TEST BANK for Contemporary Business Mathematics with Canadian Applications 12th Edition by Sieg Humm by StudyGuide

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 1 Review of Arithmetic 1) Simplify: (28 - 4)/3 Answer: (28 - 4)/3 = 24/3 = 8 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 2) Simplify: 8 + 6 ∗ 2 Answer: 8 + 12 = 20 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 3) Simplify: 5(4 + 3) Answer: 5 ∗ 7 = 35 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 4) Simplify: Answer: 5/20 = .25 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 5) Simplify: Answer: 40/20 = 2 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 6) Simplify: 9(8 - 5) + 5(6 + 4) Answer: 9 ∗ 3 + 5 ∗ 10 = 27 + 50 = 77 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 7) Evaluate: Answer: 268/(4400 ∗ .4262295) = 268/1875.4098 = .1429021 Diff: 2 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations.

8) Evaluate and round to 3 decimal places: 125(6 + 0.35 ∗ 142/678) Answer: 125 ∗ (6 + 0.35 ∗ 0.2094395) = 125 ∗ (6 + 0.0733038) = 125 ∗ (6.0733038) = 759.163 Diff: 2 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 9) Evaluate: 400(1 + .10 ∗ 100/365) Answer: 400 ∗ (1 + .10 ∗ .2739726) = 400 ∗ (1 + .02739726) = 400 ∗ (1.02739726) = 410.959 Diff: 2 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 10) Evaluate and round to one decimal place: 9210(5 - 1.38 ∗ 169/420) Answer: 9210 ∗ (5 - 1.38 ∗ 0.40238095) = 9210 ∗ (5 - 0.5552857) = 9210(4.444714) = 40,935.8 Diff: 2 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 11) Evaluate: Answer: 2424/(1 + .2 ∗ .4547945) = 2424/(1 + .0909589) = 2424/1.0909589 = 2221.899 Diff: 2 Type: SA Page Ref: 6-11 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 12) Evaluate and round to three decimal places:

Answer: 3140/(2 + 0.35 * 0.2123288) = 3140/(2 + 0.0743151) = 2910/2.074315068 = 1513.753 Diff: 2 Type: SA Page Ref: 7-13 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 13) Evaluate: Answer: 5000/(1 + .1 ∗ .5) = 5000/(1 + 0.05 ) = 5000/1.05 = 4761.90 Diff: 2 Type: SA Page Ref: 6-11 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa.

14) Spade Realty sold lots for $23 240 per hectare. What is the total sales value if the lot sizes, in hectares, were 2 , 3 , 4 ? Answer: 23240 ∗ (

+4 )

= 23240 ∗ (2 10/20 + 3 5/20 + 4 4/20) = 23240 ∗ (9 19/20) = 23240 ∗ 9.95 = $231238 Diff: 2 Type: SA Page Ref: 6-11 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 15) Three mechanics worked 15 , 14 , 18

hours respectively. What was the total cost of labour if the

mechanics were paid $14.75 per hour? Answer: Total Hours = 15

+ 14

+ 18

= 15.5 + 14.75 + 18.125 = 48.375 Total cost of labor = 48.375 ∗ 14.75 = $713.53 Diff: 2 Type: SA Page Ref: 6-11 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 16) Ana, Aamir and Charlotte worked 11 , 14

, and 22

hours respectively. What was the total cost of

labour if they were paid $18.00 per hour? Answer: Total Hours = 11

+ 14

+ 22

= 11.75 + 14.65 + 22.80 = 49.20 Total cost of labor = 49.20 ∗ 18.00 = $885.60 Diff: 2 Type: SA Page Ref: 7-13 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 17) A retailer returned 300 defective items to the manufacturer and received a credit for the retail price of $0.75 less a discount of 1/3 of the retail price. What was the amount of the credit received by the retailer? Answer: Retail value = 300($0.75) = $225 Credit = (1-1/3 )∗ $225 = (2/3) ∗ $225 = $150 Diff: 2 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.

18) Complete the following inventory sheet and find the total value. Item Quantity Cost per Unit 1 69 $.85 2 111 16 2/3 cents 3 155 $2.75 4 350 $1.66 Answer: 1161616161669 × 0.85 = 111 × 0.16 2/3 = 330 × 0.1666667 = 155 × 2.75 = 350 × 1.66 =

Total

$58.65 18.50 426.25 581.00 $1084.40

Diff: 1 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 19) Extend each of the following and determine the total. Quantity 48

Unit Price $2.45

$0.83

$2.12

$1.33

Answer: Quantity Unit Price Value 48 $2.45 $117.60 48 0.83 1/8 39.90 16 2.12 33.92 60 1.33 1/6 79.90 Total: $271.32 Diff: 1 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.

20) Purchases of packs of printing paper during the last accounting period were as follows: Number of items 8 4 15 10

Unit price $13.00 $12.00 $10.00 $10.50

What was the weighted average price per item? Answer: Number of items Unit price Weighted value 8 × $13.00 = 104.00 4 × $12.00 = 48.00 15 × $10.00 = 150.00 10 × $10.50 = 105.00 Total: 37 407.00 Average price was 407/37 = $11.00 Diff: 2 Type: SA Page Ref: 17-24 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 21) Purchases of an inventory item during last month were as follows: Number of items 5 10 8 15

Unit price $5.00 $8.00 $6.00 $3.00

What was the weighted average price per item? Answer: Number of items Unit price Weighted value 5 × $5.00 = 25.00 10 × 8.00 = 80.00 8 × 6.00 = 48.00 15 × 3.00 = 45.00 Total: 38 198.00 Average price was 198/38 = $5.21 Diff: 2 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.

22) Noriko's final mark for her Financial Mathematics course was based on four tests with different weightings. Test one counted for 10% of the final grade, test two for 20%, test three for 30% and test four for 40%. If Clara received 70% on test one, 85% on test two, 64% on test three and 72% on test four, calculate her final mark. Answer: = 70(0.1) + 85(0.2) + 64(0.3) + 72(0.4) = 7 + 17 + 19.2 + 28.8 = 72 Diff: 2 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 23) On a trip, a motorist purchased gasoline as follows: 66 litres at 69.0 cents per litre; 69 litres at 70.5 cents per litres; 80 litres at 71.5 cents per litre; and 57 litres at 74.5 cents per litre. a) What was the average number of litres per purchase? b) What was the average cost per litre? c) If the motorist averaged 9.75 km per litre, what was her average cost of gasoline per kilometre? Answer: a) 66 + 69 + 80 + 57 = 272 Average number of litres = 272 ÷ 4 = 68 b) Average cost per litre: Total cost = 66 × 69.0 = 45.54 69 × 70.5 = 48.645 80 × 71.5 = 57.20 57 × 74.5 = 42.465 193.85 cents Average cost = 193.86 ÷ 272 = 71.27 cents c) Average cost per km = 71.27 ÷ 9.75 = 7.3097436 cents Diff: 2 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 24) Dorian Frump invested $12 500 in a business on January 1. She withdrew $850 on April 3, reinvested $1920 on August 1, and withdrew $700 on September 1. What is Don's average monthly investment balance for the year? Answer: Weighted investment: January 1 — March 31: 12500 × 3/12 = 3125.0000 April 1 - July 31: 11650 × 4/12 = 3883.3333 August 1 - August 31: 13570 × 1/12 = 1130.8333 September 1 - December 31: 12870 × 4/12 = 4290.0000 Average investment balance = $12429.17 Diff: 2 Type: SA Page Ref: 17-24 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.

25) Tommy Hughes invested $10 000 in a business on January 1. He withdrew $1000 on March 1, reinvested $5000 on July 1, and withdrew $4000 on October 1. What is Tommy's average monthly investment balance for the year? Answer: Weighted investment: January 1 - February 28: 10000 × 2/12 = 1666.6700 March 1 - June 30: 9000 × 4/12 = 3000.0000 August 1 - August 31: 14000 × 3/12 = 3500.0000 September 1 - December 31: 10000 × 3/12 = 2500.0000 Average investment balance = $10666.67 Diff: 2 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 26) The following information is shown in your investment account for last year. The balance on January 1 was $7600.00. A withdrawal of $420.00 was made on March 1. A deposit of $1690.00 was made on May 1 and another deposit of $130.00 was made on October 1. What was the average monthly balance for the year in your account? Answer: Date Balance Months Weighted value January 1 7600 2 15 200 March 1 7180 2 14 360 May 1 8870 5 44 350 October 1 8740 3 26 220 Total: 12 100 130 Average monthly balance = 100130/12 = $8344.17 Diff: 2 Type: SA Page Ref: 17-24 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 27) Kevin Ash earns a semi-monthly salary of $1023.40 and works a regular workweek of 40 hours. a) What is Kevin's hourly rate of pay? b) If Kevin's gross earnings in one pay period were $1390.47, for how many hours of overtime was he paid at time and one-half of his regular pay? Answer: a) Semimonthly pay = $1023.40 Yearly salary = $24 561.60 Weekly gross pay = 24561.60 ÷ 52 = $472.34 Hourly rate = 472.34 ÷ 40 = $11.81 b) Gross pay = 1390.47 Regular pay = - 1023.40 Overtime pay = 367.07 Number of overtime hours = (367.07 ÷ 1.5) ÷ 11.81 = 20.72 hrs. Diff: 3 Type: SA Page Ref: 24-32 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

28) R.J. earns $11.70 an hour, with time-and-a-half for hours worked over 36 a week. His clock hours for a week are 10.5, 7.5, 11, 13, and 9.75. Determine his gross earnings for a week. Answer: Total hours = 10.5 + 7.5 + 11 + 13 + 9.75 = 51.75 Regular weekly earnings = 36 × $11.70 = $421.20 Overtime earnings = (51.75 - 36) × $11.70 × 1.5 = $276.41 Gross = Regular time + Overtime = $421.20 + $276.41 = $697.61 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 29) Florence Lamb is paid a commission of 10 3/4% on her net sales and is authorized to draw up to $900.00 a month. What is the amount due to Florence at the end of a month in which she drew $820.00, had sales of $14 660.00, and sales returns of $331.20? Answer: Gross sales = $14660.00 Less: returns = 331.20 Net sales = 14328.80 Gross commission = 14328.80 × .1075 = 1540.35 Less: drawings 820.00 Amount due $720.35 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 30) A sales representative selling computer parts receives a commission of 3.5% on net sales up to $15 000.00, 7% on the next $6000.00, and 9% on any further sales. If his sales for a month were $34 250.00 and sales returns were $1055.00, what was his commission for the month? Answer: Gross sales = $34250.00 Less: returns = 1055.00 Net sales = $33195.00 Commission: = .035 × 15000.00 = $525.00 = .07 × 6000.00 = 420.00 = .09 × 12195.00 = 1097.55 Total commission = $2042.55 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

31) A salesperson had gross earnings of $943.25 for last week on gross sales of $8320.00. If returns and allowances were 5.5% of gross sales, what is his rate of commission based on net sales? Answer: Net sales = (1 - .055) * 8320 = 7862.40 Commission rate =

= 11.997%

Diff: 3 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 32) A salesperson is paid a weekly salary of $350.00 or a commission of 14.5% of his sales, whichever is the greater. What is his earnings for a week in which his sales were a) $2480.00? b) $3780.00? Answer: a) Salary = $350.00 Commission = .145 ∗ 2480.00 = $359.60 Gross earnings = $709.60 b) Salary = $350.00 Commission = .145 ∗ 3780.00 = $548.10 Gross earnings = $898.10 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 33) Beth's annual salary is $42 120.00. Her regular work-week is 36 hours and she is paid semi-monthly. a) Calculate her gross pay period b) Calculate her hourly rate of pay c) Calculate her gross pay for a period in which she works 12 hours of overtime at time and one-half regular pay. Answer: a)

= $1755.00

= $22.50

c) Overtime = 12 × 22.50 × 1.5 = $405.00 Gross pay = 1755.00 + 405.00 = $2160.00 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

34) Scott Rae had gross earnings of $554.30 for last week. Scott earns a base salary of $350.00 on a weekly quota of $4 500.00. If his sales for the week were $6124.00, what is his commission rate? Answer: Gross earnings = $554.30 Less: base salary = 350.00 Commission: = $204.30 Sales for week = $6124.00 Quota = 4500.00 Commission sales = $1624.00 Rate of commission =

= 12.58%

Diff: 3 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 35) A.J. is paid an annual salary of $41 840.00. She is paid monthly on a 40-hour work week. What is the gross pay for a pay period in which she works 9 hours overtime at time-and-a-half regular pay? Answer: Weekly pay = Hourly pay =

= 804.6154

= $20.1154

Regular monthly earnings =

= 3486.67

Overtime earnings = 20.1154 × 9 ∗ 1.5 = 271.56 Gross = Regular time + Overtime = 3486.66 + 271.56 = $3758.22 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 36) Corrine Davis had gross earnings of $937.50 for the week. If she receives a base salary of $664.00 on a quota of $7800.00 and a commission of 6.75% on sales exceeding the quota, what were Corrine's sales for the week? Answer: Gross earnings = 937.50 Less: base salary = 664.00 Commission: 273.50 Commission sales =

= 4051.85

Sales for week = $(7800 + 4051.85) = $11 851.85 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

37) Abigail receives a commission of 4.5% on the first $1250.00 of sales during a week. On the next $4500.00 she receives a commission of 11.5%. On any additional sales, the commission rate is 13.75%. Find her gross earnings for a week during which her sales amount to $14 200.00. Answer: Commission on first $1250.00 is 0.045 × 1250.00 = 56.25 Commission on next $4500.00 is 0.115 × 4500.00 = 517.50 Commission on remainder = 0.1375 × 8450 = 1161.875 After rounding to the nearest cent, gross earnings are $1293.63. Diff: 2 Type: SA Page Ref: 24-32 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 38) Last week Dana worked 46 hours. For the regular workweek of 40 hours she is paid $12.40 per hour, and for every hour over 40 hours she is paid at time and one-half regular pay. How much did she earn last week? Answer: For the first 40 hours = 40(12.40) = 496.00 For the next 6 hours = 6(12.40 + 6.20) = 111.60 The total = 496.00 + 111.60 = $607.60 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 39) Kim Farrena earns $17.60 per hour. Overtime from Monday to Friday is paid at time and one-half regular pay for any hours over 7 1/2 per day. Overtime on weekends is paid at double the regular rate of pay. Last week Kim worked regular hours on Monday, Wednesday, and Friday, 8.5 hours on Tuesday, 11.75 hours on Thursday, and 5 hours on Saturday. Determine Kim's gross wages by each of the two methods. Answer: Method A Regular hours = 37.5 × 17.60 = 660.00 Overtime pay = 5.25 × 17.60 × 1.50 = 138.60 5 × 17.60 × 2 = 176.00 Gross earnings = $974.60 Method B At regular rate: 47.75 × 17.60 = 840.40 Overtime premium: 5.25 × 17.60 × 0.50 = 46.20 Overtime premium 5 × 17.60 × 1 = 88.00 Gross earnings = $974.60 Diff: 3 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

40) C.O. is paid a semi-monthly salary of $1 250.00. If his regular work week is 35 hours, what is his hourly rate of pay? Answer: Annual salary = 1250.00 × 24 = 30000.00 Weekly pay =

= 576.92

Hourly rate =

= $16.48

Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 41) C.O. is paid a semi-monthly salary of 2 754.30. If his regular work week is 42 hours, what is his hourly rate of pay? Answer: Annual salary = 2754.30 × 24 = 66103.20 Weekly pay =

= 1271.22

Hourly rate =

= $30.27

Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 42) An employee receives a gross pay of $750.73 for 47.25 hours of work. What is the hourly rate of pay if a regular work week is 37.5 hours and overtime is paid at time-and-a-half the regular rate of pay? Answer: Let the regular rate of pay be y. Regular weekly pay = 37.5y Overtime pay = (9.75 × 1.5)y = 14.625y Total pay = 37.5y + 14.625y = 750.7 52.125y = 750.73 y = 14.40 The regular rate of pay is $14.40 Diff: 3 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

43) Ali checked his pay stub on his employee portal and it showed gross earnings of $596.00 for 51 hours of work. What is his hourly rate of pay if the regular workweek is 40 hours and overtime is paid at time and one-half the regular rate of pay? Answer: Total hours = 51 Regular hours = 40 Overtime hours = 11 At time-and-a-half, overtime hours are equivalent to 11 × 1.5 = 16.5 regular hours Rate of pay = 596/56.5 = $10.55 Diff: 3 Type: SA Page Ref: 24-32 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 44) Barb's Home Income Tax business operates only during tax season. Last season Barb grossed $38 790 including GST. During that season she spent $9500 before GST on her paper and supply purchases. How much does Barb owe Revenue Canada for GST? Answer: Barb's revenue of $38 790 includes 5% GST. GST taxable revenue =

= 36 942.86

GST collected = 5% of 36 942.86 = 1 847.14 GST paid = 5% of 9 500 = 475.00 Barb owes Revenue Canada $1847.14 - $475 = $1372.14 Diff: 2 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 45) "Save the tax" is a popular advertising tactic. How much would you save on the purchase of a sweater with a list price of $52.00 in a Manitoba store during a "Save the PST" promotion? Answer: Savings on PST = 7% of $52.00 = 0.07(52.00) = $3.64 Diff: 1 Type: SA Page Ref: 32-36 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes.

46) A retail chain sells snowboards for $855.00 plus GST and PST. What is the price difference for consumers in London, Ontario, and Lethbridge, Alberta? Answer: Total cost in London Retail price = $855.00 HST = 13% of $855.00 = 0.13(855) $111.15 Total cost in London = $966.15 Total cost in Lethbridge Retail price = $855.00 GST = 5% of $855.00 = 0.05(855) $42.75 PST nil Total cost in Lethbridge $897.75 Difference = PST $68.40 Diff: 2 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 47) Emily's residence is assessed by the local taxation department at $249 500.00. Calculate the property taxes paid on this property if the existing mill rate is 15. Answer: 249500 ×

= $3742.50

Diff: 1 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 48) Calculate the property tax on a property located in the City of Brampton and assessed at $326 500 if the current tax rate is 1.05351%. Answer: Property tax = $326 500 × 1.05351/100 = $3439.71 Diff: 1 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 49) Sean's residence is assessed by the local taxation department at $160 000. Calculate the property taxes paid on this property if the existing mill rate is 20. Answer: 160 000 ×

= $3200.00

Diff: 1 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 50) The town of Pandora assesses property at market value. How much will the owner of a house valued at owe in taxes if this year's mill rate has been set at 21.386? Answer: Property tax = 32500

= $6950.45

Diff: 2 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes.

51) A town has an assessed residential property value of $350 000 000. The town council must meet the following expenditures: Education: General Purposes: Recreation: Public works: Police and fire protection:

$11 050 000 $2 100 000 $270 000 $670 000 $857 500

a) Suppose 70% of the expenditures are charged against residential real estate. Calculate the total property taxes that must be raised. b) What is the mill rate? c) What is the property tax on a property assessed at $235 000? Answer: a) Total expenditure = $(11050000 + 2100000 + 270000 + 670000 + 958500) = 14947500 Total residential property tax = 0.70(14947500) = $10463250 b) Residential mill rate = c) Property tax = $235000

(1000) = 29.895 = $7025.33

Diff: 2 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 52) Extend and total the following invoice. Quantity 77 208 621 414

Description Item A Item B Item C Item D

Answer: 77 × 0.65

$50.05

208 × 0.83

173.16

621 × 1.19 414 × 1.95

= =

Unit Price $0.65 $83 $1.19 $1.95 Total

Amount ________ ________ ________ ________ ________

738.99 807.30 $1769.50 Diff: 1 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.

53) Denise Jantz invested $35 000 on January 1 in a partnership. She withdrew $5000 on June 1, withdrew a further $1900 on August 1, and reinvested $6 000 on November 1. What was her average monthly investment balance for the year? Answer: January 1 - May 31: 35000 × 5 = 175000 June 1 - July 31: 30000 × 2 = 60000 August 1 - October 31: 28100 × 3 = 84300 November 1 - December 31:

34100 ×

Total

68200 387500

Average monthly investment =

= $32 291.67

Diff: 2 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 54) Jessica Hughes invested $40 000 on January 1 in a partnership. She withdrew $15 000 on June 1, withdrew a further $2000 on August 1, and reinvested $8 000 on November 1. What was her average monthly investment balance for the year? Answer: January 1 - May 31: 40000 × 5 = 200000 June 1 - July 31: 25000 × 2 = 50000 August 1 - October 31: 23000 × 3 = 69000 November 1 - December 31:

31000 ×

= 62000

Total Average monthly investment =

381000 = $31 750.00

Diff: 2 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 55) Carla is paid a semi-monthly salary of $1870.80. Her regular workweek is 40 hours. Overtime is paid at time and one-half regular pay. a) What is Carla's hourly rate of pay? b) What is Carla's gross pay if she worked 7 1/2 hours overtime in one pay period? Answer: a) Annual salary = 24 × 1870.80 = 44899.20 Weekly salary = 44899.20 ÷ 52 = 863.45 Hourly rate of pay =

= $21.58625

b) Regular semimonthly pay = 1870.80 Overtime pay = 7.5 × 21.58625 × 1.5 = 242.85 Gross earnings $2113.65 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

56) Tom is employed at an annual salary of $50 292.48. His regular workweek is 37.5 hours and he is paid semi-monthly. a) What is Tom's gross pay per period? b) What is his hourly rate of pay? c) What is his gross pay for a period in which he worked 12 1/2 hours overtime at time and one-half regular pay? Answer: a) Semimonthly pay = 50292.48 ÷ 24 = 2095.52 b) Weekly pay = 50292.48 ÷ 52 = 967.16308 Hourly rate = 967.16308 ÷ 37.5 = 25.791015 c) Regular earnings = 2095.52 Overtime pay = 12.5 × 25.791015 × 1.5 = 483.58 Gross earnings = $2579.10 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 57) Last week April worked 44 hours. She is paid $11.20 per hour for a regular workweek of 40 hours and overtime at time and one-half regular pay. a) What were April's gross wages for last week? b) What is the amount of the overtime premium? Answer: a) Regular earnings = 40 × 11.20 = 448.00 Overtime pay = 4 × 11.20 × 1.5 = 67.20 Gross earnings = $515.20 b) Overtime premium = 4 × 11.20 × 0.5 = $22.40 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 58) Nick's gross earnings for one week was $698.10. His regular rate of pay is $15.60 for a 35 hour week and overtime is paid at time and one-half regular pay. Calculate the number of hours that Nick worked. Answer: Gross earnings = 698.10 Regular earnings = 35 × 15.60 = 546.00 Overtime pay = 698.10 - 546.00 = 152.10 Overtime hours =

= 6.5 hours

Total number of hours worked = 35 + 6.5 = 41.5 hours Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

59) Mohammad is paid a weekly commission of 2% on net sales of $6000.00, 4% on the next $3 000.00 and 6.25% on all further sales. His gross sales for a week were 11 160.00 and sales returns and allowances were $120.00. a) Calculate his gross earnings for the week. b) Calculate the average hourly rate of pay for the week if he worked 40 hours. Answer: a) Net sales = Gross sales - Returns = 11 160 - 120 = 11 040 Commission: 2% of 6000 = 120.00 4% of 3000 = 120.00 6.25% of (11 040 - 9000) = 127.50 Gross earnings = 120.00 + 120.00 + 127.50 = $367.50 b) Average hourly rate = 367.50 ÷ 40 = $9.19 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 60) September is paid on a weekly commission basis. She is paid a base salary of $370.00 on a weekly quota of $9500.00 and a commission of 5.75% on any sales in excess of the quota. a) If September's sales for last week were $11 340.00, what were her gross earnings? b) What are September's average hourly earnings if she worked 35 hours? Answer: a) Base salary on quota of $9500 = 370.00 Commission = 5.75% on 1840 = 105.80 Gross earnings = $475.80 b) Hourly rate = 475.80 ÷ 35 = $13.59 Diff: 3 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 61) Bill earned a gross commission of $2551.05 during August. What were his gross sales if his rate of commission is 14.5% of net sales and sales returns and allowances for the month were 6% of his sales? Answer: Net sales = 2551.05 ÷ 0.145 = 17593.448 Net sales = Gross sales - returns 17593.448 = Gross sales - 6% of Gross sales 17593.448 = 94% of Gross sales Gross sales =

= $18 716.43

Diff: 3 Type: SA Page Ref: 16-21 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

62) Yanping receives a monthly salary of $1931.54 paid semi-monthly. The regular workweek is 38 hours. a) Calculate the hourly rate of pay. b) If the gross earnings for one pay period is 1270.75, for how many hours of overtime was Yanping paid at double-time regular pay. Answer: a) Annual gross earnings = 1931.54 × 12 = 23 178.48 Weekly gross earnings = Hourly rate of pay =

= 445.74 = $11.73

b) Regular semi-monthly gross earnings =

= 965.77

Overtime pay = 1270.75 - 965.77 = 304.98 Overtime rate = 11.73 × 2 = 23.46 Overtime hours =

= 13 hours

Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 63) Norm Bates is paid a semi-monthly salary of $792.50. Regular hours are 37 1/2 per week and overtime is paid at time and one-half regular pay. a) What is Norm's hourly rate of pay? b) How many hours overtime did Norm work in a pay period for which his gross pay was $946.30? Answer: a) Annual salary = 792.50 × 24 = 19020.00 Weekly pay = 19020.00 ÷ 52 = 365.76923 Hourly rate of pay = 365.76923 ÷ 37.5 = $9.75 b) Gross earnings = 946.30 Regular earnings = 792.50 Overtime pay = 153.80 Overtime hourly rate = 9.75 × 1.5 = 14.625 Overtime hours = 153.80 ÷ 14.625 = 10.516 hr. Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

64) Mark's gross wages for a week were $711.20. His regular workweek is 40 hours and overtime is paid at time and one-half regular pay. What is Mark's regular hourly wage if he worked 45 1/2 hours? Answer: Total hours = 45.5 Regular hours = 40.00 Overtime hours = 5.5 5.5 overtime hours are equivalent to 5.5 × 1.5 = 8.25 regular hours. Hourly rate of pay =

= $14.74

Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 65) Shaggy's Grocery Store shows sales revenue (exclusive of GST) of $235 000 for the year. Shaggy's GST taxable expenses were (exclusive of GST) $24 750. How much should he remit to the government at the end of the year? Answer: GST collected = 5% of $235000 = 0.05(235000) = 11750.00 GST paid = 5% of $24750 = 0.05(24750) = - 1237.50 GST remittance = $10512.50 Diff: 2 Type: SA Page Ref: 32-35 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 66) A store located in Penticton, B.C., sells a computer for $2975.00 plus HST. If the same model is sold at the same price in a store in Thunder Bay, Ontario, what is the difference in the prices paid by consumers in the two stores? Answer: Amount paid in Penticton, BC = Retail Price + 12% HST = 2975(1.12) = $3332.00 Amount paid in Thunder Bay, ON = Retail price + 13%HST = 2975(1.13) = $3361.75 The difference = 3361.75 - 3332 = $29.75, that is the 1% difference in the HST. Diff: 2 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes.

67) A computer store located in Oakville, Ont., sells a laptop for $1000.00 plus HST. If the same model is sold at the same price in a store in Victoria, B.C.., what is the difference in the prices paid by consumers in the two stores? Answer: Amount paid in Oakville, Ont = Retail Price + 13% HST = 1000(1.13) = $1130 Amount paid in Victoria, B.C. = Retail price + 12% HST = 1000 (1.12) = $1120 The difference = 1130- 1120 = $10.00, that is the 1% difference in the HST Diff: 2 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 68) Two people living in different communities build houses of the same design on lots of equal size. If the person in Airdrie has his house and lot assessed at $165 000 with a mill rate of 22.051 mills, will his taxes be more or less than the person in Kimberly with an assessment of $145 000 and a mill rate of 25.124 mills? Answer: Property tax in Airdrie = 165000 Property tax in Kimberly = 145000

= 3638.42 = 3642.98

The person in Kimberly pays $4.56 more in property tax. Diff: 2 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 69) Extend each of the following and determine the total. Quantity Unit Price 74 $1.35 90

70 58 Answer: 74 × 1.35

$0.885 $1.35 = 99.90

90 × 0.16

= 14.70

70 × .885 = 61.95 58 × 1.35 = 78.30 Total $254.85 Diff: 1 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.

70) Spade Realty sold lots for $17 120 per hectare. What is the total sales value if the lot sizes, in hectares, were 5 3/4, 7 1/3, 5 5/8, and 4 1/6? Answer: Total size

= (5.75 + 7.3333333 + 5.625 + 4.1666667) ha = 22.875 ha Sales value = 17120 × 22.875 = $391620.00 Diff: 1 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 71) Heart of Gold Realty sold lots for $50 000 per hectare. What is the total sales value if the lot sizes, in hectares, were 1 3/4, 2 1/3, 3 5/8, and 4 1/6? Answer: Total size

= (1.75 + 2.3333333 + 3.625 + 4.1666667) ha = 11.875 ha Sales value = 50000 × 11.875 = $593 750.00 Diff: 1 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 72) A salesperson earned a commission of $926.59 for last week on gross sales of $7880. If returns and allowances were 10.5% of gross sales, what is his rate of commission based on net sales? Answer: Net sales = 0.895 × 7880.00 = 7052.60 Commission rate =

= 13.14%

Diff: 2 Type: SA Page Ref: 16-21 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 73) Levi earns $19.60 an hour with time and one-half for hours worked over 8 a day. His hours for a week are 9.25, 8.5, 10.5, 13.5, and 6.25. Determine his gross earnings for a week. Answer: Total hours = 9.25 + 8.5 + 10.5 + 13.5 + 6.25 = 48 Regular hours = 8 + 8 + 8 + 8 + 6.25 = 38.25 Overtime hours = 1.25 + 0.5 + 2.5 + 5.5 = 9.75 Regular pay = 38.25 × 19.60 = 749.70 Overtime pay = 9.75 × 19.60 × 1.5 = 286.65 Gross pay = $1036.35 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

74) A salesperson receives a weekly base salary of $800.00 on a quota of $2900. On the next $2100, she receives a commission of 14%. On any additional sales, the commission rate is 19%. Find her gross earnings for a week in which her sales total $8455. Answer: Base salary on first $2900 = $800.00 Commission on next $2100 = 0.14 × 2100 = 294.00 Commission on additional sales = (8455 - 5000) × 0.19 = 3455 × .19 = 656.45 Gross earnings = $1750.45 Diff: 2 Type: SA Page Ref: 16-21 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 75) A clothing salesperson receives a weekly base salary of $200.00 on a quota of $3 000. On the next $1000, he receives a commission of 25%. On any additional sales, the commission rate is 40%. Find her gross earnings for a week in which her sales total $6000. Answer: Base salary on first $3000 = $200.00 Commission on next $1000 = 0.25 × 1000 = 250.00 Commission on additional sales = (6000 -4000) × 0.40 = 2000 ×. 40 = 800.00 Gross earnings = $1250.00 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 76) Esther's flower shop had sales revenue of $152 000.00 for the year. If the shop's GST taxable expenses were 29 920.00. Calculate how much Colleen should remit to the government at the end of the year. Answer: GST collected = (0.05)152 000 = $7 600 GST paid = (0.05)29 920 = $1 496 GST remittance = 7600 - 1496 = $6104 Diff: 2 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 77) Alicia Helm of Wawanesa, Manitoba, bought a ring for $5700. Since the jeweller is shipping the ring, Alicia must pay a shipping charge of $30.00. She must also pay PST and GST on the ring. Find the total purchase price of Alicia's ring. Answer: Total value $5730.00 GST 5% of $5730.00 $286.50 Manitoba PST 7% of 5730.00 401.10 687.60 Total purchase price $6417.60 Diff: 1 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes.

78) Suppose you went shopping and bought bulk laundry detergent worth $11.65. You then received a $1.50 trade discount, and had to pay a $2.10 shipping charge. Find the total purchase price of the detergent in Nova Scotia. Answer: Purchase price $11.65 Less discount 1.50 Net price 10.15 Add shipping charge 2.10 Total cost before taxes 12.25 HST 15% of $12.25 1.8375 1.84 Total cost in Nova Scotia $14.09 Diff: 1 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 79) A town has a total residential assessment of 900 million dollars. The town must meet expenditures of $45 million. a) If 90% of the expenditures are charged against residential real estate, calculate then total property taxes that must be raised. b) Calculate the mill rate. c) Calculate the property tax on a property assessed at $235 000.00 Answer: a) Total Residential property tax = 0.9(45 000 000) = $40 500 000 b) Mill rate = c)

(1000) = 45

Property tax = 235000

= $10 575.00

Diff: 2 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 80) A small town has a total residential assessment of $1.1 billion. The town must meet expenditures of $300 million. a) If 80% of the expenditures are charged against residential real estate, calculate then total property taxes that must be raised. b) Calculate the mill rate. c) Calculate the property tax on a property assessed at $500 000.00 Answer: a) Total Residential property tax = 0.8(300 000 000) = $240 000 000 b) Mill rate = c)

Property tax = 500000

(1000) = 218.18 = $109 090

Diff: 2 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes.

81) Simplify: 6 +

-4

A) 46 B) 18 C) 24 D) 4 E) 8 Answer: D Diff: 1 Type: MC Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 82) Simplify: 10 +

-5

A) 2 B) 5 C) 55 D) 7.5 E) 8 Answer: B Diff: 1 Type: MC Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 83) Calculate the final answer for the following expression: (53 + 22) ÷ 20

A) 5.45 B) 4.45 C) 6.45 D) 7.45 E) 9.45 Answer: C Diff: 1 Type: MC Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 84) Simplify the following: 7125 × A) 23592.74 B) 24592.74 C) 25592.74 D) 26592.74 E) 27592.74 Answer: B Diff: 1 Type: MC Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations.

85) Calculate the weighted-average cost of the following inventory purchases: Date May 4 May 11 May 29

Quantity Purchased 33 41 37

Cost per Unit $12.25 $13.87 $11.99

Total Amount

A) $12.76 B) $12.12 C) $22.11 D) $21.21 E) $22.22 Answer: A Diff: 1 Type: MC Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 86) You are paid a commission on the gross profit per vehicle that you sell. The commission increases with each additional vehicle that you sell. The minimum commission is 25% and it increases in increments of 1% for each additional vehicle sold. Your sales for the following month were as follows: Vehicle 1 2 3 4 5 6 7 8

Gross Profit $755 $1023 $474 $1512 $864 $1021 $1953 $1207

What is your monthly gross pay? A) $2462.69 B) $2569.42 C) $2669.69 D) $2642.69 E) $2554.60 Answer: E Diff: 1 Type: MC Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

87) Your full-time job pays you a bi-weekly salary of $2963.56. In addition to this position you have been working part-time for the last 15 months and earn $500 semi-monthly. You have other payments that total $14 400 per year. What is the maximum monthly amount that your payments can be towards a house purchase? Assume that property taxes and heating costs are included in the $14 400. A) $2844.42 B) $3488.42 C) $2488.42 D) $1961.42 E) none of the above Answer: C Diff: 1 Type: MC Page Ref: 25-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 88) You are supposed to teach 15

credit hours per week per term for a trimester school year. Anything

above is considered to be overtime and is compensated at time and one-half you regular rate of $27.13 per hour. You work the following credit hours. How much total overtime, for the entire term, does your employer owe you at the end of the third term (assume that you were not paid any overtime in terms one and two and that the term is 15 weeks long)? Term 1 2 3

Credit Hours 20 26 19

A) $9987.50 B) $10 987.50 C) $8987.5 D) $11 987.50 E) $11 187.50 Answer: B Diff: 1 Type: MC Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

89) Your gross annual pay is $19 163. Employment insurance premiums are deducted at a rate of 2.25% and Canada Pension Plan premiums are 3.75% based on total earnings. You pay income taxes at a rate of 17% on all amounts over $8131. What is your Net Pay for the year? A) $15 137.78 B) $17 137.78 C) $16 317.78 D) $16 137.78 E) $18 237.78 Answer: D Diff: 2 Type: MC Page Ref: 16-21 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 90) Calculate the final answer for the following expression: (103 + 202) ÷ 20. A) 50.00 B) 20.00 C) 70.00 D) 3.5 E) 5000 Answer: C Diff: 1 Type: MC Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 91) Calculate the weighted-average cost of the following inventory purchases: Date June 4 June 14 June 29

Quantity Purchased 30 40 30

Cost per Unit $11.50 $15.00 $10.99

Total Amount $345.00 $600.00 $329.70

A) $12.74 B) $12.75 C) $22.74 D) $22.75 E) $37.49 Answer: B Diff: 1 Type: MC Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.

92) You bought a new car in Ontario for $19 500 which included HST. What is the total amount of HST that you paid? A) $2535.00 B) $2243.36 C) $2913.79 D) $1197.37 E) $171.05 Answer: B Diff: 1 Type: MC Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 93) Phil works for a hydro company in a 35 hour work week schedule. His contract pays him $1345/week for working Monday to Friday, 7 hours a day. If he spends more time than his usual time, he is entitled to "time and a half" on time worked in excess of 7 hours per day. If he works on Sundays or statutory holidays, he is entitled to twice the time. Calculate his gross earnings for the month of October 2013, if he worked for 5 hours on Thanksgiving Day in addition to his regular hours for the month. Answer: Rate per day = Rate per hour =

= $269/day

= $38.43/hr

Total working days in October 2013 = 22 days Pay for regular days = 269 × 22 = $5918 Pay for overtime = 5 × 2 × 38.43 = $384.29 Gross earnings for October 2013 = $5918 + $384.29 = $6302.29 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 94) Isabelle works for The Brick, a furniture company. She earns a base salary $12/hr, 40 hours a week. However, she is also paid commission of 2% for all sales and an extra lump sum of $1000 incentive if her monthly sales exceed $10 000. In the month August, she won the title of employee of the month for total sales of $75 000. How much did she earn in the month of August, assuming a 4 week month? Answer: Base salary = 12 × 40 × 4 = $1920 Commission earned in August = 2% × 75000 = $1500 Total pay for August = $1920 + $1500 + $1000 = $4420 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

95) Isabelle works for The Brick, a furniture company. She earns a base salary $12/hr, 40 hours a week. However, she is also paid commission of 2% for all sales and $1000 incentive pay for every sale of $10 000. In the month August, she won the title of employee of the month for total sales of $75 000. How much did she earn in the month of August, assuming a 4 weeks month? Answer: Base salary = 12 × 40 × 4 = $1920 Commission earned in August = 2% × 75000 = $1500 Total incentive pay = 7 × $1000 = $7000 Total pay for August = $1920 + $1500 + $7000 = $10 420 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 96) Phil works for a hydro company in a 35 hour work week schedule. His contract pays him $1345/week for working Monday to Friday, 7 hours a day. If he spends more time than his usual time, he is entitled to "time and a half" on time worked in excess of 7 hours per day. If he works on Sundays or statutory holidays, he is entitled to twice the time. Calculate his gross earnings for the month of November 2013, if he worked straight time only. A) $5380 B) $4035 C) $5649 D) $8473.50 E) $11 298 Answer: C Diff: 2 Type: MC Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 97) Raj works for Honda Pickering on commission based on each unit sold. He gets $700 for every used car he sells and $975 for every new car. He sold 8 used cars and 5 new cars in January 2020. What was his gross earning in January? A) $5600 B) $4875 C) $10,475 D) $11,300 Answer: C Diff: 2 Type: MC Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

98) Abu sells mutual funds for CIBC. On mutual funds sales, CIBC charges a "front-end load" or gross commission rate of 5%. Abu is paid on a graduated commission structure. He receives 30% commission on the first $50 000 worth of mutual funds he sells in a month, 40% commission for the next $50 000 worth of mutual funds, and 75% commission on all additional sales in the same month. What is Abu's commission for a month in which he sells $150 000 worth of mutual funds? A) $3625 B) $2250 C) $2700 D) $4500 E) $5625 Answer: A Diff: 2 Type: MC Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 99) On March 1, Sam invested $90 000 in a business. On June 1, he invested another $10 000. On July 1, he withdrew $17 000 to spend on the repair of his house. On September 1, he injected another $2000 into the business. What was his average investment per month in the business during the year? Assume all months have the same length or weighting. Answer: $0 are invested for 2 months $90,000 are invested for 3 months $90000 + $10000 = $100,000 are invested for 1 month $100,000 - $17000 = $83000 are invested for 2 months $83000 + $2000 = $85000 are invested for 4 months Average investment = =

= $73 000 per month

Diff: 2 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 100) The engineering department of a consulting company has 8 junior engineers working at $24.75/hour, 5 senior engineers working at $39.49/hour, and an engineering manager working at $47.24 per hour. Calculate the weighted average hourly rate earned by the engineering department. A) $31.62 B) $37.16 C) $88.19 D) $35.09 E) $34.05 Answer: A Diff: 1 Type: MC Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.

101) In 2011, Danny invested his savings among four mutual funds as follows: 25% in bonds fund, 20% in Canadian equity fund, 40% in US equity fund, and the rest in money markets. During the past year, the rates of return on the individual funds were 7%, 3%, 9%, and -1%, respectively. What was the overall return rate on his portfolio? Answer: Contributing rate of return from bonds fund = 25% × 7% = 1.75% Contributing rate of return from Canadian equity fund = 20% × 3% = 0.6% Contributing rate of return from US equity fund= 40% × 9% = 3.6% Contributing rate of return from money market = (100% - 25% - 20% - 40%) × (-1%) = 15% × (-1%) = -0.15% Overall return rate on portfolio = 1.75% + 0.6% + 3.6% - 0.5% = 5.8% Diff: 1 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 102) In the first term, Sam's courses and grades were as follows: Course English French Math Physics Chemistry Social Science Arts

Credit 2 2 3 3 3 1 1

Grade Point Value 2.7 3.0 4.0 4.0 3.7 3.3 2.0

Calculate Sam's Grade Point Average (GPA). Answer: Course Credits GPV English 2 2.7 French 2 3.0 Math 3 4.0 Physics 3 4.0 Chemistry 3 3.7 Social Science 1 3.3 Arts 1 2.0 Total 15 Grade Point Average =

Credits × GPV 5.4 6.0 12.0 12.0 11.1 3.3 2.0 51.8

= 3.45

Diff: 1 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.

103) Kim invested $7500 in a business for 3 months. She withdrew $2500 at the start of the fourth month and kept the rest of the money in the savings account for the remaining 9 months in the year. What is Kim's average monthly investment balance for the year? A) $833 B) $1042 C) $3750 D) $5625 E) $5000 Answer: D Diff: 1 Type: MC Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 104) Certify Auto Repair (CAR) files GST returns annually. In 2012, CAR billed its customers $195 432 for labour, $24 732 for parts, and then added the GST. During this year, CAR paid $19 785 for parts, $36 000 for rent, $9767 for utilities, and $21 873 for shop repairs, plus the GST on these goods and services. What GST must be remitted (or refunded by the CRA ) for the year 2012? Answer: Total Revenue = 195432 + 24732 = $220 164 Total Costs = 19785 + 36000 + 9767 + 21873 = $87 425 Net Revenue = 220164 - 87425 = $132 739 GST remitted = $6636.95 Diff: 2 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 105) Samir plans to sell his 2400 sq. ft. bungalow in Whitby for $389 000 and buy a similar-sized house in Oshawa for $369 000. Whitby charges the property tax at the mill rate of 13.59815, whereas Oshawa charges the property tax at the mill rate of 16.18347. What will be his net property tax penalty/saving? Answer: Property Tax in Whitby = Property Tax in Oshawa =

× 389000 = $5289.68 × 369000 = $5971.70

Net property tax penalty = $682.02 Diff: 2 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes.

106) Justin is planning to buy a house. He has two options. A 1800 sq. ft. bungalow in Courtice costs $389 000. A similar house costs $369 000 in Bowmanville. Justin plans to base his decision on the cost of property tax paid. Whitby charges the property tax at the mill rate of 13.59815, whereas Oshawa charges the property tax at the mill rate of 16.18347. Where will he end up paying less taxes? A) Taxes are the same in both the cities B) In Bowmanville C) In Courtice D) There is not enough data to make a decision Answer: C Diff: 1 Type: MC Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 107) Paul is a homeowner in Whitby and his home value has been recently assessed by MPAC as $339 500. Paul's tax notice lists the following mill rates for various local services and capital developments. Calculate current year's total property tax. Town general municipal Region subtotal Education

4.19938 7.18877 2.21000

Answer: Total property tax rate = 4.19938 + 7.18877 + 2.21 = 13.59815 Total property tax =

× 339500 = $4616.57

Diff: 2 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 108) Simplify: A) 1 B) 0.24 C) 6.11 D) 10.49 E) 0.58 Answer: E Diff: 1 Type: MC Page Ref: 6-11 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa.

109) Simplify: A) -1.5 B) 1.5 C) 0 D) -1 E) 1 Answer: D Diff: 1 Type: MC Page Ref: 6-11 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 110) Simplify:

A) $7067 B) $771 C) $695 D) $765 E) $665 Answer: C Diff: 1 Type: MC Page Ref: 6-11 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 111) Simplify: 5

-9

A) 3 B) 3 C) 22 D) 3 E) 3 Answer: B Diff: 1 Type: MC Page Ref: 6-11 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa.

112) Simplify and round the answer to two decimal places: 56.929 - 36.434 A) 20.49 B) 20.50 C) 20.54 D) 20.53 E) 20.48 Answer: B Diff: 1 Type: MC Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 113) Change 0.25% into a decimal. Answer: 0.25/100 = 0.0025 Diff: 1 Type: SA Page Ref: 12-15 Topic: 1.3 Percent Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 114) Change 0.035 into a percent. Answer: 0.035 ∗ 100 = 3.5% Diff: 1 Type: SA Page Ref: 12-15 Topic: 1.3 Percent Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 115) Change 0.2% into a fraction. A) B) C) D) E) Answer: A Diff: 1 Type: MC Page Ref: 12-15 Topic: 1.3 Percent Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa.

116) Express the following as a percentage:

A) 16.67% B) 166.67% C) 1.667% D) .1667% E) .001667% Answer: A Diff: 1 Type: MC Page Ref: 12-15 Topic: 1.3 Percent Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 117) Change the following expression into a decimal: 166.67% A) 166667 B) 166.7 C) 16667 D) 1.6667 E) 0.16667 Answer: D Diff: 1 Type: MC Page Ref: 12-15 Topic: 1.3 Percent Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 118) 87 % is equal to: A)

B) 87 C) 8.75 D) E) Answer: D Diff: 1 Type: MC Page Ref: 12-15 Topic: 1.3 Percent Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa.

119) Convert 99.5% into decimals. A) 0.9 B) 0.99 C) 0.999 D) 0.995 E) 9.95 Answer: D Diff: 1 Type: MC Page Ref: 12-15 Topic: 1.3 Percent Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 120) Simplify: 2[42 - 3 × 12 + 4(7 - 9 × 50)] + (6 + 94) Answer: = 2[42 - 3 × 12 + 4(7 - 9 × 50)] + (6 + 94) = 2[16 - 3 × 12 + 4(7- 9 × 50)] + 6 + 94 = 2[16 - 36 + 4(-443)] + 6 + 94 = 2[16 - 36 - 1772] + 6 + 94 = 2[-1792] + 6 + 94

= -3584 + 6 + 94 =-3584 + 6 + 6561 = -3578 + 6561 = 2983 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 121) Simplify: 44[(9 ÷ 3) + 72 - 6 × 2.5 - 19] - [42 - 3(77 ÷ (3.5 × 2)) - 9] Answer: = 44[3 + 72 - 15 - 19] - [42 - 3(77 ÷ 7) - 9] = 44[60 - 19] - [42 - 3(11) - 9] = 44(41) - [42 - 33 - 9] = 44(41) - [9 - 9] = 1804 - 0 = 1804 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations.

122) Simplify, then round to the nearest hundredth:

Answer: =

= = = 1233.01 Diff: 1 Type: SA Page Ref: 6-11 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 123) Convert this fraction into the decimal form. If appropriate, place a dot above a decimal number to show that it repeats: Answer: 1.416 Diff: 1 Type: SA Page Ref: 6-11 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 124) Convert this fraction into the decimal form. If appropriate, place a dot above a decimal number to show that it repeats: Answer: 0.333 Diff: 1 Type: SA Page Ref: 6-11 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 125) Reduce the following fraction to the lowest term: Answer: Diff: 1 Type: SA Page Ref: 6-11 Topic: 1.2 Fractions Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa.

126) Change the following percent into a common fraction in lowest terms: 34% Answer: Diff: 1 Type: SA Page Ref: 12-15 Topic: 1.3 Percent Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 127) Change the following percent into a common fraction in lowest terms: 82.6% Answer: Diff: 1 Type: SA Page Ref: 12-15 Topic: 1.3 Percent Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 128) Express the following as a percent: 0.67 Answer: 67% Diff: 1 Type: SA Page Ref: 12-15 Topic: 1.3 Percent Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 129) Express the following as a percent: 4.55 Answer: 455% Diff: 1 Type: SA Page Ref: 12-15 Topic: 1.3 Percent Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 130) Express the following as a percent: Answer: 87.5% Diff: 1 Type: SA Page Ref: 12-15 Topic: 1.3 Percent Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa.

131) A house in Detroit was sold at $246,000, which is

of its actual value. If the realtor made a

commission of $1.25 on each $250 of the actual house value, how much commission does the realtor earn while selling the house? Answer:

of the house's actual cost is

of the cost ÷ 4.

$246 000 ÷ 4 = $61 500. of the cost or all the actual value of the house, is

of the cost × 13.

$61 500 × 13 = $799 500 of the house's price is $799 500. $799 500 ÷ $250 = $3198. $3198 × $1.25 = $3997.50 The realtor earned $3997.50. Diff: 3 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 132) A party room was rented out at $11.75 per hour. If 6 people each rented the room for 2 , 1 3 , 6 , and 2

,4 ,

hours respectively, how much did the party room owners earn in total (excluding all

taxes)? Round your answer to the nearest cent. Answer: =

× ×

= 30.55 + 21.7375 + 56.7916 + 14.6875 + 74.4166 + 31.725 = 229.91 = The party room owners were paid $229.91 in total. Diff: 2 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.

133) Complete the following statement: Quantity 88

Description Alpha

65 73 46

Bravo Charlie Delta

$0.675 $1.44

Description Alpha

Unit Price $0.37

Answer: Quantity 88 65 73 46

Bravo Charlie Delta

Unit Price $0.37 96

Total $

$0.675 $1.44

Total $ 32.56 62.92 49.275 66.24

Diff: 2 Type: SA Page Ref: 16-21 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages. 134) Oliver received a pay at the end of the month of $4 303.00 in June 2016, working 6 hours a day, and a total of 20 days in the month. Included in the pay is a bonus of $243. What is Oliver's hourly rate of pay? Round to the nearest cent. Answer: $4 303.00 – $243 = $4 100.00 $4 060 ÷ 20 = $203 $203 ÷ 6 = $33.83 Oliver's hourly rate of pay is $33.83 Diff: 2 Type: SA Page Ref: 23-29 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions.

135) An employee at a nuclear power plant in New York has an annual salary of $143 000 and is paid every two weeks. The employee works twenty days in a month, five days in a week, and eight hours in a day. (Assume 52 weeks in a year) a) What is the gross pay per pay period? b) What is the employee's hourly rate of pay? Answer: a) 52 ÷ 2 = 26 He has 26 pay periods in a year. $143 000 ÷ 26 = $5500 The employee's gross pay per pay period is $5500 b) $5500 ÷ 10 = $550 $550 ÷ 8 = $68.75 The employee earns $68.75 per hour. Diff: 2 Type: SA Page Ref: 23-28 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 136) Determine the amount of provincial sales tax in Quebec after purchasing an electric key board for $182. Answer: 9.975% of 182 = 0.09975(182) = $18.15 Diff: 1 Type: SA Page Ref: 32-35 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 137) While in Prince Edward Island, Charlotte goes to the barbershop, where she pays $24 for the haircut and $21 for the colour, both of which are subject to the 14% HST. She also tips the barber 17% of the combined cost of the haircut and colouring, excluding taxes. How much does Charlotte spend in total? Answer: 24 + 21 = 45 Tip = 17% of 45 = 0.17(45) = $7.65 HST on the haircut = 14% of 24 = 0.14(24) = $3.36 HST on the colouring = 14% of 21 = 0.14(21) = $2.94 Total = 45 + 7.65 + 3.36 + 2.94 = $58.95 Diff: 2 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes. 138) The city of Toronto had a property tax rate of approximately 1.529% on multiresidential buildings in 2015. If a multiresidential building had an estimated value of $ 12 500 000, how much property tax did the building's owners have to pay? Answer: 0.01259(12500000) = 157375 Diff: 2 Type: SA Page Ref: 30-33 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes.

139) Evaluate: Answer: (4 + 6(2)2 + 2)/15 = (4 + 6(4) + 2)/15 = (4 + 24 + 2)/15 = 30/15 = 2 Diff: 2 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 140) Simplify: 6 ÷ 2 + 15 × 2 Answer: 3 + 30 = 33 Diff: 1 Type: SA Page Ref: 5-6 Topic: 1.1 Basics of Arithmetic Objective: 1-1: Simplify arithmetic expressions using the basic order of operations. 141) Turn 78

% into a common fraction in lowest terms.

Answer: 69/88 Diff: 2 Type: SA Page Ref: 13-17 Topic: 1.3 Percent Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 142) Express 93/44 as a percent. Answer: 211.36% Diff: 1 Type: SA Page Ref: 13-17 Topic: 1.3 Percent Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 143) Compute 187% of 342. Answer: 639.54 Diff: 1 Type: SA Page Ref: 13-17 Topic: 1.3 Percent Objective: 1-2: Determine equivalent fractions, and convert fractions to decimals, decimals to percents, and vice versa. 144) Find the arithmetic average of the following data set: 2, 9,

, 34.02, 52, 1, 83 , 0.5, 6,

Answer: = (2 + 9 + 0.875 + 34.02 + 52 + 1 + 83.75 + 0.5 + 6 + 4.909)/10 = 19.405 Diff: 2 Type: SA Page Ref: 17-24 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.

145) Pooja has an annual salary of $98 351.00 and is paid biweekly. Her regular workweek is 32 hours. a) What is Pooja's gross pay per pay period? b) What is the hourly rate of pay? c) What are the gross overtime earnings for Pooja if the overtime rate is one and three quarters the regular hourly rate of pay and she works 8 extra hours? Answer: a) Yearly salary = $98 351.00 Biweekly gross pay = 98 351/26 = $3782.73 b) Biweekly gross pay = $3782.73 Weekly gross pay = 3782.73/2 = $1891.37 Hourly rate = 1891.37/32 = $59.11 c) Hourly rate = $59.11 Gross overtime earnings = 59.11(1.75)(8) = $827.54 Diff: 3 Type: SA Page Ref: 24-32 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 146) Danny receives a commission of 13.4% on the first $5000 of his net sales, 14.1% on the next $3000, and 15.0% on any additional net sales for the month, and is entitled to drawings of up to $1750 per month. During February, Danny's gross sales amounted to $11 456 and sales returns and allowances of $512. a) What is Danny's gross commission for the month? b) If Danny drew $1235 in February, what is the amount due to him? Answer: a) Gross sales = $11 456.00 Commission on first $5000 = 0.134(5000) = $670.00 Commission on next $3000 = 0.141(3000) = $423 Commission on additional sales = 0.15(11456 - 8000) = $518.40 Total gross commission = $1611.40 b) 1611.40 - 1235 = $376.40 Diff: 3 Type: SA Page Ref: 24-32 Topic: 1.5 Applications - Payroll Objective: 1-4: Determine gross earnings for employees remunerated by the payment of salaries, hourly wages, or commissions. 147) In 2018, Ashley's bike repair shop made a gross profit of $23 000, of which $4032 were GST-taxable expenses. How much did Ashley remit to the government in 2018? Answer: GST collected: 5% of $23000 = 0.05(23000) = $1150.00 GST Paid: 5% of 4032 = 0.05(4032) = - $201.60 GST Remittance: = $948.40 Diff: 1 Type: SA Page Ref: 32-36 Topic: 1.6 Applications - Taxes Objective: 1-5: Through problem solving, compute GST, HST, PST, sales taxes, and property taxes.

148) Suppose that in your Business Math course you receive the following grades (%): Assessment Quiz 1 Quiz 2 Quiz 3 Quiz 4 Quiz 5 Quiz 6 Quiz 7 Quiz 8 Quiz 9 Quiz 10 Term Test 1 Term Test 2 Final Exam

Grade Received 80% 90% 50% 40% 80% 90% 80% 50% 80% 70% 65% 78% 72%

Each assessment category counts as following towards the final grade: Assessments Quizzes (best 8 out of 10) Tests (two; each 15%) Final Exam

Course Grade 20% 30% 50%

a) What is your average quiz grade in this course? b) What is your average grade for the course overall? Answer: a) Quiz Average = (80 + 90 + 80 + 90 + 80 + 50 + 80 + 70)/8 = 77.5% b) Average course grade = [(77.5) 20 + (65)15 + (78)15 + (72)50] / (20 + 15 + 15 + 50) = (1350 + 2145 + 3600)/100 = 70.95% Diff: 3 Type: SA Page Ref: 17-24 Topic: 1.4 Applications - Averages Objective: 1-3: Through problem solving, compute simple arithmetic and weighted averages.

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 2 Review of Basic Algebra 1) Simplify: 7m - 2m - 3m Answer: 2m Diff: 1 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 2) Simplify: 8x - 5y - 14x + 5y Answer: -6x Diff: 1 Type: SA Page Ref: 46-53 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 3) Simplify: 4x - 6y - 4x - 2y Answer: -8y Diff: 1 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 4) Simplify: 2x + 0.73x Answer: 2.73x Diff: 1 Type: SA Page Ref: 46-53 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 5) Simplify: x - 0.32x Answer: 0.68x Diff: 1 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 6) Simplify: 3ax - 4x + 1 - 7 + 3x - 4ax Answer: -ax - x - 6 Diff: 1 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 7) Simplify: -(4 - 6a) - (-4 + 3a) Answer: -4 + 6a + 4 - 3a = 3a Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution.

8) Simplify: -(4 - 6a ) - (-8 + 6a ) Answer: -4 + 6a + 8 - 6a = 4 Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 9) Simplify: -(3m - 6m - 5) - (4 - 7m - 2m) Answer: -3m + 6m + 5 - 4 + 7m + 2m = 12m + 1 Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 10) Simplify: (7a - 7b ) - (-3a + 9b ) - 11b Answer: 7a - 7b + 3a - 9b - 11b = 10a - 27b Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 11) Simplify: (8a - 7b ) - (-3a + 7b ) - 11b Answer: 8a - 7b + 3a - 7b - 11b = 11a - 25b Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 12) Simplify: -9a(-5b ) Answer: 45ab Diff: 1 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 13) Simplify: -5a(-5b ) Answer: 25ab Diff: 1 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 14) Simplify: -6m(-3m) Answer: 18m2

Diff: 1 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 15) Simplify: 3a(-3b)(-4c)(-1) Answer: -36abc Diff: 1 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution.

16) Simplify: -2a(-3b)(-4c)(-5) Answer: 120abc Diff: 1 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 17) Simplify: 5(4x - 2) Answer: 20x - 10 Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 18) Simplify: -3x(4 - 2b - b) Answer: -12x + 6bx + 3bx = -12x + 9bx Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 19) Simplify: -3x(4 - 2b - 2b ) Answer: -12x + 6bx + 6bx = -12x + 12bx Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 20) Simplify: 8(9y - 4) - 2(y - 1) - (1 - 3y) Answer: 72y - 32 - 2y + 2 - 1 + 3y = 73y - 31 Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 21) Simplify: 4(9y - 4) - 2(y - 1) - (1 - 3y) Answer: 36y - 16 - 2y + 2 - 1 + 3y = 37y - 15 Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 22) Simplify: (3m - 4n)(m + 7n) Answer: 3m2 + 21mn - 4mn - 28n2 = 3m2 +17mn - 28n2

Diff: 2 Type: SA Page Ref: 46-53 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 23) Simplify: (3a - 1)(a - 3a + 1) Answer: 3a2 - 9a2 + 3a - a + 3a - 1 = -6a2 + 5a - 1

Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution.

24) Simplify: (4a - 1)(a - 3a + 1) Answer: 4a2 - 12a2 + 4a - a + 3a - 1 = -8a2 + 6a - 1

Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 25) Simplify: 2(a - 1)(7a - 3 ) - 3(6a - 2)(2a + 1) Answer: 2(7a2 - 3a - 7a + 3 ) - 3(12a2 + 6a - 4a - 2 ) = 14a2 - 6a - 14a + 6 - 36a2 - 18a + 12a + 6 = -22a2 - 26a + 12

Diff: 1 Type: SA Page Ref: 46-53 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 27) Simplify: 60xy ÷ (-5xy) Answer: -12 Diff: 1 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 28) Simplify: (-45a3b) ÷ 15a2 Answer: -3ab Diff: 1 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 29) Simplify: (-64ab ) ÷ (8ab ) Answer: -8 Diff: 1 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 30) Simplify: (-8ab ) ÷ (8a ) Answer: -b Diff: 1 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution.

31) Simplify: (21x - 36 ) ÷ (-3 ) Answer: -7x + 12 Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 32) Simplify: (-a3 - 11a2 - 3a ) ÷ (-a ) Answer: a2 + 11a + 3

Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 34) Evaluate: 4x2 - 10xy - 8y2 for x = -3, y = 5 Answer: 4x2 - 10xy - 8y2 = 4(-3)2 - 10(-3)(5 ) - 8(5)2 = 4(9) + 150 - 8(25 ) = 36 + 150 - 200 = -14 Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 35) Evaluate y: y = (3x2 - x - 1 ) - (5 - 2x - x2) for x = -3 Answer: (3x2 - x — 1 ) - (5 - 2x - x2) = [3(-3)2 - (-3) - 1] - [5 - 2(-3) - (-3)2] = (27 + 3 - 1) - (5 + 6 - 9 ) = (29 ) - (2 ) = 14.5 - 0.5 = 14.0 Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution.

36) Evaluate R: R = Answer:

for I = 83, P = 845, T =

= .163708

Diff: 3 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 37) Evaluate r: r =

, here I = 116, P = 1760, t =

Answer: r =

= .1603788

Diff: 3 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 38) Evaluate r: r =

, where I = 200, P = 800, t =

Answer: r =

= .6083333

for N = 32, C = 20, P = 1859, n = 26 =

= .0255

= 1400 [1 - .185 ∗ 7]

= 1400[1 - .1295] = 1400[.8705] = 1218.7 Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution.

41) Evaluate P: P = A(1 - RT) for A = 700, R = 0.185, T = Answer: A(1 - RT) = 700

= 1400 [1 - .185 ∗ 7]

= 700[1 - .0925] = 700[.9075] = 635.25 Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 42) Evaluate p: p = s

, where s = 3120, r = 0.123, t = 295

Answer: p = 3120

= 3120[1 - .123 ∗ .8194444]

= 3120[1 - .1007917] = 3120[.8992083] = 2805.53 Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 43) Evaluate P: P = Answer:

for A = 752, R = 0.145, T =

= 689.5129

Diff: 3 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 44) Evaluate s: s =

Answer: s =

, where p = 3411.50, r = 0.0925, t = 75

= 3347.001

, PMT = $750, i = 0.025, and n = 10.

= $8402.54

Diff: 3 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution.

46) Evaluate: 16 Answer: 1 Diff: 1 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 47) Evaluate: (-1)14 Answer: 1 Diff: 1 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 48) Evaluate: (-1)17 Answer: -1 Diff: 1 Type: SA Page Ref: 53-61 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 49) Evaluate: Answer: Diff: 1 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 50) Evaluate: (-0.1)7 Answer: -0.0000001 Diff: 1 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 51) Evaluate: (-0.1)4 Answer: 0.0001 Diff: 1 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero.

52) Evaluate: m0 Answer: 1 Diff: 1 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 53) Evaluate: (-5)-3 Answer: -

or -.008

Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 54) Evaluate: Answer:

or 3.375

Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 55) Evaluate: (3.09)0 Answer: 1 Diff: 1 Type: SA Page Ref: 53-61 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 56) Evaluate: -(288888)0 Answer: -1 Diff: 1 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 57) Evaluate: (3)2(3)4 Answer: (3)2(3)4 = 36 = 729

Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero.

58) Simplify: (-4)4 ∗ (-4) Answer: (-4)4 + 1 = (-4)5 = -1024

Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 59) Simplify: (-4)3 ∗ (-4) Answer: (-4)3 + 1 = (-4)4 = 256

Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 60) Simplify: (-3)7 ÷ (-3)4 Answer: (-3)7-4 = (-3)3 = -27

Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 61) Simplify: (m4)5 Answer: (m4)5 = m4 × 5 = m20

Diff: 1 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 62) Simplify: (m5)5 Answer: (m5)5 = m5 × 5 = m25

Diff: 1 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 63) Simplify: [(-4)4]3 Answer: (-4)4 × 3 = (-4)12 = 16777216

Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero.

64) Simplify: m13 ÷ m6 Answer: m13-6 = m7

Diff: 1 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 65) Simplify: (-1)4(-1)2(-1)5 Answer: (-1)4 + 2 + 5 = (-1)11 = -1

Diff: 1 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 66) Simplify: Answer: x3 + 5 - 4 = x4 Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 67) Simplify: Answer: x16 + 4 - 2 = x18 Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 68) Simplify:

Answer: [1/9]8-3 = [1/9]5 = 1/59049 Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 69) Simplify:

Answer: [-3/5]9 - 4 = [-3/5]5 = -243/3125 = -.07776 Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero.

70) Simplify: 1.005240 ÷ 1.005160 Answer: 1.005240 -160 = 1.00580 = 1.4903386

Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 71) Simplify: Answer: [-3/8]6 × 3 = [-3/8]18 = (-3)18/818 or .0000000215062 Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 72) Simplify: (1 - r)2(1 - r)3(1 - r) Answer: (1 - r)6

Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 73) Simplify: (1 - r)3(1 - r)4(1 - r) Answer: (1 - r)8

Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 74) Simplify: [(1 - r)51]2 Answer: (1 - r)51 × 2 = (1 - r)102

Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 75) Simplify: (5wz)3 Answer: 53w3z3 = 125w3z3

Diff: 1 Type: SA Page Ref: 53-61 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero.

76) Simplify: Answer:

Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 77) Simplify: Answer:

Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 78) Simplify: 7-5 ÷ 7-7 Answer: 7-5 - (-7) = 72 = 49

Diff: 3 Type: SA Page Ref: 53-61 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 79) Simplify: Answer: Diff: 3 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 80) Compute: Answer: = 11.74 Diff: 1 Type: SA Page Ref: 61-64 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents.

81) Compute: Answer: 1.01 Diff: 2 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 82) Compute: Answer: 1.03 Diff: 2 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 83) Compute: Answer: .5 Diff: 2 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 84) Compute: Answer: 1.010595566 Diff: 2 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 85) Compute: Answer: 1.13 Diff: 2 Type: SA Page Ref: 61-64 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 86) Compute: 19562/5 Answer: 20.727529 Diff: 2 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents.

87) Compute: 32.53/4 Answer: 13.611705 Diff: 2 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 88) Compute: 82/6 Answer: = =2 Diff: 2 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 89) Compute: Answer: 1.2423925 Diff: 2 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 90) Compute: 3.94-6/17 Answer: 0.616346 Diff: 3 Type: SA Page Ref: 51-64 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 91) Compute: Answer:

= 9.935074

Diff: 3 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 92) Compute the value of Answer:

= 15.61388

Diff: 3 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents.

93) Evaluate: 50.00 Answer: 50.00 = 50 = 50 = 50 = 50(60.40198318) = 3020.10 Diff: 2 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 94) Express in logarithmic form: 29 = 512 Answer: 9 = log2 512

Diff: 2 Type: SA Page Ref: 64-70 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 95) Express in logarithmic form: 56 = 19683 Answer: 6 = log5 15625

98) Express in exponential form: log3 Answer: log3

= -4

= -4, 3-4 =

Diff: 2 Type: SA Page Ref: 63-69 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 100) Evaluate: ln 250 Answer: ln 250 = 5.521460918 Diff: 2 Type: SA Page Ref: 64-70 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 101) Evaluate: ln 60 Answer: ln 60 = 4.094344562 Diff: 2 Type: SA Page Ref: 63-69 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 102) Evaluate: ln[400(1.177)] Answer: ln[400(1.177)] = ln 400 + ln 1.177

= ln 400 + 7(ln 1.17) = 5.9914645 + 7(.1570038) = 5.9914645 + 1.0990262 = 7.090491 Diff: 2 Type: SA Page Ref: 63-69 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms.

103) ln Answer: ln = ln = ln = ln = ln 8.270120604 = 2.112649092 Diff: 2 Type: SA Page Ref: 63-69 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 104) Solve: 9x = 63 Answer: x =

Diff: 1 Type: SA Page Ref: 72-77 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 105) Solve: 2x = 40 Answer: x =

= 20

Diff: 3 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 106) Solve: -5x = 35 Answer: x =

= -7

Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 107) Solve: -

x = 64

Answer: x = 64

= -288

Diff: 2 Type: SA Page Ref: 72-77 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division.

108) Solve: 0.04x = 37 Answer: x =

= 925

Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 109) Solve:

= 0.24

Answer: x = 0.24(4) = 0.96 Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 110) Solve:

= 0.3

Answer: x = 0.3(8) = 2.4 Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 111) Solve: -

x = 15

Answer: x = 15 ∗ -8, x = -120 Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 112) Solve: -

x = -49

Answer: x = -49 ∗

= 36.75

Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 113) Solve: -2x = 8 - 18x Answer: 16x = 8, x = 2 Diff: 1 Type: SA Page Ref: 72-77 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 114) Solve: 3x = 9 + 12x Answer: -9x = 9, x = -1 Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division.

115) Solve: 2x + 17 = 7x - 15 Answer: 17 + 15 = 7x - 2x, 32 = 5x, 6.4 = x Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 116) Solve: 2x + 17 = 8x - 3 Answer: 17 + 3 = 8x - 2x, 20 = 6x, 3.3333 = x Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 117) Solve: x - 0.23x = 2105 Answer: .77x = 2105, x =

= 2733.766

Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 118) Solve: x + 0.307x = 640.20 Answer: 1.307x = 640.20, x =

= 489.824

Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 119) Solve: 51 - 14x = -34 - x Answer: 51 + 34 = -x + 14x, 85 = 13x, 6.538462 = x LS: 51 - 14(6.538462) = 51 - 91.53846 = -40.53846 RS: -34 - 6.53846 = -40.53846 Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division.

120) Solve: 4x - 8 - 19x = 210 + 7x - 4 Answer: 4x -8 - 19x = 210 + 7x - 4 -15x - 8 = 206 + 7x -8 - 206 = 7x + 15x -214 = 22x -9.7272727 = x LS: = 4(-9.7272727) - 8 - 19(-9.7272727) = -38.909091 - 8 + 184.81818 = 137.90909 RS: = 210 + 7(-9.7272727) - 4 = 206 - 68.090909 = 137.90909 Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 121) Solve: Answer:

x= x=

xx-

520 + 1 = 55x + 12x, 521 = 67x, 7.7761194 = x Diff: 2 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 122) Solve: 10(3 - x) + 2(x - 2) = 6(2x - 2) Answer: 10(3 - x) + 2(x - 2) = 6(2x - 2) 30 - 10x + 2x - 4 = 12x - 12 -10x + 2x - 12x = -12 - 30 + 4 -20x = -38 x = 1.9 Diff: 2 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 123) Solve: 4(2x - 5) + 3 = 3(x - 4) Answer: 4(2x - 5) + 3 = 3(x - 4) 8x - 20 + 3 = 3x - 12 8x - 3x = -12 + 20 - 3 5x = 5 x=1 Diff: 2 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement.

124) Solve: Answer: (24)

- (24)

=1 =1 = (24)1

(6)3 - 3x - (4)(x + 2) = 24 18 - 3x - 4x - 8 = 24 -7x = 14 x = -2 Diff: 2 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 125) Solve: 5(2x - 4) - 3(1 - 3x) = -64 Answer: 5(2x - 4) -3(1 - 3x) = -64 10x - 20 - 3 + 9x = -64 19x = -41 x = 2.1578947 LS: 5[2(-2.1578947) - 4] - 3[1 - 3(-2.1578947)] = -64 5[-4.3157895 - 4] - 3[1 - (-6.4736842)] = -64 5[-8.3157895] - 3[7.4736842] = -64 -41.578948 - 22.421053 = -64 -64 = -64 RS: = -64 Diff: 2 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 126) Solve: 17 - 4(2x - 7) = 15x - 3(2x - 3) Answer: 17 - 4(2x - 7) = 15x - 3(2x - 3) 17 - 8x + 28 = 15x - 6x + 9 45 - 8x = 9x + 9 36 = 17x 2.117647 = x LS: 17 - 4[2(2.117647) - 7] 17 - 4(4.2353941 - 7) 17 - 4(-2.7647059) 17 + 11.058824 28.058824 RS: 15(2.117647) - 3[2(2.117647) - 3] 31.764706 - 3[4.2352941 - 3] 31.764706 - 3[1.2352941] 31.764706 - 3.7058823 = 28.058824 Diff: 2 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement.

127) Solve: y + Answer: y +

y = 308 y = 308

y = 308 y = 308 × y = 196 Diff: 2 Type: SA Page Ref: 77-81 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 128) Solve: 2 Answer: 2 -

x= x=

x+ x+

36 - 9x = 12x + 50 -14 = 21x -

Diff: 2 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 129) Solve: Answer:

(4 - 3x) + (4 - 3x) +

= =

xx-

(2x - 3) (2x - 3)

112(4 - 3x) + 23 = 28x - 15(2x - 3) 448 - 336x + 23 = 28x - 30x + 45 471 - 336x = -2x + 45 426 = 334x 1.2754491 = x Diff: 3 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement.

130) Solve: Answer:

(5x - 2) (5x - 2) -

(16x - 3) = (16x - 3) =

+ 3x + 3x

80(5x - 2) - 36(16x - 3) = 17 + 180x 400x - 160 - 576x + 108 = 17 + 180x -176x - 52 = 17 + 180x -69 = 356x -.1938202 = x Diff: 3 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 131) Solve: I = Prt for r Answer: I = Prt, r = Diff: 1 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 132) Solve: I = Prt for t Answer: I = Prt, t = Diff: 1 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 133) Solve: Answer:

for V

v(R + r) = Vr V= Diff: 2 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 134) Solve: Q = Answer: Q =

for p , 4Q = p - q, 4Q + q = p

Diff: 2 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement.

135) Solve: P = S(1 + i)-n for i Answer: P = S(1 + i)-n, = (1 + i)-n, = 1 + i, i=

= 1 + i,

-1

Diff: 3 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 136) Solve: S = P(1 + rt) for t Answer: S = P(1 + rt), S = P + Prt, S - P = Prt t= Diff: 2 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 137) Solve: D = Answer: D =

for F ,

= E + F, F =

-E

Diff: 3 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 138) Conor had to pay income taxes of $3 440.00 plus 22% of the amount by which his taxable income exceeded $36 000.00. If his tax bill was $3 684.00, calculate his taxable income. Answer: Let the taxable income (in dollars) be x. Then x - 36 000 is the amount that his income is greater than $36 000. 3440 + 0.22(x - 36 000) = 3684 3440 + 0.22x - 7920 = 3684 0.22x = 8164 x = $37 109.09 Diff: 2 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations.

139) Jane invests part of her $3 500 savings into a savings account at 4% and part into a GIC at 9.5% simple interest. If she gets $260 in interest from the two investments, calculate how much she invested at each rate. Answer: Let the amount invested in the savings account (at 4%) be $x. Then the amount invested at 9.5% is (3500 - x). 0.04x + 0.095(3500 - x) = 260 0.04x + 332.50 - 0.095x = 260 -0.055x = -72.5 x = 1318. Jane invested about $1318.18 at 4% (in her savings account). 3500 - 1318.18 = $2182.82. Jane invested about $2182.82 at 9.5% (in her GIC). Diff: 3 Type: SA Page Ref: 82-87 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 140) Bow Valley Electronics sold a mini stereo set during a sale for $776. Determine the regular selling price of the set if the price of the set had been reduced by 1/4 of the original regular selling price. Answer: Let the regular selling price be $x. Sale price = $ ∴

x = 776

4x - x = 3104 3x = 3104 x = 1034.67 The regular selling price was $1034.67. Diff: 2 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 141) After an increase of 1/9 of his current hourly wage, Pierre will receive a new hourly wage of $12.35. How much is his hourly wage before the increase? Answer: Let the original hourly wage be $x. New hourly wage = $ ∴

x = 12.35

9x + x = 111.15 10x = 111.15 x = 11.12 The hourly wage before increase was $11.12. Diff: 2 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations.

142) After a reduction of of the selling price, a VCR was sold for $470.00. Determine the regular selling price. Answer: Let the regular price be y. Then reduction in price is y. y - y = 470 y = 470 y = $506.15 The regular selling price is $506.15. Diff: 2 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 143) A rubber tube 120 cm long is cut into two pieces so that the longer piece is 30 cm longer than twice the length of the shorter piece. What is the length of the longer piece? Answer: Let the shorter piece be x cm. Length of the longer piece = (2x + 30) cm. Total length = (x + 2x + 30) cm. ∴ x + 2x + 30 = 120 3x = 90 x = 30 The longer piece is 2(30) cm + 15 cm = 75 cm. Diff: 2 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 144) A clothing store sells fancy hats at a gross margin of $3.50 each and ordinary hats at a gross margin of $5.00 each. During July, 110 hats were sold for a total gross margin of $460.00. How many fancy hats were sold? Answer: Let the number of fancy hats be y. Then the number of ordinary hats = 110 - y. The total gross margin on fancy hats = $5(y). The total gross margin on ordinary hats = $3.5(110 - y) 5y + 3.5(110 - y) = 460 5y + 385 - 3.5y = 460 1.5y = 75 y = 50 The number of fancy hats sold = 50. Diff: 2 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations.

145) An electronics company has been producing 1705 CD Players a day working two shifts. The second shift has produced 95 CD players fewer than four-fifths of the number of CD players produced by the first shift. Determine the number of CD players produced by the second shift. Answer: Let the number of CD players produced by the first shift be x. Number of CD players produced by the second shift = Total production = x + ∴

x - 95.

x - 95 = 1705

x = 1800 x = 1000 Production by the second shift is

x(1000) - 95 = 705.

Diff: 2 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 146) A 3D printer requires 3 hours to make a unit of Part A and eight hours to make a unit of Part B. Last month the printer operated for 232 hours producing a total of 74 units. How many units of Part B were produced? Answer: Let the number of units of Part A produced last month be x. Then the number of units of Part B produced last month is 74 - x, the number of hours taken to produce Part A was 3x, and the number of hours taken to produce Part B was 8(74 - x). ∴ 3x + 8(74 - x) = 232 3x + 592 - 8x = 232 -5x = -360 x = 72 Part B: 82 - 72 = 10. ∴ 10 units of Part B were produced last month. Diff: 2 Type: SA Page Ref: 82-87 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations.

147) The local amateur soccer club spent $1640 on tickets to a professional hockey game. If the club bought 2.5 times as many eight-dollar tickets than the number of twelve-dollar tickets and four fewer fifteen-dollar tickets than 7/10 the number of twelve dollar tickets, how many of each type of ticket did the club buy? Answer: Let the number of $12 tickets be x. Number of $8 tickets = 2.5x Number of $15 tickets =

x-4

Value of the $12 tickets = $12x Value of the $8 tickets = $8(2.5x) Value of the $15 tickets = $15(

x - 4)

12x + 8(2.5x) + 15 (7/10x - 4) = $1640 12x + 20x + 10.5x - 60 = 1640 42.5x = 1700 x = 40 Sales were 40 $12 tickets, 100 $8 tickets, and 24 $15 tickets. Diff: 3 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. ∴

148) Evaluate s: s = ut + at2 for u = 10, a = 24, t = 7 Answer: s = 10(7) + (24)(7)2

= 70 + 1896 = 1966 Diff: 2 Type: SA Page Ref: 46-53 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 149) Evaluate s: s = ut + Answer: s = 20(5) +

at2 for u = 20, a = 10, t = 5

(10)(5)2

= 100 + 125 = 225 Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 150) Evaluate z: z = 5x2 - 5xy - 3y2 for x = -6, y = + 5 Answer: z = 5(-6)2 - 5(-6)(5) - 3(5)2

= 5(36) + 150 - 3(25) = 180 + 150 - 75 = 255 Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution.

151) Evaluate c: c = 8(7a - 4b) - 4(5a + 3b) for a = Answer: c = 8 =8

,b=-

-4 -4

= 12 + 55 = 67

Diff: 3 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 152) Evaluate K: K =

for N = 112, C = 250, P = 2450, n = 24

Answer: For N = 12, C = 400, P = 2000, n = 24 K=

= .914286

Diff: 3 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 153) Evaluate T: T =

for I = 224, P = 6700, r = 0.11

Answer: For I = 324, P = 5400, r = 0.15 T=

= .3039349

Diff: 3 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 154) Evaluate P: P = S(1 - dt) for S = 1886, d = 0.15, t = Answer: P = 1886 = 1886(1 - .15 × .6821918) = 1886(1 - .1023288) = 1886(.8976712) = 1693.01 Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution.

155) Evaluate P: P =

for S = 1665, r = 0.14, t =

Answer: P =

= $1497.30

Diff: 3 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 156) Evaluate P: P =

for S = 1000, r = 0.1, t =

Answer: P =

= $952.38

Diff: 3 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 157) Compute: Answer: = 1.016315 Diff: 2 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 158) Compute: Answer:

= 14.93169

= 27.36

Diff: 3 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents.

160) Compute: ln 1.257 Answer: ln 1.257 = .228728 Diff: 1 Type: SA Page Ref: 63-69 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 161) Compute: ln[2.50e-0.2] Answer: ln(2.50e -0.2)

= ln 2.50 + ln e-0.2 = ln 2.50 - 0.2 ln e = 0.7162907319 Diff: 3 Type: SA Page Ref: 64-70 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 162) Solve: x - 0.26x = 8.96 Answer: x - .26x = 8.96, .74x = 8.96, x = 12.10811 Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 163) Solve: x - 0.75x = 9.00 Answer: x - .75x = 9.00, .25x = 9.00, x = 36.00 Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 164) Solve: 0.4x - 4 = 6 - 0.8x Answer: .4x - 4 = 6 - .8x, 1.2x = 10, x = 8.33333333 Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 165) Solve: (2 + 8x) - (12x — 3) = 73 Answer: (2 + 8x) - (12x — 3) = 73 2 + 8x - 12x + 3 = 73 5 - 4x = 73 -4x = 68 x = -17 Diff: 2 Type: SA Page Ref: 77-81 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement.

166) Solve: 5(8x - 2) - 5(3x + 5) = 36 Answer: 5(8x - 2) -5(3x + 5) = 36 40x - 10 -15x - 25 = 36 25x - 35 = 36 25x = 71 x = 2.84 Diff: 2 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 167) Solve: x + Answer: x +

x+ x+

+x+ +x+

x + 1 = 256 x + 1 = 256

x + .7x + 1.8x + 1.5 = 256 3.5x = 254.5 x = 72.71429 Diff: 3 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 168) After reducing the regular selling price by 1/7, Moon Electronics sold a TV set for $294. What was the regular selling price? Answer: Let the regular selling price be $x. Reduction in price + $ x x-

x = 294 x = 294

x = $343.00 Diff: 2 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 169) After reducing the regular selling price by 1/8, Sepaba Inc. sold a Stereo set for $300. What was the regular selling price? Answer: Let the regular selling price be $x. Reduction in price + $ x x-

x = 300 x = 300

x = $342.86 Diff: 2 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations.

170) The zinc department of a factory occupies 500 square metres more than 2 times the floor space occupied by the copper department. The total floor space is 9500 square metres. Determine the floor space occupied by the cooper department. Answer: Let the floor space occupied by copper be x. Floor space occupied by zinc = 2x + 500 Total floor space = x + 2x + 500 ∴ x + 2x + 500 = 9500 3x = 9000 x = 3000 The floor space occupied by copper is 3000 square metres. Diff: 2 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 171) A company employs 204 employees. There are three shifts. There are three times as many on the first shift as on the second shift, and four more on the third shift than on the second shift. Determine how many were on each shift. Answer: Let x be the number on the second shift. Then 3x is the number on the first shift. And x + 4 is the number on the third shift. x + 3x + (x + 4) = 204 5x = 200 x = 40 on the second shift 3x = 120 on the first shift x + 4 = 44 on the third shift Diff: 2 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 172) A machine requires 4 hours to make a unit of Product A and 7 hours to make a unit of Product B. The machine operated for 810 hours producing a total of 150 units. How many units of Product B were produced? Answer: Let the number of units of Product A be x. Number of units of Product B = 150 - x. Number of hours for Product A = 4x. Number of hours for Product B = 7(150 - x). ∴ 4x + 7(150 - x) = 810 4x + 1050 - 7x = 810 -3x = -240 x = 80 The number of units if Product B is 150 - 80 = 70. Diff: 2 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations.

173) Simplify: (5x - 4)(3x + 1) A) 15x2 - 7x + 4 B) 15x2 - 7x - 4 C) 15x2 + 7x - 4

D) -15x2 - 7x + 4 E) -10x2 - 7x + 4

Answer: B Diff: 1 Type: MC Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 174) Simplify: A) 20a4b4c5 B) 5a4b4c5 C) 5a3b2c2 D) 20a3b2c2

E) 55a4b4c5 Answer: B Diff: 1 Type: MC Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 175) Simplify: (3)2(3)5 A) 30 B) 90 C) 59 049 D) 2187 E) 120 Answer: D Diff: 1 Type: MC Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero.

176) Simplify: A) 16777216 B) 17666216 C) 12222617 D) 17222167 E) 17333167 Answer: A Diff: 1 Type: MC Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 177) Compute the value of (four decimal places): A) .0045 B) .5449 C) .4459 D) .0445 E) 0.044946 Answer: E Diff: 1 Type: MC Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 178) Simplify: (m2)6 A) m12 B) m-4 C) m8 D) m3 E) m2

Answer: A Diff: 1 Type: MC Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero.

179) Simplify: [(33) - 62]3 A) 279 B) -279 C) 729 D) -729 E) 972 Answer: D Diff: 1 Type: MC Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 180) Calculate the following: ln A) 7.895 B) -8.795 C) 7.588 D) -7.895 E) -7.958 Answer: D Diff: 2 Type: MC Page Ref: 64-70 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 181) Solve the following equation: 5x - 4 + 9 = -3x - 2 - 13 A) 2.5 B) -2.5 C) 20 D) -20 E) -25 Answer: B Diff: 1 Type: MC Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 182) Solve the following and check your solutions: 14x - 8(9 + 2x) = 322 A) 197 B) 125 C) -98.5 D) -197 E) -125 Answer: D Diff: 1 Type: MC Page Ref: 77-81 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement.

183) You have a lotto ticket with three numbers between zero and 49. The total of the three numbers is 93. One number is twice as large plus two as the lowest number. The second number is

the size of the

smaller number. What are the values of each of the numbers? A) 1-44, 2-28, 3-21 B) 1-41, 2-31, 3-21 C) 1-44, 2-25, 3-24 D) 1-44, 2-20, 3-15 E) 1-44, 2-15, 3-20 Answer: A Diff: 3 Type: MC Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 184) Simplify: 11.57843/7 A) 3.857 B) 4.857 C) 2.857 D) 2.587 E) 4.785 Answer: C Diff: 1 Type: MC Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 185) You have 4 colors of paper slips in a small bag - white, pink, and brown. There are 3 times minus 4 as many white papers as there are pink papers. There is 7/8 as many plus 2 brown papers as there are pink. There are a total of 622 papers in the bag. How many of each colour paper are there? A) 130 pink, 110 brown, 382 white B) 127 pink, 113 brown, 377 white C) 128 pink, 112 brown, 382 white D) 128 pink, 114 brown, 380 white E) 311 pink, 108 brown, 203 white Answer: D Diff: 2 Type: MC Page Ref: 82-87 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations.

186) Following equation is used in carbon dating of the artefacts: N = N0 × where, N = Decay rate of the artefact, expressed as disintegrations per minute (dpm) N0 = Decay rate of natural carbon = 14 dpm τ = Mean life time =

= 8267 years

t = Age of the artefact Calculate the age of the artefact, if the decay rate of the artefact is measured as 4 dpm. Answer: N = N0 ×

⇒ ⇒ ln ⇒

ln(e)

t = -τ × ln

because ln(e) = 1

Plugging the values in the above equation, we get: t = -8267 × ln ⇒ t = -8267 × -1.253 ⇒ t = 10357 Age of the artefact is 10,357 years Diff: 3 Type: SA Page Ref: 63-69 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms.

187) Following equation is used in carbon dating of the artefacts: N = N0 × where, N = Decay rate of the artefact, expressed as disintegrations per minute (dpm) N0 = Decay rate of natural carbon = 14 dpm τ = Mean life time =

= 8267 years = 8267 years

t = Age of the artefact What is the expected decay rate of the artefact, if it is known to be from Canadian New France era, 350 years ago? Answer: N = N0 × Plugging the values in the above equation, we get N = 14 ×

= 14 × 0.959 = 13.4 dpm

Diff: 2 Type: SA Page Ref: 63-69 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 188) Find b, if = 0.25 A) -0.0625 B) -1.386 C) -0.347 D) 0.347 E) -0.229 Answer: D Diff: 2 Type: MC Page Ref: 63-69 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 189) Calculate ln A) -1.16 B) 0.06 C) 6.458 D) 4.138 E) 198.84 Answer: D Diff: 2 Type: MC Page Ref: 63-69 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms.

190) Calculate ln A) 9.75 B) 17162 C) 2653.29 D) 10.52 E) 11.23 Answer: A Diff: 2 Type: MC Page Ref: 63-69 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 191) In the following equation: PVn = PMT calculate n, if PVn = 75,000; PMT = 3500; i = 0.028

Answer: Plugging the values in the above equation: 75000 = 3500 ⇒ 0.6 = 1 ⇒ = 0.4 ⇒ -n ln(1.028) = ln (0.4) ⇒ -n(0.027615) = -0.916291 ⇒ n = 33.180706 Diff: 2 Type: SA Page Ref: 63-69 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 192) Evaluate ln A) 5505 B) 8.613 C) 112.35 D) 104.66 E) 8.786 Answer: B Diff: 2 Type: MC Page Ref: 63-69 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms.

193) Simplify: A) 3 B) 0.333 C) 531,441 D) 0.00000188 E) 1 Answer: E Diff: 1 Type: MC Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 194) Simplify: A) 8.000015625 B) 8 C) 8.304 D) 11.39 E) 91.125 Answer: C Diff: 1 Type: MC Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 195) Evaluate: A) 32768 B) -32768 C) -40 D) 256 E) -30 Answer: B Diff: 1 Type: MC Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero.

196) Simplify A)

+ n3

B) C) D) E) Answer: D Diff: 1 Type: MC Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 197) Evaluate A) 341 B) 13 C) 1.24 D) 27 E) Answer: A Diff: 1 Type: MC Page Ref: 61-64 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 198) Evaluate 5000 A) 0.08 B) 5000 C) 8425.29 D) 5281.12 E) 25000 Answer: C Diff: 1 Type: MC Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents.

199) Evaluate: 1862.14 Answer: 1862.14

= 1862.14

= 1862.14 = 1862.14[11.07931]-1 = 168.07 Diff: 2 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 200) Evaluate: $5000 Answer: $5000

= $100,000

Diff: 1 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents.

201) Evaluate: Answer:

= = 199 Diff: 2 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents.

202) Evaluate:

Answer:

= = 979.84 Diff: 2 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 203) Simplify: (a + b + c)(a2 + b2 + c2 - ab - bc - ca) Answer: (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

= a(a2 + b2 + c2 - ab - bc - ca) + b(a2 + b2 + c2 - ab - bc - ca) + c(a2 + b2 + c2 - ab - bc - ca) = a3 + ab2 + ac2 - a2b - abc - ca2 + ba2 + b3 + bc2 - ab2 - b2c - abc + ca2 + b2c + c3 - abc - bc2 - ac2 = a3 + b3 + c3 - 3abc Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 204) Simplify: (a + b)3 Answer: (a + b)3 = (a + b)(a + b)2 = (a + b)(a2 + 2ab + b2) = a3 + 3a2b + 3ab2 + b3 Diff: 1 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 205) Simplify: 7x - (5x - 4y) - (6x + 8y -5x) A) x - 4y B) 11x - 12y C) x + 12y D) 3x + 12y E) 3x - 4y Answer: A Diff: 1 Type: MC Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution.

206) Simplify: x - 2y - [4x - 6y - {3x - z + 2(2x - 4y + z)}] Answer: x - 2y - [4x - 6y - {3x - z + 2(2x - 4y + z)}] = x - 2y - [4x - 6y - {3x - z + 4x - 8y + 2z}] = x - 2y - [4x - 6y - {7x + z - 8y}] = x - 2y - [4x - 6y - 7x - z + 8y] = x - 2y - [-3x + 2y - z] = x - 2y + 3x - 2y + z = 4x - 4y + z Diff: 2 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 207) Solve: 2x + 5 = 7 A) x = 1 B) x = 6 C) x = 2 D) x = 4 E) x = Answer: A Diff: 1 Type: MC Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 208) Solve: 8x - 11 = 5x + 4 A) x = B) x = C) x = 5 D) x = 15 E) x = 4 Answer: C Diff: 1 Type: MC Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 209) Solve: 0.5x - 0.75 + 9x = 5x + 1.5 A) x = 0.05 B) x = 0.155 C) x = -0.5 D) x = 0.5 E) x = 0.55 Answer: D Diff: 1 Type: MC Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division.

210) Solve: y = 192 + 0.04y A) y = 184.62 B) y = 192 C) y = 200 D) y = 320 E) y = 137.14 Answer: C Diff: 1 Type: MC Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 211) Solve and find the value of a, given 3a + 4(15-2a) = 20 A) a = B) a = 7 C) a = 5 D) a = -1 E) a = 8 Answer: E Diff: 2 Type: MC Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 212) Solve and find the value of y, given 3(1 - 2y) + 5y = 5 Answer: 3(1 - 2y) + 5y = 5 ⇒ -y + 3 = 5 ⇒ y = -2 Diff: 1 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 213) Solve:

Answer:

⇒

+ 2x

= $115.10

+ 2.11452x = $115.10

⇒ 3.06442x = $115.10 ⇒ x = $37.56 Diff: 2 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement.

214) Solve: x

+ 2x

= $2551.65

Answer: 1.04438x + 2x(1.02096)-1 = $2551.65 ⇒ 1.04438x + 1.95894x = $2551.65 ⇒ x = $849.61 Diff: 2 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 215) Solve: x(1.05)3 + $1000 +

Answer: 1.157625x + 0.710681x = $11791.38 - $1000 ⇒ x= ⇒ x = $5776.02 Diff: 1 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement.

216) Solve and find the value of x, given:

A) x = 20 B) x = -9 C) x = 9.02 D) -17.16 E) 4.33 Answer: C Diff: 2 Type: MC Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 217) Luxury furniture is promoting a no HST (13%) event for the fall sale. What is the actual selling price for the Sofa before tax shown on your receipt, if the ticket price of the sofa is $795.00? A) $795.00 B) $6,115.38 C) $103.35 D) $898.35 E) $703.54 Answer: E Diff: 1 Type: MC Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations.

218) Sobeys sells fruit trays consisting of pineapple and melons. The manager of the fruit department obtains pineapples at a wholesale price of $2.50 per kg, and melons at $1.85 per kg. He is required to produce 10 kg of the mixed fruit. What is the maximum weight of pineapple that he can put in the mix in order to have an effective wholesale cost no greater than $2.00 per kg? A) 2.31 kg B) 7.69 kg C) 4.56 kg D) 4 kg E) 5.41 kg Answer: A Diff: 2 Type: MC Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 219) Sobeys sells fruit trays consisting of pineapple and melons. The manager of the fruit department obtains pineapples at a wholesale price of $2.50 per kg, grapes at $2.15 per kg, strawberries at $1.50 per kg and melons at $1.85 per kg. He is required to produce 10 kg of the mixed fruit. He is required to add 2 kg of strawberries and 3 kg of grapes. What is the maximum weight of pineapple that he can put in the mix in order to have an effective wholesale cost no greater than $2.00 per kg? Answer: Cost of 2 kg of strawberries = $1.50 × 2 = $3.00 Cost of 3 kg of grapes = $2.15 × 3 = $6.45 Total cost of the mixed fruit = $2 × 10 = $20 Cost of pineapple and melons = $20 - ($3 + $6.45) = $10.55 ⇒ 2.5p + 1.85m = 10.55 Maximum weight of pineapple and melons = 10 - (2 + 3) = 5kg ∴ 2.5p + 1.85(5-p) = 10.55 ⇒ 0.65p + 9.25 = 10.55 ⇒ p=

∴ Maximum weight of pineapple = 2 kg Diff: 2 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 220) ABC financial has introduced a stock option incentive program. For a total of 301,375 options, each of the 3 executives receives twice as many options as the senior managers. Each senior manager receives twice the options as each of the middle managers and each of the middle managers receives 1.5 times the options as each of the employee. If there are 25 senior managers, 75 middle managers and 1000 employees, how many options will the each executive receive? A) 250 B) 375 C) 750 D) 1500 E) 4500 Answer: D Diff: 2 Type: MC Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations.

221) Sleep Inc. reduced the price of the Sealy mattress by 20% for a spring sale. What was the regular price of the mattress, if the sale price is $649.95? A) $129.99 B) $779.94 C) $3,249.75 D) $812.44 E) $649.95 Answer: D Diff: 1 Type: MC Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 222) Blue Mountain resort charges $39 for a single ticket for high ropes and $29 for a single ticket for low ropes. If a day's total revenue from the sale of 540 passes is $18,910, how many tickets were sold for high ropes? Answer: 39HR + 29(540 - HR) = 18910 ⇒

HR =

= 325

325 tickets were sold for high ropes Diff: 3 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 223) Age difference between my son and me is 30 years. Sum of our ages is 54. What is my age now? A) 12 B) 42 C) 38 D) 45 E) 15 Answer: B Diff: 1 Type: MC Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 224) Transpipe owns 47% ownership stake in BP. Cameo owns a 29% stake in the BP. Suppose Transpipe sells 62.5% of its stake for $47 Million. If Comeo uses that transaction as the basis for calculating the value of its own 29% position, what value will Cameo obtain? A) $18.1 Million B) $46.4 Million C) $16 Million D) $47 Million E) $29 Million Answer: B Diff: 2 Type: MC Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations.

225) Simplify: -(3a + 7ab - 5) - (12 + 6ab - 5a) Answer: = -3a - 7ab + 5 - 12 - 6ab + 5a = 2a - 13ab - 7 Diff: 4 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 226) Simplify: (x + 4)(2x2 - 9x + 3) Answer: 2x3 - 9x2 + 3x + 8x2 - 36x + 12 = 2x3 - x2 - 33x + 12

Diff: 4 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 227) Simplify: -r(4t - 8s) — 5s(12r + 2t) Answer: -4rt + 8rs — 60rs — 10st = -4rt — 52rs — 10st Diff: 3 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 228) Solve:

(6x2 + x - 1) -

(2x2 -12x + 4)

Answer:

[6(7)2 + 7 - 1] -

[2(7)2 -12(7) + 4]

(294 + 7 – 1) -

(98 - 84 + 4)

(300) -

for x = 7

(18)

= 91 Diff: 3 Type: SA Page Ref: 44-49 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 229) Simplify: 29 ÷ 23 × 24 A) 22 B) 28 C) 27

D) 210 Answer: D Diff: 2 Type: MC Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero.

230) Simplify: Answer:

÷ =

Diff: 3 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 231) Expand: (3xyz)3 Answer: 33x3y3z3 = 27x3y3z3

Diff: 3 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 232) Solve: 635241(1093)0 Answer: 635241(1093)0

= 635241 × 1 = 635241 Diff: 2 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 233) Simplify: Answer: = (x + 2)2/ x2 = (x2 + 4x + 4)/ x2

= 1 + (4/x) + (4/x2) Diff: 4 Type: SA Page Ref: 51-58 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 234) Calculate: Answer: 67 Diff: 2 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents.

235) Calculate: Answer: 6 Diff: 3 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 236) Calculate up to 4 decimals: Answer: 3.3065 Diff: 3 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 237) Calculate: (5489031744)1/6 Answer: 42 Diff: 4 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 238) Calculate with final answer in fractional form: 390625-1/8 Answer: Diff: 4 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 239) Calculate up to 4 decimals: 48268091/6 Answer: 13 Diff: 4 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 240) Express in logarithmic form: 8-3 = Answer: log8

= -3

Diff: 3 Type: SA Page Ref: 63-67 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms.

241) Express in logarithmic form: e4x = 20 Answer: 4x = ln 20 x = ln 20 Diff: 4 Type: SA Page Ref: 63-67 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 242) Express in exponential form: log3(243) = 5

Answer: 35 = 243 Diff: 2 Type: SA Page Ref: 63-67 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 243) Express in exponential form: log-4(4096) = 6

Answer: (-4)6 = 4096 Diff: 3 Type: SA Page Ref: 63-67 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 244) Calculate up to 4 decimal places: ln 16 Answer: 2.7726 Diff: 3 Type: SA Page Ref: 63-67 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 245) Calculate up to 4 decimal places: ln [400(

)]

Answer: 2.5903 Diff: 5 Type: SA Page Ref: 63-67 Topic: 2.4 Logarithms - Basic Aspects Objective: 2-4: Write exponential equations in logarithmic form and use an electronic calculator equipped with a natural logarithm function to determine the value of natural logarithms. 246) Solve for x: x + 14 = -7 Answer: x = -7 - 14 = -21 Diff: 2 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division.

247) Solve for x: -5x = 8 - 4x Answer: -4x + 5x = 8 x=8 Diff: 2 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 248) Solve for x: 2x - 1.5x = 410 Answer: 0.5x = 410 x = 820 Diff: 3 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 249) Solve for x: Answer: (-7) -

x=8

x = 8(-7)

x = -56 Diff: 3 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 250) Solve for x: -6x = 486 + 3x Answer: -9x = 486 x = -54 Diff: 3 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division. 251) Solve for x and check your answer: 8x + 14 = 12x + 22 Answer: 14 - 22 = 12x - 8x -8 = 4x -2 = x Check: 8(-2) + 14 = 12(-2) + 22 -16 + 14 = -24 + 22 -2 = -2 Diff: 4 Type: SA Page Ref: 71-75 Topic: 2.5 Solving Basic Equations Objective: 2-5: Solve basic equations using addition, subtraction, multiplication, and division.

252) Solve for y: -2y + 8(2y - 3) = -5(4 + y) - (-12y -10) Answer: -2y + 16y - 24 = -20 - 5y + 12y + 10 14y - 24 = 7y - 10 7y = 14 y=2 Diff: 2 Type: SA Page Ref: 77-81 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 253) Solve for x:

Answer: (8) x - (8) x = (8) 4(5x) - 2(3x) = 14 14x = 14 x=1 Diff: 3 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 254) Solve for x: x +

x = 22

Answer: (8)x + (8) x = (8)22 11x = 176 x = 16 Diff: 2 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 255) Solve for x (answer will be a decimal) Answer: (24) (x + 2) - (24)

(x + 2) -

(2x + 2) =

(2x + 2) = (24)

(4)(5)(x + 2) – 8(2x + 2) = 3(2) 20(x + 2) – (16x - 16) = 6 4x = -50 x = 12.5 Diff: 4 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement.

256) Rearrange for r: A = πr2 Answer: = r2 =r Diff: 3 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 257) Rearrange for b: a2 + b2 = c2 Answer: b2 = c2 + a2 b= Diff: 3 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement.

258) Rearrange for c: x = Answer: 2ax = -b ± 2ax + b = (2ax + b)2 = b2 - 4ac

4a2x2 + 4abx + b2 = b2 - 4ac 4a2x2 + 4abx + b2 - b2 = -4ac -4a2x2 - 4abx = 4ac -

-ax2 - bx = c Diff: 5 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement.

259) A smartphone was being sold for $660 while on sale. If the price dropped by 20% for the sale, what was the original price of the smartphone? Answer: Let x represent the original smartphone's price Since a 20% drop in price equates to x-

, (x -

x) can represent the sale price

x = 660

x = 660 x=

(660)

x = 825 Diff: 3 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 260) Carlos went to the mall to buy toys for his children and nephews with a total of $1300. After he was done shopping, he had $400 left in his account. Carlos's expenditure on doll sets was $100 less than four times the amount he spent on toy cars. How much money did Carlos spend on doll sets? Answer: Let x represent Carlos's expenditure on toy cars. Consequently, the expenditure on doll sets can be represented with the expression 4x - 100. Since Carlos started with $1300 and was left with $400, we can assume he spent $900 at the mall. From this info, we can create the equation: x + 4x - 100 = 900 5x = 1000 x = 200 Diff: 4 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 261) Usually, Bob's Burgers sells x burgers for p dollars each. But today, the price was lowered by 22% and as a result, 60% more burgers were sold. Which of the below represents the revenue for today's burger sales, in dollars? A) 0.6x - 0.22p B) 1.15xp C) 1.248xp D) 1.455xp Answer: C Diff: 2 Type: MC Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations.

262) Sony has been producing 1710 VR headsets a day through two work shifts. The second shift has produced 90 fewer headsets than seven-fifths of the number of headsets produced by the first shift. What is the number of VR headsets produced by the second shift in a day? Answer: Let x represent the number of VR headsets produced by the first shift. Consequently, the number of VR headsets made by the second shift can be represented through the expression

x - 90.

Since 1675 headsets are produced daily, the equation formed is: x+

x - 90 = 1710 x = 1620

(1620)

x = 675 Diff: 4 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 263) Old Navy had $940 worth of orders for shirts. The prices for the shirts were categorized by style: $10 for tank tops, $14 for graphic tees, and $22 for polos. The number of polos ordered was one more than half the number of graphic tees, and the number of graphic tees ordered was eight less than twice the number of tank tops. Determine the number of each type of shirt ordered. Answer: Let x represent the number of tank tops, let 2x - 8 represent the number of graphic tees sold, and let

(2x - 8) + 1 represent the number of polos sold. Taking the price and the value of the combined

orders into consideration, the equation is as follows: 10x + 14(2x - 8) + 22[ (2x - 8)] = 940 10x + 28x - 112 + 22[x - 4] = 940 38x + 22x - 88 = 1052 60x = 1140 x = 19 2(19) - 8 = 30 Graphic Tees 2(19) - 8] + 1 = 16 Polos Diff: 5 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations.

264) Zain began with three times more money than Max. After Zain gave Max $18, Zain still had $6 more than Max. How much money do they have altogether? Answer: Let x represent the amount of money Brock has to start (3x represents Zain) 2x - 18 = x + 6 + 18 2x - 18 = x + 24 2x - x = 42 x = 42 3x = 3(42) = 126 Diff: 3 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 265) The rapper Drake hosts a concert at the Air Canada Center, where the ticket price is $240. However, the price of the individual ticket includes the 13% HST. What is the original price of the ticket, excluding the HST? Answer: Let x represent the price of a concert ticket, excluding HST 1.13x = 240 x= x = 212.39 Diff: 2 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 266) At the beginning of the Toronto FC game, spectators left, leaving only

of the stadium's seats were full. At halftime, 1000

of the seats full. In total, how many seats are in the stadium?

Answer: Let x represent the total number of seats in the stadium x - 1000 = xx-

x = 1000 x = 1000

x = 1000 x = 21 000 Diff: 3 Type: SA Page Ref: 81-84 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations.

267) Calculate the answer in fractional form: 625-1/2 Answer:

Diff: 2 Type: SA Page Ref: 59-62 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents. 268) Shannon's Bookstore is having a "Save the Tax" (no HST 13%) event for the fall sale. What is the final sale price of a book (rounded to the nearest cent), if, including HST, the book would be sold for $17.50? A) $15.49 B) $14.59 C) $19.78 D) $15.48 E) $2.28 Answer: A Diff: 1 Type: MC Page Ref: 82-87 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 269) Cedar Park Resort charges $13 for a one adult general admission ticket (excluding large waterslides), and $20 for one waterslides admission ticket. If the total revenue for one day in which 354 tickets were sold is $4728, how many waterslides admission tickets were sold? Answer: Let w represent the number of waterslide admission tickets that were sold. 20w + 13(354 - w) = 4728 20w + 4602 - 13w = 4728 7w = 126 w = 18 18 large waterslides admission tickets Diff: 2 Type: SA Page Ref: 82-87 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations. 270) Simplify: -8(4a) Answer: -32a Diff: 1 Type: SA Page Ref: 46-53 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 271) Simplify: -10(8a - b) + 3(-2a + 11b) Answer: -80a + 10b - 6a + 33b = -86a + 43b Diff: 2 Type: SA Page Ref: 46-53 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution.

272) Evaluate y: y = (1/8)(4x2 - 2x - 5) + Answer: = (1/8)[4(-2)2 - 2(-2) + 12] + = (1/8)(16 + 4 + 12) + = (1/8)(32) +

(6 + 5x + 6x2) for x = -2

[6 + 5(-2) + 6(-2)2]

(6 - 10 + 24)

(20)

=4+5 =9 Diff: 2 Type: SA Page Ref: 46-53 Topic: 2.1 Simplification of Algebraic Expressions Objective: 2-1: Simplify algebraic expressions using fundamental operations and substitution. 273) Evaluate: (-4)-3 Answer: -1/64 Diff: 2 Type: SA Page Ref: 53-61 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 274) Simplify: Answer: x11 / x6 = x5 Diff: 2 Type: SA Page Ref: 53-61 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 275) Simplify: 2.0079320 ÷ 2.0079307 Answer: 2.0079320 - 307 = 2.007913 = 8622.77465

Diff: 2 Type: SA Page Ref: 53-61 Topic: 2.2 Integral Exponents Objective: 2-2: Simplify and evaluate powers with positive exponents, negative exponents, and exponent zero. 276) Compute the value of Answer:

= 222.2101776

Diff: 3 Type: SA Page Ref: 61-64 Topic: 2.3 Fractional Exponents Objective: 2-3: Use an electronic calculator to compute the numerical value of arithmetic expressions involving fractional exponents.

277) Express in exponential form: log8

= -3

= -3 → 8-3 =

Answer: log8

= =

+2 +2

) = 6( ) + 6(2)

18x + 4x = x + 12 18x + 4x - x = 12 x = 0.57 Diff: 2 Type: SA Page Ref: 76-79 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement. 279) Solve: Answer:

for a

c(a - b + c) = ad ac - bc + c2 = ad c2 - bc = ad - ac c2 - bc = a(d - c) a= Diff: 3 Type: SA Page Ref: 77-81 Topic: 2.6 Solving Equations Involving Algebraic Simplification Objective: 2-6: Solve equations involving algebraic simplification and formula rearrangement.

280) After a reducing the price of a TV by

of the original selling price, the TV was sold for $810.00.

Determine the original selling price. Answer: Let the regular price be x. Then reduction in price is x-

x = 810 x = 810 x = 810

x = 895.26 The original selling price was $895.26 Diff: 2 Type: SA Page Ref: 82-87 Topic: 2.7 Solving Word Problems Objective: 2-7: Solve word problems by creating and solving equations.

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 3 Ratio, Proportion, and Percent 1) Set up a ratio to represent the value following: 8 quarters to 7 dimes to 13 nickels Answer: 8 quarters = 200 cents. 7 dimes = 70 cents. 13 nickels = 65 cents. 200 : 70 : 65 40 : 14 : 13w Diff: 1 Type: SA Page Ref: 96-101 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 2) Set up a ratio to represent the following: one-half of an hour to 18 minutes Answer: 30 : 18 5:3 Diff: 1 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 3) At Upscale Corporation, direct selling expense amounted to $3500 while sales volume was $97 500 for last month. What is the ratio of direct selling expense to sales volume? Answer:

Diff: 1 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 4) The cost of a unit is made up of $120.00 direct material cost, $80.00 direct labour cost, and $20.00 overhead. What is the ratio that exists between the three elements of cost? Answer: Direct material : direct labour : overhead = $120.00 $80.00 $20.00 = 120 : 80 : 20 =6:4:1 Diff: 1 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 5) Departments A, B, and C occupy floor space of 310, 120 and 475 square meters respectively. If the total rental for the space is $38 225.00 per month, how much rent should be paid by Department B? Answer: Price per square meter =

= $42.237569

Amount paid by B = 120 × price per square meter = 120 ∗ 42.237569 = $5068.51 Diff: 2 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

6) The cost of operating the Marketing Department is allocated to four departments based on the floor space each occupies, in m2. Department A occupies 1400 m2, Department B occupies 820 m2, Department C occupies 560 m2, and Department D occupies 390 m2. If the cost of operating for March was $19 000, how much of the cost should each department of the Marketing Department take up? Answer: Department A : Department B : Department C : Department D = 1400 : 820 : 560 : 390 Cost per square meter =

= 5.993690852

Allocation: Department A: 1400(5.993690852 ) = $8391.17 Department B: 820(5.993690852 ) = $4914.83 Department C: 560(5.993690852 ) = $3356.47 Department D: 390(5.993690852 ) = $2337.54 Diff: 2 Type: SA Page Ref: 96-101 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 7) Manufacturing overhead is charged to three operating divisions on the basis of capital investment in the three divisions. If the investment is $10.8 million in the Eastern Division, $8.4 million in the Southern Division, and $17.4 million in the Far Eastern Division, how should manufacturing overhead of $888 000 be allocated to the three divisions? Answer: Eastern : Southern : Far Eastern = $10.8 : $8.4 : $17.4 = 108 : 84 : 174 Let the common part be x ∴ 10.8x + 8.4x + 17.4x = 888,000 x = 24262.295 Allocation: Eastern Division: 10.8 ∗ 24262.295 = 262032 Southern Division: 8.4 ∗ 24262.295 = 203803 Far Eastern: 17.4 ∗ 24262.295 = 422164 Diff: 2 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

8) Carol, Clara and Nancy decide to buy LOTTO 649 tickets, agreeing to share any winnings in a ratio of :

, which is in the same ratio as they spent on tickets. If they end up winning $1147.00, calculate

how much each should receive from the winnings. Answer: Convert into fractions with the same denominators :

the ratio is 10 : 15 : 12 total number of parts = 10 + 15 + 12 = 37 the value of each part is 1147 ÷ 37 = 31 Carol's share = 31 × 10 = $310.00 Clara's share = 31 × 15 = $465.00 Nancy's share = 31 × 12 = $372.00 Diff: 2 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 9) Sean, Paul and Wallis decide to buy a property, agreeing to share profit in a ratio of

, which is

in the same ratio as they spent on the property. If they end up making $20 000.00, calculate how much each should receive from the sale of the property. Answer: Convert into fractions with the same denominators. :

the ratio is 20 : 15 : 12 total number of parts = 20 + 15 + 12 = 47 the value of each part is 20 000 ÷ 47 = 425.53 Sean's share = 425.53 × 20 = $8510.64 Paul's share = 425.53 × 15 = $6382.98 Wallis's share = 425.53 × 12 = $5106.38 Diff: 2 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

10) Lorraine, Estelle, and Frances own a business jointly and share profits and losses in the same proportion as their investments. How much of a profit of $14 500 will each receive if their investments are $14 000, $16 000, and $15 000 respectively? Answer: Total number of parts = 14000 + 16000 + 15000 = 45000 Value of each part = 14 500/45000 = .3222222 Distribution of profit: To Lorraine: 4 000 ∗ 0.3222222 = 4511.11 To Estelle: 6 000 ∗ 0.3222222 = 5155.56 To Frances: 15 000 ∗ 0.3222222 = 4833.33 Diff: 2 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 11) Departments, A, B, and C occupy floor space of 60 m,90 m, and 300 m, respectively. If the total rental for the space is $125 200 per month, how much rent should Department C pay? Answer: Total number of square meters = 60 + 90 + 300 = 450 Price per square meter =

= 278.2222222

Amount paid by department C = 300 ∗ 278.2222222 = 83466.67 Diff: 2 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 12) Four beneficiaries are to divide an estate of $289 000 in the ratio

. How much should

each receive? Answer: Ratio:

= 16 : 6 : 21 : 11

Let the constant part be x ∴ 16x + 6x + 21x + 11x = 289000 54x = 289000 x = 5351.851852 Distribution of estate: First beneficiary: 16 ∗ 5351.85182 = $85629.63 Second beneficiary: 6 ∗ 5351.85182 = 32111.11 Third beneficiary: 21 ∗ 5351.85182 = 112388.90 Fourth beneficiary: 11 ∗ 5351.85182 = 58870.37 Diff: 3 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

13) Four beneficiaries are to divide an estate of $500 000 in the ratio

. How much should

each receive? Answer: Ratio:

=8:6:3:1

Let the constant part be x ∴ 8x + 6x + 3x + 1x = 500000 18x = 500000 x = 27777.77778 Distribution of estate: First beneficiary: 8 ∗ 27777.77778 = $22222.22 Second beneficiary: 6 ∗ 27777.77778 = 166666.67 Third beneficiary 3 ∗ 27777.77778 = 83333.33 Fourth beneficiary: 1 ∗ 27777.77778 = 27777.78 Diff: 3 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 14) Students, A, B, and C occupy floor space of 50 m, 70 m, and 80 m, respectively. If the total rental for the space is $1000 per month, how much rent should student A pay? Answer: Total number of square metres = 50 + 70 + 80 = 200 Price per square metre

= 5.00

Amount paid by student A = 50 ∗ 5.00 = $250.00 Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 15) Solve: n : 6 = 24 : 42 Answer: n : 6 = 24 : 42 42n = 24 ∗ 6 n=

= 3.42857

Diff: 1 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 16) Solve: 7 : n = 35 : 21 Answer: 7 : n = 35 : 21 = 21 ∗ 7 = 35n n= n = 4.2 Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

17) Solve: 7 : 5 = x : 40 Answer: 7 : 5 = x : 40 5x = 40 ∗ 7 x=

= 56

Diff: 1 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 18) Solve: 4.2 : x = 4.375 : 25.6 Answer:

x= x = 24.576 Diff: 1 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 19) Solve: 2.18 : 1.61 = k : 4.4 Answer: 2.18 : 1.61 = k : 4.4 1.61k = 4.4 ∗ 2.18 k=

= 5.957764

Diff: 1 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 20) Solve: 3.80 : m = 10.8 : 8.10 Answer: 3.80 : m = 10.8 : 8.10 10.8m = 3.80 ∗ 8.10 m= m = 2.85 Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 21) Solve: 2.50 : 1.25 = k : 4.4 Answer: 2.50 : 1.25 = k : 4.4 1.25k = 4.4 ∗ 2.5 k=

= 8.8

Diff: 1 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

22) Solve: 3.1 : m = 1.4 : 1.9 Answer: 3.1 : m = 1.4 : 1.9 1.4m = 1.9 ∗ 3.1 m=

= 4.207143

Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 23) Solve: 1.15 : 2.30 = k : 1.30 Answer: 1.15 : 2.30 = k : 1.30 2.3 ∗ k = 1.15 ∗ 1.30 k= k = 0.65 Diff: 1 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 24) Solve:

Answer:

∗

t = 0.4902 Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 25) Solve:

:t=

Answer:

:t=

∗

= 1.0666667

Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

26) Solve: x :

Answer: =

= 0.666666667

Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 27) Solve:

Answer:

t= t=

∗ ∗

= .200

∗

Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 28) Solve:

Answer:

t= t=

∗ ∗

∗

= .464

Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

29) Solve: 1.15 : 0.95 = k : 1.19 Answer: 1.15 : 0.95 = k : 1.19 0.95k = 1.15 ∗ 1.19 k= k = 1.44 Diff: 1 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 30) Solve: y :

Answer: y :

∗

∗ = 2.142857

Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 31) A clothing company has found that, based on past experience, $17.50 must be spent on labour costs for every $13.25 spent on materials. If a budget of the year 2005 is set at $231 875.00 for materials, calculate the amount that must be budgeted for labour. Answer: Let the labour cost for 2005 be $x. = x= x = $306250.00 Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 32) The community of Oakville sets a property tax rate of $25.50 per $1000 assessed valuation. What is the assessment if a tax of $554 is paid on a property? Answer: Let the tax assessment for a tax of $554 be $x. = 25.5x = 554000 x = $21725.49 Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

33) A manufacturing process requires $68.5 supervision cost for every 64 labour hours. At the same rate, how much supervision cost should be budgeted for 18 000 labour hours? Answer: Let the supervision cost for 18000 hours be $x. = 64x = 68.5 ∗ 18000 64x = 1233000 x = $19265.63 Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 34) Seven-eighths of Jim Whyte's inventory was destroyed by fire. He sold the remaining part, which was slightly damaged, for one-third of its value and received $1400. a.) What was the value of the destroyed part of the inventory? b.) What was the value of the inventory before the fire? Answer: Let the slightly damaged part be $x. 1 : 3 = 1400 : x x = 4200 Let the total value be $y. = a)

y = 33600 (33600) = 29400

b) $33600 Diff: 3 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 35) Five-eighths of Jim Bernard's inventory was destroyed by water damage. He sold the remaining part, which was slightly damaged, for one-third of its value and received $5000. a) What was the value of the destroyed part of the inventory? b) What was the value of the inventory before the fire? Answer: Let the slightly damaged part be $x. 1 : 3 = 5000 : x x = 15000 Let the total value be $y. = a)

y = 40000 (40000) = 25000

b) $40000 Diff: 3 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

36) Gross profit for March was three-fourths of net sales, and net income was four-fifteenth of gross profit. Net income was $16400. a) What was the gross profit for March? b) What were net sales for March? Answer: a)

Net income =

of gross profit

G = 16400 G = 16400 /

= $61500

Gross profit is $61 500 b)

Gross profit =

of net sales

N = 61500 N = 61500 /

= $82000

Net sales are $82 000 Diff: 2 Type: SA Page Ref: 101-106 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 37) A gas tank with a capacity of 245 L can be drained in twenty-five minutes. At the same rate, how many minutes will it take to drain a tank containing 176 L? Answer: Let the number of minutes for the 176 L tank be x. = = 245x = 25(176) x= x = 17.95918 It will take 17.95918 minutes to heat the 176 L tank. Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

38) Mr. Duckworth owned 3/8 of a store. He sold 2/3 of his interest in the store for $37 000. What was the value of Mr. Duckworth's share of the store? Answer: Let Duckworth's interest in the store be $x; then the amount sold by him is $ x. x = 37000 x = 37000 ∗ x = 55500 Let the value of the store be $y; then the interest held by Duckworth before selling is $ y. y = 55500

∴

y = 55500 ∗ y = 148,000 The value of the store is $148,000. Diff: 3 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 39) Overhead cost of a fan belt is three-eighths of total cost and labour cost is one-third of overhead cost. If labour cost is $15, what is the total cost of the fan belt? Answer: Let overhead cost be $x. =

, x = 45

Let the total cost be $y. =

, 3y = 8(45), y = 120

Total cost is $120 Diff: 3 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 40) $36 is what percent of $14? Answer: R =

= 257%

Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems.

41) 142% of what amount is $355? Answer: 142% of x = 355 1.42x = 355 x = 250 Diff: 1 Type: SA Page Ref: 107-115 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 42) 225% of what amount is $310? Answer: 225% of x = 310 2.25x = 310 x = 137.78 Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 43) 160% of what amount is $200? Answer: 160% of x = 200 1.60x = 200 x = 125.00 Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 44) $170 is what percent of $80? Answer: R =

= 212.5%

Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 45) $189 is what percent of $4150? Answer: R =

= 4.5542%

Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 46) 27.5% of what amount is $28.63? Answer: 27.5% of x = 28.63 .275x = 28.63 x = $104.11 Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems.

47) $901 is 31.2% of what amount? Answer: 901 = 31.2 of x 901 = .312x x = 2887.82 Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 48) $750 is 33.3% of what amount? Answer: 750 = 33.3 of x 750 = .333x x = 2252.25 Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 49) What is

% of $14200?

Answer: x =

% of 14200

x = 0.00375(14200) x = $53.25 Diff: 1 Type: SA Page Ref: 107-115 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 50) 63 is what percent of 45? Answer: Rate =

= 1.4 = 140%

Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 51) What is 3/4% of $164.00? Answer: x =

% of 164.00

x = .0075 ∗ 164.00 x = $1.23 Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 52) What percent of 0.14 is 0.035? Answer: Rate =

= 0.25 = 25%

Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems.

53) What percent of $62.50 is $11.25? Answer: Rate =

= .18 = 18%

Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 54) $14.35 is 1/3% of what amount? Answer: 14.35 =

of x

14.35 = $4305.00 = x Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 55) Overhead content in an article is 37 1/2% of total cost. How much is the overhead cost if the total cost is $72? Answer: Let labour cost be $x. x = 37 % of 72 x=

× 72

x = 27 Diff: 2 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 56) If total deductions on a yearly salary of $28 600 amounted to 16 2/3%, how much was deducted? Answer: Let the deduction be $x. x = 16 % of 28600 =

× 28600

= $4766.67 Diff: 2 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems.

57) The Canada Pension Plan premium deducted from an employee's wages was $131.59. If the premium rate is 2.925% of gross wages, how much were the employee's gross wages? Answer: Let gross wages be $x. 2.925% of x = 131.59 0.02925x = 131.59 x = $4498.80 Diff: 2 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 58) A firm's debts of $8215 were 1 4/5% of sales. What were the firm's sales? Answer: Debts = 1 4/5% of sales Let s represent the firm's sales in $. 8215 = 0.018s s = 456 388.89 Sales are $456 388.89. Diff: 1 Type: SA Page Ref: 107-115 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 59) 350% of $15.00 is what amount? Answer: 350% of 15.00 = x x = 3.50 ∗ 15 = $52.50 Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 60) $275 is 87 1/2% of what amount? Answer: 275 = 87 % of x x = 275 x = 275 ∗ x = $314.29 Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems.

61) $90 is 37 % of what amount? Answer: 90 = 37 % of x x = 90 x = 90 ∗ x = $240.00 Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 62) 2 % of what amount is $19.90? Answer: 2 % of x = 19.90 0.0275x = 19.90 x = $723.64 Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 63) $32 is what percent of $26? Answer: Rate =

= 1.230769 = 123.0769%

Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 64) $75 is what percent of $50? Answer: Rate =

= 1.50 = 150.00%

Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems.

65) 166 2/3% of what amount is $280? Answer: 166 % of x = 280 x = 280 x = 280 x = 280 ∗ x = $168 Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 66) The Canada Pension Plan premium deducted from an Peter's wages was $650.25. If the premium rate is 2.925% of gross wages, how much were Peter's gross wages? Answer: Let gross wages be $x. 2.925% of x = 650.25 0.02925x = 650.25 x = $22230.77 Diff: 2 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 67) A property was sold for 250% of what the vendors originally paid. If the vendors sold the property for $218 000, how much did they originally pay for the property? Answer: Let the original cost be $x. 250% of x = 218000 2.5x = 218000 x = 87200 Diff: 2 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 68) What is the rate percent if the base is 88 and the percentage is 65? Answer: Rate =

= .738636 = 73.8636%

Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 69) What is the rate percent if the base is 50 and the percentage is 35? Answer: Rate =

= .7000 = 70.00%

Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems.

70) In the last provincial election, 71 1/2% of the population of 1 284 600 was eligible to vote. Of those eligible, 68 1/3% voted. a) What was the number of eligible voters? b) How many voted? Answer: a) Number eligible = 71 % of 1 284 600 = 0.715(1284600) = 918 489 b) Number voting = 68 % of 918 489 = 0.68333(918489) = 627631.09 Diff: 2 Type: SA Page Ref: 107-115 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 71) What is 870 decreased by 21%? Answer: x = 870 - 21% of 870 = 870 - 0.21 ∗ 870 = 870 - 182.70 = 687.3 Diff: 1 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 72) How much is $32.60 increased by 312%? Answer: x = 32.60 + 312% of 32.60 = 32.6 + 3.12(32.6) = 134.312 Diff: 1 Type: SA Page Ref: 115-119 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 73) What amount is 16 2/3% less than $69? Answer: x = 69 - 16 % of 69 = 69 -

∗ 69

= 69 - 11.5 = $57.50 Diff: 1 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems.

74) What amount is 8 1/3% less than $84? Answer: x = 84 - 8 % of 84 = 84 -

∗ 84

= 84 - 7 = $77.00 Diff: 1 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 75) What percent less than $160 is $120? Answer: Decrease = $40 R = 40/160 = 0.25 = 25% Diff: 1 Type: SA Page Ref: 115-119 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 76) The amount of $123 is what percent less than $165? Answer: Decrease = 42 R=

= .2545 = 25.45%

Diff: 1 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 77) The amount of $2250 is what percent more than $2100? Answer: Increase = 150 R=

= .0714 = 7.14%

Diff: 1 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 78) The number 675 is 37 1/2% more than what number? Answer: 675 = x + 37.5% of x x+

x = 675 = x + .375x x = 675 = 1.375x x = x = 490.91

Diff: 2 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems.

79) The number 500 is 62 1/2% more than what number? Answer: 500 = x + 62.5% of x x+

x = 500 = x + .625x x = 500 = 1.625x

x = 307.69 Diff: 2 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 80) Sales at Pat's Flower Shop decreased in September 12.5% from August. If September sales amounted to $25 200.00, calculate the August sales. Answer:

= x=

x = $28800.00 Diff: 2 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 81) What sum of money decreased by 16 2/3% will equal $475? Answer: x - 16 % of x = 475 x-

x = 475 x = 475 x = 475 ∗

= 570

Diff: 2 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 82) After increasing a dollar amount, x, by 12%, the new amount became $940. What was the original amount? Answer: x + 12% of x = 940 1.12x = 940 x = 839.29 Diff: 1 Type: SA Page Ref: 115-119 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems.

83) A buyer bought merchandise for $2900. If he sells the merchandise at 66 2/3% above cost, how much gross profit does he make? Answer: Let the gross profit be $x. x = 66 % of 2900 =

∗ 2900

= $1933.33 Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 84) A storekeeper bought merchandise for $1798. If she sells the merchandise at 33 1/3% above cost, how much gross profit does she make? Answer: Let the gross profit be $x. x = 33 % of 1798 =

∗ 1798

= $599.33 Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 85) Your gas bill for March is $274.40. If you pay after the due date, a late payment penalty of $10.72 is added. What is the percent penalty? Answer: Penalty =

= 3.9067%

Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 86) The price of an article increased from $15.50 to $17.50. Calculate the percent change in price. Answer: % change =

= 0.129032258 = 12.9% increase

Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems.

87) Royal Collection Agency retains a collection fee of 20% of any amounts collected. How much did the agency collect on a bad debt if the agency forwarded $2520 to a client? Answer: Let the amount collected be $x. 80% of x = 2520 .8x = 2520 x = $3150 The amount collected was $3150. Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 88) A residential property is assessed for tax purposes at 45% of its market value. The residential property tax rate is 3 2/3% of the assessed value and the tax is $1300. a) What is the assessed value of the property? b) What is the market value of the property? Answer: a) Let the assessed value be $x. 3 % of x = 1300 .0366666x = 1300 x = 35 4 54.55 Assessed value is $35 454.55 b)

Let the market value be $y. 45% of y = 35454.55 0.45y = 35454.55 y = 78 787.89 Market value is $78 787.89 Diff: 3 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 89) A wholesale outlet is offered a discount of 2 % for payment in cash of an invoice of $950. If it accepted the offer, how much was the cash payment? Answer: Let the cash payment be $x. x = 950 - 2 % of 950 = 950 - 23.75 = $926.25 Cash payment is $926.25. Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems.

90) A retail outlet is offered a discount of 5 1/2% for payment in cash of an invoice of $2000. If it accepted the offer, how much was the cash payment? Answer: Let the cash payment be $x. x = 2000 - 5 % of 2000 = 2000 - 110 = $1890.00 Cash payment is $1890.00. Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 91) From March 1991 to March 2001, the price of gasoline increased 170%. If the price in 1991 was 33.0 cents per litre, what was the price per litre in 2001? Answer: Let the 2001 price be $x. x = 0.33 + 170% of 0.33 = 0.33 + .561 = .891 The 2001 price was 89.1 cents per litre. Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 92) Ms. Rose pays 37 1/2% of her monthly salary as rent on a townhouse. If the monthly rent is $750, what is her monthly salary? Answer: Let the monthly salary be $x. 37 % of x = 750 x = 750 x = 750 ∗

= $2000.00

The monthly salary is $2000.00. Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 93) A discount brokerage house charges a fee of 2 3/4%. If its fee on a stock purchase was $757, what was the amount of the purchase? Answer: Let the amount of the purchase be $x. 2 % of x = 757 0.0275x = 757 x = 27527.27 The amount of the purchase was $27527.27. Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems.

94) An article was sold for $567.50. The amount paid included a sales tax of 6%. Find the amount of sales tax on the article. Answer: Price paid = 106% of selling price 567.50 = 1.06 S S=

= 535.38

Sales tax = 567.50 - 535.38 = $32.12 Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 95) An article was sold for $212.00. The amount paid included a sales tax of 6%. Find the amount of sales tax on the article. Answer: Price paid = 106% of selling price 212.00 = 1.06 S S=

= 200.00

Sales tax = 212.00 - 200.00 = $12.00 Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 96) A boat owner listed his sailboat for 125% more than he paid for it. The owner eventually accepted an offer 5 1/2% below his asking price and sold the property for $50 000. How much did the owner pay for the sailboat? Answer: 94 % of asking price = sale price .945 = 50 000 A = $52910.05 List = C + 125% of C $52910.05 = 1.25C C = 42328.04 Cost was $42328.04 Diff: 3 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 97) A wage earner's hourly rate of pay was increased from $13.25 to $14.55. What was the percent raise? Answer: Increase = 14.55 - 13.25 = $1.30 Rate = 1.30/13.25 = 0.098113 = 9.8113% The raise was 9.8113%. Diff: 1 Type: SA Page Ref: 120-127 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems.

98) An article originally advertised at $210.00 was reduced to $178.75 during a sale. By what percentage was the price reduced? Answer: Reduction = 210.00 - 178.75 = 31.25 Rate of reduction =

= .14881 = 14.881%

Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 99) The Bank of Nova Scotia reduced its annual lending rate from 11% to 8.75%. What is the percent reduction in the lending rate? Answer: Reduction = 11% - 8.75% = 2.25% Rate =

= .204545 = 20.4545%

The reduction in rate is 20.4545%. Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 100) Calculate the cost of a $312.00 suit if the price is increased by 2 %. Answer: New value is 312.00 + 312.00(0.025) = $319.80 Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 101) After a reduction of 33 % of the marked price, an article was sold for $412.00. What was the marked price? Answer: Let the marked price be $x. Then the reduction in price is $ x x-

x = 412 x = 412

x = 618 The marked price was $618.00. Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems.

102) After a reduction of 66 % of the marked price, an article was sold for $150.00. What was the marked price? Answer: Let the marked price be $x. Then the reduction in price is $ x x-

x = 150 x = 150

x = 450 The marked price was $450.00. Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 103) A special purpose index has increased 155% during the last ten years. If the index is now 310, what was the index ten years ago? Answer: Let the index ten years ago be x. x + 155% of x = 310 x + 1.55x = 310 2.55x = 310 x = 121.56863 The index ten years ago was 121.56863. Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 104) Sales in June increased 16 2/3% over May sales. If June sales amounted to $27 535, what were May sales? Answer: Let May sales be $x. x + 16 % of x = 27535 x+

x = 27535

x = 23601.43 May sales were $23 601.43. Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems.

105) After real estate fees of 7.5% had been deducted from the proceeds of a property sale, the vendor of the property received $188 090. What was the amount of the realtor's fee? Answer: Let the proceeds be $x. x - 7.5% of x = 188090 0.925x = 188090 x = 203340.54 Fees = 7.5% of 203340.54 = $15 250.54 Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 106) A property owner listed his property for 172% more than he paid for it. The owner eventually accepted an offer 11 1/2% below his asking price and sold the property for $201 100. How much did the owner pay for the property? Answer: 88 % of asking price = sale price .885 = 201100 A = $227 231.6384 List = C + 172% of C $227 231.6384 = 1.72C C = 132 111.42 Cost was $132 111.42. Diff: 3 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 107) The direct labour cost of manufacturing a product is $123.95, direct material cost is $56.20, and overhead is $57.75. a) What is the percent content of each element of cost in the product? b) What is the overhead percent rate based on direct labour? Answer: a) Labour cost = $123.95 52.1% Material cost = 56.20 23.6% Overhead = 57.75 24.3% Total = $237.90 100.00% b) Overhead rate =

= 46.59%

Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems.

108) A sales representative's orders for January were 83 1/3% less than her December orders, which amounted to $20 000. a) How much were the sales rep's orders in January? b) By what amount did her orders decrease? Answer: a) Maximum order = 20000 - 83 % of 20000 = 20000 -

∗ 20000

= 20000 - 16666.67 = 3333.33 b) $16666.67 Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 109) In Victoria, a car was sold for $19 621.60 including 12% HST. How much was the harmonized sales tax on the car? Answer: Let the price of the car be $x. x + 12% of x = 19621.60 x + 0.012x = 19621.60 1.12x = 19621.60 x = 17 519.29 Sales tax = 7% of 17519.29 = $1226.35 Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 110) In Manitoba, a car was sold for $21 500.55 including 5% GST and 7% PST. How much was the provincial sales tax on the car? Answer: Let the price of the car be $x. x + 5% of x + 7% of x = 21500.55 x + 0.05x + 0.07x = 21500.55 1.12x = 21500.55 x = 19196.92 Sales tax = 7% of 19 196.92 = $1 343.78 Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems.

111) A sales representative's orders for July were 16 2/3% less than her June orders, which amounted to $151 120. a) How much were the sales rep's orders in July? b) By what amount did her orders decrease? Answer: a) Maximum order = 151120 - 16 % of 151120 = 151120 -

∗ 151120

= 151120 - 25186.67 = 125 933.33 b) $25 186.67 Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 112) The price of a stock a week ago was $36.25 per share. Today the price per share is $31.75. a) What is the percent change in price? b) What is the new price as a percent of the old price? Answer: a) $ change = 31.75 - 36.25 = -4.50 change =

- .1241 = -12.41%

b) New price as a percent of old price =

= .8759 = 87.59%

Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems.

113) After a reduction of 8 1/3% of the marked price, a snowboard sold for $400.00. What was the marked price? Answer: Let the marked price be $x; then the price reduction is 8 % of x. x-

x = 400 x = 400 x = 400 ∗ x = 436.36

The marked price is $436.36. Diff: 2 Type: SA Page Ref: 188-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 114) Alanna Morrisette's hourly rate of pay was increased from $19.00 to $22.54. What was the percent raise? Answer: Raise = 22.54 - 19.00 = $3.54 Percent raise =

= .186316 = 18.6316%

Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 115) An MP3 player originally advertised at $320.00 is reduced to $279.00 during a sale. By what percent was the price reduced? Answer: Price reduction = 320.00 - 279.00 = $41.00 % reduction =

= .128125 = 12.8125%

Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 116) Calculate the cost of a $14 500.00 machine after a decrease of 15% in price. Answer: New value is 14 500.00(0.85) = $12 325.00 Diff: 2 Type: SA Page Ref: 120-127 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems.

117) After a reduction of 16 2/3% of the marked price, a pair of boots sold for $60.00. What was the marked price? Answer: Let the marked price be $x; then the price reduction is 16 % of x. x-

x = 60 x = 60 x = 60 ∗ x = 72

The marked price is $72.00. Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 118) D&L Corporation's profit and loss statement showed a net income of 9 3/4% or $41 250. Twenty-one percent of net income was paid in corporation tax and 65% of the net income after tax was paid out as dividends to Cindi and Kelsey, who hold shares in the ratio 2 to 3. a) What was the revenue of D&L Corporation? b) How much was the after-tax income? c) How much was paid out in dividends? d) What percent of net income did Kelsey receive as a dividend? Answer: a) Let revenue be $x. ∴ 9 % of x = 41250

0.0975x = 41250 x = 423,076.92 After-tax income = 41250 - 21% of 41250 = .79(41250) = 32,587.50 Dividend paid = 65% of 32,587.50 = .65 ∗ 32,587.50 = 21,181.88

= ∴

= x = $12,709.13

Kelsey's dividend

= .6 = 60%

Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems.

119) How many Canadian dollars can you buy for $850.00 U.S. dollars if one Canadian dollar is worth $0.9641 U.S. dollars? Answer: Let the number of Canadian dollars be x. = = x= x = 881.65 $850.00 U.S. will buy C$881.65. Diff: 2 Type: SA Page Ref: 124-127 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency. 120) What is the price of gasoline per litre in Canadian dollars if a U.S. gallon of gasoline costs $2.90 U.S.? One U.S. dollar is worth $1.0372 Canadian, and one U.S. gallon is equivalent to 3.8 litres. Answer: Let the cost per U.S. gallon be C$x. =

= x = 1.0372(2.90) U.S. gallon costs C$3.00788 Cost per litre =

= C$0.791547

Diff: 2 Type: SA Page Ref: 124-127 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency.

121) Suppose it costs $363.50 Canadian to purchase $350.00 U.S. a) What is the exchange rate? b) How many U.S. dollars will you receive if you convert $925 Canadian into U.S. dollars? Answer: a) Exchange rate U.S$ per C$ =

= .962861

C$1 = US$0.962861. b) C$925.00 = 925.00(0.962861) = US$890.65. Diff: 2 Type: SA Page Ref: 124-127 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency. 122) If one Canadian dollar is equivalent to $0.9749 U.S., how much do you need in Canadian funds to buy $850.00 U.S.? Answer: Let the number of Canadian dollars be x. = = x=

= 871.88

$850.00 U.S. costs $871.88 CDN. Diff: 2 Type: SA Page Ref: 124-127 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency. 123) If one Canadian dollar is equivalent to $0.9421 U.S., how much do you need in Canadian funds to buy $1000 U.S.? Answer: Let the number of Canadian dollars be x. = = x=

= 1061.46

$1000.00 U.S. costs $1 061.46 Canadian. Diff: 2 Type: SA Page Ref: 124-127 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency.

124) Use Table 3.2 on page 120 of your textbook to convert C$900.00 to euros. Answer: From the table, the exchange rate for Canadian dollars to euros is 0.7294. Conversion = 900 × 0.7294 = C$656.46 Diff: 1 Type: SA Page Ref: 124-127 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency. 125) Mr.Singh needs to buy an airline ticket from Air Canada for C$1 480. How much would it cost him in Indian rupees? Use Table 3.2 on page 120 to answer the question. Answer: Let the number of rupees be x. From the Table x × 0.0226 = 1480 x=

= 65 486.73

Mr. Singh has to pay 65 486.73 Indian rupees for his C$1 480 ticket. Alternative solution is possible. Diff: 2 Type: SA Page Ref: 124-127 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency. 126) Cindi's annual incomes for 2002, 2006, and 2009 were $25 000, $51 000, and $42 800 respectively. Given that the Consumer Price Index for the three years 100.0, 109.1, and 114.4 respectively, compute Cindi's real income for 2006 and 2009. Answer: Cindi's real income in 2006 = Cindi's real income in 2009 =

(100) = $46 746.10

(100) = $37 412.59

Diff: 2 Type: SA Page Ref: 128-130 Topic: 3.7 Applications - Index Numbers Objective: 3-5: Use index numbers and the Consumer Price Index to compute purchasing power of the Canadian dollar. 127) The Canadian CPI was 109.1 in 2006 and 114.4 in 2009 relative to the base year of 2002. Determine the purchasing power of the Canadian dollar for the two years, and interpret the meaning of the results? Answer: Purchasing power of the dollar for 2006 relative to 2002 =

(100) = 0.916590

Purchasing power of the dollar for 2009 relative to 2002 =

(100) = 0.874126

The dollar in 2006 could purchase only 91.7% of what it could purchase in 2002. In 2009, the dollar could purchase even less–about 87.4% of what it could purchase in 2002. Diff: 2 Type: SA Page Ref: 128-130 Topic: 3.7 Applications - Index Numbers Objective: 3-5: Use index numbers and the Consumer Price Index to compute purchasing power of the Canadian dollar.

128) Use the 2010 federal income tax brackets and rates listed below to answer the following question. Taxable Income (income tax brackets) $40 970 or less $40 970 to $81 941 $81 941 to $127 02 Over $127 021

Tax rates 15% of taxable income less than or equal to $40 970; plus 22% of taxable income greater than $40 970 and less than or equal to $91 941; plus 26% of taxable income greater than $81 941 and less than or equal to $127 021; plus 29% of taxable income greater than $127 021

In early 2010, Juan's gross pay increased from $35 000 per year to $43 000 per year. a) What was the annual percent increase in Juan's pay before federal income taxes? b) What was the annual percent increase in Juan's pay after federal income taxes were deducted? Answer: a) Increase in gross pay = 43 000 - 35 000 = $8 000 Percent increase =

(100%) = 22.8571%

b) Federal tax on $35 000 is 15% of $35000 Net pay after federal tax = 35000 - 5250 = 29 750 Federal tax on 43 000 On first $40970 On excess 22% of 2030

$5 250

$6145.50 446.60 $6 592.10

Net pay after federal tax = $43 000 - $6592.10 = $36 407.90 Increase in net pay after federal tax = 36 407.90 - 29 750 = 6 657.90 Percent increase =

(100%) = 22.3795%

Diff: 2 Type: SA Page Ref: 131-132 Topic: 3.8 Applications - Personal Income Taxes Objective: 3-6: Use federal income tax brackets and tax rates to calculate federal income taxes.

129) Use the 2010 federal income tax brackets and rates listed below to answer the following question. Taxable Income (income tax brackets) $40 970 or less $40 970 to $81 94 $81 941 to $127 021 Over $127 021

Tax rates 5% of taxable income less than or equal to $40 970; plus 22% of taxable income greater than $40 970 and less than or equal to $91 941; plus 26% of taxable income greater than $81 941 and less than or equal to $127 021; plus 29% of taxable income greater than $127 021

a) Dana had a taxable income of $39 500 in 2010. How much federal income tax should she report? (assuming tax rates remain the same) b) Dana expects her taxable income to increase by 15% in 2011. How much federal tax would she expect to pay in 2011(assuming tax rates remain the same). Answer: a) Federal tax on $39 500 is 0.15 × 39 500 = $5 925 b) Taxable income is 1.15 × 39 500 = $45 425 Federal tax on $40 970 is $6 145.50 On excess is 0.22 × (45 425 - 40 970) = $980.10 Total Federal tax is 6145.50 + 980.10 = $7 125.60 Diff: 2 Type: SA Page Ref: 131-132 Topic: 3.8 Applications - Personal Income Taxes Objective: 3-6: Use federal income tax brackets and tax rates to calculate federal income taxes. 130) Betty calculated her 2011 taxable income to be $133 450. How much federal income tax should she report if she is in the top tax bracket, which is taxed at $26 879.92 plus 29% of income over $127 021? Answer: Federal tax on first $127 021 $26 879.92 Federal tax on excess 29% of $(133 450 - 127 021) = 0.29(6429) $1864.41 Federal tax $28 744.33 Diff: 2 Type: SA Page Ref: 131-132 Topic: 3.8 Applications - Personal Income Taxes Objective: 3-6: Use federal income tax brackets and tax rates to calculate federal income taxes. 131) Calculate:

% of $832.00

A) $1.04 B) $10.40 C) $104.00 D) $66.56 E) $6.66 Answer: A Diff: 1 Type: MC Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems.

132) $165.00 is what percentage less than $247.50? A) 66 % B) 33 % C) 50% D) 82.5% E) 85% Answer: B Diff: 1 Type: MC Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 133) How much is $18.00 increased by 250%? A) $72.00 B) $22.50 C) $63.00 D) $45.00 E) $54.00 Answer: C Diff: 1 Type: MC Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 134) 12% of what number is 144? A) 17.28 B) 112 C) 1200 D) 136 E) 14.4 Answer: C Diff: 1 Type: MC Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 135) 21 is what percent of 200? A) 0.00105% B) 0.105% C) 10.5% D) 9.52% E) 21% Answer: C Diff: 1 Type: MC Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems.

136) A law firm has 3 partners. The first partner owns business and the third partner owns

of the business, the second partner owns

of the

of the business. The business is worth $1.5 million. A new

partner is introduced who contributes an additional $300,000 and owns

of the total value of the new

business. What are the new values for the first to the third original partners? A) 1-$476000, 2-$540000, 3-$424000 B) 1-$526000, 2-$590000, 3-$424000 C) 1-$576000, 2-$540000, 3-$324000 D) 1-$576000, 2-$450000, 3-$414000 E) 1-$576000, 2-$490000, 3-$424000 Answer: C Diff: 4 Type: MC Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 137) Calculate the value of x: x=

A) 38/288 B) 72/288 C) 38/144 D) 35/144 E) 63/20 Answer: E Diff: 2 Type: MC Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 138) You bought a car and got a 7.35% discount from the original price. If you purchase price is $24063, what was the original price? A) $25971.94 B) $24971.94 C) $26971.74 D) $27971.94 E) $27971.74 Answer: A Diff: 1 Type: MC Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems.

139) Your earnings of $967.34 for the week represent 4.168% of your entire years earnings. What are your annualized earnings? A) $23899.88 B) $23208.73 C) $28308.73 D) $22208.73 E) $28308.88 Answer: B Diff: 1 Type: MC Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 140) You sell your house at 95.7% of the list price. What was the amount of sales commission if the commission percentage is 3.75% and the list price of the house was $182 500? A) $6449.47 B) $6449.74 C) $6149.47 D) $6149.74 E) $6549.47 Answer: E Diff: 1 Type: MC Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 141) Using the Table 3.2 on page 120 of your text convert 3100 British pounds to Canadian dollars. A) $4249.78 B) $4623.65 C) $4747.32 D) $2024.30 E) $4747.65 Answer: E Diff: 1 Type: MC Page Ref: 124-127 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency. 142) On April 4, 2001, the Dow Jones Industrial (DJIA) average increased to 9918.05. This represented a one-day increase of 327.63 points. What percentage increase occurred on this day? A) 3.3% B) 2.3% C) 4.3% D) 2.8% E) 6.8% Answer: A Diff: 1 Type: MC Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems.

143) Use the income tax table (Table 3.3 on page 125 of your text) to calculate. What is your federal tax payable if you had taxable income of $134 763? A) $26 879.92 B) $29 125.10 C) $33 345.47 D) $35 270.64 E) $39 083.01 Answer: B Diff: 3 Type: MC Page Ref: 131-132 Topic: 3.8 Applications - Personal Income Taxes Objective: 3-6: Use federal income tax brackets and tax rates to calculate federal income taxes. 144) A private post-secondary corporation had a 1-year low stock price of $18.69 and a high of $40.47. Assuming that you bought it at the lowest price and sold it at its highest price, what is the simple annual rate of return on your stock purchase? A) 107.812% B) 106.712% C) 107.612% D) 116.533% E) 126.712% Answer: D Diff: 1 Type: MC Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 145) Calculate:

% of $100.00

A) $250.00 B) $25.00 C) $2.50 D) $0.25 E) $0.03 Answer: D Diff: 1 Type: MC Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 146) How much is $25.00 increased by 25%? A) $625.00 B) $30.00 C) $31.25 D) $27.50 E) $26.25 Answer: C Diff: 1 Type: MC Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems.

147) Calculate the value of x: :x=

A) 3/2 B) 8/15 C) 5/6 D) 6/5 E) 12/10 Answer: C Diff: 1 Type: MC Page Ref: 101-106 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 148) A private post-secondary corporation had a 1-year low stock price of $5.00 and a high of $7.50. Assuming that you bought it at the lowest price and sold it at its highest price, what is the simple annual rate of return on your stock purchase? A) 0.5% B) 5% C) 50% D) 500% E) 1/2 % Answer: C Diff: 1 Type: MC Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 149) Muki and Mushi invested $21 000 and $37 000, respectively. If Muki invests a further $9000, what amount should Mushi contribute to maintain their investments in the original ratio? A) $5108.11 B) $7297.30 C) $6750 D) $17 027.03 E) $15 857.14 Answer: E Diff: 1 Type: MC Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

150) Lakeridge Health Oshawa is a 423 beds hospital with 463 doctors and 4058 staff comprised of nurses and technologists. It is planning to open another 35-bed wing. Assuming the same proportionate staffing levels, how many more doctors will need to be hired? A) 39 B) 501 C) 32 D) 495 E) 4 Answer: A Diff: 1 Type: MC Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 151) Mary paid $456 for an annual contract to an ISP on July 1, 2012. She terminated her contract with the ISP on March 8. The contract requires to refund a prorated amount to the exact number of days remaining in the period of coverage. A $50 service charge is then deducted. What refund will Mary receive? A) $142.42 B) $92.42 C) $54.94 D) $50 E) $0 Answer: B Diff: 2 Type: MC Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 152) Franco, Maria and Gomez entered into a partnership to buy a steel manufacturing plant. They agreed to distribute profits in the same proportion as their respective capital investments in the partnership. How will recent period's profit of $287 800 be allocated to Franco, if Franco's capital investment shows a balance of $34 million, Maria's shows $49 million and Gomez's shows $54.5 million? Answer: Franco : Maria : Gomez = 34 : 49 : 54.5 Total Investment = 34 + 49 + 54.5 = $137.5 Million Franco's share in profit =

× $287 800

= $71 165 Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

153) Franco, Maria and Gomez entered into a partnership to buy a steel manufacturing plant. They agreed to distribute profits in the same proportion as their respective capital investments in the partnership. How will recent period's profit of $287 800 be allocated to Maria, if Franco's capital investment shows a balance of $34 million, Maria's shows $49 million and Gomez's shows $54.5 million? A) $75 165 B) $102 561 C) $114 074 D) $257 800 E) $110 000 Answer: B Diff: 2 Type: MC Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 154) SDM sells eggs at $2.80/dozen. How many eggs can be bought for $2.10? A) None B) 6 C) 9 D) 12 E) 15 Answer: C Diff: 1 Type: MC Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 155) In a remote army camp, 10 soldiers have the ration for 21 days. If 3 soldiers leave the camp, for how many days will the ration be sufficient? A) 14 B) 15 C) 21 D) 60 E) 30 Answer: E Diff: 2 Type: MC Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 156) If 35 digging machines dig 805 cubic metres of earth in 5 hours, how much earth will 30 digging machine dig in 6 hours? Answer: Number of digging machines is directly proportional to amount of earth dug. Time is also directly proportional to the amount of earth dug. ∴

⇒ x = 828 30 digging machines will dig 828 cubic metres of earth in 6 hours. Diff: 3 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

157) ABC Charity is distributing charity among the refugee camps in a remote town in Somalia. The distribution formula states that $800 is sufficient for a family of 4 members for 40 days. For how many days $1500 last for a family of 5 members? Answer: Dollars are directly proportional to the number of days. Number of family members is inversely proportional to the number of days. ∴

⇒ x = 60 $1500 will last for 60 days for a family of 5 members. Diff: 3 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 158) In an army barracks in Afghanistan the mess consumes 1.2 kg of flour per person. They have enough flour for 4200 soldiers having regular meals for 32 days. Some soldiers are reassigned to a different location and the same amount of flour becomes sufficient for 42 days at a rate of 1.6 kg per person . How many soldiers were reassigned? Answer:

⇒ x = 2400 The number of soldiers reassigned = 4200 - 2400 = 1800 Diff: 3 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 159) After spending a week in the United States, you are travelling back to Canada. Before departure from JFK, you convert your remaining US$ to C$ at the exchange rate of 1.0287 per US$. How many C$ will you receive for $321 if the currency shop charges a 5% commission on this transaction? A) $321.21 B) $330.21 C) $313.70 D) $296.44 E) $312.04 Answer: C Diff: 1 Type: MC Page Ref: 124-127 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency.

160) Using an exchange rate of ¥88.313 per C$, calculate the number of yen that C$900 could be purchased after giving away a commission of 2.5%. Round your answer to nearest yen. A) ¥77 495 B) ¥79 482 C) ¥81 469 D) ¥81 520 E) ¥77 483 Answer: A Diff: 2 Type: MC Page Ref: 124-127 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency. 161) A perfume is sold at Target Canada for $79.99 plus HST (13%). The same perfume is sold at Target USA for $74.99 plus 8% Value Added Tax (VAT). By what percentage (based on Canadian price) is perfume cheaper in the United States? Use the exchange rate of US$0.966 per C$. A) 10.4% B) 7.2% C) 6.3% D) 2.95% E) 5.9% Answer: B Diff: 2 Type: MC Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 162) In order to encourage Canadians to avoid cross border shopping, Walmart Canada is striving to keep the price of electronics in Canada equivalent to the price of electronics in its US stores before taxes. If a gaming console is sold for $649 in US stores, what should be the sale price of the same console in Canadian Walmart locations? Use the exchange rate of US$0.7660 per C$1. A) $497.13 B) $649.00 C) $727.58 D) $847.26 Answer: D Diff: 2 Type: MC Page Ref: 127-130 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency.

163) Gas is sold for C$1.293 per litre in Windsor and US$3.91 per gallon in Detroit. By what percentage (based on Windsor price) is gas cheaper in Detroit? Use the exchange rate of US$0.966 per C$. Note 1 US gallon = 3.785 litres? Answer: Fuel price in detroit in C$ = Covert price to per litre: Price difference percentage =

= C$4.04762 per gallon

= C$1.0694 per litre × 100 = 17.29%

Diff: 2 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 164) A project was tendered in 2009 for US$3.5 million, when the exchange rate was US$1.0315 per $C. It started building in 2010 and the money was borrowed from the Canadian bank when the exchange rate fell to US$0.971 per $C. What is the increase in the amount borrowed? A) $104 531 B) $106 883 C) $110 250 D) $5719 E) $211 750 Answer: A Diff: 2 Type: MC Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 165) The price of milk was $2.79 in 2006 for a 4 L bag and $3.47 in 2016 for the same bag. Compare the two prices to create an index number. Answer: Index number =

= 1.24 × 100 =124

Diff: 1 Type: SA Page Ref: 131-134 Topic: 3.7 Applications - Index Numbers Objective: 3-5: Use index numbers and the Consumer Price Index to compute purchasing power of the Canadian dollar. 166) Asia started working as a professional engineer in 2002 earning $64 500 per annum. In 2013, she is earning $88 100 per annum. The CPI in 2013 is 131 based on year 2002. What is the actual increase in Asia's pay in terms of 2002 dollars? A) $2752 B) $23 600 C) -$2450 D) $19 995 E) $3605 Answer: A Diff: 2 Type: MC Page Ref: 128-130 Topic: 3.7 Applications - Index Numbers Objective: 3-5: Use index numbers and the Consumer Price Index to compute purchasing power of the Canadian dollar.

167) Salem purchased a shirt for US$20 from the Niagara outlet mall in Buffalo on his credit card. The actual transaction on the statement shows a value of C$22.45. It also includes 2.5% commission. Calculate the exchange rate used by the bank on the statement? A) US$1 = C$1.1225 B) C$1 = US$0.891 C) US$1 = C$1.0944 D) C$1 = US$0.9111 E) US$1 = C$1.0976 Answer: C Diff: 2 Type: MC Page Ref: 124-127 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency. 168) Chris works for Ontario Hydro and his 2012 compensation was $134 532. Using Table 3.3 in the book, compute his federal tax. Answer: Federal tax = 15% × 42707 + 22% × (85414 - 42707) + 26% × (132406 - 85414) + 29% × (134532 132406) = $28 636.05 Diff: 2 Type: SA Page Ref: 131-132 Topic: 3.8 Applications - Personal Income Taxes Objective: 3-6: Use federal income tax brackets and tax rates to calculate federal income taxes. 169) Daniel started working for Oracle Canada 2 years ago. In 2012, his compensation was $78 624. Compute his federal tax using Table 3.3 in the book. A) $11 793.60 B) $6406.05 C) $14 307.79 D) $17 297.28 E) $15 744.47 Answer: C Diff: 1 Type: MC Page Ref: 131-132 Topic: 3.8 Applications - Personal Income Taxes Objective: 3-6: Use federal income tax brackets and tax rates to calculate federal income taxes. 170) Lakynn started working for Telus in 2008. His compensation in 2017 was $158 844, double the compensation he received in 2008. Using Table 3.3 (page 134) in the book, calculate his change in federal income tax, assuming that this table was also applicable in 2008. A) $11 986.59 B) $21 334.23 C) $47 758.61 D) $20 322.14 E) $20 001.06 Answer: D Diff: 2 Type: MC Page Ref: 134-135 Topic: 3.8 Applications - Personal Income Taxes Objective: 3-6: Use federal income tax brackets and tax rates to calculate federal income taxes.

171) Don works for OPG in a management position. Due to the economic turndown, provincial government has frozen management pays since 2009. In 2010, federal tax brackets and tax rates were as follows: Taxable Income ≤ $40 970 $40 970 to $81 941 $81 941 to $127 021 > $127 021

Tax rate 15% 22% for > $40 970 but ≤ $81 941 26% for > $81 941 but ≤ $127 021 29% for > $127 021

Using the above table and Table 3.3 in the book, compare his federal tax payed in 2010 versus that payed in 2012 on an annual compensation of $134 595. Answer: Federal tax in 2010 = 15% × 40970 + 22% × (81491 - 40970) + 26% × (127021 - 81941) + 29% × (134595 - 127021) = $28,977.38 Federal tax in 2012 = 15% × 42707 + 22% × (85414 - 42707) + 26% × (132406 - 85414) + 29% × (134595 132406) = $28,654.32 Decrease in federal income tax between 2010 and 2012 = $28977.38 - $28654.32 = $323.06 Percentage decrease =

× 100 = 1.115%

Diff: 3 Type: SA Page Ref: 131-132 Topic: 3.8 Applications - Personal Income Taxes Objective: 3-6: Use federal income tax brackets and tax rates to calculate federal income taxes. 172) Jon sold 40 cars in the month of July. His target for August is to sell 10% more cars than he sold in July. How many cars should he sell in August to meet his target? A) 44 B) 36 C) 45 D) 35 E) 40 Answer: A Diff: 3 Type: MC Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 173) Sleep Country advertises to beat the lower price of the competitor by 5% on similar mattress. Leon's recently slashed its prices on Sealy Posturepedic Proback Titanium mattresses by 20%. If the original sale price of this mattress was $1460, at what price would the customer get that mattress at Sleep Country? A) $1460 B) $1168 C) $1095 D) $1387 E) $1110 Answer: E Diff: 2 Type: MC Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems.

174) Total 2013 residential tax rate in Toronto is 0.7457653%. It is expected to increase by 3% in 2014. What will be the 2014 residential tax rate in Toronto? Answer: 3% of 0.7457653 = 0.02237296 2014 residential tax rate in Toronto = 0.7457653 + 0.02237296 = 0.768138% Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 175) John received 72% marks in Thermodynamics. On recounting his marks were increased to 82%. What is the percentage increase in his marks? Answer: Percent increase = (82 – 72)/72 = 0.138889 = 13.89% Diff: 1 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 176) Reduce to lowest terms: 9 to 240 Answer: 3 : 80 Diff: 1 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 177) Set up a ratio and reduce to lowest terms: $102 per half a dozen Answer: 102 : 6 17 : 1 Diff: 1 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 178) Use equivalent ratios in higher terms to eliminate the fractions: Answer: (60)

to (60)

to to (60)

25 : 36 : 32 Diff: 1 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 179) A high school has 64 teachers, 18 support staff, and 902 students. Determine the ratio between teachers, staff members, and students, and then reduce to lowest terms. Answer: 64 : 18 : 902 32 : 9 : 451 Diff: 1 Type: SA Page Ref: 96-101 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

180) Ali's company earned $576 000 and will allocate all of these earnings to its three branches: sales, customer service, and R&D. The departments earn

and

, respectively. How much of the total

earnings will each department get? Answer: :

5 + 4 + 3 = 12 $576 000 ÷ 12 = 48 000 Sales' share of earnings = 48 000 × 5 = $240 000 Customer Service's share of earnings = 48 000 × 4 = $192 000 R&D's share of earnings = 48 000 × 3 = $144 000 Diff: 3 Type: SA Page Ref: 95-98 Topic: 3.1 Ratios Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 181) Find the unknown term: 7 : 13 = 84 : x Answer: 7x = 84 × 13 7x = 1092 x = 156 Diff: 1 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 182) Find the unknown term and round to 4 decimal places: 4 : 17 = x : 61 Answer: 17x = 4 × 61 17x = 244 x = 14.3529 Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 183) Use proportions to solve the problem: Rosa drove her motorcycle 312 kilometers and used 24 liters of fuel. She wants to know how many kilometers (k) she can drive with 10 liters of fuel. Assuming that the relationship between kilometers and fuel is proportional, determine the proportion that can be used to model this scenario and the distance she can travel with 10 liters of fuel. Answer: 24 : 312 = 10 : k 24k = 10 × 312 24k = 3120 k = 130 Rosa can drive 130 kilometres with 10 litres of fuel. Diff: 1 Type: SA Page Ref: 101-106 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems.

184) David makes candles that are 7cm tall. Each candle burns 3 hours before going out. He is wondering how many hours a 21cm tall candle can burn for. Assuming that the relationship between the height of a candle and number of hours it burns is proportional; determine the proportion that can be used to model this scenario and how long a 21cm tall candle can burn for. Answer: 7 : 3 = 21 : x 7x = 21 × 3 7x = 63 x=9 Diff: 2 Type: SA Page Ref: 100-104 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 185) Compute: 25% of 170 Answer: 0.25 × 170 = 42.5 Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 186) Use fractional equivalents to compute: 62.5% of 696 Answer:

× 696 = 435

Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 187) Find the rate percent of the following: base $576; percentage $792 Answer: Rate = Rate = 1.375 = 137.5% Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 188) $81 is 20% of what amount? Answer: 0.2x = 81 x = 405 Diff: 1 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 189) 28% of Wendy's monthly salary is deducted for withholding. If those deductions total $455, what is her salary? Answer: 0.28x = 455 x = 1625 Diff: 2 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems.

190) Zara sells $4,000 worth of goods in a month and earns $280 in commission. What is her commission rate? Answer: 4000x = 280 x = 0.07 or 7% Diff: 2 Type: SA Page Ref: 105-111 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 191) What is 140 increased by 30%? Answer: 140 × 1.30 = x 182 = x Diff: 1 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 192) What amount is 6 % less than 352? Answer: x = 352 - 6 % of 352 = 352 -

(352)

= 352 - 22 = 330 Diff: 2 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 193) $49.10 is what percent more than $18.32 (round to nearest integer)? Answer: Rate of increase = =

= 1.6801 = 168%

Diff: 2 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 194) What percent less than $145.04 is $98.67 (round to nearest integer)? Answer: Rate of increase = =

= 0.3197 = 32%

Diff: 2 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems.

195) Justin took out 41% of all the money in his bank account. Following this withdrawal, Justin only had $173.91 left in his account. How much money did he have in the account prior to the withdrawal? Answer: Let x represent the original sum of money x - 41% of x = 173.91 x - 0.41x = 173.91 0.59x = 173.91 x = 294.76 Diff: 2 Type: SA Page Ref: 113-117 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 196) Arlene is getting her house renovated, and total cost is estimated to be around $23 000. However, due to unforeseen issues, the total cost of the renovation increased to $37 260. What percent increase did the total cost receive? Answer: Rate of change = =

= 0.62 = 62%

Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 197) Zhang confronted the teacher about marking his exam wrong. After going through it, his teacher bumped his mark up by 25% to an 81/84. What was his original mark? Answer: Let x represent his original mark x + 0.25x = 81 1.25x = 81 x = 64.8 Diff: 2 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems. 198) In a local school, 450 of the students are male, which makes up 62 % of the entire school population. How many students go to this school? Answer:

x = 450

x = 720 Diff: 1 Type: SA Page Ref: 118-122 Topic: 3.5 Problems Involving Percent Objective: 3-2: Find percents and percent bases to solve business problems.

199) Mia is travelling to Buffalo from Niagara Falls to purchase a pair of shoes that are only available in America. The shoes are being sold for $345 in US dollars. Calculate the price of the shoes in Canadian dollars if one Canadian dollar is equal to 0.763359USD. Answer: 1USD = 1.3100CAD 345 × 1.31 = 451.95 The shoes cost $451.95 in CAD. Diff: 3 Type: SA Page Ref: 127-130 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency. 200) Ryan bought some chocolate from Switzerland in 2007. The chocolate costed €34 in Euros. What is the price of the chocolate in CAD if one Canadian dollar is worth 0.68 Euros? Answer:

= x = 34 ÷ 0.68 = 50 Diff: 2 Type: SA Page Ref: 124-127 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency. 201) A Canadian tech company planning on building a call center got a quote from an Indian company for Rs. 5.5 million with an exchange rate of 1 CAD to 51.9600 INR (Indian rupees). The tech company also got a quote from a Pakistani company for Rs. 8 million with an exchange rate of $1CAD to 81.0756 PKR (Pakistani rupees). Which quote is cheaper, in CAD? Answer:

← INR to CAD

= x = 5500000 ÷ 51.96 = $105 850.65 =

← PKR to CAD

= x = 8 000 000 ÷ 81.0756 = $98 673.34 Diff: 4 Type: SA Page Ref: 124-127 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency.

202) The price of a 40" HDTV in 2007 was $1600 while the price in 2016 was $370. Compare the two prices to create an index number. Answer:

= 0.23125 = 23.125%

The index number is 23.125 Diff: 1 Type: SA Page Ref: 128-130 Topic: 3.7 Applications - Index Numbers Objective: 3-5: Use index numbers and the Consumer Price Index to compute purchasing power of the Canadian dollar. 203) Andre earned $94 000 in 1999, when the CPI was 92.9. In 2015, when the CPI was at 126.6, how much does Andre need to earn to keep with inflation (the base year is 2002)? Answer: Real income in 1999 (100) = $101 184 Real income in 2015 (100) = $101 184 = $1011.84 x = $128 099 Diff: 2 Type: SA Page Ref: 128-130 Topic: 3.7 Applications - Index Numbers Objective: 3-5: Use index numbers and the Consumer Price Index to compute purchasing power of the Canadian dollar. 204) Using the 2016 federal income tax brackets and rates mentioned below in Figure 3.1, how much federal income tax should Anya report if her 2016 taxable income was $67 325? Answer: Federal tax for Anya = 15% × 45 282 + 20.5% × (67 325 - 45 282) = 0.15(45 282) + 0.205(22 043) = 6792.30 + 4518.82 = $11 315.12 Diff: 2 Type: SA Page Ref: 131-132 Topic: 3.8 Applications - Personal Income Taxes Objective: 3-6: Use federal income tax brackets and tax rates to calculate federal income taxes.

205) If Biyu earns $113 420 and $34 615 from her first and second job, respectively, calculate the amount of federal income tax she should report, based off of the 2016 federal income tax brackets and rates mentioned below in Figure 3.1. Answer: 113 420 + 34 615 = 148 035 Federal tax for Biyu = 15% × 45 282 + 20.5% × 45 281 + 26% × 49 825 + 29% × (148 035 - 140 388) = $31 247.04 Figure 3.1 • 15% on the first $45,282 of taxable income or less, plus; • 20.5% on the portion of taxable income over $45,282 up to $90,563, plus; • 26% on the portion of taxable income over $90,563 up to $140,388, plus; • 29% on the portion of taxable income over $140,388 up to $200,000, plus; • 33% of taxable income over $200,000. Diff: 3 Type: SA Page Ref: 130-131 Topic: 3.8 Applications - Personal Income Taxes Objective: 3-6: Use federal income tax brackets and tax rates to calculate federal income taxes. 206) Solve: 14 : 3 = 42 : x Answer: 14 : 3 = 42 : x = 14x = (42)(3) x = 126/14 x=9 Diff: 1 Type: SA Page Ref: 101-106 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 207) Fresh Co. sells large eggs at $3.15 per dozen. How many eggs can be bought for $2.80? A) None B) 4 C) 8 D) 10 E) 13 Answer: D Diff: 1 Type: MC Page Ref: 101-106 Topic: 3.2 Proportions Objective: 3-1: Use ratios and proportions to solve allocation and equivalence problems. 208) $84 is what percent of $39? Answer: Rate = 84/39 = 2.153846154 = 215.3846154% Diff: 1 Type: SA Page Ref: 107-115 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems.

209) The R.E.S.P. premium deducted from LeeAnn's wages was $720.30. If the premium rate is 3.014% of gross wages, how much were LeeAnn's gross wages? Answer: Let gross wages be $x (round to the nearest cent). 3.014% of x = 720.30 0.03014x = 720.30 x = $2402.12 Diff: 2 Type: SA Page Ref: 107-115 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 210) 18% of what number is 222? A) 3996.40 B) 39.96 C) 1233.33 D) 261.96 E) 240.00 Answer: C Diff: 1 Type: MC Page Ref: 107-115 Topic: 3.3 The Basic Percentage Problem Objective: 3-2: Find percents and percent bases to solve business problems. 211) What percent less than $55 is $24? Answer: Decrease = 31 R = 31/55 = 0.56363636 = 56.36% Diff: 1 Type: SA Page Ref: 115-119 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 212) Alyssa sold 38 furniture sets during August. Her target for September is to sell 15% more furniture sets than she sold in August. How many sets should she sell in September to meet her target? A) 44 B) 32 C) 45 D) 43 E) 80 Answer: A Diff: 2 Type: MC Page Ref: 115-119 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems.

213) $224.00 is what percentage less than $784.00? A) 66 % B) 28.57% C) 71.43% D)

E) 35% Answer: C Diff: 1 Type: MC Page Ref: 115-119 Topic: 3.4 Problems Involving Increase or Decrease Objective: 3-3: Find rates and original quantities for increase and decrease problems. 214) If it costs $265.50 CAD to purchase $248.00 USD, a) What is the exchange rate (from CAD to USD)? b) How much USD would you receive if you converted $840 Canadian dollars into U.S. dollars? Answer: a)

Exchange rate USD$ per CAD$ =

= 0.9340866

C$1 = US$0.9340866. b) 840.00(0.9340866) = US$784.632744. Diff: 2 Type: SA Page Ref: 127-130 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency. 215) After travelling to the USA, Alyssa converts her remaining USD to CAD at the exchange rate of C$1.235 per US$1. How many C$ will she receive for $422 if the currency shop charges a 5.5% commission on this transaction? A) $521.17 B) $341.70 C) $547.23 D) $324.62 E) $492.51 Answer: E Diff: 2 Type: MC Page Ref: 127-130 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency.

216) Laurel brought back a wheel of cheese from France. The cheese costed €68 in Euros. What is the price of the cheese in CAD if one Canadian dollar is worth 0.69 Euros? Answer: Let c be the price of the cheese wheel in CAD. = 1(68) = 0.69c 98.55 = c The cheese wheel is worth C$98.55. Diff: 2 Type: SA Page Ref: 127-130 Topic: 3.6 Applications - Currency Conversions Objective: 3-4: Use proportions and currency cross rate tables to convert currency. 217) Chidi, a moral philosophy professor, earned $64 700 per annum in 2012. Since the CPI in 2012 was 121.7, and in 2018 was 133.4, what did Chidi have to earn in 2018 to keep up with inflation, assuming his real income did not increase? Answer: Let x represent Chidi's expected income for 2018. = x = 70 920.13 Chidi should have earned $70 920.13 in 2018 to keep up with inflation. Diff: 2 Type: SA Page Ref: 121-134 Topic: 3.7 Applications - Index Numbers Objective: 3-5: Use index numbers and the Consumer Price Index to compute purchasing power of the Canadian dollar. 218) Mackaenleigh began working at Ren's Animal Clinic as a veterinarian in 2002, and earned $71050 per annum there. In 2017, she became their head doctor and began earning $153 200 per annum. In 2017, the CPI was 130.4 (with the base year of 2002). What is the real increase in Mackaenleigh's pay in terms of 2017 dollars? A) $245 849.20 B) $60 550.80 C) -$24 300.50 D) $92 699.2 E) $46 434.66 Answer: B Diff: 2 Type: MC Page Ref: 131-134 Topic: 3.7 Applications - Index Numbers Objective: 3-5: Use index numbers and the Consumer Price Index to compute purchasing power of the Canadian dollar.

219) Jane earned $76 500 at her firm in 2003, when the CPI was 102.8. In 2018, the CPI was at 133.4. How much must Jane be earning in 2018 to keep up with inflation, assuming she is earning the same amount (in real income) as she was in 2003? Answer: Let x represent Jane's income in 2018. = x = 99 271.40 Jane must be earning $99 271.40 in 2018 to keep up with inflation. Diff: 2 Type: SA Page Ref: 131-134 Topic: 3.7 Applications - Index Numbers Objective: 3-5: Use index numbers and the Consumer Price Index to compute purchasing power of the Canadian dollar. 220) In 2005, the Canadian Consumer Price Index was 107, and 9 years later, it was 125.2. Determine the purchasing power of the dollar for those 2 years, relative to the base year of 2002. Answer: Purchasing power for the dollar in 2005: =

(100) = 0.934579 = 93.46% of the purchasing power in the base year.

Purchasing power for the dollar in 2014: =

(100) = 0.798722 = 79.87% of the purchasing power in the base year.

Diff: 2 Type: SA Page Ref: 131-134 Topic: 3.7 Applications - Index Numbers Objective: 3-5: Use index numbers and the Consumer Price Index to compute purchasing power of the Canadian dollar. 221) Mackaeylaah is the head of the PTA board, and this year she calculated her 2017 taxable income to be $186 980. How much federal income tax should she pay if she is in the uppermost tax bracket, which is taxed at $28 106.05 plus 31% of income over $135 224? Answer: $28 106.05 + 0.31(186 980 - 135 224) = 44 150.41 Mackaeylaah has to pay $44 150.41 in federal income tax. Diff: 2 Type: SA Page Ref: 134-135 Topic: 3.8 Applications - Personal Income Taxes Objective: 3-6: Use federal income tax brackets and tax rates to calculate federal income taxes. 222) Use the 2018 income tax table (Table 3.3 on page 134 of your textbook) to calculate your federal tax payable if you had a taxable income of $126 984? A) $25 327.17 B) $25 326.13 C) $7762.61 D) $29 877.43 E) $32 083.01 Answer: B Diff: 3 Type: MC Page Ref: 134-135 Topic: 3.8 Applications - Personal Income Taxes Objective: 3-6: Use federal income tax brackets and tax rates to calculate federal income taxes.

223) Iesha's 2018 taxable income was $92 104. Using Table 3.3 (page 134) in the book, compute her federal tax. Answer: Federal tax = 0.15(46605) + 0.22(92104 - 46605) = $17 000.53 Diff: 2 Type: SA Page Ref: 134-135 Topic: 3.8 Applications - Personal Income Taxes Objective: 3-6: Use federal income tax brackets and tax rates to calculate federal income taxes.

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 4 Linear Systems 1) Solve the system of equations:

2x + 6y = -12 2x - 5y = 10

Answer: 2x + 6y = -12 2x - 5y = 10 11y = -22 y = -2 2x + 6(-2) = -12 2x - 12 = -12 2x = 0 x=0 Diff: 2 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 2) Solve the system of equations: Answer: x + 7y = 0 x + 4y = 8 3y = -8

x + 7y = 0 x + 4y = 8

y= x+7 x-

=0 =0

x= Diff: 2 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination.

3) Solve the system of equations:

2x + 8y = -18 2x - 4y = 10

Answer: 2x + 8y = -18 2x - 4y = 10 4y = -8 y = -2 2x + 8(-2) = -18 2x - 16 = -18 2x = -2 x = -1 Diff: 2 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 4) Solve the system of equations:

5x = 10 - 2y 5y = 3x - 30

Answer: 5x = 10 - 2y 5y = 3x - 30 5x + 2y = 10 -3x + 5y = -30 15x + 6y = 30 -15x + 25y = -150 31y = -120 y = -3.87 5x + 2(-3.87) = 10 5x - 7.74 = 10 5x = 17.74 x = 3.548 Diff: 2 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination.

5) Solve the system of equations:

4x + 3y = 36 2x + 9y = -18

Answer: 4x + 3y = 36 4x + 18y = -36 -15y = 72 y = -4.8 4x + 3(-4.8) = 36 4x - 14.4 = 36 4x = 50.4 x = 12.6 Diff: 2 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 6) Solve the system of equations:

3x - 8y + 72 = 0 7x = 4y - 8

Answer: 3x - 8y + 72 = 0 7x = 4y - 8 3x - 8y = -72 7x - 4y = -8 3x - 8y = -72 14x - 8y = -16 -11x = -56 x = 5.0909 3(5.0909) - 8y = -72 15.2727 - 8y = -72 -8y = -87.2727 y = 10.909 Diff: 2 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 7) Solve the system of equations:

4x + 18y = 36 2x + 9y = -18

Answer: 4x + 18y = 36 4x + 18y = -36 0 = 72 no solution, lines are parallel Diff: 2 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination.

8) Solve the system of equations:

8x + 12y = 72 6x + 7y = 84

Answer: 8x + 12y = 72 6x + 7y = 84 48x + 72y = 432 48x + 56y = 672 16y = -240 y = -15 8x + 12(-15) = 72 8x - 180 = 72 8x = 252 x = 31.5 Diff: 2 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 9) Solve the system of equations: Answer: 0.8x + 0.3y = 1.3 0.3x + 0.2y = 1.1

0.8x + 0.3y = 1.3 0.3x + 0.2y = 1.1

8x + 3y = 13 3x + 2y = 11 16x + 6y = 26 9x + 6y = 33 7x = -7 x = -1 3(-1) + 2y = 11 -3 + 2y = 11 2y = 14 y=7 Diff: 2 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination.

10) Solve the system of equations:

6.6x + 3.5y = 110 2.9x + 4.7y = 106

Answer: 6.6x + 3.5y = 110 2.9x + 4.7y = 106 66x + 35y = 1100 29x + 47y = 1060 1914x + 1015y = 31900 1914x + 3102y = 69960 -2087y = -38060 y = 18.236703 2.9x + 4.7(18.236703) = 106 2.9x + 85.712506 = 106 2.9x = 20.287494 x = 6.9956876 Diff: 2 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 11) Solve the system of equations:

2.2x + 0.85y = 2.25 1.5x + 1.75y = 2.75

Answer: 2.2x + 0.85y = 2.25 1.5x + 1.75y = 2.75 ∗ 100 220x + 85y = 225 ∗ 100 150x + 175y = 275 ÷5 44x + 17y = 45 ÷ 25 6x + 7y = 11 ∗7 308x + 119y = 315 ∗ 17 102x + 119y = 187 206x = 128 x=

= 0.6213592

2.2(0.6213592) + .85y = 2.25 0.85y = 2.25 - 1.3669903 0.85y = 0.8830097 y = 1.0388345 Diff: 2 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination.

12) Solve the system of equations:

Answer: +

+ =

multiply the equation by 20 to eliminate the denominators

multiply the equation by 72 to eliminate the denominators

*20 *72 ÷2

36x + 30y = 94 32x + 45y = 10 18x + 15y = 47 32x + 45y = 10 *3 54x + 45y = 141 32x + 45y = 10 22x = 131 x = 5.954545 18(5.954545) + 15y = 47 15y + 107.1818184 = 47 15y = -60.181818 y = -4.012121 Diff: 3 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 13) Solve the system of equations:

6x + 5y = 30 4x - 3y = 24

Answer: 6x + 5y = 30 4x - 3y = 24 ∗4 24x + 20y = 120 ∗6 24x - 18y = 144 38y = -24 y = -.631579 6x + 5(-.631579) = 30 6x - 3.1578947 = 30 6x = 33.157895 x = 5.5263158 Diff: 2 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination.

14) Solve the system of equations:

= +

Answer: +

∗ 84 ∗ 36 ∗ 24 ∗ 21

21x + 48y = -16 24x + 54y = -7 504x + 1152y = -384 504x + 1134y = -147 18y = -237 y = -13.166667 21x + 48(-13.166667) = -16 21x - 632 = -16 21x = 616 x = 29.3333333 Diff: 3 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 15) Solve the system of equations:

y = 2.5

0.2x -

y = 1.2

Answer: x+

y = 2.5

0.2x -

y = 1.2

2x + 3y = 5 x - 2y = 6 2x + 3y = 5 2x - 4y = 12 7y = -7 y = -1 x - 2(-1) = 6 x+2=6 x=4 Diff: 2 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination.

16) Find the slope and y-intercept: 4x + 5y = 31 Answer: 4x + 5y = 31 5y = -4x + 31 y= Slope, m = -

, y-intercept, b =

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 17) Find the slope and y-intercept: 3x + 2y = 10 Answer: 3x + 2y = 10 2y = -3x + 10 y= Slope, m = -

, y-intercept, b =

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 18) Find the slope and y-intercept: 1 -

y = 4x

Answer: 1-

y = 4x

y = 4x - 1

y = -8x + 2 Slope, m = -8; y-intercept, b = 2 Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 19) Find the slope and y-intercept: 2x - y = 8 Answer: 2x - y = 8 -y = -2x + 8 y = 2x - 8 Slope, m = 2; y-intercept, b = -8 Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

20) Find the slope and y-intercept: 4 -

y = 2x

Answer: 4-

y = 2x

8 - y = 4x -y = 4x - 8 y = -4x + 8 Slope, m = -4; y-intercept, b = 8 Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 21) Find the slope and y-intercept: 10 -

y = 8x

Answer: 10 -

y = 8x

40 - y = 32x -y = 32x -40 y = -32x + 40 Slope, m = -32; y-intercept, b = 40 Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 22) Find the slope and y-intercept: 3y - 5 = 2(x - 2) Answer: 3y - 5 = 2(x - 2) 3y - 5 = 2x - 4 3y = 2x + 1 y=

Slope, m =

; y-intercept, b =

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

23) Find the slope and y-intercept: 1.5x + 3y = 2.4 Answer: 1.5x + 3y = 2.4 15x + 30y = 24 30y = -15x + 24 y=-

y=- x+ Slope, m = - ; y-intercept, b = Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 24) Find the slope and y-intercept: 4.5x + 9y = 81 Answer: 4.5x + 9y = 81 9x + 18y = 162 18y = -9x + 162 y=

y=- x+ Slope, m = -

; y-intercept, b =

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

25) Graph: x + 4y = 8 Answer: x 0 8 y 2 0

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

26) Graph: 2x - 4y = 10 Answer: Reduce the equation by 2: x - 2y = 5 x y

0 -2.5

5 0

1 -2

Diff: 1 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

27) Graph: y = 3 - 2x Answer: x 0 -3 y 3 9

3 -3

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

28) Graph: 5x + 4y = 20 Answer: x 0 4 y 5 0

Diff: 1 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

29) Graph: x = 5y Answer: x 5 y 0

5 1

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 30) Graph: 3x + 4y = 12 Answer: x 0 4 y 3 0

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

31) Graph: 7x + 4y = 0 Answer: x 0 4 y 0 -7

Diff: 1 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

32) Graph: x = 7 Answer:

Diff: 1 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

33) Graph: y = -6 Answer:

Diff: 1 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 34) Graph: x = -4 Answer:

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

35) Graph: y = 8 Answer:

Diff: 1 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

36) Graph: y = -4x + 16 Answer: For y = -4x + 16 Slope, m = -4 y-intercept, b = 16

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

37) Graph: -2y = -x + 6 Answer: For -2y = -x + 6 y = 0.5x - 3 Slope, m = 0.5 y-intercept, b = -3 x -2 0 6 10

y -4 -3 0 2

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

38) Graph: 3x - 2y = 18 Answer: x y 0 -9 2 -6 4 -3 6 0 8 3

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

39) Solve graphically: 5x - 2y = 10 and y = 5 Answer: (4, 5) is the solution.

x -2 0 6

y=5 y 5 5 5

y -10 -5 10

5x - 2y = 10

Diff: 2 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

40) Solve graphically: x - y = 4 and x + y = 6 Answer: (5,1) is the solution.

Diff: 2 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

41) Solve graphically: 2x + 3y = -4 and x - 2y = 5 Answer: 2x + 3y = -4 x 4 -2 1 y -4 0 -2

x - 2y = 5 x 0 y -2.5

5 0

3 -1

Solution is (1,-2) which can be confirmed by elimination: 2x + 3y = -4 x - 2y = 5 2x + 3y = -4 2x - 4y = 10 7y = -14 y = -2 2x + 3y = -4 2x + 3(-2) = -4 2x - 6 = -4 2x = 2 x=1

Diff: 2 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

42) Solve graphically: x = y, 2y = x - 200 Answer: Solution is (-200,-200). x=y x y

0 0

400 400

-400 100

2y = x - 200 x 0 y -100

400 100

-400 -300

Diff: 2 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

43) Solve graphically: 2y = x + 200 and 3x + 2y = 240 Answer: Solution is (10,105). 2y = x + 200 x 0 y 100

20 110

40 120

3x + 2y = 240 x 0 y 120

20 90

40 60

Diff: 2 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

44) Solve graphically: y = -x - 5 and x - y = 6 Answer: Solution is (0.5,-5.5)

Diff: 2 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

45) Solve graphically: 4x = -2y and x = 6 Answer: Solution is (6,-12).

Diff: 2 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

46) Solve graphically: 4x + 3y = 12 and 3x - 4y = -24 Answer: 4x + 3y = 12 x 0 3 1 y 4 0 3 3x - 4y = 24 x 0 y 6

-8 0

4 -3

4x + 3y = 12 3x - 4y = -24 12x + 9y = 36 12x - 16y = -96 25y = 132 y= 4x + 3y = 12 4x + 3 ∗

= 12

4x +

= 12

4x = 12 100x = 300 - 396 100x = -96 x=

(-24/25, 132/25) is the solution.

Diff: 2 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

47) Solve graphically: 2x = -5y and 3x + y = 9 Answer: 2x = -5y 2x + 5y = 0 x 0 -5 5 y 0 2 -2 3x + y = 9 x y

0 9

3 0

-3 18

2x + 5y = 0 3x + y = 9 2x + 5y = 0 15x + 5y = 45 -13x = -145 x= 3∗

+y=9 +y=9

y=913y = 117 - 135 y=

(45/13, -18/13) is the solution.

Diff: 2 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

48) Solve graphically: 4y = -5x and x = -6 Answer: 4y = -5x x y

0 0

5 -4

-5 4

(-6,15/2) is the solution. 4y + 5x = 0 4y + 5(-6) = 0 4y - 30 = 0 4y = 30 y = 7.5 x = -6

Diff: 2 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

49) Solve the system graphically. 3x - 7 = y 6x -2 y = 14 Answer: Infinitely many solutions, lines coincide. Diff: 1 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables. 50) A theatre holds 400 seats. Two types of seats are available: orchestra and balcony. The cost of an orchestra seat is $60.00 and the cost of a balcony seat is $45.00. If all 400 seats are sold the theatre would collect $22 500.00. How many of each type of seat is available? Answer: Let the orchestra seat be x and the balcony seat be y. x + y = 400 60x + 45y = 22 500 60x + 60y = 24 000 60x + 45y = 22 500 15y = 1500 y = 100 x + 100 = 400 x = 300 Diff: 2 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 51) A tavern holds 50 seats. Two types of seats are available the Friday night jazz festival: stage and dining. The cost of a stage seat is $15.00 and the cost of a dining seat is $5.00. If all 50 seats are sold the tavern would collect $400.00. How many of each type of seat is available? Answer: Let the stage seat be x and the dining seat be y. x + y = 50 15x + 5y = 400 15x + 15y = 750 15x + 5y = 400 10y = 350 y = 35 x + 35 = 50 x = 15 Diff: 2 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables.

52) Twice the larger number is greater than three times the smaller number by 19. The sum of the larger number plus one-third of the smaller number is 26. Find the two numbers. Answer: Let the larger number be x and the smaller number be y. 2x - 3y = 19 x+

y = 26

2x - 3y = 19 3x + y = 78 2x - 3y = 19 9x + 3y = 234 11x = 253 x = 23 2(23) - 3y = 19 46 - 3y = 19 -3y = -27 y=9 Diff: 2 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 53) The difference between seven times the first number and three times the second number is 12. The sum of three-fourths of the first number and two-thirds of the second number is 21. Find the two numbers. Answer: Let the first number be x and the second number be y. 7x - 3y = 12 x+

y = 21

∗ 12 ∗9 ∗7

9x + 8y = 252 63x - 27y = 108 63x + 56y = 1764 -83y = -1656 y = 19.951807 7x - 3(19.951807) = 12 7x - 59.855422 = 12 7x = 71.855422 x = 10.26506 Diff: 2 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables.

54) Hamidah's sales last week were $251 less than three times Colleen's sales. Together they sold $1123. Determine how much each person sold last week. Answer: Let Hamidah's sales be $x and Colleen's sales be $y. x + y = 1123 x = 3y - 251 Rearrange x - 3y = -251 x + y = 1123 x - 3y = -251 4y = 1374 y = 343.50 x + 343.50 = 1123 x = $779.50 Diff: 2 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 55) The Oak Store has been producing 4640 chairs a day working two shifts. The second shift has produced 120 chairs fewer than four-thirds of the number of chairs produced by the first shift. Determine the number of chairs each shift has produced. Answer: x + y = 4640 y=

x - 120

3y = 4x - 360 -4x + 3y = -360 x + y = 4640 -4x + 3y = -360 ∗3 3x + 3y = 13920 -4x + 3y = -360 7x = 14280 x = 2040 2040 + y = 4640 y = 2600 Diff: 2 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. Rearrange

∗3

56) A bakery has been baking 4000 donuts a day for distribution. The second shift has produced 1000 donuts fewer than two-thirds of the number of donuts produced by the first shift. Determine the number of donuts each shift has produced. Answer: x + y = 4000 y=

x - 1000

3y = 2x - 3000 -2x + 3y = -3000 x + y = 4000 -2x + 3y = -3000 ∗3 3x + 3y = 12000 -2x + 3y = -3000 5x = 15000 x = 3000 3000 + y = 4000 y = 1000 Diff: 2 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. Rearrange

∗3

57) A machine requires five hours to make a unit of Product A and four hours to make a unit of Product B. Last month the machine operated for 440 hours producing a total of 100 units. How many units of each type of product did it produce? Answer: Let the number of units of Product A be x and the number of units of Product B be y. x + y = 100 5x + 4y = 440 ∗4 4x + 4y = 400 5x + 4y = 440 -x = -40 x = 40 40 + y = 100 y = 60 Diff: 2 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables.

58) A machine requires 3 hours to make a unit of Product A, and 6 hours to make a unit of Product B. The machine operated for 195 hours producing a total of 40 units. How many hours were used to manufacture units of Product A? Answer: Let the number of units of Product A be x. Let the number of units of Product B be y. x + y = 40 3x + 6y = 195 ∗3 3x + 3y = 120 3x + 6y = 195 -3y = -75 y = 25 x + 25 = 40 x = 15 Diff: 2 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 59) A machine requires 1 hours to make a unit of Product A, and 4 hours to make a unit of Product B. The machine operated for 195 hours producing a total of 80 units. How many hours were used to manufacture units of Product A? Answer: Let the number of units of Product A be x. Let the number of units of Product B be y. x + y = 80 x + 4y = 200 -3y = -120 y = 40 x + 40 = 80 x = 40 Diff: 2 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables.

60) Solve the system of equations: Answer:

28 - 7x = 4y 36 - 4y = 3x

28 - 7x = 4y 36 - 4y = 3x Rearrange 28 = 7x + 4y 36 = 3x + 4y -8 = 4x x = -2 36 = 3(-2) + 4y 36 = -6 + 4y 42 = 4y y = 10.5 Diff: 2 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 61) Bonny and Clyde divide a profit of $15 700. If Clyde is to receive $4200 more than one-fifths of Bonny's share, how much will Bonny receive? Answer: Let Clyde's share of the profit be $x, and Bonny's share of the profit be $y. x + y = 15700 x=

y + 4200

∗5 Rearrange

5x = y + 21000 5x - y = 21000 x + y = 15700 5x - y = 21000 6x = 36700 x = 6116.67 6116.67 + y = 15700 y = $9583.33 Diff: 2 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables.

62) Solve for the following:

2x - 3y = -24 3x + 5y = 45

A) x = 0.789, y = 8.526 B) x = .489, y = 8.526 C) x = 4.789, y = 9.526 D) x = 2.789, y = 7.526 E) x = 0.2789, y = 0.7526 Answer: A Diff: 2 Type: MC Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 63) Solve for x and y:

2x + 1y = 5 4x + 2y = 20

A) x = 2, y = 1 B) x = -2, y = -1 C) x = 4, y = 2 D) x = -4, y = -2 E) no solution, the lines are parallel Answer: E Diff: 1 Type: MC Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 64) Find the y-intercept for the following equation: 5y + 7 = 19 A) 2.6 B) 0 C) 2.4 D) 2.8 E) 3.0 Answer: C Diff: 1 Type: MC Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 65) Find the y-intercept for the following equation: 8x = 2 + y A) 8 B) -8 C) 2 D) -2 E) 4 Answer: D Diff: 1 Type: MC Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

66) Find the y-intercept for the following: 3x - 5y = 60 A) 12 B) -12 C) 60 D) -60 E) -5 Answer: B Diff: 1 Type: MC Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 67) Find the slope of the line: 2y = 4(y + 2) A) 0 B) 2 C) 4 D) 0.5 E) 1 Answer: A Diff: 2 Type: MC Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 68) Find the slope of the following: 2x + 5y = 50 A) .4 B) 2.5 C) -2.5 D) -.4 E) 10 Answer: D Diff: 1 Type: MC Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 69) Find the slope of the following: 2y - 3x = 6 A) 3 B) 1.5 C) -1.5 D) -3 E) 6 Answer: B Diff: 1 Type: MC Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

70) Solve for x and y:

4x + 9y = 72 6x + 15y = 90

A) x = 45, y = -12 B) x = -45, y = 12 C) x = 45, y = 12 D) x = -45, y = -12 E) x = -45, y = -15 Answer: A Diff: 2 Type: MC Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 71) It takes a snowboarder 46 minutes to descend from an elevation of 9200 metres to an elevation of 4508 metres. What is the rate of descent (slope) per minute? A) 120 B) 220 C) 10.2 D) 102 E) 12.0 Answer: D Diff: 2 Type: MC Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 72) A couple is about the get married. They decide to start a joint bank account to save for a house. They want to have an initial investment of $9000. The first partner contributes $350 more than three-quarters what the second partner contributes. How much does each partner contribute? A) P1 = 4557.14, P2 = 4442.86 B) P1 = 3557.14, P2 = 5452.86 C) P1 = 4057.14, P2 = 4942.86 D) P1 = 5057.14, P2 = 3942.86 E) P1 = 5857.14, P2 = 3992.86 Answer: C Diff: 3 Type: MC Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables.

73) Find the slope of the line: 4y = 5(y + 3) A) 15 B) -15 C) 0 D) -5 E) 3 Answer: C Diff: 2 Type: MC Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 74) Solve for x and y:

7x + 3y = 42 4x + 6y = 36

A) x = -4.8, y = 2.8 B) x = 4.8, y = 2.8 C) x = -4.8, y = -2.8 D) x = 4.8, y = -2.8 E) x = 4.8, y = -5.6 Answer: B Diff: 3 Type: MC Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 75) Find the slope of the following: x + 5y = 10 A) 2 B) 0.5 C) 5 D) -0.2 E) 1 Answer: D Diff: 1 Type: MC Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 76) Find the slope of the following: 20y - 4x = 5 A) 0.8 B) 1.25 C) 0.2 D) 5 E) 4 Answer: C Diff: 1 Type: MC Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

77) Find the slope of the line: 2y = 5(x + 2) A) 2 B) 5 C) 1 D) 2.5 E) 0.4 Answer: D Diff: 2 Type: MC Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 78) Solve for x and y:

x + 9y = 18 x + 15y = 42

A) x = 18, y = 4 B) x = 18, y = -4 C) x = -18, y = -4 D) x = -18, y = 4 E) x = 4, y = -18 Answer: D Diff: 2 Type: MC Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 79) It takes a car 10 seconds to increase its speed from from 60 k/hr to 90 k/hr. What is the rate of acceleration (slope) in k/hr/sec? A) 6 B) 9 C) 3 D) 1/6 E) 1/9 Answer: C Diff: 1 Type: MC Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables.

80) It takes your company 6 labour hours to manufacture a simple hot tub and 9 labour hours to manufacture a deluxe hot tub. Last week your company used a total of 795 labor hours to manufacture a total of 105 hot tubs. How many of each type of hot tub did you manufacture? A) Simple 40, Deluxe 65 B) Simple 55, Deluxe 50 C) Simple 75, Deluxe 30 D) Simple 50, Deluxe 55 E) Simple 65, Deluxe 50 Answer: D Diff: 3 Type: MC Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 81) Two brothers decide to invest in a vacant lot which cost $51 000. The first brother contributes $1000 more than one-quarter of what the second brother contributes. How much does each brother contribute? A) B1 = 11000, B2 = 40000 B) B1 = 40000, B2 = 11000 C) B1 = 12500, B2 = 38500 D) B1 = 50000, B2 = 10000 E) B1 = 1000, B2 = 50000 Answer: A Diff: 2 Type: MC Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 82) Solve and find the value of a, given 3a + 4b = 20 and 2a + b = 15 A) a = B) a = 7 C) a = 5 D) a = -1 E) a = 8 Answer: E Diff: 1 Type: MC Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination.

83) Solve and find the value of x, given:

and x + 5y = 38

A) x = 20 B) x = -9 C) x = 9.02 D) x = -17.16 E) x = 4.33 Answer: C Diff: 1 Type: MC Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 84) Blue Mountain resort charges $39 for a single ticket for high ropes and $29 for a single ticket for low ropes. If a day's total revenue from the sale of 540 passes is $18 910, how many tickets were sold for high ropes? Answer: 39HR + 29LR = 18910 HR + LR = 540 ⇒ 29HR + 29LR = 29 × 540 HR =

= 325

325 tickets were sold for high ropes Diff: 2 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 85) The age difference between my son and me is 30 years. In 3 years from today, I will be thrice as old as my son. What is my age now? A) 12 B) 42 C) 38 D) 45 E) 15 Answer: B Diff: 2 Type: MC Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables.

86) Solve and find the value of y, given 3x + 5y = 5 and x + 2y = 1 Answer: 3x + 5y = 5 x + 2y = 1 ⇒ 3x + 6y = 3 ⇒ y = -2 Diff: 1 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 87) Sobeys sells fruit trays consisting of pineapple and melons. The manager of the fruit department obtains pineapples at a wholesale price of $2.50 per kg, grapes at $2.15 per kg, strawberries at $1.50 per kg and melons at $1.85 per kg. He is required to produce 10 kg of the mixed fruit. He is required to add 2 kg of strawberries and 3 kg of grapes. What is the maximum weight of pineapple that he can put in the mix in order to have an effective wholesale cost no greater than $2.00 per kg? Answer: Cost of 2 kg of strawberries = $1.50 × 2 = $3.00 Cost of 3 kg of grapes = $2.15 × 3 = $6.45 Total cost of the mixed fruit = $2 × 10 = $20 Cost of pineapple and melons = $20 - ($3 + $6.45) = $10.55 ⇒ 2.5p + 1.85m = 10.55 Maximum weight of pineapple and melons = 10 - (2 + 3) = 5kg ⇒ p + m = 5 2.5p + 1.85m = 10.55 1.85p + 1.85m = 5 × 1.85 0.65p = 10.55 - 9.25 ⇒ p=

∴ Maximum weight of pineapple = 2 kg Diff: 2 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 88) An electrician charges a flat fee of $90 for a home service call. In addition he charges $25 for every 20 minutes as labour cost. If one writes an equation to calculate total charges in terms of hours of labour, what will be the slope of the line? A) 25 B) 90 C) 50 D) 75 E) 100 Answer: D Diff: 1 Type: MC Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

89) Solve the equation for x and find slope and x-intercept of 2x + 4y = 8 A) -2, 4 B) - , 2 C) -4, 8 D) -2,8 E) 4, -2 Answer: A Diff: 1 Type: MC Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 90) Solve graphically: 0.2y = 0.1x + 2 and 0.3x + 0.2y = 2.4 Answer: Solution is (1,10.5). 0.2y = 0.1x + 2 x 0 y 10

1 10.5

4 12

0.3x + 0.2y = 2.4 x 0 y 12

1 10.5

4 6

Diff: 2 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

91) Operations costs increases linearly with the number of staff and are represented by the equations OC = 12000N + 3000 and 3OC - 7000N = 252000. Graphically calculate the operations cost and number of staff. Answer: Solution is (9,$111,000).

Diff: 2 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables. 92) Solve the system of equations:

y-8=

(x - 8)

2y - x = 8 A) 10,9 B) 20,14 C) 12,10 D) -4,2 E) Unique answer is not possible Answer: E Diff: 2 Type: MC Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination.

93) Calculate the x-intercept and y-intercept for the following equation: -x + 2y = 8 A) x-intercept does not exist; y-intercept = 4 B) -2; C) -8; 8 D) -8; 4 E) 2; Answer: D Diff: 2 Type: MC Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 94) My son and daughter were born two years apart. 9 years ago, my son was twice the age of my daughter. What are their ages now? A) 12, 10 B) 4, 2 C) 16, 14 D) 22, 20 E) 13, 11 Answer: E Diff: 2 Type: MC Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables.

95) Graph: x = Answer: x -3.33 y 0

-4 2

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 96) Graph: y = -(0.22x + 7) Answer: x 0 5 y -7 -8.1

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

97) 1000 tickets were sold for Cirque du Soleil. Adult tickets cost $8.50, children's cost $4.50, and a total of $7300 was collected. How many tickets were sold to adults? A) 1000 B) 300 C) 700 D) 500 E) Not enough data to answer the question Answer: C Diff: 2 Type: MC Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 98) Solve the system of equations: -

3x + 4y = 12 Answer: 9x - 4y = 72 3x + 4y = 12 12x = 84 x=7 3(7) + 4y = 12 4y = -9

=6 (× 12)

y= Diff: 4 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 99) How many solutions does the following system of linear equations have? x = 3y + 4 3x - 9y = 13 Answer: 3x - 9y = 13 3x = 9y + 13 x = 3y +

(÷ 3)

x = 3y + 4 Both linear equations have the same slope but different y-intercepts, making them parallel. As a result, there are no solutions to this set. Diff: 2 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination.

100) Solve the system of equations: 3x + y = 13 x - 6y = 7 Answer: x - 6y = 7 3x - 18y = 21 (× 3) 3x + y = 13 -17y = 34 y = -2 x – 6(-2) = 7 x + 12 = 7 x = -5 Diff: 3 Type: SA Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 101) Solve for x and y: 34x + 8y = 42 2y =

A) x = 2, y = -2 B) x = 0, y = -3 C) x = 1, y = 2 D) infinite solutions, lines coincide E) no solutions, lines are parallel Answer: D Diff: 2 Type: MC Page Ref: 141-147 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination. 102) Find the slope and y-intercept: -6y - 24 = 5x Answer: -6y = 5x + 24 Y=

x-4

Slope, m =

, y-intercept, b = 4

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

103) Graph: y =

x-4

Answer:

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables. 104) Find the slope and y-intercept: 43 + 7y = 16x Answer: 7y = 16x - 43 y=

Diff: 2 Type: SA Page Ref: 148-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

105) Solve graphically: 4y - 20 = 6x and x = 2 Answer: the POI (x,y) = (2, 8)

Diff: 2 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

106) Solve graphically: 4x = -3y and 42 - 7y = 3x Answer: 4x = -3y 4x - 3y = 0 x 0 -3 3 y 0 4 -4 42 - 7y = 3x -3x - 7y = -42 x 0 y 6

(-6,

-7 9

7 3

) is the solution

Diff: 2 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

107) Graph the system x = 2y and -x - y + 3 = 0 for all values of x from x = -3 to x = 3 Answer: x = 2y x=y -x - y + 3 = 0 -y = x - 3 y = -x + 3

(2,1) is the solution Diff: 2 Type: SA Page Ref: 160-163 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

108) Alex speaks 150 words per minute on average in French, and 190 words per minute on average in Spanish. She once gave cooking instructions in French, followed by cleaning instructions in Spanish. Alex spent 5 minutes total giving both instructions, and spoke 270 more words in Spanish than in French. How long did Alex speak in French, and how long did she speak in Spanish? Answer: Let x represent the minutes spoken in French and let y represent the minutes spoken in Spanish x+y=5 y = -x + 5 190y - 150x = 270 190(-x + 5) - 150x = 270 -190x + 950 - 150x = 270 -340x = -680 x=2 2+y=5 y=3 Diff: 3 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 109) The town of Whitby held a painting day for adults and kids alike. Each child made 15 paintings and each adult made 7 paintings. A total of 208 paintings were made by the group, and there were 3 times as many kids as there were adults. How many kids were there, and how many adults were there? Answer: Let x represent the number of kids at the event, and let y represent the number of adults x = 3y 15x + 7y = 2236 15(3y) + 7y = 2236 45y + 7y = 2236 52y = 2236 y = 43 x = 3(43) x = 129 Diff: 3 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables.

110) Mrs. Aamir's English exam has 20 questions worth 100 points. Short answer questions are out of 3 and long answer questions are marked out of 11. How many of each type of question exist on this quiz? Answer: Let x represent short answer questions and let y represent long answer questions x + y = 20 3x + 3y = 60 3x + 11y = 100 8y = 40 y=5 x + 5 = 20 x = 15 Diff: 2 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 111) The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended? Answer: Let x represent the number of adults and let y represent the number of kids x + y = 2200 y = -x + 2200 4x + 1.5y = 5050 4x + 1.5(-x + 2200) = 5050 4x - 1.5x + 3300 = 5050 2.5x = 1750 x = 700 700 + y = 2200 y = 1500 Diff: 3 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 112) The difference of two numbers is 3. Their sum is 13. Find the numbers. Answer: Let x represent the first number and y represent the second number x + y = 13 x-y=3 2x = 16 x=8 8 + y = 13 y=5 Diff: 1 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables.

113) Xavier and Yannic went to an arcade where the machines took tokens. Xavier played 4 games of skee ball and 3 games of pinball, using a total of 17 tokens. At the same time, Yannic played 4 games of skee ball and 4 games of pinball, using up 20 tokens. How many tokens does each game require? Answer: Let x represent the number of tokens used for skee ball, and let y represent the number of tokens used for pinball 4x + 3y = 17 4x + 4y = 20 -y = -3 y=3 4x + 3(3) = 17 4x = 8 x=2 Diff: 2 Type: SA Page Ref: 165-169 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 114) Graph: y > -2x Answer:

The entirety of the plane to the right of the line is the answer. Diff: 1 Type: SA Page Ref: 170-176 Topic: 4.5 Graphing Inequalities Objective: 4-5: Solve systems of linear inequalities in two variables using graphical methods.

115) Graph: 6x - 2y < 14 Answer: -2y < -6x + 14 y < 3x - 7

The entirety of the plane to the left of the line is the answer. Diff: 2 Type: SA Page Ref: 170-176 Topic: 4.5 Graphing Inequalities Objective: 4-5: Solve systems of linear inequalities in two variables using graphical methods.

116) Graph the region defined by x > 0, 6y + 12x < 30, and 6x - 2y > 10 Answer: 6y < -12x + 24 y < -2x + 5 -2y > -6x + 8 y > 3x + 5

The region within triangle ABC formed by the three lines is the answer. Diff: 4 Type: SA Page Ref: 170-176 Topic: 4.5 Graphing Inequalities Objective: 4-5: Solve systems of linear inequalities in two variables using graphical methods.

117) Construct tables of values for the following equations from {x∈ I, -3 ≤ x ≤ 3}: a) 28x - 7y = 14 b)

x-2

Answer: a) y = 4x - 2 x -3 -2 -1 0 1 2 3

y -14 -10 -6 -2 2 6 10

b) y = 2x - 12 x -3 -2 -1 0 1 2 3

y -18 -16 -14 -12 -10 -8 -6

Diff: 1 Type: SA Page Ref: 143-155 Topic: 4.1 Algebraic Solution of Systems of Linear Equations in Two Variables Objective: 4-1: Solve linear systems consisting of two simultaneous equations in two variables using algebraic elimination.

118) Determine the slope and y-intercept of the lines of following equations (using algebra): a) 63x - 7y = 35 b) 6x + 24y = -36 Answer: a) 63x — 7y = 35 -7y = -63x + 35 y = 9x — 5 m = 9, b = -5. b) 6x + 24y = -36 24y = -6x - 36 y=m=-

,b=-

120) Graph the following equation for values {-12 ≤ y ≤ 8}: 20x + 4y = 32 Answer: 20x + 4y = 32 4y = -20x + 32 y = -5x + 8

121) Graph following linear system: 4x - 3y = -144 6x +

= 2x +5

Answer: 4x - 3y = -144 -3y = -4x - 144 y= 6x +

x + 48

= 2x +5 = -4x + 5 y = -8x + 10

Solution: (-4.07, 42.571) Diff: 1 Type: SA Page Ref: 155-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

122) Using the graphical method, solve the following system of linear equations: 3x - y = 7 2x + 3y = 1 Answer: 3x - y = 7 -y = -3x + 7 y = 3x — 7 2x + 3y = 1 3y = -2x + 1 y=

Diff: 1 Type: SA Page Ref: 155-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

123) Graph the following system of linear equations: -2x + y = -4x + 13 x+y=8 Answer: -2x + y = -4x + 13 y = -2x +13 x+y=8 y = -x + 8

Diff: 1 Type: SA Page Ref: 155-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

124) Solve the following linear system: x − 7y = -11 5x + 2y = -18 Answer: x - 7y = -11 -7y = -x — 11 y=

5x +2y = -18 2y = -5x - 18 y=-

-9

Diff: 1 Type: SA Page Ref: 155-159 Topic: 4.2 Graphing Linear Equations Objective: 4-2: Graph linear equalities in two variables.

125) Solve the following linear system: 18x - 6y = 24 and -9x + 3y = 15 Answer: 18x - 6y = 24 -6y = -18x + 24 y = 3x - 4 -9x + 3y = 15 3y = 9x + 15 y = 3x + 5 Both equations have the same slope (m = 3) and different y-intercepts, therefore they are parallel there is no solution to the system.

Diff: 2 Type: SA Page Ref: 155-159 Topic: 4.2 Graphing Linear Equations Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

126) Solve the following system of equations for x = 0 to x = 10 2x + 4y = -10 6x + 3y = 6 Answer: 2x + 4y = -10 4y = -2x - 10 y=-

6x + 3y = 6 3y = -6x + 6 y = -2x + 2

Diff: 1 Type: SA Page Ref: 155-159 Topic: 4.2 Graphing Linear Equations Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

127) Solve the following linear system graphically: 7x − 8y = -12 -4x + 2y = 3 Answer: 7x − 8y = -12 -8y = -7x - 12 y = 78x + 32

-4x + 2y = 3 2y = 4x + 3 y = 2x + The two equations have the same y-intercept (b =

), therefore (0,

) is the solution.

Diff: 2 Type: SA Page Ref: 155-159 Topic: 4.2 Graphing Linear Equations Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

128) Solve the following system: 3x + 9y = -6 -4x − 12y = 8 Answer: 3x + 9y = -6 9y = -3x — 6 y=-

-4x - 12y = 8 -12y = 4x + 8 y=-

Both equations are the same (have the same slope (m = -

) and y- intercept (b = -

an infinite number of solutions to the system.

Diff: 2 Type: SA Page Ref: 155-159 Topic: 4.2 Graphing Linear Equations Objective: 4-3: Graph linear systems consisting of two linear relations in two variables.

)), therefore there is

129) Solve the following system of linear equations algebraically 8x + 2y = y + 13 -3x + 4y = -6x — 6 Answer: 8x + 2y = y + 13 y = -8x + 13 3x + 4y = -6x - 6 -3x + 4(-8x + 13) = -6x - 6 -3x - 32x + 52 = -6x - 6 -29x = -58 x=2 8(2) + 2y = y + 13 16 + y = 13 y = -3 The solution to the system is (2,-3). Diff: 1 Type: SA Page Ref: 167-171 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 130) Solve the following linear system using an algebraic method: x + 3y = 12 -7x + 6y = 51 Answer: x + 3y = 12 2x + 6y = 24 -7x + 6y = 51 2x + 6y = 24 -9x = 27 x = -3 (-3) + 3y = 12 3y = 15 y=5 The solution to the system is (-3,5). Diff: 1 Type: SA Page Ref: 167-171 Topic: 4.3 Graphing Linear Systems of Equations in Two Unknowns Objective: 4-4: Solve problems by setting up systems of linear equations in two variables.

131) Ali has 124 coins made up of only dimes and nickels, and their total value sums up to $11.40. How many of each coin does Ali have? Answer: Ali has 1140 cents total. Let d be the number of dimes. Let n be the number of nickels. d + n = 124 d = 124 - n 10d + 5n = 1140 10(124 - n) + 5n = 1140 1240 - 10n + 5n = 1140 -5n = -100 n = 20 d + 20 = 124 d = 104 Ali has 20 nickels and 104 dimes. Diff: 1 Type: SA Page Ref: 167-171 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables. 132) Bulky Barnyard sells walnuts for $5.34/kg and salted almonds for $7.87/kg. The franchise mixes a total of 15 kg of the two nuts and sells the mix for $6.55/kg. Create a system of equations for finding out how much of each kind of nut they should add to the mix to be making the same revenue as if they were selling the nuts separately. Answer: Let x be the weight (in kg) of walnuts added to the 15 kg mix. Let y be the weight (in kg) of salted almonds added to the 15 kg mix. x + y = 10 5.34x + 7.87y = 98.25 Diff: 1 Type: SA Page Ref: 167-171 Topic: 4.4 Problem Solving Objective: 4-4: Solve problems by setting up systems of linear equations in two variables.

133) Graph the region represented by the following system of inequalities: y ≤ -x − 2 y ≥ -5x + 2 Answer:

Diff: 1 Type: SA Page Ref: 172-178 Topic: 4.5 Graphing Inequalities Objective: 4-5: Solve systems of linear inequalities in two variables using graphical methods.

134) Graph the region represented by the following system of inequalities: 2y ≤ -

x−2+y

-3x + y < -

x+2

Answer:

Diff: 1 Type: SA Page Ref: 172-178 Topic: 4.5 Graphing Inequalities Objective: 4-5: Solve systems of linear inequalities in two variables using graphical methods.

135) Lakynn is playing arcade games on machines that take tokens. She can afford to buy up to a maximum of 44 tokens. Arcade basketball requires 3 tokens per game and skee-ball requires 2 tokens per game. Create a linear inequality to represent the amount of times she can afford to play each game and graph it. Answer: Let b represent the number of basketball games Lakynn can play. Let s represent the number of skee-ball games Lakynn can play. 3b + 2s ≤ 44

Diff: 1 Type: SA Page Ref: 178-179 Topic: 4.6 Problem Solving Objective: 4-5: Solve systems of linear inequalities in two variables using graphical methods.

136) Paisleigh is renovating her house and must spend less than $4,600 to hire painters and carpenters. She also only has 5 hours per day available for 15 days to complete the work. Carpenters charge $64 per hour and painters charge $32 per hour. Write and graph the system of inequalities to represent the situation. Answer: Let c be the number of hours carpenters are hired for. Let p be the number of hours painters are hired for. c + p ≤ 75 64c + 32p < 4600

Diff: 1 Type: SA Page Ref: 178-179 Topic: 4.6 Problem Solving Objective: 4-5: Solve systems of linear inequalities in two variables using graphical methods.

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 5 Cost-Volume-Profit Analysis and Break-Even 1) A manufacturer plans to introduce a new type of shirt based on the following information. The selling price is $35.00; variable cost per unit is $15.00; fixed costs are $8200.00; and capacity per period is 740 units. a) Calculate the break-even point (i) in units (ii) in dollars (iii) as a percent of capacity b) Draw a detailed break-even chart. c) Calculate the break-even point (in units) if fixed costs are reduced to $7000.00 d) Calculate the break-even point (in dollars) if the selling price is increased to $40.00 Answer: a) (i) Let the volume be x units. Then equating the revenue function TR and the cost function TC we obtain: 35x = 8200 + 15x 35x - 15x = 8200 20x = 8200 x = 410 units (ii) 410($35) = $14 350.00 (iii) (410/700) × 100 = 58.57% b) X (units) 0 370 410 740

TR ($) 0 12 950 14 350 25 900

FC ($) 8200 8200 8200 8200

TC ($) 8200 5550 6150 11 100

Since new FC = $7000, then 35x = 7000 + 15x 20x = 7000 x = 350 units

d) 40x = 8200 + 15x 25x = 8200 x = 328 units Break-Even Sales = 328($40) = $13 120.00 Diff: 2 Type: SA Page Ref: 191-206 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships.

2) Priest and Sons, a local manufacturer of a product that sells for $13.50 per unit. Variable cost per unit is $7.85 and fixed cost per period is $1 220. Capacity per period is 1100 units. Perform a break-even analysis showing a) an algebraic statement of (i) the revenue function; (ii) the cost function; (iii) calculate the break-even point in units. b) a detailed break-even chart. Answer: a) Let the volume be x units (i) Revenue: TR = 13.5x (ii) Total cost: TC = 1220 + 7.85x (iii) TR = TC 13.5x = 7.85x + 1220 5.65x = 1220 x = 215.9 units = 216 units (rounded up) b) Break-even chart X 0 100 1100

TR = 13.5 X $0 $1 350 $14 850

TC = 7.85X + 1220 $1 220 $2 005 $9 855

FC = 1220 $1 220 $1 220 $1 220

Diff: 2 Type: SA Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships.

3) Last year a printing company had total sales of $37 500. The total of its variable costs was $ 15 000, and fixed costs were $18 000. Capacity is at sales maximum of $50 000. a) Calculate the break-even point in (i) dollars of sales (ii) as a percent of capacity b) Draw a detailed break-even chart Answer: a) (i) Let x represent the sales volume in dollars. Total Revenue: TR = x =

= 0.40 = 40%

Total Cost: TC = 0.40x + 18 000 For Break-even point: TR = TC x = 0.4x + 18 000 0.6x = 18 000 x = $30 000 (ii) Break-even point as a percent of sales capacity is

(100%) = 60%

b) X ($) 0 25 000 50000

% of Cap 0 50 100

TR = X $0 $25 000 $50 000

TC = 0.40X + 18000 $18 000 $28 000 $38 000

4) Olfert Greenhouses has compiled the following estimates for operations. Sales $150 000 Fixed cost $45 200 Variable costs 67 500 112 700 Net income $37 300 Capacity is a sales volume of $175 000. Perform a break-even analysis showing a) an algebraic statement of (i) the revenue function; (ii) the cost function; b) a detailed break-even chart. Answer: a) Let x represent the sales volume in dollars. i) Revenue: TR = x ii) b)

= 0.45

Total cost: TC = 45200 + 0.45x X ($) 0 100 000 175 000

% of Cap 0 57 100

TR = X $0 $100 000 $175 000

TC = 0.45X + 45200 $45 200 $90 200 $123 950

FC = 45200 $45 200 $45 200 $45 200

5) The gas division of Power-U-Up plans to introduce a new gas delivery system based on the following accounting information. Fixed costs per period are $4 236; variable cost per unit is $168; selling price per unit is $211; and capacity per period is 450 units. a) Draw a detailed break-even chart b) Compute the break-even point (i) in units; (ii) as a percent of capacity; (iii) in dollars. c) Determine the break-even point as a percent of capacity (i) if fixed costs are reduced to $3 788; (ii) if fixed costs are increases to $5 577 and variable costs are reduced to 75% of the selling price; (iii) if the selling price is reduced to $191. Answer: a) Let the number of units be x. Then, Total revenue: TR = 211x Total cost: TC = 4236 + 168x X 0 100 450

TR = 211 X 0 21100 94950

TC = 168X + 4236 4236 21036 79836

FC = 4236 4236 4236 4236

i) To break even, TR = TC 211x = 168 x + 4236 211x - 168x = 4236 43x = 4236 x = 98.51 units = 99 units (rounded up) ii) Break-even volume as a percent of capacity =

= .22 = 22%

iii) Break-even volume in dollars = 99($211) = $20 889. c)

i) If fixed costs are $3 788, then 211x = 168x + 3788 211x - 168x = 3788 x=

= 88.09

x = 89 units Output level =

= 19.78%

ii) If fixed costs are $5 577 and VC is 75% of P, then VC = 0.75($211) = $158.25 and 211x = 158.25x + 5577 211x - 158.25x = 5577 52.75x = 5577 x = 105.725 = 106 units The break-even point as a percent of capacity =

= 23.56%

iii) If P is $191, then 191x = 168x + 4236 x=

= 184.17

x = 185 units The break-even point as a percent of capacity =

= 41.11%

6) The Excellent DVD Company sells DVDs for $62 each. Manufacturing cost is $22.70 per DVD; marketing costs are $7.75 per DVD; and royalty payments are 15% of the selling price. The fixed cost of preparing the DVDs is $227 300. Capacity is 20 000 DVDs. a) Draw a detailed break-even chart. b) Compute the break-even point (i) in units; (ii) in dollars; (iii) as a percent of capacity. c) Determine the break-even point in units if fixed costs are increased by $3300 while manufacturing cost is reduced $1.65 per DVD. d) Determine the break-even point in units if the selling price is increased by 10% while fixed costs are increased by $2900. Answer: a) Let the number of DVDs be x. Total Revenue: TR = 62x VC = 22.70 + 7.75 + 0.15(62) = $39.75 Total cost: TC = 227 300 + 39.75x X 0 10000 20000

TR = 62 X 0 620 000 1 240 000

TC = 39.75X + 227300 227 300 624 800 1 022 300

FC = 227300 227 300 227 300 227 300

b) VC = $39.75 i) TR = TC 62x = 39.75x + 227 300 x=

= 10215.73 = 10216 units (rounded up)

ii) BE point in sales dollars

$62(10216) = $633 392.

iii) BE point as a % of capacity

= 51.08%

FC = 227 300 + 3 300 = $230 600, VC = (22.70 - 1.65) + 7.75 + 0.15(62) = $38.10 62x = 38.10x + 230 600 x=

= 9 648.53 = 9 649 units.

d) VC = 22.70 + 7.75 + .15(62 ∗ 1.1) = $40.68 P = 1.1($62) = $68.20 68.20x = 40.68x + 230200 x=

= 8 259.4 = 8 365 units

7) The operating budget of the Omega Twelve Company contains the following information. Sales at 92% of capacity $644 000 Fixed costs $215 000 Variable costs 373 520 588 520 Net income $55 480 a) Draw a detailed break-even chart. b) Compute the break-even point as a percent of capacity. c) Determine the break-even point in dollars if fixed costs are reduced by $11 200 while variable costs are changed to 62% of sales. Answer: a) Let x represent the sales volume in dollars. Total revenue: TR = x =

= 58% of revenue = 0.58

Total cost: TC = 215 000 + 0.58x X($) 0 350000 700000

% of Cap 0 50 100

TR = X 0 350 000 700 000

TC = 0.58X + 215 000 215 000 418 000 621 000

FC = 215 000 215 000 215 000 215 000

b) Using the contribution margin approach and assuming that P = $1.00, we obtain: CM = P - VC = 1 - 0.58 = 0.42 BE volume (in units) =

Capacity =

= 511904.76 = 511905 units

= 700 000

BE volume as a percent of capacity = c)

= 73.13%

FC = $203 800 and VC = 0.62, then CM = 1 - 0.62 = 0.38 CR =

= 0.38

BE in sales dollars =

= $536 315.79

Diff: 2 Type: SA Page Ref: 204-206 Topic: 5.2 Contribution Margin and Contribution Rate Objective: 5-3: Compute break-even values using contribution margin and contribution rate. 8) Trevor, the new owner of the vehicle accessory shop is considering buying sets of winter tires for $299 per set and selling them at $520 each. Fixed costs related to this operation amount to $3 250 per month. It is expected that 18 sets per month could be sold. How much profit will Trevor make each month? Answer: VC = 299, P = 520, FC = 3 250, then CM = P - VC = 520 - 299 = 221 If Volume per month is 18 units, then PFT = CM(X) - FC = 221(18) - 3250 = $728 Diff: 1 Type: SA Page Ref: 204-206 Topic: 5.2 Contribution Margin and Contribution Rate Objective: 5-3: Compute break-even values using contribution margin and contribution rate.

9) Victoria plans to set up a business selling software applications that she develops and supports. She believes that a price of $180 for her product including the technical support would be competitive. Her monthly fixed expenses amount to $952. Victoria would hire additional part-time staff to provide the technical support of the application paying them for 2 hours at $22 per hour for each client. a) How many clients does Victoria need to have to break even? b) If she wants to achieve a target profit of $800 monthly, how many clients does she need? Answer: P = $180, FC = $910, VC = 2($22) = $44. Then, CM = P — VC = 180 — 44 = $136 a) BE Volume (in units) =

= 952/136 = 7 units.

b) PFT = CM(X) — FC, where X is the volume in units, then 800 = 136X — 910 1710 = 136X X = 12.57 = ~13 units monthly. Diff: 2 Type: SA Page Ref: 206-210 Topic: 5.2 Contribution Margin and Contribution Rate Objective: 5-3: Compute break-even values using contribution margin and contribution rate. 10) Elaine's business budget included sales of $350 000 and fixed costs of $52 600. If the total contribution margin for the business was $125 000, what are the sales needed to break even? Answer: FC = $52 600, Total CM = $125 000 CR =

= 0.357143

Break-even Sales (in $) =

= $147 280

Diff: 1 Type: SA Page Ref: 208-210 Topic: 5.3 Effects of Changes to Cost-Volume-Profit Objective: 5-4: Compute the effects of changes to cost, volume, and profit. 11) A company that makes cell phones has the following cost structure. They have fixed costs of $160 000 per period and manufacturing costs of $43.12 per cell phone. Advertising costs are expected to be $20 000 per period and a special promotion will involve providing a free case for a cost of $6.90 per cell phone. Each cell phone sells for $92.85. What is the break-even point in the number of phones? A) 467 B) 1939 C) 3736 D) 4175 E) 4203 Answer: E Diff: 2 Type: MC Page Ref: 191-206 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships.

12) A pen manufacturer makes luxury pens. The pen case costs $7.26 each, the ink holder costs $1.26 each, the spring costs $.07 each and the velvet pen case costs $0.91 each. The plant has general and administrative costs of $55 000 and fixed selling expenses of $37 500. The pens sell of $39.95 each. Plant capacity is 4 000 pens per period. At what percentage of capacity is the break-even point? A) 75.95% B) 73.74% C) 75.77% D) 61.32% E) 30.79% Answer: A Diff: 2 Type: MC Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships. 13) A local health care facility has fixed costs per month of $187 400. They also have patient costs of $4.15 per day per patient for linen and cleaning, medication costs are $23.32 per patient per day and lab tests cost $75.61 per patient per day. The government is considering allowing the health care facility to charge each patient and amount to recover his or her costs and to make a "profit" of $15 000 per month. The health care facility averages 690 patients per month. The VP-Finance for the facility wants you to calculate the daily rate charge per patient. Your answer is: A) $462.92 B) $396.41 C) $345.73 D) $284.51 E) $455.21 Answer: B Diff: 2 Type: MC Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships.

14) A company that makes optical computer input devices has calculated their revenue and costs as follows for the most recent fiscal period: Sales Costs: Fixed Costs Variable Costs Total Costs Net Income

$522 000 $145 000 208 800

353 800 $168 200

What is the break-even point in sales dollars? A) $362 500.00 B) $589 666.67 C) $241 666.67 D) $870 000.00 E) $280 333.33 Answer: C Diff: 2 Type: MC Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships. 15) A company that makes environmental measuring devices has calculated their revenue and costs as follows for the most recent fiscal period: Sales Costs: Fixed Costs Variable Costs Total Costs Net Income

$750 000 $200 000 250 000

450 000 $300 000

What is the break-even point in sales dollars? A) $100 000 B) $200 000 C) $300 000 D) $250 000 E) $450 000 Answer: C Diff: 2 Type: MC Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships.

16) A company that makes audio computer input devices has calculated their revenue and costs as follows for the most recent fiscal period: Sales Costs: Fixed Costs Variable Costs Total Costs Net Income (Loss)

$723 000 $345 000 404 880

749 880 $(26 880)

The company has a target level of profitability of $35,000 per fiscal period. What sales dollar volume do they have to achieve in order to achieve their goal? A) $840 909.09 B) $616 071.43 C) $784 090.91 D) $863 636.36 E) $678 571.43 Answer: D Diff: 2 Type: MC Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships. 17) A company that makes basketballs has calculated their revenue and costs as follows for the most recent fiscal period: Sales Costs: Fixed Costs Variable Costs Total Costs Net Income (Loss)

$623 000 $??????? 404 880

??????? $(26 880)

What are the company's fixed costs per fiscal period? A) $245 000 B) $2 18120 C) $191 240 D) $1 054 760 E) $1 001000 Answer: A Diff: 2 Type: MC Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships.

18) A company that makes customized pens has calculated their revenue and costs as follows for the most recent fiscal period: Sales Costs: Fixed Costs Variable Costs Total Costs Net Income (Loss)

$100 000 $??????? 15 000

??????? $(20 000)

What are the company's fixed costs per fiscal period? A) $105 000 B) $115 000 C) $80 000 D) $85 000 E) $120 000 Answer: A Diff: 2 Type: MC Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships. 19) A local toolmaker makes the best hammers on the market. The head of the hammer costs $12.11 and the handle costs $4.37. It takes 1.4 minutes to assemble the hammer and the hourly cost is $90.00 for assembly time. The company has fixed operating costs of $22 310 per month. They sell the hammers for three times their total variable cost. The company wants to make a monthly profit of $5000. How many hammers must they sell? A) 796 B) 735 C) 776 D) 766 E) 768 Answer: B Diff: 2 Type: MC Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships.

20) A local restaurant has the best meals in town. The average variable cost per meal is $22.74 and the desserts are $5.24. Only half of the patrons order desserts. The restaurant has fixed operating costs of per month. They sell the meals and desserts for four times their average variable cost per meal. The company wants to make a monthly profit of $75 000. How many meals must they sell? A) 2 045 B) 2 236 C) 2 008 D) 2 803 E) 2 468 Answer: E Diff: 2 Type: MC Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships. 21) A local college hospitality restaurant has the best meals in town. The average variable cost per meal is $10.25 and the desserts are $1.25. The restaurant has fixed operating costs of $110 500 per month. They sell the meals and desserts for three times their average variable cost per meal. The college wants to make a monthly profit of $50 000. How many meals must they sell (Round up to nearest whole meal)? A) 4805 B) 6979 C) 6500 D) 8405 E) 9769 Answer: B Diff: 2 Type: MC Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships. 22) A company has variable costs that are 3/8 the value of their sales revenues. Total net income for the most recent period was a profit of $123 400 and sales were $400 000. The company has started a new marketing campaign that they hope will increase sales, but it will require additional advertising of $11 200. How many sales dollars does the company have to generate in order to remain at the same level of profitability as before the new ad campaign? A) $358 933.33 B) $215 360.00 C) $400 000.00 D) $417 920.00 E) $696 533.33 Answer: D Diff: 3 Type: MC Page Ref: 208-210 Topic: 5.3 Effects of Changes to Cost-Volume-Profit Objective: 5-4: Compute the effects of changes to cost, volume, and profit.

23) A company has variable costs that are 1/8 the value of their sales revenues. Total net income for the most recent period was a profit of $50 400 and sales were $500 000. The company has started a new marketing campaign that they hope will increase sales, but it will require additional advertising of $15 000. How many sales dollars does the company have to generate in order to remain at the same level of profitability as before the new ad campaign? A) $517 142.86 B) $387 100.00 C) 392 100.00 D) $387 100.86 E) 392 100.86 Answer: A Diff: 3 Type: MC Page Ref: 208-210 Topic: 5.3 Effects of Changes to Cost-Volume-Profit Objective: 5-4: Compute the effects of changes to cost, volume, and profit. 24) A company has variable costs that are 4/7 the value of their sales revenues. Total net income for the most recent period was a profit of $53 770 and sales were $420 000. The company has started a new marketing campaign that they hope will increase sales, but it will require additional advertising of $6400. How many sales dollars does the company have to generate in order to remain at the same level of profitability as before the new ad campaign? A) $434 933.33 B) $326 357.50 C) $420 210.00 D) $140 396.67 E) $105 297.50 Answer: A Diff: 3 Type: MC Page Ref: 208-210 Topic: 5.3 Effects of Changes to Cost-Volume-Profit Objective: 5-4: Compute the effects of changes to cost, volume, and profit.

25) Given the following chart, calculate the cost function. Interpret the function to calculate variable costs and fixed costs:

Answer: y-intercept = 840 Slope of the line =

∴ Line is presented by the following relationship: y = 4x + 840 In terms of cost this line is written as: TC = VC + FC ∴ FC = $840 and VC = $4 per attendant Diff: 2 Type: SA Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships.

26) An electrician charges a flat fee of $90 for a home service call. In addition he charges $25 for every 20 minutes as labour cost. Draw a graph to show the total charge against the time in hours. Calculate the slope of the line. Answer: Fixed cost = $90 (FC) Variable cost = $25 for every 20 minutes = $75 per hour (TVC) Total cost = TC = TVC + FC TC = $75t + $90 ∴ Slope of the line = 75 Following is the graph:

Diff: 1 Type: SA Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships. 27) Excel hardware is introducing a new product on a new product line of capacity 800 units per week at a production cost of $50 per unit. Fixed costs are $22,400 per week. Variable selling and shipping costs are estimated to be $20 per unit. Excel plan to market the new product at $110 per unit. What is the breakeven capacity per week? A) 204 B) 373 C) 560 D) 280 E) 320 Answer: C Diff: 1 Type: MC Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships.

28) Excel hardware is introducing a new product on a new product line of capacity 800 units per week at a production cost of $50 per unit. Fixed costs are $22 400 per week. Variable selling and shipping costs are estimated to be $20 per unit. Excel plan to market the new product at $110 per unit. What would be the weekly net income at 90% of the capacity? A) $9600 B) $7200 C) $6400 D) $56 800 E) $20 800 Answer: C Diff: 2 Type: MC Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships. 29) Sala pipe fittings produce pipe elbows and reducers from stainless steel. The company can process up to tonnes of stainless steel sheets in a year. The company pays the steel company $800 per tonne of stainless steel sheets and each tonne is used to manufacture $2000 worth of elbows and reducers. Variable processing costs are $470 per tonne and fixed processing costs $3.4 million per year at all production levels. Administrative overhead is $3 million per year regardless of the volume of the production. Marketing and transportation costs work out to be $230 per tonne. Determine the break-even volume in terms of percent capacity utilization. A) 30% B) 34% C) 43.8% D) 24.6% E) 64% Answer: E Diff: 2 Type: MC Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships. 30) Last year, Terrific Copying had total revenue of $475 000, while operating at 60% of capacity. The total of its variable cost is $150 000. Fixed costs were $180 000. What is Terrific's contribution rate? A) 62.1% B) 16.7% C) 20% D) 60% E) 68.4% Answer: E Diff: 1 Type: MC Page Ref: 204-206 Topic: 5.2 Contribution Margin and Contribution Rate Objective: 5-3: Compute break-even values using contribution margin and contribution rate.

31) Sala pipe fittings produce pipe elbows and reducers from stainless steel. The company can process up to tonnes of stainless steel sheets in a year. The company pays the steel company $800 per tonne of stainless steel sheets and each tonne is used to manufacture $2000 worth of elbows and reducers. Variable processing costs are $470 per tonne and fixed processing costs $3.4 million per year at all production levels. Administrative overhead is $3 million per year regardless of the volume of the production. Marketing and transportation costs work out to be $230 per tonne. In order to attain a net income of $2.4 million how many tonnes of steel must be processed this year? Answer: Total fixed cost = $3 000 000 + $3 400 000 = $6 400 000 Total variable cost = $800 + $470 + $230 = $1500 per tonne Net profit = Total revenue - Total Variable Cost - Fixed cost = Price × Volume - Variable cost per unit × volume - Fixed cost ⇒ $2 400 000 = $2000 × N - $,500 × N - $6 400 000 ⇒ N = 17 600 17 600 tonnes of steel must be processed this year to attain a net income of $2.4 million. Diff: 2 Type: SA Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships. 32) Last year, Terrific Copying had total revenue of $475 000, while operating at 60% of capacity. The total of its variable cost is $150 000. Fixed costs were $180 000. What is Terrific's break-even point expressed in dollars of revenue? A) $289 855 B) $1 077 844 C) $900 000 D) $263 158 E) $300 000 Answer: D Diff: 1 Type: MC Page Ref: 204-206 Topic: 5.2 Contribution Margin and Contribution Rate Objective: 5-3: Compute break-even values using contribution margin and contribution rate. 33) Last year, Terrific Copying had total revenue of $475 000, while operating at 60% of capacity. The total of its variable cost is $150 000. Fixed costs were $180 000. If the current selling price, variable costs, and fixed costs are the same as last year, what net income can be expected from revenue of $500 000 in the current year A) $381 388 B) $394 737 C) $1 418 216 D) $1 184 210 E) $346 260 Answer: E Diff: 1 Type: MC Page Ref: 208-210 Topic: 5.3 Effects of Changes to Cost-Volume-Profit Objective: 5-4: Compute the effects of changes to cost, volume, and profit.

34) Last year, Terrific Copying had total revenue of $475 000, while operating at 60% of capacity. The total of its variable cost is $150 000. Fixed costs were $180 000. What is the expected revenue this year at full capacity? A) $791 667 B) $250 000 C) $300 000 D) $550 000 E) $1 341 667 Answer: A Diff: 1 Type: MC Page Ref: 208-210 Topic: 5.3 Effects of Changes to Cost-Volume-Profit Objective: 5-4: Compute the effects of changes to cost, volume, and profit. 35) Calculate the contribution margin, if the variable cost per unit to produce a microwave is $39 and a contribution rate is 45%. A) $17.55 B) $1.15 C) $70.91 D) $31.91 E) $39 Answer: D Diff: 2 Type: MC Page Ref: 204-206 Topic: 5.2 Contribution Margin and Contribution Rate Objective: 5-3: Compute break-even values using contribution margin and contribution rate. 36) Sunbeam wants to sell microwaves at a unit contribution margin of $42 and a unit contribution rate of 60%. What should be the break-even volume if the total cost of production is kept below $10 000. A) 143 B) 238 C) 357 D) 714 E) 900 Answer: B Diff: 2 Type: MC Page Ref: 204-206 Topic: 5.2 Contribution Margin and Contribution Rate Objective: 5-3: Compute break-even values using contribution margin and contribution rate.

37) Frigidaire sell refrigerators at $1800 per unit. Their variable cost is estimated to be $680 per unit and their fixed costs are estimated to be $11.2 million per annum. Calculate the minimum number of units Frigidaire should produce every year to avoid incurring a loss. A) 4 517 B) 6 223 C) 10 000 D) 16 471 E) 20 000 Answer: C Diff: 1 Type: MC Page Ref: 191-206 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships. 38) Citizen sells a watch for $35 at a fixed cost of $1 million per annum. Recent recession has seen jump in material cost to $20. What is the new break-even volume of watches per year? A) 28 571 B) 50 000 C) 66 667 D) 6667 E) 7000 Answer: C Diff: 2 Type: MC Page Ref: 208-210 Topic: 5.3 Effects of Changes to Cost-Volume-Profit Objective: 5-4: Compute the effects of changes to cost, volume, and profit. 39) Citizen sells a watch for $35 at a variable cost of $15 per unit and fixed cost of $1 million per annum. Recent recession has seen jump in material cost by $7. What is the change in profit due to recession if Citizen produces and sells 100 000 watches in a year? A) $1 000 000 decrease B) $700 000 decrease C) $300 000 decrease D) $800 000 increase E) $1 800 000 increase Answer: B Diff: 2 Type: MC Page Ref: 208-210 Topic: 5.3 Effects of Changes to Cost-Volume-Profit Objective: 5-4: Compute the effects of changes to cost, volume, and profit.

40) Caroline needs to put some money in her pocket this winter, so she plans on removing snow from driveways. She will need to pay $560 for a snow-blower to make her job easier. A variable cost of $3 per job for supplies would also be required. She estimates that she could clean 40 driveways a month. What price should she charge the customers for the service in order to break even? Answer: X = 40; VC = 3.00; FC = 560.00 Let the price per unit be SP (SP × 40) = 560 + (3 × 40) (SP × 40) = 560 + 120 (SP × 40) = 660 SP = 17 Diff: 2 Type: SA Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-2: Compute break-even values using cost-volume-profit relationships. 41) Caroline needs to put some money in her pocket this winter, so she plans on removing snow from driveways. She will need to pay $560 for a snow-blower to make her job easier. A variable cost of $3 per job for supplies would also be required. She estimates that she could clean 40 driveways a month. What price should she charge the customers for the service in order to make a profit of $1200? Answer: X = 40; VC = 3.00; FC = 560.00; Profit = 1200.00 Let the price per unit be SP (SP × 40) - 560 - (3 × 40) = 1200 (SP × 40) - 560 - 120 = 1200 (SP × 40) = 680 + 1200 SP = 47 Diff: 2 Type: SA Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-2: Compute break-even values using cost-volume-profit relationships. 42) Raider Corporation made $130 000 in sales last year. They had fixed costs of $29 300 and variable costs of $26 000. What is the breakeven point if their capacity is at $160 000? Provide algebraic statements of the revenue and cost functions, as well as a breakeven point in sales dollars. Answer: TR = 1 × X =

= 0.2

TC = 29 300 + 0.2X 1.00X = 29 300 + 0.2X 0.8X = 29 300 X = 36 625 Diff: 2 Type: SA Page Ref: 189-202 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-2: Compute break-even values using cost-volume-profit relationships.

43) A new smartphone is being sold by Motorola at $650. The fixed cost per month to make these phones is and the variable cost per phone is $150. Determine the breakeven volume for Motorola using the contribution margin approach. Answer: Fixed cost = $840 000; Selling price per unit = $650; Variable cost per unit = $150; Contribution margin per unit = 650 - 150 = 500 Break-even volume =

= 1680

Diff: 1 Type: SA Page Ref: 204-206 Topic: 5.2 Contribution Margin and Contribution Rate Objective: 5-3: Compute break-even values using contribution margin and contribution rate. 44) Darby is running a flower stand, and is selling a single bouquet for $31.50. Her contribution margin is $24.57 per bouquet and the breakeven sales in dollars for Darby's flower stand are $700 per day. How much can she pay per day in fixed costs to run the stand? Answer: CM = 24.57 Contribution rate = Contribution rate = Contribution rate = 0.78 = 78% Breakeven sales = 700 = 700 × 0.78 = FC 546 = FC Diff: 4 Type: SA Page Ref: 204-206 Topic: 5.2 Contribution Margin and Contribution Rate Objective: 5-3: Compute break-even values using contribution margin and contribution rate. 45) Tarfex Tech intended to sell a new product in order to keep up with customer demands. The product was to be sold to consumers at $620, creating a contribution margin of $150 per unit. They spent $9800 on advertising and were able to sell 1,200 units on the first day. Determine the resulting profit. Answer: X = 1200; CM = 150; FC = 9800.00; SP = 620.00 Contribution margin per unit = Selling price per unit - Variable cost per unit 150 = 620 - VC VC = 620 - 150 VC = 470 (Selling price × Volume) - Fixed cost - (Variable cost per unit × Volume) = Profit (620 × 1200) - 9800 - (470 × 1200) = PFT 744 000 - 9800 – 564 000 = PFT 170 200 = PFT Diff: 5 Type: SA Page Ref: 204-206 Topic: 5.2 Contribution Margin and Contribution Rate Objective: 5-3: Compute break-even values using contribution margin and contribution rate.

46) Barb has a hot-dog stand near the ferry terminal for Centre Island. She pays $500 per month as rent and $3000 per month in wages for the hired help. Variable costs per hot dog come out to be $1.30 and she sells each hotdog for $3.00. In summer months, she is able to sell 4500 hotdogs in a month. Recently, another hot dog vendor opened his stand across the street. To attract and retain her customers, she added a free soda can which increased her variable costs by 20 cents. To make things worse, the rent was also increased by 15%. What is the change in her monthly net income due to these recent changes. Assume that the monthly sale figures remain at 4500 units. Answer: OLD numbers X = 4500; VC = 1.30; FC = 500 + 3000 = 3500; SP = 3.00 (Selling price × Volume) - Fixed cost - (Variable cost per unit × Volume) = Profit (3 × 4500) - 3500 - (1.30 × 4500) = PFT 4150 = PFT NEW numbers X = 4500; VC = 1.30 + 0.20 = 1.50; FC = 1.15(500) + 3000 = 3575; SP = 3.00 (Selling price × Volume) - Fixed cost - (Variable cost per unit × Volume) = Profit (3 × 4500) - 3575 - (1.50 × 4500) = PFT 3175 = PFT Monthly profit is reduced by = 4150 - 3175 = 975 Diff: 5 Type: SA Page Ref: 208-210 Topic: 5.3 Effects of Changes to Cost-Volume-Profit Objective: 5-4: Compute the effects of changes to cost, volume, and profit. 47) Barb has a hot-dog stand near the ferry terminal for Centre Island. She pays $500 per month as rent and $3000 per month in wages for the hired help. Variable costs per hot dog come out to be $1.30 and she sells each hotdog for $3.00. In summer months, she is able to sell 4500 hotdogs in a month. To increase her sales, she added a free soda can which increased her variable costs by 20 cents. However, it increased her monthly sales to 5000 units. What is the change in her monthly net income due to these recent changes. Answer: OLD numbers X = 4500; VC = 1.30; FC = 500 + 3000 = 3500; SP = 3.00 (Selling price × Volume) - Fixed cost - (Variable cost per unit × Volume) = Profit (3 × 4500) - 3500 - (1.30 × 4500) = PFT 4150 = PFT NEW numbers X = 5000; VC = 1.30 + 0.20; FC = 500 + 3000 = 3500; SP = 3.00 (Selling price × Volume) - Fixed cost - (Variable cost per unit × Volume) = Profit (3 × 5000) - 3500 - (1.50 × 5000) = PFT 4000 = PFT Change in monthly profit = 4000 - 4150 = -150 Diff: 4 Type: SA Page Ref: 208-210 Topic: 5.3 Effects of Changes to Cost-Volume-Profit Objective: 5-4: Compute the effects of changes to cost, volume, and profit.

48) Mary-Jane's Hair Salon has monthly costs of $7600 for rent, utilities and insurance. Mary-Jane also faces an average cost of $14.52 per customer. She charges $45.00 per customer for the package of wash, cut, and blow-dry. What is Mary-Jane's break-even point in dollars? A) $349.43 B) $249.34 C) $257.45 D) $127.69 E) $7540.48 Answer: B Diff: 2 Type: MC Page Ref: 191-206 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships. 49) Romero Co., a company that makes custom-designed stainless-steel water bottles and tumblers, has shown their revenue and costs for the past fiscal period: Revenue: Sales Costs: Fixed Costs Variable Costs Total Costs Net Income

$187 200 $104 696 $? $? $40 000

What are the company's variable costs per fiscal period? A) $41 500 B) $251 896 C) $42 504 D) $37 962 E) $122 504 Answer: C Diff: 1 Type: MC Page Ref: 191-206 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-1 and 5-2: Construct and interpret cost-volume-profit charts; Compute break-even values using cost-volume-profit relationships.

50) Beezus is planning on freelancing their tutoring services to save up for the new commercial printer. They will need to pay $126 (only once) for business cards to hand out and posters to put up around their neighborhood. There would also be a variable cost of $2 per job (e.g. notebooks, textbooks, calculators, etc.). They estimate that they could do 21 tutoring sessions per month. What price do they need to charge the customers per tutoring session (assuming they are all of equal cost and price) in order to break even? Answer: X = 21 VC = 2.00 FC = 65.00 Let the selling price per session be SP. 21SP = 126 + 21(2) 21SP = 126 + 42 SP = 168/21 SP = 8 Beezus should charge $8 per tutoring session to break even. Diff: 2 Type: SA Page Ref: 191-206 Topic: 5.1 Cost-Volume-Profit Analysis and Break-Even Charts Objective: 5-2: Compute break-even values using cost-volume-profit relationships. 51) Beiberlicious Inc. sells boxes of toaster strudels at a unit contribution margin of $3.50 and a unit contribution rate of 70%. What should the break-even sales (rounded to the nearest dollar) be if the total cost of production needs to be under $15 000? A) $21 428 B) $33 710 C) $21 429 D) $4285 E) $52 500 Answer: A Diff: 2 Type: MC Page Ref: 206-210 Topic: 5.2 Contribution Margin and Contribution Rate Objective: 5-3: Compute break-even values using contribution margin and contribution rate. 52) Attyre, a clothing company, sells novelty "Ok Boomer" sweatshirts for a selling price of $62, with a variable cost of $18 per unit and a fixed cost of $86 000 per year. A decrease in aggregate supply in the past year has caused the cost of materials for the sweatshirts to increase by $6.50. What is Attyre's change in profit due to these changes if they produce and sell 75 000 sweatshirts in one year? A) Decrease of $938 400 B) Decrease of $1 000 000 C) Increase of $487 000 D) Increase of $561 819 E) Decrease of $487 500 Answer: E Diff: 2 Type: MC Page Ref: 210-213 Topic: 5.3 Effects of Changes to Cost-Volume-Profit Objective: 5-4: Compute the effects of changes to cost, volume, and profit.

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 6 Trade Discounts, Cash Discounts, Markup, and Markdown 1) A hitch and trailer listed at $2149.00 has a net price of $1360.00. What is the rate of discount? Answer: Amount of Discount = 2149 - 1360 = $789.00 Rate of discount =

= 36.7148%

Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 2) Compute the rate of discount allowed on a rotor-tiller that lists for $441.00 and is sold for $205.00. Answer: Amount of discount = 441 - 205 = $236.00 Rate of discount = 236/441 = .535147 = 53.5147% Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 3) Compute the rate of discount allowed on a cordless drill that lists for $149.99 and is sold for $79.99. Answer: Amount of discount = 149.99-79.99 = 70.00 Rate of discount =

= .466697 = 46.670%

Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 4) What rate of discount has been allowed if an item that lists for $720.00 is sold for $681.57? Answer: Amount of discount = 720 - 681.57 = 38.43 Rate of discount =

= 5.3375%

Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 5) A store advertises a discount of $54.72 on sandals. If the discount is 33.5%, for how much were the sandals sold? Answer: 33.5% of list = $54.72 .335L = 54.72 L = $163.34 Sale price = 163.34 - 54.72 = $108.62 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts.

6) A store advertises a discount of $10.00 on a shirt. If the discount is 30%, for how much was the shirt sold? Answer: 30% of list = $10 0.30L = 10 L = $33.33 Sale price = 33.33 - 10 = $23.33 Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 7) Kaz purchases an article with a net price of $114.54 after a discount of 17% was allowed. Calculate the list price. Answer: 114.54 = L(1 - 0.17) 0.83L = 114.54 L = $138.00 Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 8) A 12 % discount on a pair of washer and dryer that Gayle purchased, amounted to $156.00. Calculate the net price. Answer: 0.125L = 156 L = $1248.00 N=L-D N = 1248 - 156 = $1092.00 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 9) A 14% discount on a dryer that Sean purchased, amounted to $70.00. Calculate the net price. Answer: 0.14L = 70 L = 500 N=L-D N = 500 - 70 = $430.00 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 10) A 47% discount allowed on an article amounts to $3.57 What is the net price? Answer: 0.47L = 3.57 L = $7.60 N = 7.60 - 3.57 = $4.03 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts.

11) A fridge was sold during a clearance sale for $657.00. If the fridge was sold at a discount of 34%, what was the list price? Answer: N = L(1 - d) 657 = L(0.76) L = $864.47 List price was $864.47 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 12) A television set was sold during a clearance sale for $299.00. If the television was sold at a discount of 70%, what was the list price? Answer: N = L(1 - d) 299 = 0.30L $996.67 = L List price was $996.67 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 13) Determine the net price of an article listed at $12 050 less 20%, 16%, 9 %. Answer: N = 12050(1 - 0.2)(1 - 0.16)(1 -

) = 12050(0.8)(0.84)

= $7 341.82

Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 14) The net price of an article is $84.62. What is the list price if a discount of 17% was allowed? Answer: 0.83L = 84.62 L = $101.95 Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 15) Determine the equivalent single rate of discount for each of the following series of discounts. a) 18 %, 10.0% b) 50%, 33 %, 5.0% Answer: a) Single rate = 1 - (1 -

)(1 - 0.1) = 1- (0.8166667)(.9) = 26.5%

b) Single rate = 1 - (1 - 0.5)(1-

)(1 - 0.05) = 1 - (0.50)( )(0.95) = 68.33333%

Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series.

16) A lathe is listed for $477.00 less 19 %, 9%, 7.7%. a) What is the net price? b) How much is the amount of discount allowed? c) What is the equivalent single rate of discount that was allowed? Answer: a) N = 477(0.8033333)(0.91)(0.923) = $321.85 b) Discount = 477 - 321.85 = $155.15 c) Single rate =

= 32.526%

Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 17) Determine the equivalent single rate of discount for each of the following series of discounts. a) 18 %, 4.5% b) 36%, 7 %, 1.5% Answer: a) Single rate = 1 - (0.8133333)(0.955) = 22.32667% b) Single rate = 1 - (0.64)(0.9266667)(0.985) = 41.58293% Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 18) A discount store lists an article for $875.00 less 25% and 20.5%. a competitor carries the same article for $847.90 less 21.5%. What further discount (correct to the nearest so that its net price will be the same as that of the discount store? Answer: Net price of discount store = 875(1 - 0.25)(1 - 0.205) = $521.72 Net price of competitor = 847.90(1 - 0.215) = 665.60 Additional discount needed = 665.60 - 521.72 = 143.88 Additional percent discount =

= 21.61681%

Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series.

of 1%) must the competitor allow

19) Uptown Goldsmith sells watches for $840.00 less 12.55%. Its competitor across the street offers the same type of watch for $860.00 less 8.5%, 6%, 3.25%. What additional rate of discount must Uptown offer to meet the competitor's price? Answer: Uptown N = 840(1 - 0.1255) = 840(0.8745) = 734.58 Competitor's N = 860 (1- 0.085)(1 - 0.06)(1 - 0.0325) = $715.65 Additional Uptown discount = 734.58 - 715.65 = $18.93 Additional Uptown discount rate = 18.93/734.58 = 2.57698% Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 20) Big Boy Clothing Store lists an article for $99.00 less 20% and 15%. A competitor carries the same article for $115.00 less 18%. What further discount (correct to the nearest 1/10 of 1%) must the competitor allow so that its net price is the same as Deep Discount's? Answer: Store's net price = 99(1 - .20)(1 - .15) = 67.32 Competitor's price = 115(1 - .18) = 94.30 Additional discount needed = 94.30 - 67.32 = 26.98 Additional percent discount =

= 28.6%

Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 21) Deep Discount Electronics lists an article for $127.00 less 12.5% and 11.5%. A competitor carries the same article for $137.00 less 15.5%. What further discount (correct to the nearest 1/10 of 1%) must the competitor allow so that its net price is the same as Deep Discount's? Answer: Store's net price = 127.00(1 - .125)(1 - .115) = 98.35 Competitor's price = 137.00(1 -.155) = 115.77 Additional discount needed = 115.77 - 98.35 = 17.42 Additional percent discount =

= 15.0471%

Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 22) What is the list price of an article that is subject to discounts of 23.25%, 11.2%, 4% if the net price is $664.88? Answer: L = L= L= L = $1015.94 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series.

23) What is the list price of an article that is subject to discounts of 40.5%, 10.4%, 5.0% if the net price is $650.00? Answer: L = L= L = $1283.41 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 24) The net price of an article is $272.20 after discounts of 21% and 16% have been allowed. What was the list price? Answer: N = L(1 - 0.21)(1 - 0.16) 272.20 = L(0.79)(0.84) L= L = $410.19 List price = $410.19 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 25) An invoice shows a net purchase price of $1809.13 after the subtraction of discounts of 12.5%, 25%, 8.3%. What was the list price? Answer: L = L = 3006.29 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 26) An invoice shows a net price of $231.88 after discounts of 17.8%, 8.6%, 6.3%. What was the list price? Answer: N = L(1 - 0.178)(1 - 0.086)(1 - 0.063) 231.88 = L(0.822)(0.914)(0.937) L= L = $329.39 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series.

27) An invoice for $6200.00, dated May 28, 3/10, 1/20, n/60, was received on May 30. What payment must be made on June 5 to reduce the debt to $4760.00? Answer: Invoice = $6 200, allow the 3% discount Amount of reduction = 6200 - 4760 = 1 440 Amount paid = 1440(0.97) = $1396.80 Diff: 2 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 28) The following invoices, all with terms 5/10, 2/30, n/60, were paid together on March 15. Invoice No. 457 dated January 30 is for $294.67; invoice No. 542 dated February 15 is for $486.50; and invoice No. 877 dated March 10 is for $911.25. What amount was remitted? Answer: Invoice Date Amount Rate of discount Net amount paid January 30 $294.64 --$294.67 February 15 486.50 2% (0.98)(486.50) 476.77 March 10 911.25 5% (0.95)(911.25) 865.69 Amount remitted $1637.13 Diff: 2 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 29) What amount will reduce the amount due on an invoice of $9410.25 by $1042.00 if the terms of the invoice are 4.5/5, n/30 and the payment was made during the discount period? Answer: Allow 4.5% discount on partial payment of $1042.00 Amount paid = 1042(0.955) = $995.11 Diff: 2 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 30) On June 18, an invoice dated June 18 for $6250.00 less 17%, 16%, terms 5/10, n/30, was received by Nova Distributors. a) What is the last day of the discount period? b) What is the amount due if the invoice is paid in full within the discount period? Answer: a) Last day of discount period is June 28 b) Net amount of invoice = 6250(0.83)0.84 = $ 4357.50 Amount paid = 4357.50(0.95) = $4139.63 Diff: 1 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount.

31) The Snow Store received an invoice for $6710.00 dated July 13, terms 5/10, 2/30, n/90, for a shipment of skis. The Ski Shop made two partial payments. a) How much was paid July 20 to reduce the unpaid balance to $4000.00? b) How much was paid on August 11 to reduce the outstanding balance by $1700.00? Answer: a) Allow 5% on partial payment of 6710 - 4000 = $2 710 Amount paid = 2710(0.95) = $2574.50 b) Allow 2% on partial payment of $1700. Amount paid = 1700 (0.98) = $1666.00 Diff: 3 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 32) The Local Dollar Store received an invoice for $8500.00 dated January 31, terms 5/10, 2/30, n/90, for a shipment of skis. The Local Dollar Store made two partial payments. a) How much was paid on February 7 to reduce the unpaid balance to $5000.00? b) How much was paid on March 1 to reduce the outstanding balance by $1500.00? c) What was the outstanding balance on the account on March 2? d) How much did the Local Dollar Store save by taking the advantage of the cash discounts? Answer: a) Amount of reduction is 8500 - 5000 = $3500. Local Dollar Store is entitled to the 5% discount on February 7. Hence, Amount paid = 3500(0.95) = $3325.00 b) Amount of reduction is $1500. The payment on March 1 is within the 2% discount period Amount paid = 1500(0.98) = $1470.00 c) Amount still owing is $5000 - 1500 = $3500.00 d) Amount saved = 0.05(3500) + 0.02(1500) = $205.00 Diff: 3 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 33) Bamboo & Teak furniture shop sells bamboo bedroom sets for a price of $3099 if the purchase is financed. The company also advertises "no payments and no interest for one year." Joan and her husband suggested to pay cash for the furniture and got an offer from the salesperson for $2883, which they took. (i) How much was the discount for paying cash? (ii) What was the rate of discount on the cash purchase? Answer: (i) Amount of discount = 3099 - 2883 = $216 (ii) d = 216/3099 = 6.97% Diff: 2 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount.

34) An invoice dated May 12, terms: 5/10, n/30 for $3000.00 was received by Bruce Distributors. a) What is the last day of the discount period? b) What is the amount due if the invoice is paid on June 5? c) What is the amount due if the invoice is paid on May 20? Answer: a) May 22 b) $3000 c) 3000(1 - 0.05) = 3000(0.95) = $2850.00 Diff: 1 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 35) Daffodil Supplies received a partial payment of $2300.50 from J&K Landscape company on June 7 on an invoice of $7539 dated May 12, terms 5/10, 3/30, n/60. a) For how much should the Daffodil Supplies credit J&K Landscape's account for the payment? b) How much does J&K Landscape still owe on the invoice? Answer: a) Partial payment = $2300.50. Payment is made within the 3% discount period. Hence, Let $x be amount credited to the account. x(1 - 0.03) = 2300.50 x = $2371.65 b) The amount still owing is 7539 - 2371.65 = $5167.35 Diff: 2 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 36) DeSomma and Brown received an invoice for $14 520 dated November 29, terms 5/20, n/45, from Happy Valley Winery for a truckload of grapes received December 17. DeSomma and Brown made a partial payment of $4955.00 on December 12. a) By how much did DeSomma and Brown reduce the amount due on the invoice? b) How much does DeSomma and Brown still owe? Answer: a) Partial Payment = $4955. Given 5% cash discount, the amount credited to the account $x is x(0.95) = 4955 x = $5215.79 The debt is reduced by $5215.79 b) Amount owing = 14 520 - 5215.79 = $9034.21 Diff: 2 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount.

37) Burke Company received an invoice dated November 18 for a shipment of goods received November 21. The invoice was for $9255.00 less 40%, 13.375% with terms 5/10, 3/20, n/45. How much must Burke pay on December 2 to reduce its debt a) by $2000.00? b) to $2000.00? c) if Burke Company makes a partial payment of $4000 on November 25, how much should be credited to the account? Answer: Net amount of invoice = 9255(0.86625)0.6 = $4810.29 December 2 is within 3% discount period: a) Payment = 2000(0.97) = $1940.00 b) Amount of reduction = 4810.29 - 2000 = 2810.29 Payment = 2810.29(0.97) = $2 725.98 c) November 25 is within 5% discount period. Let $x be the amount credited, then Payment = x(1 - 0.05) 4000 = x(0.95) x = $4210.53 Diff: 2 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount.

38) Export Ltd. received an invoice dated August 21 from Dutch Specialties of Amsterdam with terms 5/20, n/45 for: 10 wood trays at $37.45 each; 35 wood planters at $43.75 each; 50 wood bowls at $37.25 each. All items are subject to trade discounts of 33.5%, 7.125%, 4%. a) What is the last day of the discount period? b) What is the amount due if the invoice is paid in full on September 6? c) If only a partial payment is made on the last day of the discount period, what amount is due to reduce the outstanding balance to $1500.00? Answer: a) September 10 b) 10 ∗ 37.45 = $374.50 35 ∗ 43.75 = $1531.25 50 ∗ 37.25 = $1862.50 $3768.25 Net Amount of Invoice = 3768.25(.665)(.92875)(.96) = $2234.25 A 5% discount can be taken on September 6 Full Payment = 2 234.25(0.95) = $2122.54 c)

Net Amount = $2234.25 Remaining balance = 1500.00 Amount of reduction = $734.25 Payment = 734.25(1 - 0.05) = $697.54 Diff: 3 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount.

39) Sepaba Ltd. received an invoice dated November 20 from Wallis Specialties of Germany with terms 5/20, n/45 for: 10 vases at $35.00 each; 20 figurines at $50.50 each; 40 glass flutes at $80.50 each. All items are subject to trade discounts of 30%, 10%, 5%. a) What is the last day of the discount period? b) What is the amount due if the invoice is paid in full on December 8? c) If only a partial payment is made on the last day of the discount period, what amount is due to reduce the outstanding balance to $1500.00? Answer: a) December 10 b) 10 ∗ 35.00 = $350.00 20 ∗ 50.50 = $1010.00 40 ∗ 80.50 = $3220.00 $4580.00 Net Amount = 4580(0.70)(0.90)0.95 = $2741.13 Payment on December 8 is within the 5% discount period Full Payment = 2741.13(0.95) = $2 604.07 c)

Net Amount = $2741.13 Remaining balance = 1500.00 Amount of reduction = $1241.13 Discount of 5% applies Payment = 1241.13(1 - 0.05) = $1179.07 Diff: 3 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 40) The Exotic Import Company received an invoice dated January 5 for a shipment of goods received January 11. The invoice was for $5525.00 less 40%, 8% with terms 3/20 R.O.G. How much must Exotic Import pay on January 20th to reduce its debt a) by $3000.00? b) to $3000.00? Answer: Net invoice = (.60)(.92)(5525.00) = $3049.80 Last day for discount is January 31: allow 3% a) Pay = (0.97)(3000.00) = $2910.00 b) Pay = (0.97)(49.80) = $48.31 Diff: 2 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount.

41) Encino Ltd. received an invoice dated February 16 for $520.00 less 25%, 8.75%, terms 3/15 E.O.M. A cheque for $159.20 was mailed by Encino on March 15 as part payment of the invoice. a) By how much did Encino reduce the amount due on the invoice? b) How much does Encino still owe? Answer: Net invoice = (.75)(.9125)520.00 = $355.88 a) Allow 3% discount Payment = 97% of credit 159.20 = 0.97C C = $164.12 Amount due is reduced by $164.12 b) Amount owing = 355.88 - 164.12 = $191.76 Diff: 2 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 42) An invoice for $5800.00, dated May 27, terms 3/10 E.O.M., was received May 29. What payment must be made on June 8 to reduce the debt to $2945.00? Answer: Reduction needed = 5800 - 2945 = $2855 Discount allowed: 3% Payment = 2855(1 - .03) = $2769.35 Diff: 1 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 43) An invoice for $5999.00, dated June 27, terms 2/10 E.O.M., was received June 30. What payment must be made on July 10 to reduce the debt to $2000.00? Answer: Reduction needed = 5999 - 2000 = $3999.00 Discount allowed: 3% Payment = 3999(1 - 0.02) = $3919.02 Diff: 1 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 44) What is the cost of a new refrigerator that was sold for $2230.00 in order to realize a margin of 113% based on cost. Answer: C + 1.13C = 2230 2.13C = 2230 C = $1046.95 Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price.

45) A tire dealer bought tires from a wholesaler at $75.00 dollars each and sold the tires at a mark-up of 25% of cost. a) Calculate the selling price of the tires. b) Calculate the rate of markup based on the selling price. Answer: a) S = 75.00 + 0.25(75.00) S = 75.00 + 18.75 S = $93.75 b) Rate of markup based on selling price =

= 0.2 = 20%

Diff: 1 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 46) A bicycle at Janice's sport shop, costing $108.50 was marked up to reflect 40% of the cost price. Calculate the selling price. Answer: S = C + M S = 108.50 + 0.40(108.50) S = 108.50 + 43.40 S = 151.90 Diff: 1 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 47) Christine's pet shop buys cat litter for $15.00 less 20% per bag. The store's overhead is 45% of cost and the owner requires a profit of 20 % of cost. a) For how much should a bag be sold? b) Calculate the amount of markup. Answer: a) C = 15(1 - 0.2) = 15(0.8) = $12 E = 0.45 × 12 = $5.40 P = 0.2 × 12 = $2.40 S = 12 + 5.40 + 2.40 S = $19.80 b) M = 5.40 + 2.40 M = 7.80 Diff: 1 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price.

48) What is the selling price of an item bought for $1420.00 if the markup based on selling price is 59%? Answer: C + M = S 1420.00 + .59S = S 1420 = .41S S = 3463.41 The selling price is $3463.41 Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 49) An article cost $4270.00 and was sold for $9947.00. What was the percent markup based on selling price (correct to the nearest

of 1%)?

Answer: M = 9947 - 4270 = $5677.00 Rate of M on selling price =

= .57072 = 57%.

Diff: 1 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 50) A gross profit of $181.00 is made on a sale. If the gross profit is 22.5% based on selling price, what was the cost? Answer: M = 181 and represents 22.5% of selling price 181 = 0.225 S S= S = $804.44 C = 804.44 - 181.00 = $623.44 Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 51) A gross profit of $800.00 is made on a sale of services. If the gross profit is 15% based on the selling price, what was the cost? Answer: M = 800 and represents 15% of S 800 = 0.15 S S=

= $5 333.33

C = 5 333.33 - 800.00 = $4 533.33 Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price.

52) A gross profit of $96.00 is made on a sale. If the gross profit was 52% based on selling price, what was the cost? Answer: M = 0.52 S 96.00 = 0.52S S = 184.62 C = 184.62 - 96.00 = $88.62 Diff: 1 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 53) What is the regular selling price of an item that cost $1345.00 if the markup is 27% of the regular selling price? Answer: C + M = S 1345 + 0.27S = S 0.73S = 1345 Regular selling price = $1842.47 Diff: 1 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 54) Ali just sold his car for $25 000 and realized a markup of 14% based on cost. What was the cost of the car? What was the amount of the markup rounded to the nearest dollar amount? Answer: C + M = S C + .14C = S 1.14C = 25000 C = $21 929.82 M = 25 000 - 21929.82 = $3070 Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 55) Find the cost of an item sold for $4115 to realize a markup of 36% based on cost. Answer: C + M = S C + .36C = S 1.36C = 4115 C = $3025.74 Diff: 1 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price.

56) A merchant buys an item listed at $69.00 less 34% from a distributor. Overhead is 44% of cost and profit is 37.5% of cost. For how much should the item be retailed? Answer: C = 69(1 - 0.34) = $45.54, E = 0.44C = 0.44(45.54) = $20.04 P = 0.375C = 0.375(45.54) = $17.08 S = 45.54 + 20.04 + 17.08 S = 82.66 Selling price should be $82.66. Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 57) The Alpaca Wool Shoppe bought 150 scarves for $4150.00; 50 scarves were sold at a markup of 175% of cost and 30 scarves at a markup of 125% of cost; 40 of the scarves were sold during a clearance sale for $35.50 each and the remaining Scarves were disposed of at 15% below cost. Assume all scarves had the same cost. a) What was the markup realized on the purchase? b) What was the percent markup realized based on cost? c) What was the gross profit realized based on selling price? Answer: a) Cost per scarf =

= $27.67

Revenue: On 50 scarves sold at a markup of 175% of cost S = 27.67 + 1.75(27.67) = 27.67 + 48.42 = $76.09 Total revenue = 50(76.09) = $3804.50 On 30 scarves sold at a markup of 125% of cost S = 27.67 + 1.25(27.67) = 27.67 + 34.59 = $62.26 Total revenue = 30($62.26) = $1867.80 On 40 scarves sold at $35.50 each Total revenue = 40(35.50) = $1420.00 On 30 remaining scarves sold at 15% below cost S = (1 - .15) $27.67 = $23.52 Total revenue = 30(23.52) = $705.60 Total sales revenue = $7797.90 Total cost = 4150.00 Total markup = $3647.90 b) Percent markup realized based on cost =

(100%) = 87.9012%

c) Percent of gross profit realized based on selling price =

(100%) = 46.7805%

Diff: 3 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price.

58) The Nursery bought 800 plants auctioned off en bloc for $5550.00 This means that each plant has the same cost. On inspection, the plants were classified as normal quality, seconds, and substandard. The 420 normal quality plants were sold at a markup of 85% of cost, the 290 plants classified as seconds were sold at a markup of 21% of cost, and the plants classified as substandard were sold at 83% of their cost. a) What was the unit price at which each of the three classifications was sold? b) If overhead is 23% of cost, what was the amount of profit realized on the purchase? c) What was the average rate of markup based on the selling price at which the plants were sold? Answer: a)

Cost per normal =

= 6.9375

Unit prices: Normal quality = 1.85(6.9375) = $12.83 Seconds = 1.21(6.9375) = $8.39 Substandard = .83(6.9375) = $5.76 b) Total revenue = 12.83(420) + 8.39(290) + 5.76(90) = 5388.60 + 2433.10 + 518.40 = $8 340.10 Total cost = 5550 + 0.23 (5550) = 5550 + 1267.50 = $6826.50 Profit = 8340.10 - 6826.50 = $1513.60 c) Total markup realized = 8340.10 - 5550 = $2970.10 Average rate of markup based on selling price =

= 33.45%

Diff: 3 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 59) A dealer bought laptops for $2250.00 less 35%, 29%. They were sold for $1585.00. a) What was the markup as a percent of cost? b) What was the markup as a percent of selling price? Answer: C = 2250(1 - 0.29)(1 - 0.35) = $1038.38 M = 1 585 - 1038.38 = 546.62 a) Rate of M based on cost = 546.62/1038.38 = 52.6416% b) Rate of M based on selling price = 546.62/1585 = 34.4871% Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price.

60) An appliance store sells electric kettles at a markup of 23.6% of the selling price. The store's margin on a particular model is $7.78. a) For how much does the store sell the kettles? b) What was the cost of kettles to the store? c) What is the percent markup based on cost? Answer: a) M = $7.78 and represents 23.6% of S 7.78 = 0.236S S = $32.97 Selling price is $32.97 b) C = 32.97 - 7.78 = $25.19 c)

Rate of Markup based on cost =

= 30.8853%

Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 61) Sheldan Services sells oil at a markup of 46% of the selling price. If Sheldan paid $1.99 per litre of oil, a) what is the selling price per litre? b) what is the rate of markup based on cost? Answer: a) C + 0.46 of S = S 1.99 + 0.46S = S 1.99 = 0.54S S = 3.69 Selling price is $3.69 per litre. b) M = 0.46 (3.69) = 1.70 Rate of Markup based on cost =

= 85.4271%

Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 62) Chris's Photographic Supplies sells a Minolta camera for $551.83. The markup is 72% of cost. a) How much does the store pay for this camera? b) What is the rate of markup based on selling price? Answer: a) C + 0.72C = S 1.72C = $551.83 C = $320.83 Cost is $320.83 b) M = 0.72(320.83) = $231.00 Rate of Markup based on selling price =

= 41.8607%

Diff: 2 Type: SA Page Ref: 2396-242 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price.

63) A skateboard costing $16.23 was marked up to realize gross profit of 50% of the selling price. a) What was the selling price? b) What was the gross profit as a percent of cost? Answer: a) S = 16.23 + 0.5S 0.5S = 16.23 S = $32.46 Selling price was $32.46 b) Rate of Markup based on cost =

= 100%

Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 64) Sean's Hardware Supplies sells a snowblower for $750.99. The markup is 70% of cost. a) How much does the store pay for this snowblower? b) What is the rate of markup based on selling price? Answer: a) C + 0.70C = S 1.70C = $750.99 C = $441.76 Cost is $441.76 b) M = 0.70(441.76) = $309.23 Rate of Markup based on selling price =

= 41.1763%

Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 65) A bike shop reduces the price of a bike for quick sale from $455.00 to $395.00. Compute the markdown correct to the nearest

of 1%.

Answer: Markdown = 455 - 395 = 60.00 Rate of markdown =

= .13187 = 13.1868%

Diff: 1 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown.

66) Dave and Lance buy pants for $32.75 less 12.75%, 18%. The pants are priced at a regular selling price to cover expenses of 35% of selling price and a profit of 32% of selling price. For a special weekend sale, shirts were marked down 15%. a) What was the operating profit or loss on the pants sold during the weekend sale? b) What rate of markup was realized based on the sale price? Answer: a) C = 32.75(.8725)(.82) = $23.43 23.43 + .35S + .32S = S 23.43 = .33S $71.00 = S Sale price = 71(0.85) = $60.35 Total cost = 23.43 + .35(71) = 23.43 + 24.85 = $48.28 Profit = 60.35 - 48.28 = $12.07 b) Markup realized = Sale Price-C Markup realized = 60.35-23.43 = $36.92 Rate of markup realized on sale price =

= 61.1765%

Diff: 3 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown. 67) An item that cost the dealer $2331.00 less 14%, 12.25% carries a price tag at a markup of 68% based on cost. During Black Friday sale, the item was marked down by 32%. What was the sale price? Answer: C = 2331(0.86)(0.8775) = $1759.09 S = 1759.09 + 0.68(1759.09) S = 2955.27 Sale price = 2955.27(1 - 0.32) = $2009.58 Diff: 2 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown. 68) The regular selling price of merchandise sold in a store includes a margin of 62% based on selling price. During a sale, an item which cost the store $231.25 was marked down 47%. For how much was the item sold? Answer: C + 0.62S = S 231.25 = .38S $608.55 = S Sale price = 608.55(.53) = $322.53 Diff: 2 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown.

69) A furniture store sold a bed regularly priced for $990.00 for $720.00. The bed was originally purchased for $600.00 less 40%. The store's overhead is 25% of the regular selling price. a) Calculate the rate of markdown at which the bed was sold. b) Calculate the operating profit or loss at the sale price. Answer: a) Markdown = 990 - 720 = $270.00 Rate of markdown =

× 100 = 27.2727%

b) C = 600(1 - 0.4) = $360.00 Overhead E = 0.25(990) = $247.50 S=C+E+P 720 = 360 + 247.50 + P P = 720 - 360 - 247.50 P = $112.50 Diff: 2 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown. 70) A bed and bedding store sold a futon regularly priced for $330.00 for $240.00. The futon was originally purchased for $200.00 less 60%. The store's overhead is 25% of the regular selling price. a) Calculate the rate of markdown at which the futon was sold. b) Calculate the operating profit or loss at the sale price. Answer: a) Markdown = 330 - 240 = $90. Rate of markdown =

× 100 = 27.2727%

b) C = 200(1 - 0.6) = $80 Overhead E = 0.25 (330) = $82.50 S=C+E+P 240 = 80 + 82.50 + P P = 240 - 80 - 82.50 P = $77.50 Diff: 2 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown.

71) A retailer buys an appliance for resale at $2123.00 less 27%, 20%. The store marks the merchandise to cover expenses of 23% of the regular selling price and a net profit of 12.9% of the regular selling price. During a clearance sale, the appliance is sold at a markdown of 22%. What is the operating profit or loss? Answer: C = 2123(0.73)(0.8) = 1239.83 S = 1239.83 + 0.23S + 0.129S 0.641S = 1239.83 S = $1934.21 Sale price = 1934.21(0.78) = $1508.69 Total cost = 1239.83 + 0.23(1934.21) = 1239.83 + 444.88 = $1684.70 Operating profit = 1508.69 - 1684.70 = -$176.01 (net loss) Diff: 2 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown. 72) A retail store realizes a gross profit of $62.67 if it sells an article at a margin of 32% of the selling price. a) What is the regular selling price? b) What is the cost? c) What is the rate of markup based on cost? d) If overhead is 21% of cost, what is the break-even price? e) If the article is sold at a markdown of 25%, what is the operating profit or loss? Answer: a) M = 32% of S 62.67 = 0.32S S = $195.84 Regular selling price is $195.84 b) C = 195.84 - 62.67 = $133.17 c)

Rate of markup based on cost =

= .470601 = 47.0601%

d) Break-even price = C + E = C + 0.21C = 1.21C = 1.21(133.17) = $161.14 e) Sale price = 0.75(195.84) = $146.88 Profit = 146.88 - 161.14 = -$14.26 (a loss ) Diff: 2 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown.

73) A retailer paid $59.23 for a set of utensils. Overhead is 14% of the regular selling price and profit is 11% of the regular selling price. During a clearance sale, the set was sold at a markdown of 17%. What was the operating profit or loss on the sale? Answer: Regular selling price = C + E + P S = 59.23 + 0.14S + 0.11S S = 59.23 + 0.25S 0.75S = 59.23 S = $78.97 Sale price = 78.97(0.83) = $65.55 Total cost = 59.23 + 0.14(78.97) = 59.23 + 11.06 = $70.29 Profit = 65.55 - 70.29 = -$4.74 (a loss) Diff: 3 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown. 74) A retailer buys a computer for resale at $500.00 less 35%, 10%. The store marks the merchandise to cover expenses of 15% of the regular selling price and a net profit of 24% of the regular selling price. During a clearance sale, the appliance is sold at a markdown of 25%. What is the operating profit or loss? Answer: Cost = 500(0.65)(0.9) = 292.50 S = C + .15S + .24S .61S = 292.50 S = 479.51 Sale price = .75(479.51) = $359.63 Total cost = 292.50 + .15(479.51) = 292.50 + 71.93 = $364.43 Profit = 359.63 - 364.43 = -$4.80 (a loss) Diff: 2 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown.

75) Bargain City bought desks for $175.00 less 29%, 21%, 13%. The store's overhead is 65% of cost and normal profit is 27% of cost. a) What is the regular selling price of the desks? b) At what price can the desks be put on sale so that the store incurs an operating loss of no more than 30% of the overhead? c) What is the maximum rate of markdown at which the desks can be offered for sale in part b)? Answer: a) C = 175(.71)(.79)(.87) = $85.40 S = C + 0.65C + 0.27C S = 1.92C S = 1.92(85.40) S = $163.97 Regular selling price is $163.97 b) Overhead E = 0.65(85.40) = $55.51 Acceptable loss = 0.3(55.51) = $16.65 Total cost = 85.40 + 55.51 = $140.91 Profit = Sale price - Total cost - 16.65 = Sale price - 140.91 Sale price = 140.91-16.65 Sale price = $124.26 c)

Maximum rate of markdown =

= 24.2178%

Diff: 2 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown.

76) An answering machine cost a dealer $220.00 less 31.5%, 7%. It is regularly priced at $212.00. The dealer's overhead is 17% of the regular selling price and the answering machine was cleared out for $175.45. a) What is the regular markup based on selling price? b) What was the rate of markdown at which the answering machine was sold? c) What was the operating profit or loss? d) What rate of markup based on cost was realized? Answer: a) C = 220(1 - 0.315)(1 - 0.07) = $140.15 M = 212 - 140.15 = $71.85 Rate of markup based on selling price =

= 33.8915%

b) Markdown = 212.00 - 175.45 = 36.55 Rate of markdown =

= 17.2406%

Total cost = cost + expense = 140.15 + 0.17(212) = 140.15 + 36.04 = $176.19 Profit = 175.45 - 176.19 = -0.74 (a loss) d) Rate of markup realized based on cost =

= 25.1873%

Diff: 2 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown. 77) A natural gas oven cost a retailer $420.00 less 37%, 22%, 3.75%. It carries a regular selling price on its price tag at a markup of 65% of the regular selling price. During the end-of-season sale, the barbecue is marked down 42%. a) What is the end-of-season sale price? b) What rate of markup based on cost will be realized during the sale? Answer: a) Cost = 420(1 - 0.37)(1 - 0.22)(1 - 0.0375) = $198.65 Regular selling price = Cost + markup S = 198.65 + .65S .35S = 198.65 S = $567.57 Sale price = 567.57 - 0.42(567.57) = 567.57 - 238.38 = $329.19 b) Markup realized = 329.19 - 198.65 = 130.54 Rate of markup realized on cost =

= 65.7136%

Diff: 2 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown.

78) The Camping Store buys sleeping bags for $54.10 less 12% for buying more than 50 sleeping bags. The store operates on a margin of 51% of the sale price and advertises that all merchandise is sold at a discount of 25% of the regular selling price. What is the regular selling price of the sleeping bags? Answer: Cost = 54.10(.88) = $47.61 47.61 + 0.51 Sale price = Sale price 47.61 = .49 Sale price $97.16 = Sale price Regular selling price - discount = Sale price S - 0.25 of S = Sale price 0.75S = $97.16 S = $129.55 Diff: 3 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown. 79) A value-oriented store buys video recorders for $230.75 less 42%, 11.5%. Expenses are 65% of normal selling price and the required profit is 13% of normal selling price. All merchandise is marked so that the store can advertise a discount of 55%, while still maintaining its regular margin. During the annual clearance sale, the price of the unsold item is marked down 70%. What operating profit or loss is made by the store on items sold during the sale? Answer: C = 230.75(0.58)(0.885) = $118.44 S = 118.44 + 0.65S + 0.13S 0.22S = 118.44 S = $538.36 MP - 0.55MP = S 0.45MP = 538.36 MP = $1196.36 Sale price = 1196.36(0.3) = $358.91 Total cost = 118.44 + 0.65(538.36) = 118.44 + 349.93 = $468.37 Profit = 358.91 - 468.37 = -$109.46 (a loss) Diff: 3 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown.

80) Murray's Pro Shop purchased sets of golf clubs for $710.00 less 22%, 17%, and 19%. Expenses are 19% of the regular selling price and the required profit is 19% of the regular selling price. The store decided to change the regular selling price so that it could offer a 47% discount without affecting its margin. At the end of the season, the unsold sets were advertised at a discount of 60%. What operating profit or loss was realized on the sets sold at the end of the season? Answer: C = 710(0.78)(0.83)(0.81) = $372.32 S = 372.32 + 0 .19S + 0.19S S = 372.32 + 0.38S 0.62S = 372.32 S = $600.52 MP - 47% of MP = S MP - 0.47MP = 600.52 0.53MP = 600.52 MP = $1133.06 Sale price = MP(1 - 0.6) = 1133.06(0.4) = $453.22 Total cost = 372.32 + 0.19(600.52) = 372.32 + 114.10 = $486.42 Profit = 453.22 - 486.42 = -$33.20 (a loss) Diff: 3 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown.

81) RV World sells a make of motor home for $66 725. This regular selling price covers overhead of 12% of cost and a normal profit of 30% of cost. The cruisers were marked with a new regular selling price so that the company can offer a 25% discount while still maintaining its regular gross profit. At the end of the summer season, the motor home was marked down. The company made 20% of its usual profit and reduced the usual commission paid to the sales personnel by 27%. The normal commission accounts for 45% of the normal overhead. What was the rate of markdown? Answer: S = C + E+ P S = C + 0.12C + 0.3C 66725 = 1.42C C = $46 989.44 MP - 25% of MP = S 0.75MP = 66 725 MP = $88 966.67 Normal profit = 0.3(46 989.44) = $14 096.83 Required profit = 0.2(14 096.83) = $2819.37 Normal overhead = 0.12(46 989.44) = $5638.73 Normal commission = 0.45(5 638.73) = $2537.43 Reduction in commission = 0.27(2537.43) = $685.11 Overhead to be recovered = 5 638.73 - 685.11 = 4953.62 Markdown = 88 966.67 - 54 762.43 = 34 204.24 Rate of markdown =

= 38.44%

Diff: 3 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown. 82) Mike's Suits and Accessories purchased men's shoes for $97.00 less 25.6%. The store operates at a normal gross profit of 35% of regular selling price. The owner marks all merchandise with new regular selling prices so that the store can offer a 25% discount. What is the new regular selling price (round to the nearest dollar)? Answer: C = 97(0.744) = $72.17 S = 72.17 + 0.35S 0.65S = 72.17 S = $111.03 MP-25% of MP = S MP -0.25MP = 111.03 0.75MP = 111.03 MP = $148.04 New regular selling price is $148 Diff: 2 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown.

83) For Your Eyes Only Company bought plasma screen TV sets for $11 585 less 4.5%, 3.2%. Overhead is 19% of regular selling price and required profit is 31% of regular selling price. The gas plasma TV sets were marked at a new regular selling price so that the store was able to advertise a discount of 10% while still maintaining its margin. To clear the inventory, the remaining TV sets were marked down 15%. What operating profit or loss is realized at the clearance price? Answer: C = 11 585(.955)(.968) = $10 709.64 10709.64 + .19S + .31S = S 10709.64 = 0.5S $21419.28 = S MP - 10% of MP = S MP -0 .1MP = 21 419.28 0.9MP = 21 419.28 MP = $23 799.20 Sale price = 23799.20(0.85) = $20 229.32 Total cost = 10709.64 + .19(21419.28) = 10709.64 + 4069.66 = 14 779.30 Profit = 20 229.32 - 14 779.30 = $5450.02 Diff: 3 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown. 84) Eastside Pawn Shop bought items for $241.00 less 13%, 20%, 11.5%. The store's overhead is 36% of regular selling price and the profit required is 47% of regular selling price. a) What is the break-even price? b) What is the maximum rate of markdown that the store can offer to break even? c) What is the realized rate of markup based on cost if the items are sold at the break-even price? Answer: a) C = 241(0.87)(0.8)(0.885) = 148.45 S = 148.45 + .36S + .47S 0.17S = 148.45 S = $873.24 Break-even price = 148.45 + 0.36(873.24) = 148.45 + 314.37 = 462.82 b) maximum markdown = 873.24 - 462.82 = $410.42 Maximum rate of markdown =

= 46.9997%

c) Markup realized = 462.82 - 148.45 = 314.37 Rate of markup realized based on cost =

= 211.7683%

Diff: 3 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown.

85) Deep Discount Audio buys home theatre systems for $3 895 less 29%, 15%. Expenses are 30% of the regular selling price and the required profit is 41% of the regular selling price. All merchandise is marked with a new regular selling price so that the store can advertise a discount of 35% while still maintaining its regular markup. During the annual clearance sale, the new regular selling price of unsold items is marked down 40%. What operating profit or loss does the store make on items sold during the sale? Answer: C = 3895(1 - .29)(1 - .15) = $2 350.63 C + 0.3S + 0.41S = S 2350.63 + 0.71S = S 2350.63 = 0.29S 8105.62 = S MP-35% of MP = S 0.65MP = 8105.62 MP = $12 470.18 Sale price = 12470.18(1 - 0.4) = $7482.11 Total cost = 2350.63 + 0.3(8105.62) = 2350.63 + 2431.69 = $4782.32 Operating profit = 7482.11 - 4782.32 = $2699.79 Diff: 3 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown. 86) Home System Alarms buys wireless alarm systems for $4000.00 less 30%, 10%. Expenses are 25% of the regular selling price and the required profit is 20% of the regular selling price. All merchandise is marked with a new regular selling price so that the store can advertise a discount of 40% while still maintaining its regular markup. During the annual clearance sale, the new regular selling price of unsold items is marked down 25%. What operating profit or loss does the store make on items sold during the sale? Answer: C = 4000(1 - .30)(1 - .10) = $2 520 C + .25S + .20S = S 2520 + .45S = S 2520 = 0.55S S = 4 581.82 MP - 40% of MP = S MP - 0.40MP = S 0.60MP = 4 581.82 MP = 7 636.37 Sale price = 7636.37(1 - 0.25) = $5757.28 Total cost = 2520.00 + 0.25(4581.82) = 2520.00 + 1145.46 = $3665.46 Operating profit = $5757.28 - 3665.46 = $2091.82 Diff: 3 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown.

87) Shoot for the Moon Auto Sales and Service has an end of season sale on the remaining SUVs in their lot. The original price of one of these SUVs was $67 998.00. The first discount is 12% and the second discount is 13%. Calculate the final price of the vehicle. A) $50 998.50 B) $52 059.27 C) $57 798.30 D) $59 838.24 E) $61 198.20 Answer: B Diff: 1 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 88) Determine the equivalent single rate of discount for the following series of discounts: 15%, 10%, 5%. A) 27.3% B) 30.0% C) 70.0% D) 72.7% E) 99.9% Answer: A Diff: 2 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 89) Determine the equivalent single rate of discount for the following series of discounts. Discount of 17.25%, 14.7% and 16.37%. A) 40.969% B) 42.838% C) 41.223% D) 41.162% E) 43.084% Answer: A Diff: 1 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 90) Determine the equivalent single rate of discount for the following series of discounts. Discount of 20%, 15% and 5%. A) 35.4% B) 66.0% C) 2.85% D) 83.85% E) 28.5% Answer: A Diff: 1 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series.

91) A store has an end of season sale on snow blowers. The original price was $679.98. The first discount is 10% and the second discount is 14%. What is the final sale price of the snow blowers? A) $526.30 B) $517.54 C) $541.01 D) $540.86 E) $529.40 Answer: A Diff: 1 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 92) The net price of an article is $79.84. What is the list price if a discount of 23% was allowed? A) $116.09 B) $102.52 C) $103.69 D) $64.91 E) $117.41 Answer: C Diff: 1 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 93) The net price of an article is $125.46. What is the list price if a discount of 38% was allowed? A) $200.90 B) $204.13 C) $202.35 D) $90.91 E) $738.00 Answer: C Diff: 1 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 94) The net price of a tennis racket is $79.99. What is the list price if a discount of 35% was allowed? A) 27.00 B) 51.99 C) 123.06 D) 228.54 E) 51.06 Answer: C Diff: 1 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts.

95) Ward's Jewellers sells engagement rings for $2350.00 each less 20%. Its competitors across town offer the same ring for $2390.00 less 19%, 12%. What additional rate of discount must Ward's offer to meet the competitor's price? A) 15.830% B) 9.384% C) 10.900% D) 8.236% E) 9.486% Answer: B Diff: 2 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 96) Lumber King sells 2×4 studs for $3.15 each less 15%. Its competitors across town offer the same stud for $3.55 less 22%, 18%. What additional rate of discount must Lumber King offer to meet the competitor's price? A) 16.392% B) 15.198% C) 14.188% D) 16.232% E) 13.023% Answer: B Diff: 2 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 97) Lumber Queen sells 4×8 sheets of drywall for $8.00 each less 14%. Its competitors across town offer the same drywall for $12.00 less 30%, 20%. What additional rate of discount must Lumber King offer to meet the competitor's price? A) 2.33% B) 25% C) 36% D) 23% E) 233% Answer: A Diff: 2 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series.

98) A fierce price war has broken-out over gas. Super-Save offered pre-price war gas at 73.9 cents per litre with 5% Super-Save discount. Cooperative gas was 75.9 cents per litre and they offered a 3%, 10% discount. What additional discount must Super-Save offer to beat their competitor's price by 2%? A) -79.325% B) 89.513% C) 31.889% D) 7.506% E) 8.105% Answer: D Diff: 2 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 99) What amount must be paid to settle an invoice of $1200.00 dated November 28, terms 3/10, n/30 if the invoice is paid on December 3? A) $1200.00 B) $1236.00 C) $1080.00 D) $1164.00 E) $964.00 Answer: D Diff: 1 Type: MC Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 100) The following invoice was received for 30 shrubs at $3.50 each and 10 raspberry plants at $1.50 each. Terms 4/10, 1/30, n/60. If the invoice was dated May 20 and it was paid on May 30 what was the amount of the payment? A) $115.20 B) $99.75 C) $10.93 D) $126.00 E) $199.50 Answer: A Diff: 1 Type: MC Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount.

101) An invoice dated on September 30 was received on October 2. It was for 4 pairs of in-line skates at $125.00 each and two pairs of gel roller blades at $110.50 each. The terms of the invoice were 5/10, 2/30, n/60. What is the invoice payment amount on October 27? A) $707.58 B) $708.58 C) $705.58 D) $706.85 E) $706.58 Answer: E Diff: 2 Type: MC Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 102) An invoice dated on March 31 was received on April 2. It was for 30 training manuals at $25.00 each and four posters at $40.00 each. The terms of the invoice were 5/10, 2/30, n/60. What is the invoice payment amount on April 3? A) 891.80 B) 910.00 C) 750.50 D) 864.50 E) 64.50 Answer: D Diff: 1 Type: MC Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 103) A wine merchant buys a case of wine for a listed price of $144.00 less 27% from a wine distributor. Overhead is 37% of cost and profit is 51% of cost. For how much should the item be retailed? A) $159.78 B) $75.80 C) $211.35 D) $197.62 E) none of the above Answer: D Diff: 1 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price.

104) Wayne's Pet Supplies sells dog food for $84.50 per 10-kg bag. The markup is 62% of cost. How much does the store pay for this bag of dog food? A) $485.53 B) $224.74 C) $222.37 D) $114.19 E) $52.16 Answer: E Diff: 1 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 105) A data storage device that cost $20.25 was sold for $79.99. Calculate the rate of mark-up based on cost price. A) 74.68% B) 25.32% C) 395.01% D) 295.01% E) 75.32% Answer: D Diff: 1 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 106) An article that costs $542.50 was marked up to reflect 30% of the selling price. Calculate the selling price. A) $162.75 B) $705.25 C) $775.00 D) $379.75 E) $335.00 Answer: C Diff: 1 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 107) Fred's Electric Supplies sells an electric women's shaver for $112.83. The markup is 39% of cost. How much does the store pay for this electric razor? A) $184.97 B) $58.46 C) $88.37 D) $152.40 E) $81.17 Answer: E Diff: 1 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price.

108) A calculator that cost $48.50 was sold for $65.96. Calculate the rate of mark-up based on cost price. A) 64% B) 36% C) 17.46% D) 26.47% E) 73.53% Answer: B Diff: 1 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 109) A cheese merchant buys a case of cheese for a listed price of $96.00 less 16.5% from a cheese distributor. Overhead is 17% of cost and profit is 61% of cost. For how much should the item be retailed? A) $149.10 B) $102.55 C) $144.22 D) $132.45 E) $142.69 Answer: E Diff: 1 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 110) A dollar store bought 1000 pair of socks auctioned off en bloc for $2000.00. This means that each pair of socks has the same cost. On inspection, the socks were classified as normal quality, seconds, and substandard. The 600 normal quality socks were sold at a markup of 90% of cost, the 300 pair of socks classified as seconds were sold at a markup of 30% of cost, and the socks classified as substandard were sold at 60% of their cost. What was the unit price at which each of the three classifications was sold? A) $3.80, $2.60, $1.20 B) $3.80, $2.60, $3.20 C) $3.80, $3.20, $2.60 D) $3.80, $1.20, $2.60 E) $3.80, $1.20, $1.20 Answer: A Diff: 2 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price.

111) The Amusement Park bought 1200 t-shirts auctioned off en bloc for $6600.00. This means that each tshirt has the same cost. On inspection, the t-shirts were classified as normal quality, seconds, and substandard. The 640 normal quality t-shirts were sold at a markup of 95% of cost, the 330 t-shirts classified as seconds were sold at a markup of 35% of cost, and the t-shirts classified as substandard were sold at 47% of their cost. What was the unit price at which each of the three classifications was sold? A) $10.73, $7.43, $2.59 B) $10.73, $1.93, $2.59 C) $10.73, $7.43, $8.09 D) $9.75, $7.43, $2.59 E) $10.73, $2.59, $1.93 Answer: A Diff: 2 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 112) The Paintball Park bought 100 paint ball guns auctioned off en bloc for $7889.00. This means that each gun has the same cost. On inspection, the guns were classified as high quality, medium, and low quality. The 70 high quality guns were sold at a markup of 135% of cost, the 22 guns classified as medium were sold at a markup of 36% of cost, and the guns classified as low quality were sold at 73% of their cost. What was the unit price at which each of the three classifications was sold? A) $185.39, $107.29, $57.59 B) $106.50, $107.29, $57.59 C) $185.39, $102.56, $57.59 D) $185.39, $107.29, $136.48 E) $107.50, $106.29, $57.59 Answer: A Diff: 2 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 113) The Llama Wool Shoppe bought 225 wool overcoats for $16 875; 75 overcoats were sold at a markup of 160% of cost and 100 overcoats at a markup of 115% of cost; 20 of the overcoats were sold during a clearance sale for $130.00 each and the remaining overcoats were disposed of at 11% below cost. Assume all overcoats had the same cost. What was the markup realized on the purchase? A) $18 477.50 B) $18 677.50 C) $18 758.75 D) $18 215.00 E) $18 555.00 Answer: A Diff: 3 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price.

114) The Sheepskin Shoppe bought 375 sheepskin coats for $75 000; 115 coats were sold at a markup of 210% of cost and 95 coats at a markup of 145% of cost; 25 of the coats were sold during a clearance sale for $260.00 each and the remaining coats were disposed of at 14% below cost. What was the gross profit based on selling price? A) 89.154% B) 18.451% C) 49.47% D) 86.546% E) 18.154% Answer: C Diff: 3 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 115) A gas hedge trimmer cost a retailer $218.00 less 37%, 22%, 3.75%. It carries a regular selling price on its price tag at a markup of 82% of the regular selling price. During the end-of-season sale, the hedge trimmer is marked down 21%. What is the end-of-season sale price? A) $78.11 B) $150.83 C) $51.55 D) $452.54 E) $71.55 Answer: D Diff: 2 Type: MC Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown. 116) A power ice auger cost a retailer $327.00 less 13%, 17%, 2.25%. It carries a regular selling price on its price tag at a markup of 94% of the regular selling price. During the end-of-season sale, the power ice auger is marked down 30%. What is the end-of-season sale price? A) $2 692.78 B) $915.30 C) $403.23 D) $914.99 E) $803.23 Answer: A Diff: 2 Type: MC Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown.

117) A gas generator cost a retailer $400.00 less 20%, 17%, 3%. It carries a regular selling price on its price tag at a markup of 90% of the regular selling price. During the end-of-season sale, the gas generator is marked down 40%. What is the end-of-season sale price? A) $1030.53 B) $400.75 C) $458.01 D) $1545.79 E) 545.79 Answer: D Diff: 2 Type: MC Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown. 118) Barmby's Aquarium Supplies purchased sets of aquariums for $315.00 less 10%, 12%. Expenses are 21% of the regular selling price and the required profit is 42% of the regular selling price. The store decided to change the regular selling price so that it could offer a 20% discount without affecting its margin. During the Christmas season, the unsold sets were advertised at a discount of 40%. What operating profit or loss was realized on the sets sold during the Christmas season? A) $112.83 B) $114.62 C) $112.11 D) $110.85 E) $123.49 Answer: B Diff: 3 Type: MC Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown. 119) Duckworth's Bird Supplies purchased sets of cages for $160.00 less 9%, 14%. Expenses are 17% of the regular selling price and the required profit is 55% of the regular selling price. The store decided to change the regular selling price so that it could offer a 25% discount without affecting its margin. During the Christmas season, the unsold sets were advertised at a discount of 25%. What operating profit or loss was realized on the sets sold during the Christmas season? A) $22.11 B) $245.96 C) $138.06 D) $216.72 E) $238.06 Answer: B Diff: 3 Type: MC Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown.

120) Canadian Tire lists a lawn mower at $449 less 20%. What is the net price of the lawn mower? A) $89.80 B) $359.20 C) $360 D) $449 E) $409.50 Answer: B Diff: 1 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 121) At Lowe's, each contractor receives 5% discount for his purchases. A contractor building a deck for a customer pays $7144.89 for the material he bought. What is the list price of the material? A) $142 897.80 B) $357.24 C) $6787.65 D) $7502.13 E) $7520.94 Answer: E Diff: 1 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 122) At Lowe's, each contractor receives 5% discount for his purchases. A contractor buys a saw which is listed at $149 less 10%. What is the actual price paid by the contractor? A) $134.10 B) $141.55 C) $127.40 D) $128.68 E) $129 Answer: C Diff: 1 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 123) What single discount rate is equivalent to multiple discounts of 13% and 25%? Answer: Equivalent single discount rate = 1- [(1 - 0.13) × (1 - 0.25)] = 1 - 0.6525 = 0.3475 = 34.75% Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series.

124) An invoice for $1121.45 with terms 2/10, n/30 is dated March 7. The contractor received it on March 14. What payment will settle the invoice, if the contractor pays it the next day? A) $1121.45 B) $1009.31 C) $22.43 D) $1099.02 E) $1119.45 Answer: D Diff: 1 Type: MC Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 125) Alia keeps his savings of $5000 in a current account at TD Bank for 3 years in order to avoid paying any banking fee. The rate of return for the current account is 0%. Mandated discount rate in Ontario is 2.5%. What is the worth of Alia's money in 3 years? Answer: Alia's money worth in 3 years = $5000 × (1 - 0.025) × (1 - 0.025) × (1 - 0.025) = $5000 × 0.92686 = $4634.30 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 126) An invoice for $1121.45 with terms 2/10, n/30 is dated March 7. The contractor received it on March 14. What payment will settle the invoice if the contractor pays it March 20? A) $1121.45 B) $1009.31 C) $22.43 D) $1099.02 E) $1119.45 Answer: A Diff: 2 Type: MC Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 127) HON received an invoice from its supplier on May 9 for $32 479.87 dated May 7 with terms 3/10, 1/20, n/60. HON pays $15 000 on May 11, $10 000 on May 13 and $2000 on May 21. What balance is still owed after May 30? Answer: Payments of $15 000 and $10 000 are within 10 days of the invoice date. Therefore, a discount of 3% applies. Payment of $2000 is made after 10 days, but before 20 days of the invoice date. Therefore, a discount of 1% applies. Balance owed = $32479.87 -

= $4686.47

Diff: 3 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount.

128) OPG received an invoice from its supplier on May 9 for $32 479.87 dated May 7 with terms 3/10, 1/20, n/60. OPG always makes full payment on the first working day 25th of a given month. What payment will settle the account? Answer: 25th of May is 18 days after the invoice date. Therefore a discount of 1% applies. Pay required = $32 479.87 × (1 - 0.01) = $32 155.07 Diff: 1 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 129) Alpha computing is importing mother boards at a unit cost of $178. Alpha estimates that operating expenses per unit are 35% of cost. What should the selling price be if the desired operating profit per unit is 30% of the cost? A) $115.70 B) $240.30 C) $231.40 D) $293.70 E) $178 Answer: D Diff: 1 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 130) Sears is adding a new line of John Deere lawn mowers. The cost of each such lawn mower is $321.00 with an operating cost of adding this line is estimated to be 40% of the cost of each unit. If Sears matches the competition's retail price of $649 on each lawn mower, what will be the operating profit per unit? A) $328 B) $199.60 C) $128.40 D) $259.60 E) $61.40 Answer: B Diff: 1 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 131) The cost of a TV to Leons is $377.00. If Leons wants a 30% rate of markup on the selling price, determine the markup. A) $538.57 B) $161.57 C) $113.10 D) $490.10 E) $879.67 Answer: B Diff: 1 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price.

132) Kalotyre sells Bridgestone tires at $141.67 each. They purchase the tires from the factory for $103.11 each. Overhead expenses are 20% of the selling price. Determine the rate of markup on the selling price and the cost price? Answer: Cost = $103.11 Markup = $141.67 - $103.11 = $38.56 Rate of markup on selling price = Rate of markup on cost price =

× 100 = 27.22% × 100 = 37.4%

Diff: 1 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 133) Kalotyre retails its Bridgestone tires at $141.67 each. They purchase the tires from the factory for $103.11 each. Overhead expenses are 20% of the selling price. Determine the operating profit per tire. A) $38.56 B) $17.94 C) $10.23 D) $59.18 E) $73.98 Answer: C Diff: 1 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 134) National sports sets the selling price of the ice-hockey kit to include expected overhead expenses of 30% of the selling price and a desired profit of 50% of the selling price. Determine the selling price of the ice-hockey kit if it costs the store $79.81? Answer: Let selling price be S Then S = $79.81 + 0.3S + 0.5S ⇒ S - 0.3S - 0.5S = $79.81 ⇒ S=

= $399.05

Diff: 1 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price.

135) Fresh Co advertises to price match its competitors. This week Fresh Co sells grapes for $3.99 per pound. No Frills has sale on grapes and is selling at $2.69 per pound. Calculate the rate of markdown for Fresh Co for the grapes to price match competition. A) 43.8% B) 48.3% C) 32.6% D) 25.4% E) 0% Answer: C Diff: 1 Type: MC Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown. 136) A piece of jewellery costing $150 was marked up by 40% of the selling price. During the stores silver jubilee sale, the selling price was reduced to $175. What was rate of markdown during the sale? A) 40% B) 16.7% C) 14.3% D) 30% E) 42.9% Answer: B Diff: 1 Type: MC Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown. 137) Kia Soul is sold at an MRSP of $18 500. In May 2013, Kia offered two incentives to its customers. The first incentive was cash back of $2000 on the final sale price, which includes 13% HST. The other one was no HST MRSP. Which incentive resulted in lesser cost to the customer? Answer: In the first incentive, the total cost to the customer including tax = $18 500 × 1.13 - 2000 = $19 275 In the second incentive, total cost to the customer = $18 500 ∴ The second incentive results in lesser cost to the customer. Diff: 1 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown.

138) Cost of electricity to the grid is calculated in $ per MWh. On a typical day, the Darlington plant produces an average of 84 000 MWh. The OEB recently ruled that Darlington would receive $51 per MWh of electricity produced. The cost of fuel is $5 per MWh. Operations and maintenance costs are $37 per MWh. How much profit will the Darlington plant make in a year? Answer: S = $51/MWh C = $5/MWh E = $37/MWh P = S - C - E = 51 - 5 - 37 = $9/MWh Annual profit =

= $275 940 000

Diff: 1 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 139) United Furniture purchased RF42 model refrigerators from Samsung for $2751 each in 2012. The store marked up the price by 30% of the regular selling price. The store's overheads were estimated to be 10% of the regular selling price of each refrigerator in 2012. Following Christmas, Samsung introduced a new model. In order to make space for the new model, United Furniture decided to sell the RF42 model refrigerators at a markdown of 25%. How much was the operating profit or loss on each refrigerator as a result of the sale? Answer: Let Sr be the regular selling price for each refrigerator. Then Sr = $2751 + 0.3Sr ⇒ 0.7Sr = 2751 ⇒

Sr =

= $3930

Let Ss be the final sale price after the markdown Then Ss = $3930 - 0.25 × 3930 ⇒ Ss = $2947.50 Profit/Loss = Ss - C - E = $2947.50 - $2751 - 0.1 × 3930 = -$196.50 With the sale price, the operating loss was $196.50 per refrigerator. Diff: 3 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown.

140) United Furniture purchased RF42 model refrigerators from Samsung for $2751 each in 2012. The store marked up the price by 30% of the regular selling price. The store's overheads were estimated to be 10% of the regular selling price of each refrigerator in 2012. Following Christmas, Samsung introduced a new model. In order to make space for the new model, United Furniture decided to sell the RF42 model refrigerators at a markdown which results into a break-even. Calculate the markdown. Answer: Let Sr be the regular selling price for each refrigerator. Then Sr = $2751 + 0.3Sr ⇒ 0.7Sr = 2751 ⇒

Sr =

= $3930

Let Ss be the final sale price after the markdown. In order to achieve break-even Ss - C - E = 0 ⇒ Ss = C + E ⇒ Ss = $2751 + 0.1 × 3930 ⇒ Ss = $3144 Markdown =

× 100 = 20%

Diff: 3 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown. 141) Kahn Motors purchased 2008 an off-lease Nissan Altima from the Nissan dealer for $9850 in 2012. Kahn Motors marked up the price by 20% of the regular selling price. After failing to sell it for 3 months, Kahn Motors decided to sell the car at a markdown of 15%. What is the final sale price of the car? A) $12 312.50 B) $11 820 C) $10 047 D) $10 465.63 E) $10 368.42 Answer: D Diff: 3 Type: MC Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown. 142) Sonia receives an invoice for $981.50 dated September 30, with terms 5/10, 2/20, n/60. She decides to pay it in full on October 25. What is the total amount paid to settle the account? A) $981.50 B) $932.43 C) $961.87 D) $934.76 E) $962.25 Answer: A Diff: 3 Type: MC Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount.

143) Hasan receives an invoice for $103.50 dated September 30, with terms 5/10, 2/30, n/60. He decides to pay it in full on October 25. What is the total amount paid to settle the account? A) $103.50 B) $98.57 C) $100.40 D) $98.33 E) $101.43 Answer: E Diff: 3 Type: MC Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 144) A 3D printer listed at $3499.00 has a net price of $2599.00. What is the rate of discount? Answer: Amount of Discount = 3499 - 2599 = $900.00 Rate of discount =

= 25.7216%

Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 145) Overstock.com advertised a discount of $1250 on one of the 3D printers. If the discount is 30.7%, how much was the printer sold for? Round the answer to the nearest dollar value. Answer: 30.7% of list = $1250 .307L = 1250 L = $4071.6612 Sale price = 4071.6612 - 1250.00 = $2821.6612 The sale price was $2822 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 146) A 7.5% discount on a microwave at Lowes amounted to $75. Calculate the net price. Answer: 0.075L = 75 L = $1000.00 N=L-D N = 1000 - 75 = $1025.00 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts.

147) At Rona's, a shed was sold for $888.00 during the end of the season clearance sale. If the clearance sale indicated a discount of 40%, what was the list price? Answer: N = L(1 - d) 888 = L(1 - 0.40) 888 = L(0.60) L = $1480 The list price was $1480.00 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 148) Determine the net price of a heavy machinery equipment listed at $24 500 less 22%, 12.5% Answer: N = 24500(1 - 0.22)(1 - 0.125) = 24500(0.78)(0.875) = $16 721.25 Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 149) An invoice shows a net price of $375.75 after discounts of 15%, 10.6%, 5.5%. What was the list price? Answer: N = L(1 - 0.15)(1 - 0.106)(1 - 0.055) 375.75 = L(0.85)(0.894)(0.945) L= L = $522.97 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 150) A certain home appliance is listed for $699.00 less 15 %, 8.5%. a) What is the net price? b) How much is the amount of discount allowed? c) What is the equivalent single rate of discount that was allowed? Answer: a) N = 699 (1 - 0.156667)(1 - 0.085) = 699(0.843333)(0.915) = $539.38 b) Discount = 699 - 539.38 = $159.62 c) Single rate = = 22.8350% Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series.

151) Honest Ed sells a product for $225.00 less 7.5% and 5%. Factory Direct carries the same product for $209.99 less 5%. What further discount must Factory Direct allow so that its net price will be the same as that of Honest Ed? Answer: Net price at Honest Ed = 225(1 - 0.075)(1 - 0.05) = $197.71875 Net price at Factory Direct = 199.99(1 - 0.05) = 199.4905 Additional discount needed = 199.4905 - 197.7185 = 1.77175 Additional percent discount = 1.77175/209.99 = 0.8437% Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 152) An invoice for $2600.00, dated June 2, with terms 3/10, 1/30, n/60, was received on June 10. What payment must be made on June 11 to reduce the debt to $1250? Answer: Invoice = $2600 Payment made on June 11 allows for the 3% discount Amount of reduction = 2600 - 1250 = 1350 Amount paid = 1350(0.97) = $1309.50 Diff: 2 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 153) A convenience store received an invoice of $10 680 on May 3 with the terms 5/10, n/30 EOM. What amount will reduce the amount due on the invoice by $7 500.00 if the payment was made on June 3? Answer: Payment on June 3 will allow for the 5% discount on the partial payment. Amount credited = $7500 Amount paid = 7500(0.95) = $7125.00 Diff: 2 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 154) A convenience store received an invoice of $10 680 on May 3 with the terms 5/10, 2/15, n/30. What amount will reduce the amount due on the invoice by $7500.00 if the payment was made on May 20? Answer: Payment on May 20 will not qualify for any cash discounts. Amount credited = Amount paid = 7500 Diff: 2 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount.

155) The following invoices, all with terms 7.5/10, 2.5/30, n/45, were paid together on March 25. First invoice dated February 1 is for $2940.67; second invoice dated February 25 is for $4086.50; and the third invoice dated March 18 is for $9101.25. What amount was remitted? Answer: Invoice Date Inv. Amount Discount Rate Amount paid February 1 $2940.67 --$2940.67 February 25 $4086.50 2.5% (0.975)(4086.50) = 3984.34 March 18 $9101.25 7.5% (0.925)(9101.25) = 8334.48 Amount remitted $15 259.49 Diff: 2 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 156) A certain Canadian company received an invoice dated March 20 from their Cambodian supplier for the following shipment of goods that they received on April 1. 2000 jeans at $25.60 each; 3050 shirts at $15.25 each; 1000 men hats at $2.50 each; 1500 women hats at $3.00 each. The terms of the invoice stated 4/20, n/45 ROG and all items are subject to trade discounts of 15%, 5%. a) What is the last day of the discount period? b) What is the amount due if the invoice is paid in full on April 22? c) If only a partial payment is made on the last day of the discount period, what amount is due to reduce the outstanding balance to $25 000.00? Answer: a) April 20 b) 2000 ∗ 25.60 = $51 200 3050 ∗ 15.25 = $46 512.50 1000 ∗ 2.5 = $ 2 500.00 1500 ∗ 3.00 = $ 4 500.00 $104 712.50 Net Amount = 104712.5(0.85)(0.95) = $84 555.34375 Payment on April 19 is not within the 4% discount period, so full $84 555.34 is paid. c)

Net Amount = $84 555.34375 Remaining balance = 25 000.00 Amount of reduction = $59 555.34375 Discount of 5% applies Payment = 59555.34375(0.96) = $57 173.13 Diff: 3 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount.

157) What is the cost of the perfume set that is sold for $165.00 with a margin of 215% based on cost. Answer: C + 2.15C = 165 3.15C = 165 C = $52.38 Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 158) A drone at Future-is-Yours store, costing $1050.00 was marked up to reflect 70% of the selling price. Calculate the selling price. Answer: S = C + M S = 1050 + 0.70(S) S - 0.70(S) = 1050 S = 1750/0.3 S = $3500.00 Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 159) A big electronics retailer bought 15" tablets for $750.00 less 21%. He sold them for $1550.00. a) What is the rate of markup based on cost? b) What is the rate of markup based on selling price? Answer: C = 750(1 - 0.21) = $592.50 M = 1550 - 592.50 = 957.50 a) Rate of M based on cost = 957.50/592.50 = 161.6034% b) Rate of M based on selling price = 957.50/1550 = 61.7742% Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 160) A gross profit of $57.75 is made on the sale of a dress. If the gross profit is 33.75% based on cost, what was the selling price? Round your answer to the nearest dollar value. Answer: Gross profit = markup. Hence, M = 57.75 and it is 37.75% of the cost 55.75 = 0.3775 C C = 55.75/.3775 C = $147.6821 S = 147.6821 + 57.75 = $205.4321 The dress is sold for $205.00 Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price.

161) A retailer bought 10 000 units of cell phone cases for $55 000.00. On delivery, it was realized that not all the cases were of the same quality. Hence they were classified as high quality, average quality and poor quality. The 4200 high quality cases were sold at a markup of 120% of cost, the 2800 average quality cases were sold at a markup of 80% of the cost, and the remaining were sold at cost price. a) What is the unit selling price at which each of the three classifications were sold? b) If overhead is 18% of the cost, what is the amount of profit realized on the purchase? Answer: a) Cost per case = 55000/10000 = 5.5 Unit prices: High quality = 5.5 + 1.20(5.5) = $12.10 Average quality = 5.5 + 0.80(5.5) = $9.90 Poor quality = 5.5 b) Total revenue = 12.10(4200) + 9.9(2800) + 5.5(10 000 - 4200 - 2800) = 50 820 + 27 720 + 16 500 = $95 040.00 Total cost = 55 000 + 0.18 (55 000) = $64 900.00 Profit = 95 040 - 64 900 = $30 140.00 Diff: 3 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 162) A toy shop reduces the price of a popular toy from $89.99 to $69.99 to increase their sales compared their online competition. a) What is the markdown amount in this case? b) What is the rate of markdown? Answer: a) Markdown = 89.99 - 69.99 = 20.00 b) Rate of markdown = 20/89.99 = 0.222247 = 22.22% Diff: 1 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown.

163) To increase their sales compared their online competition, a toy shop marks down the regular price of $49.99 of a popular toy by 25%. a) What is the markdown amount in this case? b) What is the new (sale) price of the toy? Answer: a) Markdown ($) = 49.99(0.25) = $12.50 b) Sale Price = S - Markdown = 49.99 - 12.50 = $37.49 Diff: 1 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown. 164) A microwave oven costs a retailer $220.00. It carries a regular selling price on its price tag at a markup of 60% of the regular selling price. During a special promotion, the oven is marked down 30%. What is the sale price? Answer: Cost = 220 Regular selling price = Cost + markup S = 220 + .60S 0.40S = 220 S = $550.00 Sale price = 550 - 0.30(550) = $385.00 Diff: 2 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown. 165) A clothing retailer buys winter coats from one of its suppliers for $67.50. The regular selling price of the coats includes operational expenses of 46% of the selling price and a profit of 30% of the selling price. Due to an unexpected warm winters, the sales have been slow. The retailer decides to mark down this line of coats by 20% to increase sales. What is the operating profit or loss on the coats sold during the promotional sale? Answer: C = $67.50 Regular selling price: S = 67.5 + 0.46S + 0.30S 0.24S - 67.5 S = $281.25 Sale price = 281.25(1 - 0.20) = $225.00 Profit = Sale Price - C - E = 225 - 67.5 - 0.46(67.5) = $126.45 Diff: 2 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown.

166) A clothing retailer buys winter coats from one of its suppliers for $67.50. The regular selling price of the coats includes operational expenses of 46% of the selling price and a profit of 30% of the selling price. Due to an unexpected warm winters, the sales have been extremely slow. With spring products arriving in the warehouse, the retailer decides to mark down this line of coats by 70% to clear out the inventory. What is the operating profit or loss on the coats sold during the promotional sale? Answer: C = $67.50 Regular selling price: S = 67.5 + 0.46S + 0.30S 0.24S - 67.5 S = $281.25 Sale price = 281.25(1 - 0.70) = $84.375 Profit = Sale Price - C - E = 84.375 - 67.5 - 0.46(67.5) = - $14.175 Hence the retailer will incur a loss of $14.18. Diff: 2 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown. 167) Budget Electronics bought slightly used computers for $275.00 less 29%. The store's overhead is 23% of the cost and the desired profit is 37.5% of cost. a) What should be the regular selling price of the computers? b) For a promotional sale, a computer was sold for $214.16. What is the rate of markdown offered for this sale? Answer: a) C = 275(.71) = $195.25 Regular selling price: S = C + 0.23C + 0.375C S = 1.605(192.25) S = $308.56 b) E = 0.23(195.25) = $21.91 Total cost = C + E = 192.25 + 21.91 = $214.16 Maximum Markdown = 308.56 - 214.16 = 94.40 Maximum rate of markdown = 94.4/308.56 = 30.5945% Diff: 2 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown.

168) Ice-Time purchased hockey skates for $220.00 less 12% and 7%. Expenses are 22% of the cost and the required profit is 45% of the cost. At the end of the winter season, the unsold skates were advertised at a discount of 50%. What operating profit or loss was realized on the sets sold at the end of the season? Answer: C = 220(0.88)(0.93) = $180.0480 S = C+ 0.22C + 0.45C S = 1.67(180.048) S = $300.68 Sale price = S(1 - 0. 5) = 300.68(0. 5) = $150.34 Total cost = 180.048 + 0.22(180.048) = 1.22(180.048) = $219.66 Profit = 150.34 -219.66 = -$69.32 Hence the store incurred a loss of $69.32 in the end of season sale. Diff: 3 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown. 169) The Great Outdoors buys camping tents for $160.00 less 35% for buying them out of season from the manufacturer. The store's markup is 45% of the sale price. To attract customers, the store advertises all camping gear to be sold at a discount of 20% of the regular selling price. What is the regular selling price of the camping tents? Answer: Cost = 160.00(1 - 0.35) = $104.00 Sale price = 104 + 0.45 Sale price 0.55 Sale price = 104 Sale price = $189.09 Regular selling price - discount = Sale price S - 0.20 of S = 189.09 0.8S = 189.09 S = $236.36 Diff: 3 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown.

170) Unique Furniture bought coffee tables for $399.00 less 23%, 10%, 5.5%. The store's overhead expenses come out to be 30% of regular selling price and the target profit is 55% of regular selling price. a) What is the maximum rate of markdown that the store can offer to break even? b) What is the realized rate of markup based on cost if the items are sold at the break-even price? Answer: a) C = 399 (1 - 0.23)(1 - 0.1)(1 - 0.055) = 261.30 S = 261.3 + 0.3S + 0.55S 0.15S = 261.3 S = $1742.00 Break-even price = 261.30 + 0.3(1742) = 261.3 + 522.6 = 783.9 Maximum markdown = 1742 - 783.9 = $958.10 Max. markdown rate = 958.1/1742 = 55.0% b) Markup realized = 783.9 - 261.3 = $522.60 Rate of markup realized based on cost = 522.60/261.30 = 200.00% Diff: 3 Type: SA Page Ref: 249-254 Topic: 6.5 Integrated Problems Objective: 6-6: Solve integrated problems involving discounts, markup, and markdown. 171) A new line of backpacks had an original list price of $48.99, and were subject to a trade discount of 13%. What is the amount of discount and the net price? Answer: Amount of Discount = 48.99(0.13) = $6.37 Net price = 48.99 - 6.37 = 42.62 Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 172) Go-Go Gloves is looking to set a net price of $28.50 on their premium gloves to retain some profit. What should the list price be so Go-Go Gloves can achieve their target net price with a 17.5% discount rate? Answer: Net price = $28.50; Net price factor = 1 - 0.175 = 0.825 = 82.5% 28.50 = L(0.825) L = 28.50/0.825 = $34.55 The article should be listed at $34.55. Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts. 173) The new Abibas 4000 sneakers, listed at $655, are subject to a discount series of 8%, 5%, 10%, and 15%. What's the net price? Answer: Net price = 655(1 - 0.08)(1 - 0.05)(1 - 0.10)(1 - 0.15) = 655(0.92)(0.95)(0.90)(0.85) = $437.94 Diff: 1 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-1: Solve problems involving trade discounts.

174) A new couch sells for $1299 less 25%, 7%, 4% and 12%. What's the net price and the single equivalent rate of discount? Answer: Net price = 1299(1 - 0.25)(1 - 0.07)(1 - 0.04)(1 - 0.12) = $765.43 Single rate = 1 - [(1 - 0.25)(1 - 0.07)(1 - 0.04)(1 - 0.12)] = 1 - 0.589248 = 0.410752 Diff: 2 Type: SA Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 175) A cabinet manufacturer sells cabinet doors for $55 subject to a discount series of 2%, 6%, and 10%. What's the single equivalent rate of discount? A) 13.500% B) 17.092% C) 6.667% D) 86.500% E) 82.908% Answer: B Diff: 1 Type: MC Page Ref: 222-228 Topic: 6.1 Determining Cost with Trade Discounts Objective: 6-2: Calculate single rates of discount for a discount series. 176) If given payment terms of 4/15, n/40, what's the length of the credit period and what's the cash discount rate? A) 15 days; 4% B) 40 days; 15% C) 40 days; 4% D) 4 days; 40% E) 15 days; 40% Answer: C Diff: 1 Type: MC Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 177) On December 28, Rylee pays off four different invoices of furniture for her new apartment. Every invoice extended payment terms of 10/10, 3/25, n/90. The invoices are as follows: $288 dated December 1, $1267 dated December 5, $1810 dated December 19, and $4392 dated December 26. What was the total amount she paid? Answer: December 1: beyond 10-day discount period, beyond 25-day discount period. $288 paid. December 5: beyond 10-day discount period, within 25-day discount period. (1267)(1 - 0.03) = $1228.99 paid. December 19: within 10-day discount period. (1810)(0.90) = $1629 paid. December 26: within 10-day discount period. (4392)(0.90) = $3952.80 paid. Total paid: 288 + 1228.99 + 1629 + 3952.80 = $7098.79 paid. Diff: 2 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount.

178) Raven bought snorkeling apparatus for $225 on April 11 with credit terms 4/12 EOM. Raven pays her invoice on May 14. How much did she pay? Answer: With terms EOM, we treat the invoice date as April 30. Therefore the last day of the discount period is May 12. Since Raven pays her invoice on May 14, she pays the full $225. Diff: 1 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 179) Tarfex Industries has an outstanding invoice of $4712.44 from October 28, and on November 2, receives payment of $805.20. The credit terms on the invoice were 12/5, n/30. How much is still owed on the invoice after this partial payment? Answer: Let x represent be the allowed credit amount. 805.20 = x(1 - 0.12) 915 = x Amount still owed = 4712.44 - 915 = $3797.44 Diff: 1 Type: SA Page Ref: 230-236 Topic: 6.2 Payment Terms and Cash Discounts Objective: 6-3: Apply methods of cash discount. 180) Brian is opening a skincare business, and has purchased 400 bottles of acne spot treatment for $4400. His business overhead is 36% of his cost. What's the markup and profit when he sells the spot treatment for $18 each? Answer: S = 18; C = 4400/400 = 11; E = 0.36(11) = 3.96 18 = 11 + M 7=M 7 = 3.96 + P 3.04 = P Therefore, Brian's markup per bottle is $7, and the profit per bottle is $3.04. Diff: 2 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 181) Jen's hydroflask business incurs a cost of $5.25 a bottle. If the markup is 220% of the cost, what's the selling price? A) $16.80 B) $5.25 C) $11.55 D) $115.50 E) Not enough information. Answer: A Diff: 1 Type: MC Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price.

182) The cost to Apple of the original Airpods is $55, while the cost of the new Airpods Pro is $115. The rate of markup based on cost of the original Airpods was 298.18%, while the rate of markup on the Airpods Pro was 186.09%. If the overhead costs are 25% of the cost for both products, which product does Apple make more profit from? Answer: Original Airpods Airpods Pro 2.9818 = M/55 1.8609 = M/115 164.00 = M 214.00 = M 164.00 = 0.25(65) + P 214.00 = 0.25(115) + P 164.00 = 16.25 + P 214.00 = 28.75 + P 147.75 = P 185.25 = P Therefore, Apple earns more profit from the Airpods Pro. Diff: 3 Type: SA Page Ref: 238-244 Topic: 6.3 Markup Objective: 6-4: Solve problems involving markup based on either cost or selling price. 183) Look Book Inc. purchases a pair of glasses for $86. On Boxing Day, they marked down the glasses by 25%. They have a target profit of 20% of the selling price and their expenses were 23% of the selling price. Did Look Book earn a profit or a loss during this sale? Answer: Selling price = C + E + P S = 86 + 0.23S + 0.20S 0.57S = 86 S = 150.88 Sale price = S - 0.25S = 0.75S = 0.75(150.88) = 113.16 Profit = Sale price - Total Cost P = 113.16 - [86 + 0.23(150.88)] = 113.16 - 120.70 = -$7.54 Therefore, Look Book earned a loss during this sale. Diff: 3 Type: SA Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown. 184) Brockhampton's merchandise went on sale to celebrate the release of their new album, and the sale price for a hoodie was $69. When the merchandise is not on sale, they incur overhead costs of $7 and earn a profit of $71. Each hoodie costs $22 to make. What's the rate of markdown? A) 44.92% B) 25.00% C) 68.11% D) 2.82% E) 31.00% Answer: E Diff: 2 Type: MC Page Ref: 246-250 Topic: 6.4 Markdown Objective: 6-5: Solve problems involving markdown.

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 7 Simple Interest 1) Calculate the amount of interest that will be charged on $8500.00 borrowed for 6 months at 5.25%. Answer: P = 8500; r = 0.0525; t = I = 8500(0.0525)

= $223.13

Diff: 1 Type: SA Page Ref: 268-274 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 2) Calculate the amount of interest that will be charged on $9000.00 borrowed for four months at 9.0%. Answer: P = 9000; r = 0.09; t = I = (9000)(0.09)

= $270.00

Diff: 1 Type: SA Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 3) Determine the exact time between January 30, 2014 and May 20, 2014 by counting days. Answer: The number of days is 2 + 28 + 31 + 30 + 19 = 110 days Diff: 1 Type: SA Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 4) Determine the date represented by 100 days after March 1, 2012. Answer: DT1 is March 1, 2012 DBD is 100, then DT2 is June 9, 2012 Diff: 1 Type: SA Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 5) Determine the exact time between January 20, 2012 and May 10, 2012 by counting days. Answer: The number of days is 12 + 29 + 31 + 30 + 9 = 111 days Diff: 1 Type: SA Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 6) Determine the exact time between January 21, 2014 and September 13, 2014 by counting days. Answer: (31 - 21) + 1 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 12 = 235 Diff: 1 Type: SA Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt.

7) Compute the amount of interest on $875.00 at 11.5% p.a. from May 29, 2013 to August 13, 2013. Answer: P = 875.00; r = 0.115 Time period May 29, 2013, to August 13, 2013 = 225 - 149 = 76 days I = (875)(0.115)

= $20.95

Diff: 1 Type: SA Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 8) Compute the amount of interest on $250.00 at 8.25% p.a. from March 30, 2014 to October 28, 2014. Answer: P = 250; r = 0.0825 Time period March 30, 2014, to October 28, 2014 = 212 days I = (250)(.0825)

= $11.98

Diff: 1 Type: SA Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 9) Compute the amount of interest for $500.00 at 8.75% from June 30 to December 31. Answer: Number of days between June 30 and December 31 = 365 -181 = 184 I = 500.00 ∗ 0.0875 ∗

= 22.05

Diff: 1 Type: SA Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 10) Compute the amount of interest for $679.43 at 6.25% from May 11 to January 20. Answer: Number of days = 21 + 30 + 31 + 31 + 30 + 31 + 30 + 31 + 19 = 254 I = 679.43 ∗ 0.0625 ∗

= $29.55

Diff: 1 Type: SA Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 11) Find the amount of interest on $635.00 at 6.5% p.a. from October 2, 2015 to August 4, 2016. Answer: Number of days = 30 + 30 + 31 + 31 + 29 + 31 + 30 + 31 + 30 + 31 + 3 = 307 I = 635 ∗ 0.065 ∗

= $34.72

Diff: 1 Type: SA Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt.

12) Alex borrowed $7500 to buy a car. If interest is charged on a loan at 8.5%, how much interest would he have to pay in 120 days. Answer: P = $7500; r = 0.085; t = I = 7500(0.085)

= $209.59

Diff: 1 Type: SA Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 13) Determine the deposit that must be made to earn $52.96 in 240 days at 7.3%. Answer: I = 52.96; r = 0.073; t = P=

= $11 033.33

Diff: 1 Type: SA Page Ref: 274-279 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 14) A bank pays an interest of 4.5% for a three month term deposit. Calculate the amount that Stacy must invest to earn an interest of $100.00. Answer: r = 0.045; t = P=

= 0.25; I = $100

= $8888.89

= $1387.76

Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt

16) What principal will earn $219.89 interest at 11.25% p.a. from November 16, 2013 to February 7, 2014? Answer: I = 219.89; r = 0.1125 Time period November 16, 2013, to February 7, 2014 = 14 + 31 + 31 + 7 = 83 days P=

= $8595.43

Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 17) What principal will earn $59.99 interest at 7.75% p.a. from February 4, 2014 to July 17, 2014? Answer: Number of days = 24 + 31 + 30 + 31 + 30 + 17 = 163 P=

= $1733.33

= 2326.37

= $2310.81

Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 20) If Inez is charged an interest of $42.95 on a loan of $950.00 for seven months, calculate the rate of interest charged on the loan. Answer: I = 42.95; P = 950; t = r=

= 0.0775 = 7.75%

Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt

21) At what rate of interest must a principal of $1535.00 be invested to earn interest of $75.46 in 235 days? Answer: P = 1535.00; I = 75.46; t = r=

= 7.6354%

= 8.12%

Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 23) What rate of interest is paid if the interest on a loan of $3500.00 is $99.06 from November 14, 2013 to May 20, 2014? Answer: Number of days = 16 + 31 + 31 + 28 + 31 + 30 + 20 = 187 r=

= 5.5244%

Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 24) What rate of interest is paid if the interest on a loan of $7300.00 is $692.71 from June 11, 2014 to April 2, 2015? Answer: P = 7300; I = 692.71 Time period June 11, 2014 to April 2, 2015 = 20 + 31 + 31 + 30 + 31 + 30 + 31 + 31 + 28 + 31 + 1 = 295 days r=

= 11.74%

Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt

25) What rate of interest is required for $940.48 to earn $42.49 interest from September 30, 2013, to March 4, 2014? Answer: Number of days = 1 + 30 + 31 + 31 + 29 + 3 = 125 I = 42.49; P = 940.48; t = r=

= 13.19%

Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 26) What rate of interest is paid if the interest on a loan of $15 000.00 is $2000.00 from January 31, 2014 to May 31, 2014? Answer: Number of days between January 31 and May 31 = 120 r=

= 40.5556%

Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 27) An investment of $12 000.00 earns an interest of $480.00. Calculate the time in months if the interest rate is 6%. Answer: P = 12 000; I = 480; r = 0.06 t=

= 0. (12) = 8 months

∗ 12

= 2.222 ∗ 12 = 26.67 months = 27 months Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt

29) In how many months will $9100.00 grow to $9818.75 at 6.41% p.a.? Answer: I = 9818.75 - 9100.00 = 718.75 t=

∗ 12 = 14.79 months = 15 months

∗ 12

= 5.304197 ∗ 12 = 63.65 months = 64 months Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 31) How many months will it take $639 to grow to $795 at 8.45% p.a.? Answer: I = 795.00 - 639.00 = 156.00; P = 639.00; r = 0.0845 t(months) =

∗ 12 = 34.67 months = 35 months

∗ 12 = 7.5 months = 8 months

∗ 365 =

∗ 365

= .695652 ∗ 365 = 253.91 days = 254 days Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt

34) Determine the number of days it will take $478.00 to earn $17.09 at 7 %. Answer: P = 478; I = 17.09; r = 0.0725 t (days) =

∗ 365 =

∗ 365

= .493147 ∗ 365 = 179.998 days = 180 days Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 35) Daniel's credit card statement last month showed $560 in cash advances and $5.25 in interest charges. The interest rate on the statement was 18.8%. For how many days was Daniel charged interest? Answer: P = $560; I = $5.25; r = 0.188 t (days) =

∗ 365 = 18.201 days = 19 days

Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 36) On June 12, 2013, Frank James opened a low interest line of credit at his bank, with interest at 4.5% p.a. He used this line of credit immediately to pay $5300 of the tuition fee. On what date did Frank James honour the line of credit if he ended up paying $186.20 in interest? Answer: I = 186.20; P = 5300; r = 0.045 t=

= 0.780713 years = 0.780713∗365 = 284.96 days = 295 days

DT1: June 12, 2013 DBD: 295 DT2: April 3, 2014 Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 37) Find the maturity value of $832.00 invested at 8.8% from May 20, 2013, to November 23, 2013. Answer: P = 832; r = 0.088 t = 11 + 30 + 31 + 31 + 30 + 31 + 23 = 187 days =

or 327 - 140 = 187

S = P(1 + rt) = 832.00 = 832(1.0450849) = 869.51 Diff: 1 Type: SA Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt).

38) Find the maturity value of $1080.00 invested at 3.65% from July 24, 2014 to December 21, 2014. Answer: P = 1080; r = 0.0365; t = 150 days S = P(1 + rt) = 1080

= $1096.20

Diff: 1 Type: SA Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt). 39) What payment is required to pay off a loan of $2550.00 at 6.9% sixteen months later? Answer: P = 2550; r = 0.069; t = S = P(1 + rt) = 2250 = $2457.00 Diff: 1 Type: SA Page Ref: 279-283 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt). 40) Compute the accumulated value of $10 000.00 at 9% after six months. Answer: P = 10 000.00; r = 0.09; t =

= 0.5 years

S = P(1 + rt) = 10 000 = 10 000(1.045) = $1045.00 Diff: 1 Type: SA Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt). 41) Compute the accumulated value of $6500.00 at 8.5% after eleven months. Answer: P = 6500; r = 0.085; t = S = P(1 + rt) = 6500 = 6500(1.0779167) = $7006.46 Diff: 1 Type: SA Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt).

42) Tamiko invested $1500.00 at a bank for eight months at an interest rate of 4 %. Calculate the maturity value of the investment. Answer: P = 1500; t =

; r = 0.04625

S = P(1 + rt) = 1500

= $1546.25

Diff: 1 Type: SA Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt). 43) What is the amount to which $7250.00 will grow at 7.75% p.a. in 3 months? Answer: P = 7250.00; r = 0.0775; t =

= 0.25 months

S = 7250.00 = 7250.00(1.019375) = $7390.47 Diff: 1 Type: SA Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt). 44) Leo wishes to invest $8000.00 that he saved from his summer job. His bank offers 3.75% for a one-year term investment or 3.5% for a six-month term. Help Leo to investigate his options. a) How much will Leo receive after one year if he invests at the one-year rate? b) How much will Leo receive after one-year if he invests for six-months at a time at 3.5% each time? c) What would the one-year rate have to be to yield the same amount of interest as the investment described in part b)? Answer: P = 8000 a) r = 0.0375; t = 1 year S = 8000(1 + 0.0375∗1) = $8300 b) For the 1st six months: P = 8000; r = 0.035; t = 6 months = 0.5 years S1 = 8000 (1 + 0.035 ∗ 0.5) = $8140

For the 2nd six months: P = 8140; r = 0.035; t = 6 months = 0.5 years S2 = 8140 (1 + 0.035 ∗ 0.5) = $8282.45

c) Interest described at part b) is I = 8282.45-8000 = 282.45, t = 1 year r=

= 3.5306%

Diff: 2 Type: SA Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt).

45) What sum of money will accumulate to $539.04 in 97 days at 11.4%? Answer: S = 539.04; r = 0.114; t = P=

= $523.19

Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

46) What sum of money will accumulate to $3700.00 at 14.75% p.a. in 451 days? Answer: S = 3700; r = 0.1475; t = P=

= $3129.62

Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

47) What is the present value of $1200.00 due at 7.25% p.a. in 100 days? Answer: S = 1200.00; r = 0.0725; t = P=

= $1176.63

Diff: 1 Type: SA Page Ref: 283-285 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

48) Find the present value of a debt of $760.00 ninety-five days before it is due if money is worth 6.25%. Answer: S = 760.00; r = 0.0625; t = P=

= 747.83

Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

49) What principal will have a maturity value of $100 000 at 5% p.a. in 18 months? Answer: S = 100 000.00; r = 0.05; t = P=

= 1.5 years

= 93 023.26

Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

50) What principal will have a maturity value of $61 500.00 at 6.5% p.a. in 17 months? Answer: S = 61500.00; r = 0.065; t = P=

= 56 314.38

Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

51) What sum of money will accumulate to $1426.80 in eight months at 7.78%? Answer: S = 1426.80; r = 0.778; t = P=

= $1364.86

Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

52) What is the present value of $41 230.00 due in nine months if interest is 11.1%? Answer: S = 41 230.00; r = 0.111; t =

= 0.75 years

= $38 061.39

Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

53) Determine the amount of money that would have to be invested at 5 % to accumulate to $10 000.00, 91 days after the investment date. Answer: S = 10 000; r = 0.05375; t = P=

= $9867.77

Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

54) What principal will have a maturity value of $40 000.00 at 5.25% p.a. in 4 years? Answer: P =

= $33 057.85

Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

55) What principal will have a maturity value of $12 000.00 at 6.25% p.a. in 3 months? Answer: P =

= $11 815.38

Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

56) On April 24, 2012, Rebecca purchased a government-guaranteed short-term investment maturing on July 5, 2012. How much did Rebecca pay for the investment if she will receive $6000 on the maturity date and interest is 2.75%? Answer: S = 6000; r = 0.0275; t = 72 days P=

= $5967.63

Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

57) An appliance store advertises a stove for $747.50 with nothing down, no payments and no interest for six months. Determine the cash value the store would be willing to accept if on a six-month investment, it can earn an interest of 4%. Answer: S = 747.50; r = 0.04; t = P=

= 0.5 years

= $732.84

Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

58) A loan payment of $1700 was due 60 days ago and another payment of $1200 is due 45 days from now. What single payment 90 days from now will pay off the two obligation if interest is to be 14% and the agreed focal date is 90 days from now? Answer: Let the size of the single payment be $x. The focal date is 90 days from now. The equation of equivalence is 1700.00

+ 1200.00

1797.81 + 1220.71 = x x = 3018.52 The single payment 90 days from now is $3018.52. Diff: 2 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 59) Debt payments of $800.00 due now and $1400.00 due in five months are to be repaid by a payment of $1000.00 in three months and a final payment in eight months. Calculate the size of the final payment if interest is 6%. Answer: Use 8 months as the focal date and let $x represent the final payment. 1000

+ x = 800

+ 1400

1025 + x = 832 + 1421 x = 1228 The size of the payment is $1228.00. Diff: 2 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates.

60) Debt payments of $1610.00 due today, $725.00 due in 101 days and $670.00 due in 296 days respectively are to be combined into a single payment to be made 170 days from now. What is that single payment, if money is worth 9.5% p.a. and the agreed focal date is 170 days from now? Answer: Let the size of the single payment be $x. At the agreed focal date (170 days now): 1610

+ 725

1681.24 + 738.02 + 648.73 = x x = 3067.99 The size of the single payment is $3067.99. Diff: 2 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 61) Debt payments of $2560.00 and $1965.00 are due four months from now and ten months from now respectively. What single payment is required to pay off the debt today if interest is 5.94%? Answer: Let the size of the single payment be $x. The focal date is today. +

2510.30 + 1872.32 = x 4382.62 = x The size of the single payment is $4382.62. Diff: 2 Type: SA Page Ref: 285-296 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 62) Three debts, the first for $1720 due four months ago, the second for $1315 due in 5 months, and the third for $1640 due in 7 months, are to be paid by a single payment today. How much is the single payment if money is worth 7.5% p.a. and the agreed focal date is today? Answer: Let the size of the single payment be $x. The focal date is today. x = 1720

x = 1763.00 + 1275.15 + 1571.26 x = 4609.41 The size of the single payment is $4609.41. Diff: 2 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates.

63) Debt payments of $1000.00 due today, $500.00 due in 90 days, and $500.00 due in 120 days are to be combined into a single payment to be made 120 days from today. What is that single payment if money is worth 9.00% p.a. and the agreed focal date is 120 days from today? Answer: Let the single payment be $x. x = 1000

+ 500

x = 1029.59 + 503.70 + 500.00 x = $2033.29 Diff: 2 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 64) Debt payments of $500.00, $1000.00, $1500.00 are due on March 1, May 1, and December 1 of the same year. If interest is 8% calculate the single payment on August 1 of the same year that would replace the three payments. Answer: Use August 1 as the focal date. For the $500.00 debt: P = 500; r = 0.08; t = S = 500

= 516.77

For the $1000.00 debt: S = 1000; t = S = 1000

= 1020.16

For the $1500.00 debt: P = 1500; t = P=

= 1460.93

The single equivalent payment on August 1 is 516.77 + 1020.16 + 1440.93 = $2977.86 Diff: 2 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates.

65) Debt payments of $1925.00 due today, $510.00 due in 87 days, and $674.00 due in 337 days are to be combined into a single payment to be made 115 days from now. What is that single payment if money is worth 8.65% p.a. and the agreed focal date is 115 days from now? Answer: Let the single payment be $x. x = 1925

+ 510

x = 1977.46 + 513.38 + 640.31 x = $3131.15 Diff: 2 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 66) Debt payments of $1170.00 due two months ago and $1243.00 due today respectively are to be repaid by a payment of $1505.00 in one month and the balance in four months. If money is worth 8.4% p.a. and the agreed focal date is four months from now, what is the size of the final payment? Answer: Let the size of the final payment be $x. At the agreed focal date (4 months from now): 1170

+ 1243

= 1505

1219.14 + 1277.80 = 1536.61 + x 960.33 = x The size of the final payment is $960.33. Diff: 2 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 67) Debt obligations of $700.00 due three months ago and $1810.00 due in one month are to be repaid by a payment of $1200.00 today and the balance in six months. What is the size of the final payment if interest is 7.75% and the agreed focal date is one month from now? Answer: Let the size of the final payment be $x. The focal date is one month from now. 700

+ 1810.00 = 1200

718.08 + 1810.00 = 1207.75 + 1320.33 = .9687185x 1362.97 = x The size of the final payment is $1362.97. Diff: 3 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates.

68) Two obligations of $835 each, due 90 days ago and 35 days ago respectively, are to be settled by two equal payments to be made today and 65 days from now respectively. If interest allowed is 8.75% and the agreed focal date is today, what is the size of the equal payments? Answer: Let the size of the equal payments be $x. The focal date is today. The equation of equivalence is: 835

+ 835

=x+

853.02 + 842.01 = x + 0.9846569x 1695.03 = 1.9846569x x = 854.07 The size of the payments is $854.07. Diff: 3 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 69) Debts of $1480.00 due four months ago and $1385.00 due in one month are to be settled by two equal payments due now and nine months from now respectively. Find the size of the equal payments at 12% p.a. with the agreed focal date now. Answer: Let the size of the equal payments be $x. At the agreed focal date (now): 1480.00

=x+

1539.20 + 1371.29 = x + .9174312x 2910.49 = 1.9174312x 1517.91 = x The size of the equal payments is $1517.91. Diff: 3 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates.

70) A loan of $3325.00 is to be repaid by three equal payments due in 102 days, 157 days and 189 days respectively. Determine the size of the equal payments at 12.15% p.a. with a focal date of today. Answer: Let the size of the equal payments be $x. At the agreed focal date (today): 3325 =

3325 =

= .9671616x + .950334x + .9408102x = 2.8583058x 1163.28 = x The size of the equal payment is $1163.28. Diff: 3 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 71) A loan of $1000 taken out on January 1 requires equal payments on February 1,March 1, and April 1. If the focal date is april 1, what is the size of the equal payments at 6.0%? Answer: Let the size of the equal payments be $x. Focal date is September 30. Equation of equivalence is: 1000.00

1014.79 = x(1.00969863) + x(1.00509589) + x 1014.79 = 3.01479452x 336.60 = x The size of the equal payments is $336.60. Diff: 3 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates.

72) A loan of $1825 taken out on March 10 requires equal payments on April 30, June 19, and August 3, and a final payment of $700 on September 30. If the focal date is September 30, what is the size of the equal payments at 8.6%? Answer: Let the size of the equal payments be $x. Focal date is September 30. Equation of equivalence is: 1825

+ 700

1212.72 = x(1.0360493) + x(1.0242685) + x(1.0136658) 1212.72 = 3.0739836x 394.51 = x The size of the equal payments is $394.51. Diff: 3 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 73) Payments of $1430 due one year ago and $1175 due with interest of 6% in nine months are to be settled by three equal payments due today, seven months from now, and one year from now at 7.5%. Determine the size of the equal payments if the agreed focal date is one year from today. Answer: Let the size of the equal payments be $x. The focal date is one year from now. The maturity value of $1175 due in 9 months with 6% interest = 1175

= 1227.88

The equation of equivalence is: 1430(1 + 0.075 ∗ 2) + 1227.88

= x(1 + 0.075 ∗ 1) + x

1644.50 + 1250.90 = x(1.075) + x(1.03125) + x 2895.40 = 3.10625x 932.12 = x The size of the equal payments is $932.12. Diff: 3 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates.

74) A debt of $1410.00 due 6 months ago and a second debt of $1815.00 due in 4 months with interest at 6.16% p.a. are to be settled by two equal payments due now and 9 months from now respectively. Find the size of the equal payments at 7.37% p.a. with the agreed focal date now. Answer: Let the size of the equal payments be $x. Maturity value of 2nd debt = 1815.00 1410 [1 + 0.0737 ∗ 0.5] +

= 1852.27 =x+

1461.96 + 1807.86 = x + .9476203x 3269.82 = 1.9476203x 1678.88 = x Diff: 3 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 75) Payments of $5100 each due in five, ten, and fourteen months from now are to be settled by three equal payments due today, seven months from now, and eleven months from now. What is the size of the equal payments if interest is 8.75% and the agreed focal date is today? Answer: Let the size of the equal payments be $x. The focal date is today. +

=x+

4920.60 + 4753.40 + 4627.60 =

14301.60 = x + .951437x + .9257474x 14301.60 = 2.8771844x 4970.69 = x The size of the equal payments is $4970.69. Diff: 3 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 76) A loan of $10 000.00 is to be repaid in four equal payments due in three months, six months, nine months, and one year respectively after the date of the loan. Calculate the size of the equal payments if interest is 8.5%. Answer:

= 10 000

0.979192166x + 0.959232614x + 0.940070505x + 0.921658986x = 10 000 3.800154272x = 10 000 x = $2631.47 Diff: 3 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates.

77) A loan of $4100 is to be repaid in three equal installments due 110, 197, and 311 days respectively after the date of the loan. If the focal date is the date of the loan and interest is 6.89% p.a., find the size of the installments. Answer: Let the size of the installments be $x. The focal date is the date of the loan. 4100.00 =

4100.00 =

4100.00 = .979658x + .9641462x + .9445488x 4100.00 = 2.888353x 1419.49 = x The size of the installments is $1419.49. Diff: 3 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 78) A debt of $7120 is to be settled by two equal payments due today, and three years from now respectively. Determine the size of the equal payments if money is worth 8.73% and the agreed focal day is today. Answer: Let the size of the equal payments be $x. The focal date is today. 7120 = x + 7120 = x + 7120 = x + .7924558x 7120 = 1.7924558x x = 3972.20 The size of the equal payments is $3972.20. Diff: 3 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 79) The exact number of days between January 25, 2012 and March 25, 2012 is? A) 59 days B) 60 days C) 61 days D) 62 days E) 63 days Answer: B Diff: 1 Type: MC Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt.

80) The exact number of days between January 5 and March 17 during the next leap year is? A) 72 days B) 74 days C) 73 days D) 71 days E) 75 days Answer: A Diff: 1 Type: MC Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 81) Calculate the amount of interest if $2000.00 is invested at 6% for 200 days. A) $65.75 B) $120.00 C) $54.25 D) $6575.34 E) $240.00 Answer: A Diff: 1 Type: MC Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 82) Calculate the amount of interest if $700.00 is invested at 5.5% for two years and nine months. A) $111.65 B) $105.88 C) $10 587.50 D) $1058.75 E) $158.65 Answer: B Diff: 1 Type: MC Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 83) How much interest will you pay on a loan of $15 500 if you are paying the loan off in 9 months? Your loan rate is 7.125%. A) $828.28 B) $838.28 C) $848.28 D) $818.28 E) $858.28 Answer: A Diff: 1 Type: MC Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt.

84) Calculate the amount of interest if $3000.00 is invested at 6.25% from November 30, 2014 to May 15, 2015. A) $843.75 B) $86.82 C) $85.27 D) $70.31 E) $70.75 Answer: C Diff: 1 Type: MC Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 85) What rate of interest did you receive over a period of 90 days if your principal was $5000 and it has a maturity value of $6000? A) 81.11% B) 8.11% C) 0.811% D) 0.0811% E) 811.11% Answer: A Diff: 1 Type: MC Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 86) What rate of interest did you receive over a period of 67 days if your principal was $7444 and it has a maturity value of $7601? A) 11.94% B) 11.49% C) 11.69% D) 11.29% E) 11.99% Answer: B Diff: 1 Type: MC Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 87) What rate of interest did you earn for a period of 2417 days if you invested $175 and you earned $222 in interest? A) 16.19% B) 17.19% C) 29.19% D) 21.19% E) 19.16% Answer: E Diff: 1 Type: MC Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt

88) How many days did it take for your investment of $4300 to accrue $147 in interest at a rate of 8.12%? A) 163.36 days B) 135.36 days C) 157.36 days D) 153.67 days E) 163.67 days Answer: D Diff: 1 Type: MC Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 89) You have $6500 invested in a 30-day savings certificate at an interest rate of 2.00%. How much money will you have when the certificate matures? A) $6510.83 B) $6510.68 C) $5985.12 D) $10.68 E) $6600.68 Answer: B Diff: 1 Type: MC Page Ref: 279-283 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt). 90) You have $4447 invested in a 180-day savings certificate at an interest rate of 3.65%. How much money will you have when the certificate matures? A) $5427.05 B) $4572.05 C) $5472.05 D) $4527.05 E) $4257.05 Answer: D Diff: 1 Type: MC Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt). 91) How much money do you have to invest in order to accumulate a total of $10 000 in 365 days if you are able to earn 5.00% on your money? A) $9533.81 B) $9523.81 C) $10000.81 D) $8523.81 E) $5923.81 Answer: B Diff: 1 Type: MC Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

92) How much money do you have to invest in order to accumulate a total of $4761 in 327 days if you are able to earn 9.52% on your money? A) $3485.86 B) $4358.86 C) $4386.85 D) $3458.86 E) $3845.86 Answer: C Diff: 1 Type: MC Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

93) You currently owe $1400 today and $2317 in 195 days at a rate of interest of 7%. What is the single equivalent payment 150 days from today? A) $2297.18 B) $2297.81 C) $2279.18 D) $2279.81 E) $3737.45 Answer: E Diff: 1 Type: MC Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 94) You owe $3572.75 in 147 days and have decided to pay off your loan early. Your interest rate is 6.52%. What is the size of the cheque that you write today to pay off all of your debt? A) $4381.33 B) $3481.33 C) $3581.33 D) $3381.33 E) $3814.33 Answer: B Diff: 1 Type: MC Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates.

95) You owed $700.00 35 days ago, and you missed the payment. You also owe an additional $600 in 5 months. What single equivalent value is due if your focal date is today and you are paying at an interest rate of 8.46% on your debt? A) $1403.83 B) $1304.83 C) $1204.83 D) $1503.83 E) $1285.25 Answer: E Diff: 2 Type: MC Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 96) What amount of interest will be charged on $20 000 in 1 month for a simple interest of 5.99%? A) $1198 B) $119.80 C) $99.83 D) $9.98 E) $0 Answer: C Diff: 1 Type: MC Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 97) If a 6 month term deposit at a bank pays a simple interest of 3%, how much will have to be deposited to earn $18 000? A) $600 000 B) $120 000 C) $1 200 000 D) $270 E) $540 Answer: C Diff: 1 Type: MC Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 98) John buys a car for $20 000 on Kia finance with a simple interest of 5.99%. After one month he pays off the Kia finance using his secured line of credit which charges him 4% simple interest. He pays off the line of credit in 11 months. How much did he pay in total? A) $20 737 B) $20 837 C) $20 733 D) $20 800 E) $21 198 Answer: B Diff: 2 Type: MC Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt

99) Interest of $429.48 was charged on a loan of $9500 over a period of 7 months. What simple rate of interest was charged on the loan? A) 4.52% B) 2.64% C) 7.75% D) 3.48% E) 4.38% Answer: C Diff: 1 Type: MC Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 100) The interest earned on a $6000 investment was $240. What was the term in months if the interest rate was 6%? A) 4 months B) 7 months C) 8 months D) 9 months E) 12 months Answer: C Diff: 1 Type: MC Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 101) The simple interest rate being charged on a $10 000 is 0.5% per month. What is the total amount paid, if the entire loan is paid in 9 months? Answer: P = $10 000 r = 0.5% per month t = 9 months S = P (1 + rt) = $10 000 (1 + 0.005 × 9) = $10 450 Diff: 1 Type: SA Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt).

102) Principal of $259 000 on a house is mortgaged at a rate of 3.7% amortized over a period of 25 years resulting into a total interest of $137 178.49. The bank also offers simple interest for the customers, who do not want to pay compound interest. If bank wishes to earn the same amount of interest over the period of 25 years, what rate will the bank charge to its customers, if the loan is paid in equal annual installments? Answer: Principal to be paid every year = Interest to be paid every year =

= $10 360 = $5487.14

Total annual installment = $15 847.14 $15,847.14 = $10 360(1 + r × 1) ⇒ r = 0.5296 ⇒ r = 52.96% Diff: 3 Type: SA Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt). 103) Didzo took a loan of $14 600 on January 10, 2013 at a rate of 9.25%. He repaid the loan on June 4, 2013. What is the interest due on repayment date? A) $529.10 B) $532.80 C) $536.50 D) $540.20 E) $543.90 Answer: C Diff: 1 Type: MC Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 104) What amount of money would have to be invested at 4.25% to grow to $15 000 after 150 days? A) $14 742.51 B) $15 969.12 C) $12 364.69 D) $14 973.85 E) $14 000.00 Answer: A Diff: 1 Type: MC Page Ref: 283-285 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

105) Sally put $9500 in term deposit on May 22. It matured on September 4 at $9588.82. What interest rate did she earn on the term deposit? Answer: P = $9500 S = $9588.82 t = 10 + 30 + 31 + 31 + 3 = 105 days ⇒ t= r=

= 0.2877 years =

= 0.0325

⇒ r = 3.25% Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

106) Dion is supposed to pay Kristy $3500 on April 20, but delays the payment until July 1. What amount should Dion expect to pay on July 1, if Kristy can earn 8.25% on a low-risk investment? A) $3788.75 B) $3557.75 C) $3556.96 D) $3556.17 E) $3500 Answer: C Diff: 1 Type: MC Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt). 107) Leons advertises a bed room set for $1495 with nothing down, no payments and no interest for 6 months. What cash price should the store be willing to accept if, on a 6 month investment, Leons can earn a rate of 9%? A) $1629.55 B) $1562.28 C) $1371.56 D) $1430.62 E) $1495 Answer: D Diff: 1 Type: MC Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

108) Misra purchases an airplane ticket to Calgary today on the airline's early bird sale for $559, instead of buying it for $599 four months from now. Had he not bought the ticket today, he would have invested the money in bonds yielding 9% return. What is his true savings? Answer: Future value from bond = S = P(1 + rt) = $559

= $575.77

Future value of the airline ticket = $599 Misra's true savings = $599 - $575.77 = $23.23 Diff: 1 Type: SA Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt). 109) A $10 000 loan advanced on February 15 at a 10% interest rate requires equal payments after 61 days, 122 days and 183 days from the day of the loan. What should be the value of the equal payments? Answer: Let x be the value of equal payment. Present value for a payment made after 61 days =

Present value for a payment made after 122 days =

Present value for a payment made after 61 days =

= $10 000

⇒ x = $3444 Diff: 3 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 110) What principal will have a future value of $640, at 12.1% in 6 months? A) $600 B) $603.50 C) $678.72 D) $370.80 E) $800 Answer: B Diff: 1 Type: MC Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

111) Ed plans to receive $2000 interest in 200 days on his investment of $40 000. At what interest rate should he plan to invest? A) 9.1% B) 9.6% C) 34.7% D) 10% E) 20% Answer: A Diff: 2 Type: MC Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

112) Hydro One plans to complete a modification with the project start date of March 12. The procurement activities begin on July 6 worth $3.5 million. Calculate the amount the project manager obtains on March 12 to pay for all the procurement, if money is worth 2.5% per year Answer: P =

= $3 472 175

Diff: 2 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

113) Kennedy owed $925 to his friend a year ago and owed $600 to another friend 6 months ago. If he clears his debt today, what total payment is he expected to make, if the simple interest of 20% is applicable? A) $1770 B) $1110 C) $660 D) $1525 E) $1800 Answer: A Diff: 2 Type: MC Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 114) Danny bought a camera from Future Shop for $385 at 3% p.a. rate and 4 equal quarterly payments. Determine the size of the quarterly payments if the agreed focal date is the date of the loan. A) $98.05 B) $96.25 C) $99.14 D) $93.45 E) $128.33 Answer: A Diff: 2 Type: MC Page Ref: 261-270 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates.

115) On Christmas day, Sebastian deposited the $265 he got in a savings account that had a simple interest rate of 1.7% p.a. By next Christmas, how much interest was earned and paid into his account? Answer: P = $265.00; r = 1.7% = 0.017; t = 1 I = Prt = (265)(0.017)(1) = $4.51 Diff: 1 Type: SA Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 116) CentSmart Inc. deposited $83 215 into an account that had a simple interest rate of 2.3% on February 27th, 2015. On New Year's Eve, how much interest had been earned on that deposit? Answer: The starting date is February 27, 2016 and the ending date is December 31, 2015. The number of days between dates is 307. P = $83 125.00; r = 2.3% = 0.023; t = I = Prt = (83 125)(0.023)

= $1608.07

Diff: 2 Type: SA Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 117) Bilal had got his first monthly paycheck of $5800 from his job and decided to deposit it into a new account with a 1.2% simple interest rate p.a. After half a year, how much interest had this initial deposit garnered? Answer: P = $5800.00; r = 1.2% = 0.012; t = 0.5 I = Prt = (5800)(0.012)(0.5) = $34.80 Diff: 2 Type: SA Page Ref: 266-271 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 118) To buy a house, Noah borrowed $55 230 from the bank. If interest was charged on the loan at a 1.12% simple interest rate p.a. how much interest would Noah have to pay after half a year? Answer: P = $55 230; r = 0.0112; t = I = Prt = (55 230)(0.0112)

= $309.29

Diff: 2 Type: SA Page Ref: 268-274 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt.

119) Determine the principle amount Bruce Wayne must invest if $1485 of interest was earned over 10 months with a rate of 0.6%. Answer: P =

= $297 000

= $7542.86

= 0.0089 = 0.89%

Diff: 1 Type: SA Page Ref: 272-276 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 122) Approximately how many months would it take for $9520 to earn $17.00 of interest on a 0.55% rate? Answer: P = 9520; I = 17; r = 0.0055 t= t = 0.32 yrs # of months = 0.32(12) = 3.84 ≈ 4 Diff: 2 Type: SA Page Ref: 274-279 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt

123) Bart deposited $4300 into a savings account with a rate of 1.4% p.a. After 200 days, what is the maturity value of the deposit? Answer: P = 4300; r = 1.4% = 0.014; t = S = (4300) = (4300)[1 + (0.0077)] = (4300)[1.0077] = 4333.11 Diff: 1 Type: SA Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt). 124) Pablo invested $1965 in a 3-month term deposit at 2.2% p.a. What is the maturity value of this deposit? Answer: P = 1965; r = 2.2% = 0.022; t = S = (1965) = (1965)[1 + (0.0055)] = (1965)[1.0055] = 1975.81 Diff: 1 Type: SA Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt). 125) Genevieve won $42 016 from the lottery and deposited all of it into an RESP account for her children that had an interest rate of 0.7% for a 1-year term. What is the maturity value of the deposit? Answer: P = 42 016; r = 0.7% = 0.007; t = 1 S = (42 016)[1 + (0.007)(1)] = (42 016)[1 + (0.007)] = (42 016)[1.007] = 42 310.11 Diff: 1 Type: SA Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt).

126) CloudTrust Limited has a monthly payroll of $6 million. They choose to make a short-term investment in an account with a 0.9% rate. At the end of the month, what is the maturity value of the deposit? Answer: P = 6 000 000; r= 0.9% = 0.009; t = S = (6 000 000) = (6 000 000)[1 + (0.00075)] = (6 000 000)[1.00075] = 6 004 500 Diff: 1 Type: SA Page Ref: 277-279 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt). 127) Five months after Nadir makes an investment into an account with a rate of 0.65%, the maturity value of his deposit is $1822. What was the initial value of his investment? Answer: S = 1822; r = 0.65% = 0.0065; t = P=

P= P= P = 1817.09 Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

128) What principal will have a future value of $7325 at a 1.1% rate in 350 days? Answer: S = 7325; r = 1.1% = 0.011; t = P=

P= P= P = 7248.89 Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

129) On New Years' Day in 2014, Legend's Hair Emporium borrowed a sum of money, and promised to

pay $11 781 back on October 28, 2014. This loan included interest at 1.4%. How much money was initially borrowed on New Years' Day by Legends Hair Emporium? Answer: Time between dates = 301 days S = 11781; r = 1.4% = 0.014; t = P=

P= P= P = 11646.48 Diff: 2 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

130) Louis took out a loan to pay off his new car, and four months later, has to pay back $8752. This figure also includes a 0.93% interest rate. How much did Louis originally borrow? Answer: S = 8752; r = 0.93% = 0.0093; t = P=

P= P= P = 8724.95 Diff: 1 Type: SA Page Ref: 280-282 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

131) Jake borrowed $5590 from Gina 8 months ago and Jake also borrowed $1210 from Rosa five months ago. If Jake wanted to repay both of these debts today, determine the single payment that would fully repay both debts. Allow for simple interest at 0.9%. Answer: A focal date of today. The dated value of Jake's debt to Gina at the focal date: S = 5590[1 + (0.009) S = 5623.54 The dated value of Jake's debt to Charles at the focal date: S = 1210 S = 1214.54 Single payment required now = 5623.54 + 1214.54 = 6838.08 Diff: 2 Type: SA Page Ref: 285-296 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates. 132) Eighty days ago, Emma was supposed to pay $870 to Regina. She also will owe Regina $1230 in 120 days. However, Emma is unable to pay these amounts; so instead, Regina has agreed to accept two equal payments due after three months and six months, respectively. If the interest rate is 1.25%, determine the size of the quarterly payments. Answer: Dated value of the original scheduled payments: S = 870

= 870[1 + 0.00274] = 872.38

= 1224.97

Dated value of replacement payments at focal date: P=

= 0.99688x

= 0.99379x

0.99688x + 0.99379x = 872.38 + 1224.97 1.99067x = 2097.35 x = 1053.59 Diff: 3 Type: SA Page Ref: 284-293 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates.

133) Calculate the amount of interest that will be charged on $3500.00 borrowed for three months at 4.8%. Answer: P = 3500. r = 0.048. t =

I = 3500(0.048) I = $42.00 Diff: 1 Type: SA Page Ref: 268-274 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 134) Calculate the amount of interest that will be charged on $8500.00 borrowed for half a year at 6.2%. Answer: P = 8500. r = 0.062. t =

I = 8500(0.062) I = $263.50 Diff: 1 Type: SA Page Ref: 268-274 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 135) Ricky borrowed $11 350.00 for 8 months at a rate of 4.245%. How much interest was accumulated in his borrowing period? Answer: P = 11350. r = 0.04245. t =

I = 11350(0.04245) I = 321.21 Ricky's loan accumulated $321.21 in interest. Diff: 1 Type: SA Page Ref: 268-274 Topic: 7.1 Finding the Amount of Simple Interest Objective: 7-1: Compute the amount of simple interest using the formula I = Prt. 136) Calculate the principal that would earn $252.98 at 5.755% in 10 months. Answer: I = 252.98. r = 0.05755. t = P=

= $5275.00

Diff: 1 Type: SA Page Ref: 274-279 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt

137) Calculate the annual interest rate that would accumulate $105.00 of interest on $1000.00 in 14 months. Answer: P = 1000.00. I = 105.00. t = r=

= 0.09 = 9%.

Diff: 1 Type: SA Page Ref: 274-279 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 138) Determine the number of months required for a principal of $1200 to earn $67.50 simple interest at 7.5% per annum. Answer: P = 1200. I = 67.50. r = 0.075 t=

= 0.75

0.75(12) = 9 ∴ It would take 9 months. Diff: 1 Type: SA Page Ref: 274-279 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 139) Chuck's repair shop received $175.69 interest on a 150-day term deposit of $9600. At what rate of interest was the deposit invested? Answer: P = 9600. I = 175.69. t = r=

= 0.0445 = 4.45%

Diff: 1 Type: SA Page Ref: 274-279 Topic: 7.2 Finding the Principal, Rate, or Time Objective: 7-2: Compute the principal, interest rate, or time using variations of the formula I = Prt 140) Determine the future value of an investment of $1500 earning 5.6% per annum for 5 months. Answer: P = 1500. r = 0.056. t = S = 1500

= $1535.

Diff: 1 Type: SA Page Ref: 279-283 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt).

141) Flynn invested $3400 in a two-year deposit, earning 6.35% per annum. What is the future value of Flynn's investment? Answer: P = 3400. r = 0.0635. t = 2 S = 3400[1 + 0.0635(2)] = $3831.80 Diff: 1 Type: SA Page Ref: 279-283 Topic: 7.3 Computing Future Value (Maturity Value) Objective: 7-3: Compute the maturity value (future value) using the formula S = P(1 + rt). 142) Compute the principal value of an investment that earns interest at 5.5% per annum and has a maturity value of $690 after 10 months. Answer: S = 690. r = 0.055. t = P=

= $659.76

Diff: 1 Type: SA Page Ref: 283-285 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

143) What amount of money must have been invested on March 25, 2019, to amount to $8400 on December 15, 2019, at 7.2% per annum? Answer: Time period in days = 265. S = 8400. r = 0.072. t = P=

= $7982.71

Diff: 1 Type: SA Page Ref: 283-285 Topic: 7.4 Finding the Principal (Present Value) Objective: 7-4: Compute the principal (present value) using the formula P =

144) Two debts have to be paid today. One debt is $720 from two years ago and the other was $610 from five months ago. Determine the single payment today that would fully repay the debts if both debts accumulated simple interest at 1.15% per annum Answer: S = P(1 + rt) for both debts. i. P = 720. r = 0.0115. t = 2. S = 720[1 + 0.0115(2)] = $736.56 ii.

P = 610. r = 0.0115. t = S = 610

= $612.92

$736.56 + $612.92 = $1349.48. Diff: 2 Type: SA Page Ref: 285-296 Topic: 7.5 Computing Equivalent Values Objective: 7-5: Compute equivalent or dated values for specified focal dates.

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 8 Simple Interest Applications 1) For the following promissory note, determine the amount of interest due at maturity. $1250.00 Ottawa, Ontario January 30, 2019. 10 months after date we promise to pay to the order of Yonk Steel Company * * * * * * * *EXACTLY* * * * * * * 1250.00* * * * * * * * * * *DOLLARS at Yonk Steel Company for value received with interest at 9.05% per annum. Due ___________________ (Seal) ____________________ (Seal) ____________________ Answer: The term ends November 30, 2019 Legal due date is December 2, 2019, because November 30 falls on a weekend (Saturday). Interest period January 30, 2019 to December 2, 2019 is 306 days. Interest = (1250)(0.0905)

= $94.84

Diff: 2 Type: SA Page Ref: 302-309 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 2) For the following promissory note, determine the amount of interest due at maturity. $3000.00 Oakville, Ontario January 15, 2013. Six months after date we promise to pay to the order of Hughes Pet Supply Company * * * * * * * *EXACTLY* * * * * * * 3000.00* * * * * * * * * * *DOLLARS at Hughes Pet Supply Company for value received with interest at 9.00% per annum. Due ____________________ (Seal) ____________________ (Seal) ____________________ Answer: Legal due date is July 18, 2013. Interest period January 15, 2013 to July 18, 2013 is 184 days. Interest = 3000(0.09)

= 136.11

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes.

3) A note for $665 dated March 22, 2014, with interest at 7.34% per annum, is issued for 128 days. Determine a) the legal due date; b) the interest period (in days); c) the amount of interest; d) the maturity value. Answer: a) DT1: March 22, 2014; DBD: 128; DT2: July 28, 2014 the legal due date is July 31, 2014 b) Number of days = 131 c) I = 665.00 ∗ 0.0734 ∗

= $17.52

d) Maturity value = 665.00 + 17.52 = $682.52 Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 4) A note for $800 dated June 4, 2010, with interest at 8.25% p.a., is issued for 10 days. Determine a) the legal due date; b) the interest period (in days); c) the amount of interest; d) the maturity value. Answer: a) June 17, 2010 b) Number of days = 13 c) I = 800.00 ∗ 0.0825 ∗

= $2.35

d) Maturity value = 800.00 + 2.35 = $802.35 Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 5) Find the maturity value of a $473 note issued on October 4 at 8.5% for 190 days. Answer: Number of days = 193 P = 473.00; r = .085; t = S = 473.00

= 473.00(1.0449452) = $494.26

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes.

6) Find the maturity value of a six-month, $642 note dated November 1, 2013, earning interest at 7.5%. Answer: Due date is May 4, 2014 Number of days = 184 P = 642.00; r = .075; t = S = 642

= 642(1.0380137) = $666.27

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 7) Determine the maturity value of a 120-day note for $2260.00 dated May 9 and bearing interest at 5.66%. Answer: P = 2260; r = .00566; t = S = P(1 + rt) = 2260

= 2260(1.0190734) = 2303.11

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 8) Find the maturity date and the maturity value of a $1415.00, 5.25%, 220-day note dated February 25, 2012. Answer: Legal due date is October 5, 2012. Interest period February 25, 2013 to October 2, 2012 = 223 days. Maturity value = 1415

= 1415.00(1.0320753) = 1460.39

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 9) Find the maturity value of a 60-day, 4% note for $3000 note dated February 25, 2011. Answer: Legal due date is April 19, 2011. Interest period February 25, 2011 to April 19, 2011 = 63 days. Maturity value = 3000

= 3000(1.00690411) = $3020.71

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 10) Find the maturity value of a $1190, 7.275%, 120-day note dated February 5, 2013. Answer: Interest period: 123 days S = 1190

= 1190.00(1.0245158) = $1219.17

Diff: 1 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes.

11) The maturity value of a 155-day 7.5% note dated March 14 is $1721.74. Compute the face value of the note. Answer: Number of days = 158 S = 1721.74; r = 0.075; t = P=

= $1667.60

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 12) The maturity value of a five-month promissory note issued May 31, 2013, is $2134.00. What is the present value of the note on the date of issue if money is worth 6.3%? Answer: Legal due date: November 3, 2013 Interest period: May 31, 2013 - November 3, 2013 is 156 days P=

= $2078.05

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 13) What is the face value of a three-month promissory note dated July 30, 2013, with interest at 6.5 percent if its maturity value is $1742.84? Answer: Legal due date: November 2, 2013 Interest period: July 30- November 2 = 95 days S = 1742.84; r = 0.065; t = P=

= $1713.85

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 14) Compute the present value of a 150-day non-interest-bearing promissory note for $5000 dated June 15, if the note was sold on August 21 and money is worth 6.5%. Answer: There are 153 days until the legal due date, November 15. From August 21 to November 15 is 86 days. Present value: P =

= $4924.58

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes.

15) A 60-day non-interest-bearing promissory note for $10 000 is dated June 1, 2013. Compute the present value of the note on June 14, 2013, if money is worth 5%. Answer: There are 63 days until the legal due date, August 3. From June 14 to August 3 is 50 days. Present value: P =

= $9931.97

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 16) Find the present value of a non-interest-bearing seven-month promissory note for $1800 dated August 27, 2012, on December 4, 2012, if money is then worth 8.375%. Answer: Due date: March 30, 2013 Maturity value: $1800.00 Discount period: December 4, 2012 - March 30, 2013 28 + 31 + 28 + 29 = 116 days P=

= $1753.33

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 17) Determine the present value on January 17, 2014 of a non-interest-bearing promissory six-month note for $17 000.00 dated October 15, 2013 if money is worth 4.15%. Answer: Legal due date is April 18, 2014 Interest period January 17, 2014 to April 18, 2014 contains 91 days. Present value: P =

= $16 825.91

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes.

18) Find the present value on June 1, 2014 of a non-interest-bearing note for $950 issued February 2, 2014, for 210 days if money is worth 8.31%. Answer: Due date is 213 days after February 2, 2014, i.e., September 3, 2014. Interest period is June 1 - September 3, i.e., 94 days. S = 950.00; r = 0.0831; t = P=

= $930.09

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 19) Find the present value and the amount of discount for a four-month non-interest-bearing note for $9 180 issued December 2, 2014, if money is worth 7.2% on February 28, 2015. Answer: Legal due date is April 5, 2015. the interest period is February 28 - April 5 = 36 days S = 9180; r = 0.072; t = P=

= $9115.27 Present value

Discount = 9180 - 9115.27 = $64.73 Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 20) A 4-month, 7.26% percent promissory note dated June 10, 2013 has a maturity value of $6231.34. What is the face value of the note? Answer: Legal due date is October 13, 2013. Interest period June 10, 2001 to October 13, 2001 contains 125 days. Face value: P =

= 6080.17

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes.

21) Find the proceeds of a six-month note for $966 dated September 16, 2012, with interest at 5.45% if money is worth 5.75% on November 4, 2012. Answer: Maturity value: Legal due date: March 19, 2013 Interest period: September 16, 2012 - March 19, 2013 is 184 days P = 966; r = 0.0545; t = S = P(1 + rt) = 966

= $992.54

Proceeds: Discount period: November 4, 2012 - March 19, 2013 is 135 days S = 986.44; r = .0545; t = P=

= $971.87

Proceeds = $971.87 Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 22) A $2850, five-month promissory note with interest at 6.15% is issued on June 1. Compute the proceeds of the note on August 13, when money is worth 7.5%. Answer: Maturity value: Legal due date: November 4 Interest period: June 1 - November 4 = 156 days S = P(1 + rt) = 2850

= $2924.91

Proceeds: Interest period: August 13 - November 4 = 83 days P=

= $2875.86

Proceeds = $2875.86 Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes.

23) Find the present value of a seven-month note for $3940 dated April 1 with interest at 7.45% if money is worth 6%, on August 15. Answer: Maturity value: The legal due date is November 4. Number of days = 217 P = 3940.00; r = 0.0745; t = S = 3940

= 3940(1.044292) = $4114.51

Present value on August 15 Interest period is August 15 to November 4 is = 81 days S = 4114.51; r = 0 .06; t = P=

= $4060.44

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 24) A six-month promissory note dated November 8, 2013 is made at 6% for $2900.00. What is the present value of the note thirty-eight days later if money is worth 7.2%? Answer: Legal due date is May 11, 2014. Interest period November 8, 2013 to May 11, 2014 is 184 days S = 2900

= $2987.72

Discount period is 184 - 38 = 146 days Present value: P =

= $2904.08

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 25) A 7-month note dated May 1, 2014 is made at 3.75% for $3705. What is the present value of the note on September 4, 2014, if money is worth 6.87%? Answer: Legal due date is December 4, 2014. Interest period May 1, 2014 to December 4, 2014 is 217 days. Maturity value S = 3705

= 3787.60

Discount period September 4, 2014 to December 4, 2014 is 91 days. Present value: P =

= $3723.82

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes.

26) On July 12, 2013 a 140-day note promissory note for $9 175 with interest at 5.75% was issued. Find the proceeds of the note on September 30, 2013 if money is worth 7%. Answer: Maturity value: Legal due date is December 2, 2013. Interest period is 143 days. P = 9175; r = 0.0575; t = S = 9175.00

= 9175.00(1.0225274) = 9381.69

Proceeds: Discount period is September 30 - December 2 is 63 days S = 9381.69; r = 0.07; t = P=

= $9269.69

Proceeds = $9269.69 Discount = 9381.69 - 9269.69 = $112.00 Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 27) What is the price of a 91-day, $25 000 Government of Canada treasury bill that yields 2.45% per annum? Answer: S = 75 000; r = 0.0245; t = P=

= $24 848.22

Diff: 2 Type: SA Page Ref: 309-312 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills.

28) An investment dealer bought a 182-day Government of Canada treasury bill at the price required to yield an annual rate of return of 3.38% a) What was the price paid by the investment dealer if the T-bill has a face value of $1 000 000? b) Later the same day, the investment dealer sold this T-bill to a large corporation to yield 3.25%. What was the investment dealer's profit on this transaction? Answer: a) S = 1 000 000; r = 0.0338; t = P1 =

= $983 425.64

The investment dealer paid $983 425.64 for the T-Bill. b) S = 1 000 000; r = 0.0325; t = P2 =

= $984 052.95

Profit = P2-P1 = 984 052.95 - 983 425.64 = $627.31 The investment dealer profit was $627.31. Diff: 2 Type: SA Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills. 29) Tracey bought a 182-day Government of Canada treasury bill at the price to yield an annual rate of return of 4.68%. a) What was the price paid by Tracey if the T-bill has a face value of $100 000? b) Later the same day, Tracey sold this T-bill to a large corporation to yield 4.48%. What was Tracey's profit on this transaction? Answer: a) S = 100 000.00; r = 0.0468; t = P1 =

= $97 719.63

The investment dealer paid $97 719.63 for the T-Bill. b) S = 100 000.00; r = .0448; t = P2 =

= $97 814.95

Profit = 97 814.95 - 97 719.63 = $95.32 Tracey's profit was $95.32 Diff: 2 Type: SA Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills.

30) Bradley purchased a 91-day, $25 000 T-bill on its issue date for $24 875.42. What was the original yield of the T-bill? Answer: S = 25 000.00; P = 24 875.42; t = Amount of yield = 25000 - 24875.42 = 124.58 Yield rate =

= 0.020088 = 2.0088%

The original yield on the T-Bill is 2.01%. Diff: 2 Type: SA Page Ref: 309-312 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills. 31) Sean purchased a 182-day, $10 000 T-bill on its issue date for $9754.25. What was the original yield of the T-bill? Answer: S = 10 000.00; P = 9754.25; t = Amount of yield = 10000.00 - 9754.25 = 245.75 Yield rate =

= 0.050526 = 5.05267%

The original yield on the T-Bill is 5.05267%. Diff: 2 Type: SA Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills.

32) Darren purchased $250 000 in 364-day T-bills 315 days before maturity to yield 2.86%. After holding it for 120 days, Darren sold the T-bill for a yield of 3.25%. a) How much did Darren pay for the T-bills? b) For how much did Darren sell the T-bills? c) What rate of return (per annum) did Darren realize on the investment? Answer: a) S = 250 000.00; r = 0.0286; t = P=

= $243 978.09

b) Time to maturity at date of sale is 315 - 120 = 195 days S = 250 000; r = 0.0325; t = P=

= $245 733.33

c) I = 245733.33 - 243978.09 = $1 755.24 Yield rate =

= 0.021883 = 2.1883%

Diff: 2 Type: SA Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills.

33) An investor purchased $250 000 in 91-day T-bills on the issue date for $248 157.56. After holding the T-bills for 37 days, she sold them for a yield of 3.25%. a) What was the original yield of the T-bills? b) For how much did the investor sell the T-bills? c) What rate of return (per annum) did the investor realize while holding this T-bill? Answer: a) S = 250000; P = 248157.56; t = Amount of yield = 250 000 - 248 157.56 = $1842.44 Yield rate =

= 0.029779 = 2.9779%

The original yield on the T-Bill is 2.9779%. b) Time to maturity at date of sale is 91 - 37 = 54 days S = 250 000; r = 0.0325; t = P=

= $248 803.70

The investor sold the T-bills for is $248 803.70. c) I = 248 803.70 - 248 157.56 = $646.14 Yield rate =

= 0.025686 = 2.5686%

The rate of return realized on the T-Bill is 2.5686%. Diff: 2 Type: SA Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills.

34) Meher purchased a $100 000.00, 182-day T-bill on April 4, 2012. a) If the annual rate of return is 4.2%, calculate the purchase price. b) Calculate the selling price of the bill if it is sold on June 20, 2012 to yield 3.19%. c) Calculate the rate of interest realized. Answer: a) S = 100 000.00; r = 0.042; t = P=

= $97 947.71

b) Due date is Oct. 3, 2012. Discount period: June 20 to Oct. 3 is 105 days S = 100 000.00; r = 0.0319; t = P=

= $99 090.67

c) Interest I = 99 090.67 - 97 947.71 = 1 142.96 r=

= 0.055314 = 5.5314%

Diff: 2 Type: SA Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills.

35) A government of Ontario 364-day T-bills with a face value of $50 000 were purchased on January 2 for . The T-bills were sold on September 28 for $48 999.99. a) What was the market yield rate on January 2? b) What was the yield rate on September 28? c) What was the rate of return realized? Answer: a) S = 50 000; P = 48 000.76; t = Amount of yield = 50 000 - 48 000.76 = $1999.24 Yield rate =

= 4.1765%

b) Time period, January 2 to September 28 is 269 days. The time to maturity = 364 - 269 = 95 days S = 50 000; P = 48 999.99; t = Amount of yield = 50 000 - 48 999.99 = $1001.01 Yield rate =

= 7.8490%

c) Gain from January 2 to September 28 = 48999.99 - 48000.76 = 999.23 Rate of return =

= 2.82%

Diff: 2 Type: SA Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills.

36) Government of Manitoba 364-day T-bills with a face value of $2 000 000 were purchased on April 17 for . The T-bills were sold on May 25 for $1 946 340. a) What was the market yield rate on April 17? b) What was the yield rate on May 25? c) What was the rate of return realized? Answer: a) S = 2 000 000; P = 1 945 970; t = Amount of yield = 2000000 - 1945970 = $54 030.00 Yield rate =

= 2.7841%

b) Time period, April 17 to May 25 is 38 days. The time to maturity = 364 - 38 = 326 days S = 2 000 000; P = 1 946 340; t = Amount of yield = 2000000 - 1946340 = $53 660.00 Yield rate =

= 3.0868%

c) Gain from April 17 to May 26 = 1 946 340 - 1 945 970 = 370.00 Rate of return =

= .1826%

Diff: 2 Type: SA Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills.

37) Linda borrowed $19 000.00 on August 17. She paid $4500.00 on November 11, $5500.00 on December 8, and the balance on February 21. The rate of interest on the loan was 8.5%. How much did Linda pay on February 20? Use the declining balance method. Answer: August 17 Original balance $19 000.00 Nov. 11 Partial payment $4500.00 Less interest 19 000(0.085) Dec. 8

Feb. 20

380.52

Balance Partial payment Less interest

$5500.00

14 880.52 (0.085)

93.56

4 119.48 $14 880.52

5 406.44

Balance Interest due

$9 474.08

9474.08 (0.085)

163.27

Final payment $9 310.81 Diff: 3 Type: SA Page Ref: 310-313 Topic: 8.3 Demand Loans Objective: 8-3: Compute interest and balances for demand loans.

38) Dirk Propp borrowed $14 300.00 for investment purposes on May 19 on a demand note providing for a variable rate of interest and payment of any accrued interest on December 31. He paid $1300.00 on June 28, $1450 on September 25, and $4200.00 on November 15. How much is the final payment on December 31 if the rate of interest was 11.5% on May 19, 8.21% effective August 1, and 6.35% effective November 1? Use the declining balance method. Answer: May 19 Original balance $14 300.00 June 28 Partial payment $1300.00 Less interest 14 300.00 (0.115) Sept. 25

Nov. 15

180.22

Balance Partial payment Less interest

1119.78 1450.00

13180.22 (0 .115)

137.04

13180.22 (0.0821)

163.06

Balance Partial payment Less interest

300.10 4200.00

12030.32 (0.0821)

100.12

12030.32 (0.0635)

29.30

$13 180.22

$12 030.32

129.42

Balance Dec. 31 incl. Interest

$7 959.74

7 959.74 (0.0635 )

63.70

Payment due $8 023.44 Diff: 3 Type: SA Page Ref: 310-313 Topic: 8.3 Demand Loans Objective: 8-3: Compute interest and balances for demand loans.

39) Amertech borrowed $32 000.00 from Balzac Credit Union on May 17 at 12%. The interest rate was changed to 14.11% effective July 1 and to 13.27% effective October 1. The loan was repaid by payments of $17 000.000 on July 15 and the balance, including the accumulated interest, on November 20. How much did the loan cost? Answer: Interest period Principal Rate Interest May 17 - June 30

$32000.00

0.12

32000.00 ∗ .12 ∗

= 473.42

$32000.00

0.1411

32000.00 ∗ .1411 ∗

= 173.19

$15000.00

0.1411

15000.00 ∗ .1411 ∗

= 452.29

$15000.00

0.1327

15000.00 ∗ .1327 ∗

= 272.67

inclusive July 1 - July 14 inclusive July 15 - September 30 inclusive October 1 - November 20

$1371.57 Diff: 3 Type: SA Page Ref: 310-313 Topic: 8.3 Demand Loans Objective: 8-3: Compute interest and balances for demand loans.

40) The West Lake Country Club arranged short-term financing of $36 200.00 on August 27 with the Bank of Commerce and secured the loan with a demand note. The club repaid the loan by payments of $11 000.00 on September 25, $9000.00 on November 15, and the balance on December 29. Interest calculated on the daily balance and charged to the club's current account on the last day of each month, was at 8.4% per annum on July 27. The rate was changed to 8.6% effective September 1 and to 9.05% effective December 1. Determine the total interest cost. Answer: Interest Period Principal Rate Interest August 27-August 31 inclusive $36200.00 0.084 36200(0.084)(5/365) = $41.65 September 1-September 24 inclusive $36200.00 0.086 36200(0.086)(24/365) = $204.70 September 25-September 30 inclusive $25200.00 0.086 25200(0.086)(6/365) = $35.63 October 1-October 31 inclusive $25200.00 0.086 25200(0.086)(31/365) = $184.06 November 1-November 14 inclusive $25200.00 0.086 25200(0.086)(14/365) = $83.13 November 15-November 30 inclusive $16200.00 0.086 16200(0.086)(16/365) = $61.07 December 1-December 29 $16200.00 0.0905 16200(0.0905)(28/365) = $112.47 Total interest = $722.71 Diff: 3 Type: SA Page Ref: 312-315 Topic: 8.3 Demand Loans Objective: 8-3: Compute interest and balances for demand loans.

41) The owner of Easy Clips borrowed $8800.00 from Red Deer Community Credit Union on June 17. The loan was secured by a demand note with interest calculated on the daily balance and charged to the store's account on the 5th day of each month. The loan was repaid by payments of $2500.00 on July 25, $2300.00 on October 17, and $3500.00 on December 30. The rate of interest charged by the credit union was 6.5% on June 17. The rate was changed to 6.65% effective July 1 and to 6.95% effective November 1. Determine the total interest cost on the loan, up to and including Dec. 30. Answer: DATE INTEREST PERIOD PRINCIPAL INTEREST June 17 8800.00 July 5 June 17 - June 30 incl. 8800.00 8800.00 * .065 * 14/365 21.94 July 1 - July 4 incl. 8800.00 8800.00 * .0665 * 4/365 6.41 28.35 Aug 5 July 5 - July 24 incl. 8800.00 8800.00 * .0665 * 20/365 32.07 July 25 - Aug 4 incl. 6300.00 6300.00 * .0665 * 11/365 12.63 44.70 Sept 5 Aug 5 - Sept 4 incl. 6300.00 6300.00 * .0665 * 31/365 35.58 Oct 5 Sept 5 - Oct 4 incl. 6300.00 6300.00 * .0665 * 30/365 34.43 Nov 5 Oct 5 - Oct 16 incl. 6300.00 6300.00 * .0665 * 12/365 13.77 Oct 17 - Oct 31 incl. 4000.00 4000.00 * .0665 * 15/365 10.93 Nov 1 - Nov 4 incl. 4000.00 4000.00 * .0695 * 4/365 3.50 98.21 Dec 5 Nov 5 - Dec 4 incl. 4000.00 4000.00 * .0695 * 30/365 22.85 Dec 30 Dec 5 - Dec 29 incl. 4000.00 4000.00 * .0695 * 25/365 19.04 December 30 500.00 500.00 * .0695 * 1/365 .10 41.99 Total Interest $213.25 Diff: 3 Type: SA Page Ref: 310-313 Topic: 8.3 Demand Loans Objective: 8-3: Compute interest and balances for demand loans.

42) Determine the missing information for the following line of credit. Harold has a line of credit secured by the equity in his home. The limit on his line of credit is $85 000. Transactions for the period May 1 to September 30 are shown below. Harold owed $45 967.06 on his line of credit on May 1. Principal Principal Interest Date Withdrawal Payment Payment ----------------------------------------------------------------May 01 30 1200.00 ? June 23 5050.00 30 2200.00 ? July 10 6700.00 ? Aug 05 11 700.00 31 8200.00 ? Sept 30 6200.00 ?

Balance -45 967.06

Note: "-" indicates a negative balance. Overdraft interest is 28.8% p.a. The line of credit interest is variable. It was 6.15% on May 1, 6.50% effective June 20, and 6.55% effective September 10. a) Calculate the interest payments on May 31, June 30, July 31, August 31, and September 30. b) What is the account balance on September 30? Answer: Balance May 1 -45967.06 Balance May 1 -45967.06 + 1200.00 = -44767.06 May 30 -44767.06 Interest payment May 31

45967.06(.0615)

+ 44767.06(.0615)

= 224.61 + 15.09 = $239.70 Balance June 1 Balance June 1 June 23 June 30 Interest payment June 30 44767.06(.0615)

-44767.06 -44767.06 - 5050.00 = -49817.06 -49817.06 + 2200.00 = -47617.06 -47617.06

+ 44767.06(.065)

) + 49817.06(.065)

= 143.32 + 23.92 + 70.97 = 238.21 Balance July 1 -47617.06 Balance July 1 -47617.06 + 6700.00 = -40917.06 July 10 -40917.06 Interest payment 47617.06(.065)

+ 40917.06(.065)

= 76.32 + 160.31 = 236.63

Balance August 1 Balance August 1 August 5 August 31 Interest payment 40917.06(.065)

+ 52617.06(.065)

= 29.15 + 252.99 = 282.14 Balance Sept. 1 Balance Sept 1 Sept 30 Interest payment 44417.06(.065)

-40917.06 -40917.06 - 11700.00 = -52617.06 -52617.06 + 8200.00 = - 44417.06 -44417.06

-44417.06 -44417.06 + 6200.00 = -38217.06 -38217.06

+ 44417.06(.0655)

= 71.19 + 167.39 = 238.58 a) Interest payments

May 31 June 30 July 31 Aug 31 Sept 30

$239.70 238.21 236.63 282.14 238.58

b) Balance Sept 30 -$38217.06 Diff: 3 Type: SA Page Ref: 314-317 Topic: 8.4 Lines of Credit and Credit Card Loans Objective: 8-4: Compute interest and balances for lines of credit and credit card loans. 43) Sean borrowed $3000.00 from Sepaba Savings and Loan. The line of credit agreement provided for repayment of the loan in three equal monthly payments plus interest at 6.00% per annum calculated on the unpaid balance. Determine the total interest cost. Answer: Monthly payment = 3000.00 ÷ 3 = $1000.00 Monthly rate of interest = .0600 ÷ 12 = .005 Month 1 2 3

Loan Amount owing 3000.00 3000 -1000 = 2000 2000 -1000 = 1000

Interest for Month 15.00 10.00 5.00 Total Interest = $30.00

Diff: 2 Type: SA Page Ref: 314-317 Topic: 8.5 Loan Repayment Schedules Objective: 8-5: Construct repayment schedules for loans using blended payments.

44) Raymond borrowed $3900.00 from Airdrie Regional Savings. The line of credit agreement provided for repayment of the loan in four equal monthly payments plus interest at 9.56% per annum calculated on the unpaid balance. Determine the total interest cost. Answer: Monthly payment = 3900.00 ÷ 4 = $975.00 Monthly rate of interest = .0956 ÷ 12 = .00796666667 Month 1 2 3 4

Loan Amount Owing Interest for Month $3,900.00 $31.07 3900.00 - 975.00 = 2925.00 23.3 2925.00 - 975.00 = 1950.00 15.51 1950.00 - 975.00 = 975.00 7.77 Total Interest $77.68 Diff: 2 Type: SA Page Ref: 319-322 Topic: 8.5 Loan Repayment Schedules Objective: 8-5: Construct repayment schedules for loans using blended payments. 45) FIGURE 8.1 Basic Design of a Loan Repayment Schedule (2) (1) Balance (3) (4) (5) Payment before Amount Interest Principal number payment paid paid repaid

(6) Balance after payment

Use the design shown in Figure 8.1 to construct a complete repayment schedule including the totaling of the Amount Paid, Interest Paid, and Principal Repaid columns for the following loan. On April 22, Tim borrowed $2900.00 from Keewatin Credit Union at 6.5% per annum calculated on the daily balance. He gave the Credit Union four cheques for $535.00 dated the 15th of each of the next four months starting May 15 and a cheque dated October 15 for the remaining balance to cover payment of interest and repayment of principal. Answer: Repayment schedule

Payment number 0 1 2 3 4 5 Totals

Balance before payment 2900.00 2376.88 1855.00 1329.91 802.25

Amount paid

Interest paid

Principal repaid

535.00 535.00 535.00 535.00 806.68 2946.68

11.88 13.12 9.91 7.34 4.43 46.68

523.12 521.88 525.09 527.66 802.25 2900.00

Balance after payment $2900.00 2376.88 1855.00 1329.91 802.25

Diff: 3 Type: SA Page Ref: 319-322 Topic: 8.5 Loan Repayment Schedules Objective: 8-5: Construct repayment schedules for loans using blended payments.

46) You borrowed the amount indicated in the Balance after payment column of the following schedule from your friendly credit union. You agreed to repay the loan in monthly payments equal to

of the

original loan, including interest due and principal. Interest is charged at a rate of 12.2% per annum computed on the monthly balance. Repayment Schedule Balance Balance Payment before Amount Interest Principal after number payment paid paid repaid payment 0 1200.00 1 2 3 4 5 6 Totals Required: Complete the repayment schedule (this includes totaling of the Payment, Interest paid and Principal repaid columns to check the accuracy of your work). a) What is the loan balance after the third payment? b) What is the total amount needed to repay the loan? Answer: Repayment Schedule Payment Balance before number payment 0 1 1200.00 2 972.20 3 742.08 4 509.62 5 274.80 6 37.59 Totals

Amount paid

Interest paid

Principal repaid

240.00 240.00 240.00 240.00 240.00 37.97 1237.97

12.20 9.88 7.54 5.18 2.79 .38 37.97

227.80 230.1 232.46 234.82 237.21 37.59 1200.00

Balance after payment 1200.00 972.20 742.08 509.62 274.80 37.59

a) $509.62 b) $1237.97 Diff: 2 Type: SA Page Ref: 319-322 Topic: 8.5 Loan Repayment Schedules Objective: 8-5: Construct repayment schedules for loans using blended payments.

47) You borrowed $5200 at 8.7% per annum calculated on the unpaid monthly balance and agree to repay the principal together with interest in monthly payments of $950 each. Construct a complete repayment schedule. Answer: Repayment Schedule Balance Balance Payment before Amount Interest Principal after number payment paid paid repaid payment 0 5200 1 5200 950 37.70 912.30 4287.70 2 4287.7 950 31.09 918.91 3368.79 3 3368.79 950 24.42 925.58 2443.21 4 2443.21 950 17.71 932.29 1510.92 5 1510.92 950 10.95 939.05 571.87 571.87 576.02 4.15 571.87 0.00 Totals 5326.02 126.02 5200.00 Diff: 3 Type: SA Page Ref: 319-323 Topic: 8.5 Loan Repayment Schedules Objective: 8-5: Construct repayment schedules for loans using blended payments. 48) Calculate the legal due date of a $600.00, 120 day note with interest at 5% dated March 2, 2014. A) July 5 B) July 2 C) June 30 D) July 3 E) July 31 Answer: D Diff: 1 Type: MC Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 49) Calculate the legal due date of a $10 000, 120-day note with interest at 4.56% dated March 31, 2012. A) July 28 B) July 29 C) August 1 D) August 2 E) August 3 Answer: C Diff: 1 Type: MC Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes.

50) A promissory note has a face value of $1725 and it has a date of issue of October 1 this year. The term is for 4 months. The rate of interest is 6.13%. What is the legal due date of the promissory note? A) Feb. 1 B) Feb. 2 C) Feb. 3 D) Feb. 4 E) Feb. 5 Answer: D Diff: 1 Type: MC Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 51) A promissory note has a face value of $5175.00 and it has a date of issue of April 2 this year. The term is for 5 months. The rate of interest is 6.75%. What is the maturity value of the note? A) $5324.30 B) $5234.30 C) $5342.30 D) $5334.20 E) $5444.30 Answer: A Diff: 1 Type: MC Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 52) A promissory note has a face value of $6000 and it has a date of issue of June 1 this year. The term is for 6 months. The rate of interest is 8.00%. What is the maturity value of the note? A) $6242.60 B) $6246.60 C) $6200.60 D) $6248.60 E) $6244.60 Answer: E Diff: 1 Type: MC Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 53) Calculate the maturity value of a 180-day note for $4000 dated August 18 if the rate of interest is 7.5%. A) 4147.95 B) 4150.41 C) 5479.45 D) 5504.11 E) 4440.41 Answer: B Diff: 1 Type: MC Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes.

54) A promissory note has a face value of $4500 and it carries an interest rate of 8.73% for a period of 4 months (including the period of grace). It is sold 3 months before the legal due date. What is the present value of the note on the date of sale if money is worth 8.2%? A) $4573.92 B) $4537.92 C) $4357.92 D) $4592.37 E) $4777.92 Answer: B Diff: 2 Type: MC Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 55) A promissory note has a face value of $5000 and it carries an interest rate of 5% for a period of 6 months (including the period of grace). It is sold 3 months before the legal due date. What is the present value of the note on the date of sale if money is worth 4%? A) $5047.26 B) $5070.26 C) $5074.26 D) $5125.00 E) $5125.26 Answer: C Diff: 2 Type: MC Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 56) A non-interest bearing promissory note has a $2400 maturity value and it matures in 90 days. You decide to sell the note 12 days before the legal due date. How much money do you sell it for if money is worth 4.84%? A) $3096.32 B) $2396.23 C) $2323.96 D) $2347.42 E) $2500.00 Answer: B Diff: 1 Type: MC Page Ref: 302-309 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes.

57) You purchase a 182-day treasury bill for $240 000 at a rate of 3.546%. What did you sell it for 111 days before maturity if the new interest rate is 3.778%? A) $238 220.20 B) $238 320.20 C) $238 200.20 D) $238 100.20 E) $241 469.24 Answer: E Diff: 1 Type: MC Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills. 58) You bought a $100 000 91-day T-bill for $99 453.67 61 days before maturity. What discount rate was used? A) 3.29% B) 2.39% C) 3.49% D) 4.39% E) 3.99% Answer: A Diff: 1 Type: MC Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills. 59) You bought $150 000 in 364-day T-bills. The T-bills were discounted at a rate of 4.432%. If you paid $148 811.24 for the T-bills, how many days before maturity did you buy it? A) 6.527 days B) 605.27 days C) 56.27 days D) 52.67 days E) 65.79 days Answer: E Diff: 2 Type: MC Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills. 60) You bought a $100 000 364-day T-bill. The T-bill was discounted at a rate of 5%. If you paid $99 900.00 for the T-bill, how many days before maturity did you buy it? A) 7.3 days B) 3.7 days C) 37 days D) 73 days E) 1 day Answer: A Diff: 2 Type: MC Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills.

61) You borrow $2500 on September 3rd this year. Your demand loan carries an interest rate of 7.46%. You make partial payments of $500 on October 15th and $1575 on November 17th. You want to make a final payment to pay off the remaining outstanding balance on November 30th. What is the size of your final payment? Use the declining balance method. A) $851.11 B) $399.76 C) $449.07 D) $450.02 E) $612.84 Answer: D Diff: 2 Type: MC Page Ref: 312-315 Topic: 8.3 Demand Loans Objective: 8-3: Compute interest and balances for demand loans. 62) You lend a friend $1300 on May 11th. The demand loan rate is 10.12%. Your friend makes a partial payment on May 26th for $550 and on June 19th for $675. You demand full repayment of the outstanding balance on July 17th. What is the final payment amount to? Use the declining balance method. A) $68.10 B) $86.10 C) $88.10 D) $82.10 E) $96.10 Answer: B Diff: 2 Type: MC Page Ref: 310-313 Topic: 8.3 Demand Loans Objective: 8-3: Compute interest and balances for demand loans. 63) You take out a demand loan on February 17th, in a non-leap year, with a local financing company. The loan is for $4500 at an interest rate of 12.15%. The interest rate rises to 12.75% on March 18th and then goes down to 12.25% on April 29th. You make partial payments of $900 on March 2nd and $1700 on May 14th. The loan is demanded in full on June 21st. What is the size of the final payment? A) $2063.37 B) $2043.37 C) $2083.37 D) $2036.37 E) $2116.37 Answer: D Diff: 2 Type: MC Page Ref: 310-313 Topic: 8.3 Demand Loans Objective: 8-3: Compute interest and balances for demand loans.

64) Average rate of return or yield on a 6-month Government of Canada treasury bills sold on June 18, 2013 was 1.04% (http://www.bankofcanada.ca/rates/interest-rates/t-bill-yields/). At this yield, what price was paid for a T-bill with a face value of $50 000? A) $50 260 B) $49 740 C) $49 741.35 D) $260 E) $258.65 Answer: C Diff: 1 Type: MC Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills. 65) Average rate of return or yield on 180-day Government of Canada treasury bills sold on June 18, 2013 was 1.04%. The client sold the $50 000 T-bill after 39 days. What rate of return (per annum) did the client realize while holding the T-bill, if the short term interest for this maturity had risen to 1.13% by the date of sale? Answer: Present value of $50,000 discounted at 1.04% for 180 days =

Days remaining to maturity = 180 - 39 = 141 days Selling price =

= $49 782.69

Effective interest on initial investment = $49 782.69 - $49 744.87 = $37.82 Rate of return =

= 0.00712 = 0.712%

Diff: 3 Type: SA Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills.

= $49 744.87

66) PC Financial approved a $75 000 line of credit on a demand basis to Little Blessings day care to finance equipment. Interest at the rate of prime plus 1% is charged to the account at the bank on the 15th of each month. The initial advance was $45 000 on July 15, when the prime rate stood at 3%. There were further advances of $8000 on August 20 and $10 000 on September 10. Payments of $15 000 each were applied against the principal on October 1 and November 1. What was the total interest accumulated on the loan for the period July 15 to November 15? Answer: Construct a repayment schedule, listing chronological order of all dates on which a transaction or an event affecting the loan occurs. Use I = Prt to calculate interest at an interest rate of 4%

Date July 15 August 15 August 20 September 10 September 15 October 1 October 15 November 1 November 15

Number of Days 0 31 5 21 5 15 14 17 14

Interest 0 $152.88 $24.66 $121.97 $34.52 $110.47 $73.64 $89.42 $50.63

Payment / (Advance) ($45000) 0 ($8000) ($10000) 0 $15,000 0 $15,000 0

Balance $45000 $45000 $53000 $63000 $63000 $48000 $48000 $33000 $33000

Total interest accumulated = $658.19 Diff: 3 Type: SA Page Ref: 310-313 Topic: 8.3 Demand Loans Objective: 8-3: Compute interest and balances for demand loans. 67) CIBC approved a $10 000 personal line of credit on a demand basis to Tara, who takes an advance of $2240 on 1st of every month to settle her monthly budget. She settles her loan when she receives her pay on 15th of every month. Interest at the rate of prime (3%) plus 2.75% is charged to the account at the bank on the 15th of each month. What is Tara's payment each month to settle the loan? A) $2245.29 B) $5.29 C) $2240 D) $2368.80 E) $2263.63 Answer: A Diff: 2 Type: MC Page Ref: 310-313 Topic: 8.3 Demand Loans Objective: 8-3: Compute interest and balances for demand loans.

68) Islamic banking is being introduced in Oman, such that no interest is given on the promissory notes. Compute the present value on the date of issue of a non-interest-bearing, worth $850 000, three-month promissory note dated September 1, 2013 (plus 3 days of grace), if money is worth 12.5% in Oman. A) $824 311 B) $847 272 C) $877 363 D) $823 490 E) $26 510 Answer: D Diff: 2 Type: MC Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 69) Zahid purchased a large-screen TV at a local store in December, that had advertised, "No payment for 6 months." Amount to be paid after 6 months is $1495 plus HST (13%) plus an administration fee of $79. If money is worth 2.5%, what is the actual cost of the TV to Zahid on the day of the purchase? A) $1746.52 B) $1790.45 C) $1476.54 D) $1554.57 E) $1768.35 Answer: A Diff: 2 Type: MC Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 70) Lee is planning to buy furniture worth $7000 from Leons. He can buy on his MasterCard and pay it within 7 days following the grace period of 21 days. His second option is to buy on the personal (unsecured) line of credit and pay it back after 6 months. His third option is to use secured line of credit and pay it back in 9 months. MasterCard charges an interest rate of 19.9%. Unsecured line of credit charges a rate of prime (3%) plus 3% and secured line of credit charges a rate of prime plus 0.5%. What is his best option? A) MasterCard B) Unsecured line of credit C) Secured line of credit D) All options are the same E) Either secured or unsecured line of credit. Answer: A Diff: 3 Type: MC Page Ref: 314-317 Topic: 8.4 Lines of Credit and Credit Card Loans Objective: 8-4: Compute interest and balances for lines of credit and credit card loans.

71) Leonard buys a painting on his credit card for $14 990. He pays his credit card in full within the grace period of 11 days using his secured line of credit, which charges him prime (3%) plus 1%. He repays his loan in 168 days. If his credit card company charges him a rate of 28% after the grace period, what is the total amount of interest paid on the purchase of the painting? A) $0 B) $206.99 C) $275.98 D) $126.49 E) $402.47 Answer: C Diff: 2 Type: MC Page Ref: 314-317 Topic: 8.4 Lines of Credit and Credit Card Loans Objective: 8-4: Compute interest and balances for lines of credit and credit card loans. 72) Caprice buys a painting on his credit card for $14 990. She pays her credit card in full 3 days after the grace period of 11 days using her secured line of credit, which charges her prime (3%) plus 1%. She repays her loan in 168 days. If her credit card company charges her a rate of 28% after the grace period, what is the total amount of interest paid on the purchase of the painting? Answer: Interest paid to the credit card = $14 990 × 0.28 ×

= $160.99

Total payment to the credit card using secured line of credit = $14 990 + $160.99 = $15 150.99 Interest paid to the secured line of credit = $15 150.99 × 0.04 ×

= $278.94

Total interest paid on the purchase of the painting = $160.99 + $278.94 = $439.93 Diff: 2 Type: SA Page Ref: 314-317 Topic: 8.4 Lines of Credit and Credit Card Loans Objective: 8-4: Compute interest and balances for lines of credit and credit card loans. 73) Caprice buys a painting on his credit card for $14 990. She pays her credit card in full 3 days after the grace period of 11 days using her secured line of credit, which charges her prime plus 1%. She repays her loan in 168 days. The prime rate is 2.5% on the day of repayment of credit card loan and increases to 3% 90 days after that day. If her credit card company charges her a rate of 28% after the grace period, what is the total amount of interest paid on the purchase of the painting? Answer: Interest paid to the credit card = $14 990 × 0.28 ×

= $160.99

Total payment to the credit card using secured line of credit = $14 990 + $160.99 = $15 150.99 Interest rate for first 90 days = 3.5% Interest rate for (168 - 90) = 78 days = 4% Interest paid to the secured line of credit = $15 150.99 × 0.035 ×

+ $15 150.99 × 0.04 ×

Total interest paid on the purchase of the painting = $160.99 + $260.26 = $421.25 Diff: 3 Type: SA Page Ref: 314-317 Topic: 8.4 Lines of Credit and Credit Card Loans Objective: 8-4: Compute interest and balances for lines of credit and credit card loans.

= $260.26

74) Marty took a $5000 loan from a financial institute at a rate of 6%, which should be repaid in two equal installments made every 4 months. What should be the value of the installment? Answer: Interest paid after first 4 months = $5000 × 0.06 ×

= $100

Let p be the first installment, then balance after first four months = $5000 - p + 100 = $5100 -p Interest paid after next 4 months = ($5100 - p) × 0.06 ×

= $102 - 0.02p

Total interest paid = $100 + $102 - 0.02p = 202 - 0.02p Balance after 8 months = $5000 - 2p + 202 -0.02p = 0 ⇒ 5202 - 2.02p = 0 ⇒ p=

= $2575.25

Diff: 3 Type: SA Page Ref: 319-322 Topic: 8.5 Loan Repayment Schedules Objective: 8-5: Construct repayment schedules for loans using blended payments. 75) Marty took a $5000 loan from a financial institute at a rate of 6%, which should be repaid in two equal installments of $2575.25 made every 4 months. How much more interest would have been paid, had Marty paid it in a single installment after 8 months? A) $75.25 B) $150.50 C) $200 D) $49.50 E) $0 Answer: D Diff: 3 Type: MC Page Ref: 319-322 Topic: 8.5 Loan Repayment Schedules Objective: 8-5: Construct repayment schedules for loans using blended payments. 76) Excel Electronics is Eastern Alberta accepted a 6-month promissory note from a customer for the $3950 balance owed on the purchase of a refrigerator. The note was dated November 8, 2015 and charged interest at 6.99%. The store's proprietor sold the promissory note 38 days later to Standard Union finance company at a price that would yield Standard Union 19% rate of return on its purchase price. What price did Standard Union pay to the proprietor of Excel Electronics? A) $4089.94 B) $4087.67 C) $3797.12 D) $3799.22 Answer: D Diff: 4 Type: MC Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes.

77) Saint Mary's Corporation issued a 150-day non-interest bearing note for $25 300 to Carlos Services on June 15. On August 21, the note was sold to Bearing Collections at a price reflecting a discount rate of 12.5%. What were the proceeds of the note? Answer: There are 153 days from the issue date until the legal due date. By August 21, 67 of the days have passed and 86 days remain. Since it is a non-interest bearing note, the future value remains the same as the present value. Hence 86 days earlier, at a discount rate of 12.5%: Proceeds =

= $24 576.18

Diff: 2 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 78) Leon's advertised a "No payment, No interest" plan for one year on all their dining set. Hab bought a dining set for $4520 plus 13% HST and an administration fee of $99 on 14 May 2016. On the same day, Sally bought the same dining set, paid the amount in full, but convinced the general manager of the store to give her a discount based on Canada's inflation rate, which is translated into a money worth of 2%. What did Sally pay? Answer: S = $4520; r = 2%; t = P=

= $4431.37

Amount paid by Sally = $4431.37 1.13 = $5007.45 Diff: 1 Type: SA Page Ref: 301-307 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 79) Syed bought 168 day Government of Canada T-Bills on 30 April 2016, yielding 0.6%. The face value of the bills was $55 000. He immediately sold it to a client at a higher price that presented a yield to the client of 0.54%. What profit did Syed make? A) $54 848.53 B) $54 863.64 C) $0 D) $15.11 Answer: D Diff: 2 Type: MC Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills.

80) Syed bought 168 day Government of Canada T-Bills on 30 April 2016, yielding 0.6%. The face value of the bills was $55 000. He immediately sold it to a client, Zeb, at a higher price that presented a yield to the client of 0.54%. Zeb sold the T-bills after 99 days, when the short term maturity had risen to 0.75%. What profit did Zeb make? A) $15.11 B) $58.49 C) $0 D) $24.70 Answer: B Diff: 4 Type: MC Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills. 81) Arazasa Equipment Inc. (AEI) is planning to setup a small business importing and supplying medical equipment to the doctor's offices in British Columbia. The bank has setup a line of credit allowing AEI to import the equipment. The company received an initial advance of $100 000 on its revolving demand loan on April 1, 2016. On the 27th of each month, interest is calculated up to but not including 27th of the month. It is then deducted from AEI's deposit account. Because of the lack of credit history, the bank charged an interest rate of 9.75% but dropped it to 9.5% on April 17. On May 1, another $100 000 was drawn on the line of credit. How much interest did AEI pay by May 25? A) $427.40 B) $364.38 C) $1353.42 D) $2145.21 Answer: D Diff: 3 Type: MC Page Ref: 310-313 Topic: 8.3 Demand Loans Objective: 8-3: Compute interest and balances for demand loans. 82) Meridian credit union approved a $500 000 line of credit on a demand basis to Hard Steel to finance the import of raw material. Interest at the rate of prime plus 1% is charged to Hard Steel's chequing account at the bank on the 1st of each month. The initial advance was $300 000 on 1st of June, 2016, when the prime rate stood at 3%. There were further advances of $40 000 on July 13 and $100 000 on July 28. Payments of $70 000 each were applied against the principal on August 15 and September 15. What was the total interest paid on the loan for the period June 1 to October 1? A) $4590.68 B) $1380.82 C) $2807.67 D) $4064.66 Answer: A Diff: 3 Type: MC Page Ref: 310-313 Topic: 8.3 Demand Loans Objective: 8-3: Compute interest and balances for demand loans.

83) Bana received Scotiabank American Express (Amex) gold card, where the interest is charged at 18.9% per annum. The bill is issued on 16th of every month and is due on 4th of next month. She used her Amex card for the following transactions in the month of July: July 1 July 4 July 10 July 15

Purchased some fireworks for Canada Day for a total of $67.35 Across the US border, took a cash advance worth $124.79 Canadian Booked a return flight to Calgary for $587 Purchased some food at the airport for $12.97

The minimum payment due is greater of $10 or 2% of the outstanding balance. She intends to pay the balance in full on the 29th of July. How much must she pay? A) $667.32 B) $792.11 C) $124.79 D) $793.73 Answer: D Diff: 2 Type: MC Page Ref: 314-317 Topic: 8.4 Lines of Credit and Credit Card Loans Objective: 8-4: Compute interest and balances for lines of credit and credit card loans. 84) Bana received Scotiabank American Express (Amex) gold card, where the interest is charged at 18.9% per annum. The bill is issued on 16th of every month and is due on 4th of next month. She used her Amex card for the following transactions in the month of July: July 1 July 4 July 10 July 15

The minimum payment due is greater of $20 or 2% of the outstanding balance on the billing day. She intends to pay the minimum amount on August 4 and pay the rest on August 11. How much must she pay on August 11? A) $793.73 B) $794.11 C) $774.11 D) $776.92 Answer: D Diff: 3 Type: MC Page Ref: 314-317 Topic: 8.4 Lines of Credit and Credit Card Loans Objective: 8-4: Compute interest and balances for lines of credit and credit card loans.

85) Joan buys a snow blower on her credit card for $1499. She pays her credit card in full a day after the grace period of 21 days using her secured line of credit, which charges her prime (3.25%) plus 1%. She repays her loan in 15 days. If her credit card company charges her a rate of 28% after the grace period, what is the total amount of interest paid on the purchase of the snow blower? A) $25.30 B) $1524.30 C) $2.66 D) $27.96 Answer: D Diff: 3 Type: MC Page Ref: 314-317 Topic: 8.4 Lines of Credit and Credit Card Loans Objective: 8-4: Compute interest and balances for lines of credit and credit card loans. 86) Aslam borrowed $23 000 from Meridian Credit Union at 4% p.a. and agreed to repay the loan in monthly instalments of $3000 each, such payments to cover interest due and repayment of principal. Construct a complete repayment schedule. Answer: Payment Balance Before Amount Principal Balance After No. Payment Paid Interest Paid Repaid Payment 0 $23,000.00 1 $23,000.00 $3,000.00 $76.67 $2,923.33 $20,076.67 2 $20,076.67 $3,000.00 $66.92 $2,933.08 $17,143.59 3 $17,143.59 $3,000.00 $57.15 $2,942.85 $14,200.73 4 $14,200.73 $3,000.00 $47.34 $2,952.66 $11,248.07 5 $11,248.07 $3,000.00 $37.49 $2,962.51 $8,285.56 6 $8,285.56 $3,000.00 $27.62 $2,972.38 $5,313.18 7 $5,313.18 $3,000.00 $17.71 $2,982.29 $2,330.89 8 $2,330.89 $2,338.66 $7.77 $2,330.89 $Diff: 3 Type: SA Page Ref: 319-322 Topic: 8.5 Loan Repayment Schedules Objective: 8-5: Construct repayment schedules for loans using blended payments.

87) Aymen shopped at Canadian Tire using line of credit and purchased merchandise worth $1450.55. She is budgeting a planning to pay $200 every month back to her bank. The bank charges 6.5% p.a. on unpaid balances. Construct a complete repayment schedule for Aymen What is the total interest paid to the bank? Answer: Balance Payment Before Principal Balance After No. Payment Amount Paid Interest Paid Repaid Payment 0 $1,450.55 1 $1,450.55 $200.00 $7.86 $192.14 $1,258.41 2 $1,258.41 $200.00 $6.82 $193.18 $1,065.22 3 $1,065.22 $200.00 $5.77 $194.23 $870.99 4 $870.99 $200.00 $4.72 $195.28 $675.71 5 $675.71 $200.00 $3.66 $196.34 $479.37 6 $479.37 $200.00 $2.60 $197.40 $281.97 7 $281.97 $200.00 $1.53 $198.47 $83.50 8 $83.50 $83.95 $0.45 $83.50 $0.00 Diff: 3 Type: SA Page Ref: 319-323 Topic: 8.5 Loan Repayment Schedules Objective: 8-5: Construct repayment schedules for loans using blended payments. 88) For the following promissory note, determine the amount of interest due at maturity. $5500.00 Toronto, Ontario April 20, 2019. Six months after date we promise to pay to the order of The Orange SuperStore * * * * * * * *EXACTLY* * * * * * * 5500.00* * * * * * * * * * *DOLLARS at The Orange SuperStore for value received with interest at 8.50% per annum. Due ____________________ (Seal) ____________________ Answer: Legal due date is October 21, 2013. Interest period April 20, 2019 to October 21, 2019 is 184 days. Interest = 5500(0.085)

= 235.67

Diff: 2 Type: SA Page Ref: 302-309 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes.

89) A note for $890 dated February 12, 2019, with interest at 6.95% p.a., is issued for 128 days. Determine a) the interest period (in days); b) the maturity value Answer: a) Since the legal due date is June 21, the interest period is 131 days. b) S = 890

= $912.20

Diff: 2 Type: SA Page Ref: 302-309 Topic: 8.1 Promissory Notes Objective: 8-1: Compute maturity value and present value for promissory notes. 90) What was the price of a 91-day, $100 000 Government of Canada treasury bill that yields 3.05% per annum? Answer: S = 100 000; r = 0.0305; t = P=

= $99 245.33

Diff: 2 Type: SA Page Ref: 309-312 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills. 91) Rosa purchased $75 000 in 182-day T-bills 115 days before maturity to yield 3.06%. How much did she pay for the T-bills? Answer: S = 75 000.00; r = 0.0306; t = P=

= $74 283.82

Diff: 2 Type: SA Page Ref: 308-309 Topic: 8.2 Treasury Bills - Present Value Objective: 8-2: Compute present values for treasury bills.

92) Sam borrowed $3200 on July 23rd this year. Their demand loan carries an interest rate of 6.96% per annum. They made partial payments of $600 on August 12th and $1050 on September 5th. Jane wants to make a final payment to pay off the remaining outstanding balance on September 15th. What is the total amount of interest she will have paid? Answer: Interest Period Principal Rate Time Interest July 23 - July 31 3200.00 0.0696 0.02465753 5.49 August 1 - August 11 3200.00 0.0696 0.030136986 6.71 August 12 - August 31 2600.00 0.0696 0.054794521 9.92 September 1 - September 4 2600.00 0.0696 0.010958904 1.98 September 5 - September 15 1550.00 0.0696 0.071232877 7.68 Total interest = $31.79 Diff: 3 Type: SA Page Ref: 312-315 Topic: 8.3 Demand Loans Objective: 8-3: Compute interest and balances for demand loans. 93) George opened a line of credit for purchases worth $9100.00. She is budgeting a planning to pay $1250 every month back to her bank. The bank charges 6.9% p.a. on unpaid balances. Construct a complete repayment schedule for Aymen What is the total interest paid to the bank? Answer: Payment Balance before Amount Principal Balance after number payment paid Interest paid repaid payment 0 9100 1 9100 1250 52.33 1197.67 7902.33 2 7902.33 1250 45.44 1204.56 6697.76 3 6697.76 1250 38.51 1211.49 5486.28 4 5486.28 1250 31.55 1218.45 4267.82 5 4267.82 1250 24.54 1225.46 3042.36 6 3042.36 1250 17.49 1232.51 1809.86 7 1809.86 1250 10.41 1239.59 570.26 8 570.26 573.54 3.28 570.26 0.00 Totals 9323.54 223.54 9100.00 Diff: 3 Type: SA Page Ref: 319-323 Topic: 8.5 Loan Repayment Schedules Objective: 8-5: Construct repayment schedules for loans using blended payments.

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 9 Compound Interest - Future Value and Present Value 1) Find the compound amount of $5750.00 at 13.2% p.a. for seven years compounded monthly. Answer: Use PV = 5700.00; i = 0 .132/12 = 0.011; n = 84 FV = 5750(1 + 0 .011)84 = $114413.35 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

2) Determine the accumulated value of $4100.00 compounded semi-annually at 8% p.a. for seven years. Answer: PV = 4100.00; i = 4%; n = 14 FV = 4100(1 + .04)14 = 4100(1.7316764) = $7099.87 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

3) Calculate the accumulated value of $3000.00 at 8% compounded quarterly for fifteen years. How much of the amount is interest? Answer: PV = 3000; i =

= 2% = 0.02 ; n = 15(4) = 60; I/Y = 8; P/Y = C/Y = 4

FV = 3000(1 + 0.02)60 = 9843.09 Programmed solution:

Interest = 9843.09 - 3000.00 = $6843.09 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 4) How much will a registered retirement savings deposit of $11 750.00 be worth in 10 years at 8.44% compounded semi-annually? How much of the amount is interest? Answer: PV = 11750.00; i = 8.44%/2 = 4.22%; n = 10*2 = 20 FV = 11750(1 + 0.0422)20 = 11750(2.285711) = 26857.11 I = 26857.11 - 11750.00 = 15107.11 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

5) How much will a registered retirement savings deposit of $10 000.00 be worth in 15 years at 6.00% compounded quarterly? How much of the amount is interest? Answer: PV = 10000.00; i = 6.0%/4 = 1.5%; n = 15 * 4 = 60 FV = 10000(1 + .015)60 = 10000(2.4432198) = 24 432.20 I = 24432.20 - 10000.00 = 14 432.20 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

6) Darcy's parents made a trust deposit of $3500.00 on October 31, 2002, to be withdrawn on Darcy's eighteenth birthday on July 31, 2016. To what will the deposit amount on that date at 13.48% compounded quarterly? Answer: PV = 3500.00; i =

= 0.0337

October 31, 2002 - July 31, 2016 contains 13 years, 9 months. n = 13 * 4 + 9/12 * 4 = 52 + 3 = 55 FV = 3500(1 + 0.0337)55 = 3500(6.1901104) = 21 665.39

Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 7) To what future value will a principal of $7100.00 amount in three years at 7.6% p.a. compounded: a) annually b) semi-annually c) quarterly d) monthly Answer: P V = 7100.00 a) m = 1, i = 7.6%, n = 3 FV = 7100.00(1 + .076)3 = 7100.00(1.245767) = 8844.95 b) m = 2; i =

= 0.038; n = 3 * 2 = 6

FV = 7100.00(1 + .038)6 = 7100.00(1.2507892) = 8880.60 c) m = 4; i =

= 0.019; n = 3 * 4 = 12

FV = 7100.00(1 + .019)12 = 7100.00(1.2534015) = 8899.15 d) m = 12; i =

= 0.0063; n = 3 * 12 = 36

FV = 7100.00(1 + .0063333)36 = 7100.00(1.2551836) = 8911.80 Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

8) Find the future value of and the compound interest on $4575.00 invested at 12.24% compounded monthly for: a) 3.5 years b) 7 years c) 12.5 years Answer: PV = 4575.00; m = 12; i =

= .0102

a) n = 3.5 * 12 = 42

FV = 4575.00(1 + .0102)42 = 4575.00(1.5314728) = 7006.49 Interest = 7006.49 - 4575.00 = 2431.49 b) n = 7 * 12 = 84 FV = 4575.00(1 + .0102)84 = 4575.00(2.345409) = 10 730.25

Interest = 10730.25 - 4575.00 = 6155.25 c) n = 12.5 * 12 = 150 FV = 4575.00(1 + .0102)150 = 4575.00(4.5825227) = 20 965.04

Interest = 20965.04 - 4575.00 = 16390.04 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 9) Leridian Credit Union expects an average annual growth rate of 16% for the next four years. If the assets of the credit union currently amount to $16.3 million, what will the forecasted assets be in four years? Answer: PV = 16300000; m = 1; i = 16%; n = 4 FV = 16300000(1 + .16)4 = 16300000(1.81063936) = 29 513 421.57 Forecasted assets will amount to $29 513 421.57. Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

10) GBC Credit Union expects an average annual growth rate of 10% for the next 10 years. If the assets of the credit union currently amount to $50 million, what will the forecasted assets be in ten years? Answer: PV = 50 000 000; m = 1; i = 10%; n = 10 SFV = 50000000(1 + .10)10 = 50000000(2.59374246) = 129 687 123 Forecasted assets will amount to $129 687 123. Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

11) Use the exact method to determine the accumulated value of $3875.00 due in 61 months compounded annually at 9.75% p.a. Answer: PV = 3875.00; i = 9.75%; n =

= 5.0833333

FV = 3875.00(1.0975)5.0833333 = 3875.00(1.6046846) = $6218.15 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

12) Suppose $4320.00 is invested for five years, eight months at 8.25% compounded annually. What is the compounded amount? Answer: PV = 4320.00; i = 8.25% = .0825; n =

= 5.6666667

FV = 4320.00(1.0825)5.6666667 = 4320.00(1.5670811) = $6769.79 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 13) Find the maturity value of a promissory note for $1400.00 dated February 29, 2016, and due on August 29, 2021, if interest is 7.64% compounded quarterly. Answer: Interest period 2016-02-29 to 2021-08-29 contains 6 years 6 months. PV = 1400.00; i = .0191; n = 6 * 4 + 0.5 * 4 = 24 + 2 = 26 Maturity value = 1400.00(1.0191)26 = 1400.00(1.635448) = $2289.63 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

14) A $5725.00 investment matures in three years, seven months. Find the maturity value if interest is 9.16% p.a. compounded quarterly. Answer: PV = 5725.00; i = .0229; n = 4

= 14.3333333

FV = 5725.00(1.0229)14.3333333 = 5725.00(1.3833823) = $7919.86 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 15) A $8000.00 investment matures in five years, three months. Find the maturity value if interest is 12% p.a. compounded quarterly. Answer: PV = 8000.00; i = .03; n = 4

= 21

FV = 8000.00(1.03)21 = 8000.00(1.860294572) = $14 882.36 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 16) An investment of $21 700 is accumulated at 5.24% compounded quarterly for three and one-half years. At that time the interest rate is changed to 6.12% compounded monthly. How much is the investment worth two years after the change in interest rate? Answer: Balance after 3.5 years FV1 = 21700.00(1.0131)14 = 21700.00(1.1998651) = $26 037.07 Balance 2 years later FV2 = 26037.07(1.0051)24 = 26037.07(1.1298546) = $29 418.10 Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

17) Determine the maturity value of $5400 due in 91 months compounding annually at 8.75%. Answer: n =

= 7.5833333

FV = 5400.00(1.0875)7.5833333 = 5400.00(1.8891014) = $10 201.15 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 18) Determine the maturity value of $2000 due in 63 months compounding annually at 10%. Answer: n =

= 5.25

FV = 2000.00(1.1)5.25 = 2000.00(1.649345337) = $3298.69 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 19) At Alpha Financial Services, Michael is offered three-year term deposits at 9.125% compounded annually. Beta Bank offers such deposits at 9.0% compounded semi-annually. If he has $3000 to invest, what is the maturity value of Michael's deposit? a) at Alpha Financial Services? b) at Beta Bank? Answer: a) FV = 3000.00(1.09125)3 = 3000.00(1.299490) = $3898.47 b) FV = 3000.00(1.045)6 = 3000.00(1.302260) = $3906.78 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

20) Alternative Savings offers five-year term deposits at 10% compounded annually while your credit union offers such deposits at 9.6% compounded quarterly. If you have $1000 to invest, what is the maturity value of your deposit a) at Alternative Savings? b) at your credit union? Answer: a) FV = 1000.00(1.10)5 = 1000.00(1.61051) = $1610.51 b) FV = 1000.00(1.024)20 = 1000.00(1.606938044) = 1606.94 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

21) The Rob U Blind Bank advertises capital savings at 7.128% compounded semi-annually while Take Your Money Trust offers premium savings at 7.1% compounded monthly. Suppose you have $4400.00 to invest for two years. a) Which deposit will earn more interest? b) What is the difference in the amount of interest? Answer: a) Bank: PV = 4400.00; m = 2; i =

= .03546; n = 2 * 2 = 4

FV1 = 4400.00 (1 + .03564)4 = 4400.00(1.150364) = 5061.60 Trust: PV = 4400.00; m = 12; i =

= .0059167; n = 2 * 12 = 24

FV2 = 4400.00(1 + .0059167)24 = 4400.00(1.1520945) = 5069.22 The Take Your Money Trust investment is preferable. b) Difference in interest = 669.22 - 661.60 = $7.62 Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 22) Boston Holdings offers a savings account at 1.2% compounded monthly while Albany Secure Savings offers premium savings at 1.236% compounded yearly. Suppose you have $8100.00 to invest for two years. a) Which deposit will earn more interest? b) What is the difference in the amount of interest? Answer: a) Boston: PV = 8100.00; m = 12; i = 1.2/12 = 0.001; n = 2 * 12 = 24 FV1 = 8100.00 (1 + .001)24 = 8100.00(1.024278) = 8296.65 Albany: PV = 8100.00; m = 1; i = 1.236% = 0.01236; n = 2 * 1 = 2 FV2 = 8100.00(1 + .01236)2 = 8100.00(1.024873) = 8301.47

The Albany Secure Savings bank investment is preferable. b) Difference in interest = 8301.47 - 8296.65 = $4.82 Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 23) Two years after Sean deposited $5000 in a savings account that earned interest at 6% compounded monthly, the rate of interest was changed to 6.4% compounded semi-annually. How much was in the account fifteen years after the deposit was made? Answer: Balance after 2 years FV = 5000.00(1.005)24 = 5000.00(1.127159776) = 5635.80 Balance after a total of 15 years FV = 5635.80(1.032)26 = 5635.80(2.268151865) = $12 782.85

Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

24) Betty deposited $17 150.00 in an RRSP on March 1, 2010, at 6.4% compounded quarterly. Subsequently the interest rate was changed to 6.6% compounded monthly on September 1, 2012, and to 6.8% compounded semi-annually on June 1, 2014. What was the value of the RRSP deposit on December 1, 2016, if no further changes in interest were made? Answer: At Sept. 1, 2002: FV = 17150(1.016)10 = 17150.00(1.1720256) = $20 100.24 Further: FV = 20100.24(1.0055)21 = 20100.24(1.1220794) = $22 554.07 (at June 1, 2014) FV = 22554.07(1.034)7 = 22554.07(1.2636994) = 28 501.56 (at Dec. 1, 2016 ) $28501.56 Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

25) Fahira invested $5000.00 at 4.8% compounded quarterly on January 1, 2012. On July 1, 2013 the rate changed to 5.6% p.a. compounded semi-annually. Calculate the amount Fahira will have on January 1, 2005. Answer: Step 1 PV = 5000; i =

= 1.2% = 0.012;n =

(4) = 6; I/Y = 4.8; P/Y = C/Y = 4

FV = 5000(1 + 0.012)6 = 5370.97 Programmed solution:

Step 2 PV = 5370.97; i = n=

= 2.8% = 0.028;

(2) = 3 ; I/Y = 5.6; P/Y = C/Y = 2

FV = 5370.97(1 + 0.028)3 = $ 5834.88 Programmed solution:

Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

26) An investment of $2500.00 accumulates interest at 9.25% compounded quarterly. After 18 months the rate changed to 9.75% compounded semi-annually. Calculate the accumulated value three years after the initial investment. Answer: Step 1 PV = 2500.00; i =

= 2.3125% = 0.023125; n =

(4) = 6; I/Y = 9.25

FV = 2500(1 + 0.023125)6 = 2867.56 Programmed solution:

Step 2 PV = 2867.56; i =

= 4.875% = 0.04875; n =

(2) = 3; I/Y = 9.75

FV = 2867.56(1 + 0.04875)3 = 3307.72 Programmed solution:

Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 27) An investment of $4300.00 earns interest at 9.96% p.a. compounded monthly for three and a half years. At that time the interest rate is changed to 9% compounded semi-annually. How much will the accumulated value be six years after the change? Answer: Balance after 3.5 years: PV = 4300.00; m = 12; i =

= .0083; n = 3.5 * 12 = 42

FV = 4300.00(1 + .0083)42 = 4300.00(1.415045) = 6084.69 Balance 6 years later: PV = 6084.69; m = 2; i =

= .045; n = 6 * 2 = 12

FV = 6084.69(1 + .045)12 = 6084.69(1.695881) = 10318.91 The accumulated value 6 years after the change is $10318.91. Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

28) An investment of $5000.00 earns interest at 6% p.a. compounded monthly for two years. At that time the interest rate is changed to 8% compounded semi-annually. How much will the accumulated value be three and a half years after the change? Answer: Balance after 2 years: PV = 5000.00; m = 12; i =

= .005; n = 2 * 12 = 24

FV = 5000.00(1 + .005)24 = 5000.00(1.127159776) = 5635.80 Balance 3.5 years later: PV = 5635.80; m = 2; i =

= .04; n = 3.5 * 2 = 7

FV = 5635.80(1 + .04)7 = 5635.80(1.315931) = 7416.33 The accumulated value 2.5 years after the change is $7416.33. Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 29) Six years after Mr. Robertson deposited $7110.00 in a savings account that earned interest at 6.8% p.a., compounded semi-annually, the interest was changed to 7.2% p.a., compounded quarterly. How much was in the account ten years after the deposit was made? Answer: After 6 years: PV = 7110.00; i = .034; n = 12 FV = 7110.00(1 + .034)12 = 7110.00(1.4936418) = 10 619.79 4 years later (i.e., 10 years after the initial deposit): PV = 10619.79; i = .018; n = 16 FV = 10619.79(1 + .018)16 = 10619.79(1.3303455) = 14 127.99

Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 30) Accumulated $1720.00 at 8.4% p.a. compounded monthly from March 1, 2011, to July 1, 2013, and thereafter at 8.88% p.a. compounded quarterly. What is the amount on April 1, 2015? Answer: Balance July 1, 2013: PV = 1720.00; m = 12; i =

= .007

March 1, 2011 - July 1, 2013 contains 2 years 4 months: n = 28 FV = 1720.00(1 + .007)28 = 1720.00(1.2156965) = 2091.00 Balance April 1, 2005:

PV = 2091.00; m = 4; i =

= .022

July 1, 2003 - April 1, 2015 contains 1 year 9 months: n = 1.75 * 4 = 7 FV = 2091.00(1 + .022)7 = 2091.00(1.164545) = 2435.06

The balance on April 1, 2015 is $2435.06. Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

31) You borrowed $1700.00 at 12.36% p.a. compounded monthly, and repaid $800.00 after three years and $950.00 after five years. How much do you owe at the end of the nine years? Answer: After 3 years: PV = 1700.00; i = .0103; n = 36 FV = 1700.00(1 + .0103)36 = 1700.00(1.4461789) = 2458.45 After 3 years: PV = 2458.45 - 800.00 = 1658.45; n = 24 After 5 years FV = 1658.45(1 + .0103)24 = 1658.45(1.2788172) = 2120.85 After 5 years: PV = 2120.85 - 950.00 = 1170.85; n = 48 FV = 1170.85(1 + .0103)48 = 1170.85(1.6353735) = 1914.78 Diff: 3 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

32) Allison started an RRSP on March 1, 2011, with a deposit of $2470.00. She added $1900.00 on December 1, 2012, and $1850.00 on September 1, 2013. What is the accumulated value of her account on December 1, 2015, if interest is 7.24% compounded quarterly? Answer: m = 4; i =

= .0181

Balance on March 1, 2011 PV = 2470.00; n = 1.75 * 4 = 7 FV = 2470.00(1 + .0181)7 = 2470.00(1.1337911) = $2800.46

Balance on December 1, 2012 PV = 2800.46 + 1900.00 = $4700.46; n = 0.75 * 4 = 3 FV = 4700.46(1 + .0181)3 = 4700.46(1.05528876) = $4960.34 Balance on September 1, 2013 PV = 4960.34 + 1850.00 = 6810.34; n = 2.25 * 4 = 9 FV = 6810.34(1 + .0181)9 = 6810.34(1.1752058) = $8003.55

The accumulated value of the RRSP account on December 1, 2015 is $8003.55 Diff: 3 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 33) Find the compound interest earned by $1910 invested at 7.5% compounded semi-annually for seven years. Answer: PV = 1910.00; i = 0.0375; n = 14 FV = 1910.00(1.0375)14 = 1910.00(1.6743008) = $3197.91 Compound interest = 3197.91 - 1910.00 = $1287.91 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

34) Find the compound interest earned by $5000 invested at 3.2% compounded semi-annually for 18 years. Answer: PV = 5000.00; i = 0.016; n = 36 FV = 5000.00(1.016)36 = 5000.00(1.770816) = 8854.080995 Compound interest = 8854.08 - 5000.00 = $3854.08 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

35) Six years after Ellen deposited $4500 in a savings account that earned interest at 4.68% compounded monthly, the rate of interest was changed to 6.4% compounded semi-annually. How much was in the account thirteen years after the deposit was made? Answer: Balance after 6 years FV = 4500.00(1.0039)72 = 4500.00(1.3234657) = $5955.60 Balance after a total of 13 years FV = 5955.60(1.032)14 = 5955.60(1.5542317) = $9256.38

Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 36) Sean started an RRSP on March 1, 2007, with a deposit of $2000.00. He added $1500.00 on March 1, 2008. What is the accumulated value of his account on December 1, 2008, if interest is 6% compounded quarterly? Answer: m = 4; i =

= .015

Balance on March 1, 2007 PV = 2000.00; n = 1 * 4 = 4 FV = 2000.00(1 + .015)4 = 2000.00(1.061363551) = $2122.73 Balance on March 1, 2008 PV = 2122.73 + 1500.00 = $3622.73; n = 0.75 * 4 = 3 FV = 3622.73(1 + .015)3 = 3622.73(1.045678375) = $3788.21

Balance on December 1, 2008. Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

37) You have a line of credit loan with the Bank of Hong Kong. The initial loan balance was $7000.00. Payments of $3000.00 and $2500.00 were made after four months and ten months respectively. At the end of one year, you borrowed an additional $4250.00. Seven months later, the line of credit loan was converted into a collateral mortgage loan. What was the amount of the mortgage if the line of credit interest was 8.52% compounded monthly? Answer: m = 12; i =

= .0071

Balance after 4 months: PV = 7000.00; n = 4 FV = 7000.00(1 + .0071)4 = 7000.00(1.0287039) = 7200.93 Balance after 10 months: PV = 7200.93 - 3000.00 = 4200.93 FV = 4200.93(1 + .0071)6 = 4200.93(1.0433633) = 4383.10

Balance after 1 year: PV = 4383.10 - 2500.00 = 1883.10 FV = 1883.10(1 + .0071)2 = 1883.10(1.01425041) = 1909.93 Balance after 19 months: PV = 1909.93 + 4250.00 = 6159.93 FV = 6159.93(1 + .0071)7 = 6159.93(1.05077123) = 6472.68

The mortgage was for $6472.68 Diff: 3 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 38) A variable rate demand loan showed an initial balance of $20 000.00, payments of $5000.00 after six months, $10000.00 after one year, and a final payment after two years. Interest was 6% compounded semi-annually for the first year and 12% compounded monthly for the remaining time. How much was the size of the final payment? Answer: Balance after 6 months: PV = 20000.00; m = 2; i =

= 0.03; n = 1

FV = 20000.00(1 + .03)1 = 20000.00(1.03) = 20 600.00 Balance after 6 months: 20600-5000 = 15 600 Balance after 12 months PV = 15600.00; m = 2; i =

= .003; n = 1

FV = 15600.00(1 + .03)1 = 15600.00(1.03) = 16 068.00 Balance after 12 months: 16068 -10000 = 6068 Balance after 24 months PV = 6068.00; m = 12; i =

= 0.01; n = 12

FV = 6068.00(1 + .01)12 = 6068.00(1 + .01)12 = 6837.57 The final payment is $6837.57. Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

39) What is the present value of $7800.00 payable in six years if the current interest rate is 7.6% p.a., compounded quarterly? Answer: FV = 7800.00; i = 1.9%; n = 24 PV = 7800.00(1 + .019)-24 = 7800.00(.636531) = 4964.94 Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money.

40) Determine the sum of money that will grow to $18 530 in four years, eleven months at 5.2% compounded quarterly. Answer: n = (4 +

)4 = 19.6666667

i = 0.052/4 = 0.013

PV = 18530.00(1.013)-19.6666667 = 18530.00(.77567699) = $14 373.29 Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 41) Determine the proceeds of $14 000 three years and three months before the due date if interest is 7.6% compounded quarterly. Answer: n = (3 +

)4 = 13

i = 0.076/4 = 0.019

PV = 14000.00(1.019)-13 = 14000.00(.782953) = $10 961.34 Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 42) Find the present value and the compound discount of $6100.00 due in 7.5 years if interest is 8.26% compounded semi-annually. Answer: PV = 6100.00(1.0413)-15 = 6100.00(.5449567) = $3324.24 Discount = 6100.00 - 3324.24 = $2775.76 Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money.

43) Determine the proceeds of $5000 four years and nine months before the due date if interest is 8% compounded semi-annually. Answer: n = (4 +

)2 = 9.5

i = 0.08/2 = 0.04

PV = 5000.00(1.04)-9.5 = 5000.00(.6889429759) = $3444.71 Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money.

44) Calculate the present value of $12 500.00 due in two years and nine months if interest is 7.8% p.a. compounded semi-annually. Answer: FV = $12 500.00; n = (2 +

)(2) = 5.5; i =

= 3.9% = 0.039; I/Y = 7.8; P/Y = C/Y = 2

PV = FV(1 + i)-n = 12500(1 + 0.039)-5.5 = $10 128.02 Programmed solution:

Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 45) Find the present value and the compound discount of $6 600.00 due in seven years, three months, if interest is 7.2% compounded quarterly. Answer: FV = 6600.00; m = 4; i =

= 1.8% = 0.018; n = 7.25 * 4 = 29

PV = 6600.00(1.018)-29 = 6600.00(.5960936) = $3934.22 Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 46) What sum of money invested at 5.95% p.a., compounded semi-annually, will grow to $28 000.00 in 19.5 years? Answer: FV = 28000.00; i = 2.975%; n = 39 PV = 28000.00(1.02975)-39 = 28000.00(.318757) = $8925.20 Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money.

47) What sum of money invested at 5.2% p.a., compounded biweekly, will grow to $10 000.00 in 12.5 years? Answer: FV = 10000.00; i = 0.2%; n = 325 PV = 10000.00(1.002)-325 = 10000.00(.522385) = 5223.85 Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money.

48) What sum of money will grow to $23 000.00 in seven years at 9.612% compounded monthly? Answer: FV = 23000.00; m = 12; i =

= .0081; n = 7 * 12 = 84

PV = 23000.00(1 + .00801)-84 = 23000.00(.5116269) = $11 767.42 Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 49) How much would you have to deposit in an account today to have $37 000.00 in a three-year term deposit at maturity if interest is 7.775% compounded annually? Answer: FV = 37000.00; m = 1; i = 7.775% = .07775; n = 3 PV = 37000.00(1 + .07775)-3 = 37000.00(.7988144) = $29 556.13 Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money.

50) Compute the discounted value of $5125.00 due in three years, eight months if money is worth 8.24% compounded quarterly. Answer: n = (3 +

)4 = 14.6666667

P V = 5125.00(1.0206)-14.6666667 = 5125.00(.7415124) = $3800.25 Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 51) Find the principal which will grow to $7258.00 at 6.58% compounded semi-annually in four years and five months. Answer: n = (4 +

)2 =

= 8.8

PV = 7258.00(1.0329)-8.83 = 7258.00(.7513089) = $5453.00 Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 52) Richard wants to have $114 150.00 in six years, seven months from now. How much would he have to invest right now to achieve his goal, if he is getting 7.6% compounded quarterly rate of return on his investment? Answer: n = (6 +

)4 = 26.3333333

PV = 114150.00(1.019)-26.3333333 = 114150.00(.6091812) = $69 538.03 Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money.

53) Calculate the proceeds of $8956.00 due in seven years, eleven months discounted at 7.5% compounded semi-annually. Answer: n = (7 +

)2 = 15.8333333

PV = 8956.00(1.0375)-15.8333333 = 8956.00(.5582838) = $4999.99 Diff: 1 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 54) A $41 200.00, non-interest-bearing note due August 1, 2003, discounted on March 1, 2001, at 7.32% compounded monthly. Find the proceeds. Answer: Discount period 2001-03-01 to 2003-08-01 contains 2 years 5 months FV = 41200.00; m = 12; i =

= .0061; n = 2 * 12 + 5 = 29

PV = 41200.00(1 + .0061)-29 = 41200.00(.8383139) = $34 538.53 Diff: 1 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 55) An eleven-year promissory note discounted after six years at 9.2% compounded quarterly has a maturity value of $71 500. Find the proceeds. Answer: Discount period = 11 - 6 = 5 years FV = 71500.00; m = 4; i =

= .023; n = 5 * 4 = 20

PV = 71500.00(1 + .023)-20 = 71500.00(.6345814) = $45 372.57 Diff: 1 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 56) A twenty-year promissory note discounted after seven years at 10% compounded quarterly has a maturity value of $40 000. Find the proceeds. Answer: Discount period = 20-7 = 13 years FV = 40000.00; m = 4; i =

% = .025; n = 13 * 4 = 52

PV = 40000.00(1 + .025)-52 = 40000.00(.276923) = $11 076.92 Diff: 1 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes.

57) A $5000.00, six-year note bearing interest at 8.24% compounded quarterly, discounted three and a half years after the date of issue at 6.6% compounded monthly. Find the proceeds. Answer: Maturity value: PV = 5000.00; m = 4; i =

= .0206; n = 6 * 4 = 24

FV = 5000.00(1 + .0206)24 = 5000.00(1.6312989) = $8156.49 Proceeds: Discount period = 6 - 3.5 = 2.5 years FV = 8156.49; m = 12; i =

= .0055; n = 2.5 * 12 = 30

PV = 8156.49(1 + .0055)-30 = 8156.49(.8482771) = $6918.96 Diff: 2 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 58) A ten-year promissory note dated April 1, 2001, with a face value of $4700.00 bearing interest at 7.2% compounded semi-annually, discounted seven years later when money was worth 9.92% compounded monthly. Find the proceeds. Answer: Maturity value: PV = 4700.00; m = 2; i =

= .036; n = 10 * 2 = 20

FV = 4700.00(1.036)20 = 4700.00(2.0285939) = $9534.39 Proceeds: Discount period 10 - 7 = 3 years FV = 9534.39; i =

= .0082667; m = 12; n = 3 * 12 = 36

PV = 9534.39(1.0082667)-36 = 9534.39(.7435064) = $7088.88 Diff: 2 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 59) A seven-year, non-interest-bearing note for $6532.00 is discounted three years and two months before its due date at 9.12% compounded quarterly. Find the proceeds of the note. Answer: n = (3 +

)4 = 12.6666667

PV = 6532.00(1.0228)-12.6666667 = 6532.00(.7515956) = $4909.42 Diff: 1 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes.

60) Use the exact method to compute the proceeds of a non-interest-bearing note for $5640.00 six years and seven months before the due date, if money is worth 7.25% p.a. compounded annually. Answer: n = (6 +

)1 = 6.5833333

PV = 5640.00(1.0725)-6.5833333 = 5640.00(.6307895) = $3557.65 Diff: 1 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 61) A $4300.00 promissory note issued without interest for nine years on September 30, 2001, is discounted on July 31, 2006, at 8.32% compounded quarterly. Find the compound discount. Answer: Due date: 2010-09-30 Discount period: 2006-07-31 to 2010-09-30: 4 years 2 months n = (4 +

)4 = 16.6666667 i = 8.32%/4 = 0.0208

Proceeds = 4300.00(1.0208)-16.6666667 = 4300.00(.7095598) = 3051.11 Discount = 4300.00 - 3051.11 = $1248.89 Diff: 2 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 62) Four years and five months after its date of issue, a 7-year promissory note for $8900.00 bearing interest at 5.04% compounded monthly is discounted at 6.5% compounded semi-annually. Find the proceeds of the note using the exact method. Answer: Maturity value: PV = 8900.00; i = 0.0042; n = 84 FV = 8900.00(1.0042)84 = 8900.00(1.4219955) = 12 655.76 Proceeds: = (2 +

)2 = 5.1666667

= 6.5%/2 = 0.0325

P = 12655.76(1.0325)-5.1666667 = 12655.76(.8476854) = $10 728.10 Diff: 2 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 63) On July 28, 2018, a ten-year note dated January 28, 2015, is discounted at 8.2% compounded quarterly. If the face value of the note is $4000.00 and interest is 7.2% compounded quarterly, find the compound discount. Answer: Due date: 2025-01-28 Maturity Value = 4000.00(1.018)40 = 4000.00(2.0413201) = 8165.28 Discount period n = (6 + 0.5)4 = 26 PV = 8165.28(1.0205)-26 = 8165.28(.590013) = $4817.623389

Discount = 8165.28 — 4817.62 = $3347.66 Diff: 2 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes.

64) A ten-year note for $2220.00 bearing interest at 6.6% compounded monthly is discounted at 8.92% compounded quarterly four years and four months before maturity. Find the proceeds of the note. Answer: P V = 2220.00; i = 0.0055; n = 120 Maturity value = 2220.00(1.0055)120 = 2220.00(1.9312967) = 4287.48 = (4 +

)4 = 17.3333333

PV = 4287.48(1.0223)-17.333333 = 4287.48(.6822995) = $2925.35 Diff: 2 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 65) A twenty-year note for $10 000.00 bearing interest at 12% compounded monthly is discounted at 8% compounded quarterly four years and six months before maturity. Find the proceeds of the note. Answer: PV = 10000.00; i = 0.01; n = 240 Maturity value = 10000.00(1.01)240 = 10000.00(10.89255365) = 108 925.54 = (4 +

)4 = 18

= 0.08/4 = 0.02

PV = 108925.54(1.02)-18 = 108925.54 (.700159375) = $76 265.24 Diff: 2 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 66) A seven-year promissory note for $3100 dated May 2, 2016, bearing interest at 8% compounded quarterly, is discounted on September 2, 2019, at 8.5% compounded semi-annually. Determine the proceeds of the note. Answer: Due date: 2009-05-02 FV = 3100.00(1.02)28 = 3100.00(1.7410242) = 5397.18 n = (3 +

)2 = 7.3333333

Proceeds = PV = 5397.18(1.0425)-7.3333333 = 5397.18(.7369571) = $3977.49 Diff: 2 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes.

67) A note dated May 1, 2011 promises the payment of $5660.00 with interest at 6.5% p.a. compounded semi-annually on November 1, 2015. Find the proceeds of the sale of the note on May 1, 2013 if money was then worth 7.2% p.a. compounded monthly. Answer: Interest period: May 1, 2011 to November 1, 2015 = 4.5 years. Maturity value: PV = 5660.00; m = 2; i =

= .0325; n = 4.5 * 2 = 9

FV = 5660.00(1.0325)9 = 5660.00(1.3335538) = $7547.91 Discount period: Nov. 1, 2015 to May 1, 2013 FV = 7547.91; m = 12; i =

= .006; n = 2.5 * 12 = 30

PV = 7547.91(1.006)-30 = 7547.91(.8357196) = $6307.94 Diff: 3 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 68) Calculate the proceeds of $5000.00 due in five years, nine months discounted at 8.0% compounded semi-annually. Answer: N = (5 +

)2 = 11.5

PV = 5000.00(1.04)-11.5 = 5000.00(.6369665) = $3184.83 Diff: 1 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 69) A debt can be repaid by payments of $4125.00 today, $3770.00 in two years and $5600.00 in five years. What single payment would settle the debt three years from now if money is worth 9.88% p.a. compounded quarterly? Answer: Let the single payment be $x. Let the focal date be 3 years from now. i = 0.0988/4 = 0.0247 x = 4125.00(1.0247)12 + 3770.00(1.0247)4 + 5600.00(1.0247)-8 x = 4125.00(1.3401729) + 3770.00(1.025212) + 5600.00(.8226709) x = 5528.21 + 4156.50 + 4606.96 x = $14 291.67 Diff: 2 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

70) A debt can be repaid by payments of $1000.00 today, and $3000.00 in two years. What single payment would settle the debt four years from now if money is worth 1.6% p.a. compounded semi-annually? Answer: Let the single payment be $x. Let the focal date be 4 years from now. i = 0.008. x = 1000.00(1.008)8 + 3000.00(1.008)4 x = 1000.00(1.065821) + 3000.00(1.032386) x = 1065.82 + 3097.16 x = $4162.98 Diff: 2 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

71) Loans of $1600.00, $8300.00, and $12 100.00 are due now, in four years, and in seven years respectively. What is the equivalent single sum of money due three and a half years from now if interest is 10.8% compounded monthly? Answer: Let the focal date be three and a half years from now. i = 0.108/12 = 0.009 x = 1600.00(1.009)42 + 8300.00(1.009)-6 + 12100.00(1.009)-42 x = 1600.00(1.4568974) + 8300.00(.947661) + 12100.00(.6863901) x = 2331.04 + 7865.59 + 8305.32 = $18501.95 The single payment 3.5 years from now is $18 501.95. Diff: 2 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

72) Determine the discounted value now of $7700.00 due in forty-four months at 7.5% compounded quarterly. Answer: n =

= 14.6666667

PV = 7700.00(1.01875)-14.6666667 = 7700.00(.7615087) = $5863.62 Diff: 1 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

73) Debts of $400.00, $450.00 and $500.00 are due in one year, eighteen months and thirty months from now respectively. Determine the single payment now that would settle the debts if interest is 8% p.a. compounded quarterly. Answer: Let the focal date be now . For the $400.00 debt: FV = 400.00; n = 1(4) = 4; i =

= 2% = 0.02; I/Y = 8; P/Y = C/Y = 4

PV = FV(1 + i)-n = 400(1 + 0.02)-4 = 369.54 Programmed solution:

For the $450.00 debt: FV = 450.00; n =

(4) = 6; I/Y = 8; P/Y = C/Y = 4

PV = FV(1 + i)-n = 450(1 + 0.02)-6 = 399.59 Programmed solution:

For the $500.00 debt: FV = $500.00; n =

(4) and I/Y = 8 and P/Y = C/Y = 4

PV = FV(1 + i)-n = 500(1 + 0.02)-10 = 410.17 Programmed solution:

Payment = debts = 369.54 + 399.59 + 410.17 = $1179.30 Diff: 2 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

74) Scheduled debt payments of $1500.00 due seven months ago, $1200.00 due two months ago, and $1800.00 due in five months are to be settled by two equal payments now and three months from now respectively. Determine the size of the equal replacement payments at 9% p.a. compounded monthly. Answer: Let the size of the equal payments be $x, and the focal point now. For the $1500.00 debt: PV = 1500.00; n = 7; i =

= 0.75% = 0.0075 ; I/Y = 9; P/Y = C/Y = 12

FV = PV(1 + i)n = 1500(1 + 0.0075)7 = 1580.54 Programmed solution:

For the 1200.00 debt the only difference is PV = 1200 and n = 2. FV = PV(1 + i)n = 1200(1 + 0.0075)2 = 1218.07 Programmed solution: For the $1800.00 debt FV = $1800.00 n = 5 and I/Y = 9 and P/Y = C/Y = 12 PV = FV(1 + i)-n = 1800(1 + 0.0075)-5 = 1733.99 Programmed solution: For the first equal payment ($x) now, = $x For the second equal payment ($x) in three months FV = x and n = 3 . PV = FV(1 + i)-n = x(1 + 0.0075)-3 = 0.977833328 x Programmed solution: Payments = debts x + 0.977833328x = 1580.54 + 1218.07 + 1733.99 1.977833328x = 4532.60 x = 2291.70 The size of the two equal payments is $2291.70. Diff: 3 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

75) A debt of $5000.00 is to be repaid by payments of $2000.00 after two years, $2500.00 after three years and a final payment after five years. Determine the size of the final payment if interest is 10% p.a. compounded semi-annually. Answer: Let the final payment be $x and the focal date at five years. For the $5000.00 debt: PV = 5000.00; n = 5(2) = 10; i =

= 5% = 0.05; I/Y = 10; P/Y = C/Y = 2

FV = PV(1 + i)n = 5000(1 + 0.05)10 = 8144.47 Programmed solution:

For the $2000.00 payment: PV = 2000.00; n = 3(2) = 6; i =

= 5% = 0.05; I/Y = 10; P/Y = C/Y = 2

FV = PV(1 + i)n = 2000(1 + 0.05)6 = 2680.19 Programmed solution:

For the $2500.00 payment: PV = 2500.00; n = 2(2) = 4; i =

= 5% = 0.05; I/Y = 10; P/Y = C/Y = 2

FV = PV(1 + i)n = 2500(1 + 0.05)4 = 3038.77 Programmed solution:

Payments = debt 2680.19 + 3038.77 + x = 8144.47 x = $2425.51 Diff: 2 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

76) Debts of $2800.00 due three months from now and $4600.00 due twenty-one months from now are to be settled by two equal payments due in three months and nine months from now respectively. Determine the size of the equal replacement payments if interest is 5.5% p.a. compounded quarterly. Answer: Let the size of the equal payments be $x, and the focal point be at three months from now. For the $2800.00 debt: 2800 For the 4600 debt: FV = 4600.00; n =

(4) = 6 ; i =

= 1.375% = 0.01375 ; I/Y = 5.5; P/Y = C/Y = 4

PV = FV(1 + i)-n = 4600(1 + 0.01375)-6 = 4238.11 Programmed solution:

For the first equal payment ($x) at 3 months, = $x For the second equal payment ($x) in nine months FV = x and n =

(4) = 2.

PV = FV(1 + i)-n = x(1 + 0.01375)-2 = 0.973056965x Programmed solution:

Payments = debts x + 0.973056965x = 2800.00 + 4238.11 1. 973056965x = 7038.11 x = 3567.11 The size of the two equal payments is $3567.11. Diff: 3 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

77) Dorian owes his friend Katrina after she helped him pay for a used car. She originally asked Dorian to pay $1600.00 due today, and a second payment of $7200.00 due in 15 months with interest at 9.6% p.a. compounded quarterly. Dorian's financial position is currently disastrous, however, so he renegotiated with Katrina and now owes $3400.00 nine months from now and one more final payment in 21 months. What's the size of Dorian's final payment if the money is worth 12.12% p.a. compounded monthly? Answer: Let the focal day be in 21 months. Payment of $7200: FV = 7200.00(1.024)5 = 7200.00(1.1258999) = 8106.48 At focal date: Let the final payment be $x. i = 12.12%/12 = 0.0101 1600.00(1.0101)21 + 8106.48(1.0101)6 = 3400.00(1.0101)12 + x

1600.00(1.2349569) + 8106.48(1.0621509) = 3400.00(1.1286456) + x 1975.93 + 8610.31 = 3835.76 + x $6750.48 = x Dorian's final payment will be $6750.48 Diff: 2 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 78) Debts of $850 due in six months, $700 due in sixteen months, and $1100 due in three years are to be settled by a single payment one year from now. What is the size of that single payment if interest is 7.5% compounded monthly? Answer: Let the size of the payment be $x and the focal date be one year from now. x = 850.00(1.00625)6 + 700.00(1.00625)-4 + 1100.00(1.00625)-24 x = 850.00(1.0380908) + 700.00(.9753858) + 1100.00(.8611099) x = 882.38 + 682.77 + 947.22 x = $2512.37 Diff: 2 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

79) A debt payment in the amount of $2000.00 due today, is to be settled by a payment of $1500.00 nine months from now and a final payment in 18 months. Determine the size of the final payment if the money is worth 12% p.a. compounded quarterly. Use the Banker's Method. Answer: In 9 months: FV = 2000.00(1.03)3 = 2000.00(1.092727) = 2185.45 Balance: 2185.45 - 1500 = 685.45

Final payment: FV = 685.45(1.03)3 = 749.01 Diff: 1 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

80) A loan of $2500.00 made today is to be repaid in three equal installments due in one year, two years, and four years respectively. What is the size of the equal installments if money is worth 8.4% compounded monthly? Answer: Let the size of the equal payments be $x and the focal date be now. 2500.00 = x(1.007)-12 + x(1.007)-24 + x(1.007)-48 2500.00 = x(.9197004) + x(.8458487) + x(.7154601) 2500.00 = 2.4810092x $1007.65 = x The size of the equal payments is $1007.65. Diff: 3 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

81) An obligation of $5150.00 due in 2.5 years is to be settled by four equal payments due today, nine months from now, 21 months from now, and 27 months from now, respectively. What is the size of the equal payment at 8.88% p.a. compounded monthly? Answer: Let the size of the equal payments be $x. Let the focal date be 2.5 years from now. i = 0.0074 5150.00 = x(1.0074)30 + x(1.0074)21 + x(1.0074)9 + x(1.0074)3 5150.00 = x(1.247551) + x(1.167457) + x(1.068606) + x(1.022365) 5150.00 = 4.505979x $1142.93 = x Diff: 3 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

82) You invest $10 000 in a savings account that pays interest of 8% compounded monthly. What is the value of your account after 14 months? A) $10 970.88 B) $10 904.88 C) $10 074.88 D) $10 004.88 E) $10 974.88 Answer: E Diff: 1 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

83) Karen started a registered retirement savings plan on February 1, 2000, with a deposit of $2210. She added $2000 on February 1, 2001, and $1475 on February 1, 2004. What is the accumulated value of her RRSP account on August 1, 2005, if interest is 10.44% compounded quarterly? Answer: Balance on Feb. 1, 2001 FV1 = 2210.00(1.0261)4 = 2210.00(1.1085588) = 2449.91 2449.91 + 2000.00 = 4449.91 Balance on Feb. 1, 2004 FV2 = 4449.91(1.0261)12 = 4449.91(1.362311) = 6062.16

6062.16 + 1475.00 = 7537.16 Balance on Aug. 1, 2005 FV3 = 7537.16(1.0261)6 = 7537.16(1.167180775) = $8797.23

Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 84) Jenni started a registered retirement savings plan on January 1, 2008, with a deposit of $2000. She added $3000 on January 1, 2009, and $2000 on January 1, 2010. What is the accumulated value of her RRSP account on July 1, 2010, if interest is 12% compounded quarterly? Answer: Balance on Jan. 1, 2009 FV1 = 2000.00(1.03)4 = 2000.00(1.12550881) = 2251.02 2251.02 + 3000 = 5251.02 Balance on Jan. 1, 2010 FV2 = 5251.02(1.03)4 = 5251.02(1.362311) = 5910.07 5910.07 + 2000 = 7910.07 Balance on July 1, 2010 FV3 = 7910.07(1.03)2 = 7910.07(1.0609) = $8391.79

Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 85) Find the present value and the compound discount of $7500 due in five years and six months if interest is 4.8% compounded quarterly. Answer: PV = 7500.00(1.012)-22 = 7500.00(.7691813) = $5768.86 Discount = 7500.00 - 5768.86 = $1731.14 Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money.

86) Find the proceeds of a non-interest-bearing promissory note for $175 500 discounted 54 months before maturity at 9.75% compounded semi-annually. Answer: FV = 175000.00; i = 0.04875; n =

*2=9

PV = 17500.00(1.04875)-9 = 175000.00(.6515567) = $114 022.42 Diff: 2 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 87) A three-year, $12 000 promissory note bearing interest at 11.6% compounded quarterly is discounted two years after the date of issue at 9.64% compounded semi-annually. What are the proceeds of the note? Answer: Maturity value = 12000.00(1.029)12 = 12000.00(1.4092385) = $16 910.86 Proceeds = 16910.86(1.0482)-2 = 16910.86(.91049473) = $15 391.37 Diff: 2 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 88) An obligation of $17 320 is due two years from now with interest at 9.2% compounded semi-annually. The obligation is to be settled by a payment of $7250 in seven months and a final payment in sixteen months. What is the size of the second payment if interest is now 10.8% compounded monthly? Answer: Let the second payment be $x and the focal date be 16 months from now. = 7250.00(1.009)9 + x = 7250.00(1.083978) + x = 7858.84 + x 19299.47 = 7858.84 + x $11 440.63 = x Diff: 2 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 89) Debt payments of $1400.00 due today, $5100.00 due in twenty-one months, and $1900.00 due in 4.5 years are to be combined into a single payment due three years from now. What is the size of the single payment if interest is 8.04% p.a. compounded quarterly? Answer: Let the single payment be $x and the focal date be 3 years from now. x = 1400.00(1.0201)12 + 5100.00(1.0201)5 + 1900.00(1.0201)-6 x = 1400.00(1.2697346) + 5100.00(1.1046221) + 1900.00(.8874492) x = 1777.63 + 5633.57 + 1686.15 x = $9097.35 Diff: 1 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

90) A loan of $8000.00 taken out two years ago is to be repaid by three equal installments due now, three years from now, and six years from now respectively. What is the size of the equal installments if interest on the debt is 7.32% p.a. compounded monthly? Answer: Let the size of the equal payments be $x and the focal date be now. 8000.00(1.0061)24 = x + x(1.0061)-36 + x(1.0061)-72 8000.00(1.1571444) = x + .8033756x + .64541235x 9257.16 = 2.44878795x $3780.30 = x Diff: 3 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

91) A $17 200 debt will accumulate for six years at 11.6% compounded semi-annually. For how much will the debt sell four years after it was incurred if the buyer of the debt charges 9% compounded quarterly? Answer: Maturity value: PV = 17200.00; i = 11.6%/2 = 5.8%; n = 12 FV = 17200.00(1.058)12 = 17200.00(1.9671071) = $33 834.24 Proceeds:

i = 0.09/4 = 0.0225 N = 2 × 4 = 8 PV = 33834.24(1.0225)-8 = 33834.24(.8369383) = $28 317.17

Diff: 2 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 92) Determine the compound discount on $18 875 due in 7.75 years if interest is 2.84% compounded monthly. Answer: PV = 18875.00(1.002367)-93 = 18875.00(0.802647) = $15149.966740 Compound discount = 18875.00 — 15149.97 = $3725.03 Diff: 1 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes.

93) A note dated July 1, 2000, promises to pay $9000 with interest at 12.4% compounded quarterly on January 1, 2007. Find the proceeds from the sale of the note on July 1, 2002, if money is then worth 8.64% compounded semi-annually. Answer: Balance on Jan. 1, 2007 Maturity Value = 9000.00(1.031)28 = 9000.00(2.3509459) = $21 158.51 Discounted value on July 1, 2002 PV = 21158.51(1.0216)-9 = 21158.51(.8250344) = $17 456.50

Diff: 2 Type: SA Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes.

94) A debt of $8125 due today is to be settled by three equal payments due three months from now, 18 months from now, and 39 months from now respectively. What is the size of the equal payments at 6.8% compounded quarterly? Answer: Let the size of the three equal payments be $x and the focal date be now. 8125.00 = x(1.017)-1 + x(1.017)-6 + x(1.017)-13 8125.00 = .9832842x + .903804x + .8032072x 8125.00 = 2.6902954x $3020.11 = x Diff: 3 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

95) A debt of $4000 due today is to be settled by two equal payments due three months from now, and 9 months from now respectively. What is the size of the equal payments at 8% compounded quarterly? Answer: Let the size of the two equal payments be $x and the focal date be now. 4000.00 = x(1.02)-1 + x(1.02)-3 4000.00 = .9803921569x + .9423223345x 4000.00 = 1.922714491x $2080.39 = x Diff: 3 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

96) Payments of $10 200.00 due one year ago and $13 000.00 due nine months ago are to be replaced by a payment of $5800.00 now, a second payment of $10 000.00 fifteen months from now, and a final payment twenty-four months from now. What is the size of the final payment if interest is 9.2% compounded quarterly? Answer: Focal date is 24 months from now. Let the final payment be $x. 10200.00(1.023)12 + 13000.00(1.023)11 = 5800.00(1.023)8 + 10000.00(1.023)3 + x 10200.00(1.3137345) + 13000.00(1.2841979) = 5800.00(1.1995133) + 10000.00(1.0705992) + x 13400.09 + 16694.56 = 6957.18 + 10705.99 + x 30094.65 = 17663.17 + x $12 431.48 = x The size of the final payment is $7816.89. Diff: 2 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 97) Find m for the investment of $1000.00 for 12 years at 4% compounded quarterly. A) 48 B) 3 C) 1% D) 16% Answer: A Diff: 1 Type: MC Page Ref: 337-343 Topic: 9.1 Basic Concepts and Computations Objective: 9-1: Calculate interest rates and the number of compounding periods.

98) Find i if an amount is invested at 6.00% compounded every 3 months. A) 3.00% B) 2.00% C) 6.00% D) 1.50% E) 0.02 Answer: D Diff: 2 Type: MC Page Ref: 337-343 Topic: 9.1 Basic Concepts and Computations Objective: 9-1: Calculate interest rates and the number of compounding periods. 99) Determine n if an amount is invested for 3.5 years at 2.25% compounded quarterly. A) 14 B) 5 C) 10.5 D) 0.5625% E) 56.25 Answer: A Diff: 1 Type: MC Page Ref: 337-343 Topic: 9.1 Basic Concepts and Computations Objective: 9-1: Calculate interest rates and the number of compounding periods. 100) You invest $8500 in a savings account that pays interest of 4.8% compounded monthly. What is the value of your account after 19 months? A) $9169.79 B) $9669.79 C) $9119.79 D) $9219.79 E) $9999.79 Answer: A Diff: 1 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 101) If $3000.00 is invested for seven years and seven months at 6% p.a. compounded quarterly, calculate the maturity value. A) $3763.18 B) $3776.32 C) $4745.43 D) $4712.57 E) $4882.57 Answer: D Diff: 1 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

102) Calculate the accumulated value of $1000.00 at 8% compounded monthly for 12 years. A) $2603.39 B) $1603.39 C) $2600.39 D) $2003.39 E) $603.39 Answer: A Diff: 1 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 103) Calculate the future value of $4200 if it is invested at an interest rate of 8.6% compounded quarterly for 3 years and 10 months. A) $5919.75 B) $5891.75 C) $5819.75 D) $5875.91 E) $5888.75 Answer: C Diff: 1 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 104) Calculate the future value of $5000 if it is invested at an interest rate of 16% compounded quarterly for 5 years and 6 months. A) $11 849.59 B) $11 809.59 C) $11 049.59 D) $10 849.59 E) $5849.59 Answer: A Diff: 1 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 105) You invest $6780 in a floating rate guaranteed investment certificate. For the first 30 months you earn 4.9% compounded semi-annually. For the next 8 months you earn 4.32% compounded monthly. What is the maturity value of the certificate? A) $7775.44 B) $7785.44 C) $7857.44 D) $7875.44 E) $7888.44 Answer: D Diff: 2 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

106) You invest $8000 in a floating rate guaranteed investment certificate. For the first 24 months you earn 6% compounded semi-annually. For the next 6 months you earn 12% compounded monthly. What is the maturity value of the certificate? A) $9558.02 B) $9888.00 C) $9558.00 D) $9585.00 E) $8558.00 Answer: C Diff: 2 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 107) You have an investment that will mature in 20 months with the value of $2500. You need some quick cash and decide to sell it today at a discount rate of 10% compounded quarterly. What is the cash value? A) $659.23 B) $2009.63 C) $2120.54 D) $2299.63 E) $2219.63 Answer: C Diff: 1 Type: MC Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 108) You want to retire with $370 000 in the bank and you are able to earn 4.84% compounded quarterly for the next 25 years. You currently have $175 000 saved up. How much more do you have to invest in 15 years in order to achieve your goal? A) $100 649.58 B) $110 649.58 C) $120 649.58 D) $130 649.58 E) $0.00 Answer: E Diff: 3 Type: MC Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money.

109) Find the principal that will grow to $11 019.45 at 6% compounded semi-annually in 5 years and five months. A) $8661.65 B) $8000.00 C) $7980.32 D) $1630.33 E) $788.57 Answer: B Diff: 1 Type: MC Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 110) You want to retire with $400 000 in the bank and you are able to earn 6% compounded quarterly for the next 25 years. How much money do you have to invest today in order to achieve your goal? A) $90 251.78 B) $90 251.77 C) $90 001.78 D) $90 201.77 E) $90 001.77 Answer: A Diff: 1 Type: MC Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 111) What is the cash value of $5000 if it is discounted for 5 years at 10% compounded yearly? A) $3100.61 B) $3000.61 C) $3143.28 D) $3228.43 E) $3104.61 Answer: E Diff: 1 Type: MC Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 112) You have an investment that will mature for $6825 in 57 months. You sell the investment 21 months before maturity. The discount rates used are 5.6% compounded quarterly for the first nine months of the discount period (from the date of maturity) and then 4.92% compounded monthly for the remaining discount period. How much did you sell the investment for? A) $6262.54 B) $6232.54 C) $6223.54 D) $6632.54 E) $6666.54 Answer: B Diff: 2 Type: MC Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes.

113) Calculate the cash value of a bond that will mature with a value of $16 500 in 7 years and 5 months. The bond is discounted at 5.8% compounded semi-annually. A) $10 797.48 B) $10 979.48 C) $10 997.48 D) $10 779.48 E) $11 117.48 Answer: A Diff: 1 Type: MC Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 114) A 9-month non-interest bearing promissory note is sold 2 months after it was issued. The face value of the note is $8500 and it is discounted at a rate of 5.2% compounded annually. What are the proceeds? A) $8252.33 B) $8525.33 C) $8255.33 D) $8522.33 E) $8552.33 Answer: A Diff: 1 Type: MC Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 115) You invest $12 300 of your money in a locked-in savings account that earns 4.92% compounded monthly for 7 months. You have a family emergency and need to withdraw your funds 2 months early. You sell the savings account at a 5.6% compounded quarterly interest rate. How much did you sell the account for? A) $11 569.61 B) $12 956.61 C) $12 540.61 D) $12 659.61 E) $12 666.61 Answer: C Diff: 2 Type: MC Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes.

116) You owe $4510 due in 7 months. In addition you owe $3780 due in 13 months and $5125 due in 21 months. You are paying 8.64% compounded monthly on your loan. What single amount three months from now will pay off the entire loan of the three future payments? A) $12 440.86 B) $12 040.86 C) $12 404.86 D) $12 444.86 E) $12 884.86 Answer: C Diff: 3 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 117) You lend a friend $800 and they agree to make quarterly payments for 1 year. You charge your friend 8.52% compounded quarterly. What is the size of the payments? A) $120.76 B) $210.76 C) $220.76 D) $200.76 E) $212.76 Answer: B Diff: 3 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 118) You borrow $4750 at a rate of 6.72% compounded monthly. You make a partial payment of $1400 in 3 months. You also agree to make two equal payments 3 and 5 months after the partial payment. What is the size of the equal payments? Use a focal date of 8 months. A) $1573.84 B) $1754.02 C) $1735.84 D) $1784.35 E) $1788.02 Answer: C Diff: 3 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

119) You will need three amounts of $14 200 in each year for four years in order to go to school. You are planning on going to school starting in 5 years and ending in 8 years (years 5, 6, 7, 8). You are able to earn 9.64% compounded quarterly. How much money do you have to have today in order to be able to go to school? A) $30 574.35 B) $30 457.35 C) $30 754.35 D) $30 475.34 E) $33 354.35 Answer: C Diff: 2 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 120) You owe $4510 due in 7 months. In addition you owe $3780 due in 13 months and $5125 due in 21 months. You are paying 8.64% compounded monthly on your loan. What single amount three months from now will pay off the entire loan of the three future payments? A) $12 440.86 B) $12 040.86 C) $12 404.86 D) $12 444.86 E) $12 884.86 Answer: C Diff: 3 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 121) You have an investment that will mature in 33 months and have a value of $4300. You need some quick cash and decide to sell today at a discount rate of 8.2% compounded quarterly. What is the cash value? A) $4339.74 B) $3439.74 C) $3339.74 D) $3539.74 E) $3669.74 Answer: B Diff: 1 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

122) Determine the nominal rate of interest if the periodic rate is 1.75% per quarter. A) 1.75% B) 3.5% C) 5.25% D) 7% E) 21% Answer: D Diff: 1 Type: MC Page Ref: 337-343 Topic: 9.1 Basic Concepts and Computations Objective: 9-1: Calculate interest rates and the number of compounding periods. 123) Albert made investments of $1000 each for two years. The first one at an interest of 4.8% compounded monthly and the second at 4.9% compounded annually. Calculate the total value of his investment in two years. A) $1100.55 B) $1100.40 C) $2000 D) $2200.95 E) $2100.55 Answer: D Diff: 1 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 124) Valeri invested $5000 at 9.25% compounded quarterly. After 18 months, the rate changed to 9.75% compounded semi-annually. What amount will Valeri have 3 years after the initial investment? A) $5735.12 B) $6578.31 C) $6652.76 D) $6615.44 E) $6754.33 Answer: D Diff: 2 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

125) Keith took a loan of $20 000 to buy a car at an interest of 4% compounded quarterly, with no monthly payments. On first, second and third anniversary of the loan, he made payments of $5000. What payment on the fourth anniversary will eliminate the debt? Answer: i =

= 0.01

n=1×4=4 After 1st year: FV = PV(1 + i)n = $20000(1 + 0.01)4 = $20 812.08

After $5000 payment, PV after 1st year = $15 812.08 After 2nd year: FV = PV(1 + i)n = $15812.08(1 + 0.01)4 = $16 454.11

After $5000 payment, PV after 2nd year = $11 454.11 After 3rd year: FV = PV(1 + i)n = $11,454.11(1 + 0.01)4 = $11 919.20 After $5000 payment, PV after 3rd year = $6919.20

Payment required on the 4th anniversary to eliminate the debt = $6919.201(1 + 0.01)4 = $7200.14 Diff: 3 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 126) Nader makes semi-annual payments of $3000 to pay off his loan. Because of unemployment, he defaulted on his last two payments. What total payment is he expected to make during his third payment, if the defaulted payments are compounded monthly at a rate of 6%? Answer: i =

= 0.005

n1 = 1 × 12 = 12 n2 = 1 × 6 = 6

= $3000 + $3000(1+ 0.005)12 + $3000(1+ 0.005)6 = $9276.17 Diff: 2 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. Total payment during 3rd payment

127) Shaun wants to invest his money such that he accumulates $10 000 after 3

years at a rate of 4%

compounded monthly? How much money should he invest today? A) $8717.33 B) $9615.38 C) $9884.20 D) $8695.61 E) $11 500.06 Answer: D Diff: 2 Type: MC Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 128) Alex's property taxes are $4768.00 for his house valued at $349 900 in Whitby due on July 1 of 2014. How much discount should city give, if Alex pays his property tax 8 months in advance, when the city can earn 3% compounded monthly on the surplus fund? Answer: i

= 0.0025

n=8 PV =

= $4673.70

Discount = $4768 - $4673.70 = $94.30 Diff: 2 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 129) Two payments of $49 000 each must be made 3 year and 5 year from now. If money can earn 4.9% compounded monthly, what single payment 3 years from now would be equivalent to the two scheduled payments? A) $93 435 B) $98 000 C) $88 869 D) $105 742 E) $101 177 Answer: A Diff: 2 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

130) Two payments of $49 000 each must be made 3 year and 5 year from now. If money can earn 4.9% compounded monthly, what single payment 5 years from now would be equivalent to the two scheduled payments? A) $98 117 B) $98 000 C) $103 034 D) $105 742 E) $101 177 Answer: C Diff: 2 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 131) Two payments of $49 000 each must be made 3 year and 5 year from now. If money can earn 4.9% compounded monthly, what single payment 4 years from now would be equivalent to the two scheduled payments? Answer: i =

= 0.004083

For 3rd year payment made in 4th year PV = $49 000; n = 12 FV = PV(1 + i)n = $49000(1+ 0.004083)12 = $51 455.66 For 5th year payment made in 4th year FV = $49 000; n = 12 PV = FV(1 + i)-n = $49000(1+ 0.004083)-12 = $46 661.53 Single payment made in 4th year = $51 455.66 + $46 661.53 = $98 117.19 Diff: 2 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 132) Doris borrowed $5000 from a finance company at an interest of 8% quarterly. The loan is to be repaid in 3 equal payments, annually. What is the amount of the equal payment? A) $1948.69 B) $1666.67 C) $1419.51 D) $1787.42 E) $1856.77 Answer: A Diff: 2 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

133) What periodic payment does Imran receive from a $100 000, 3-year, monthly payment GIC earning a nominal rate of 2.25% payable monthly? A) $180.50 B) $187.50 C) $191.75 D) $185.50 E) $208.33 Answer: B Diff: 2 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 134) ICICI quotes nominal annual interest rate of 6.6% compounded annually, 6.5% compounded semiannually and 6.4% compounded monthly on 5 year compound interest GICs. What rate should an investor choose? A) 6.6% compounded annually B) 6.5% compounded semi-annually C) 6.4% compounded monthly D) Any, as all alternatives are same E) Insufficient data to make a decision Answer: B Diff: 2 Type: MC Page Ref: 337-343 Topic: 9.1 Basic Concepts and Computations Objective: 9-1: Calculate interest rates and the number of compounding periods. 135) A $100 000 face value strip bond has 6 years remaining until maturity. If the current market return rate is 4% compounded semi-annually, what is the fair market value of the strip bond? A) $78 849 B) $79 031 C) $88 797 D) $100 000 E) $126 824 Answer: A Diff: 2 Type: MC Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes.

136) A five year promissory note with a face value of $5000, bearing interest at 6% compounded semiannually, was sold 18 months after its issue date to yield the buyer 4% compounded quarterly. What amount was paid for the note? A) $6719.58 B) $5845.79 C) $5463.64 D) $4753.16 E) $4349.82 Answer: B Diff: 2 Type: MC Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 137) Ontario's recommended discount rate is 2.5% per year, which is the same as rate of inflation in Ontario. What annual income will be needed 17 years from now, to have the same purchasing power as a $134 000 annual income today? Answer: PV = $134 000 n = 17 i = 0.025 FV = $134 000 × 1.02517 = $203 897 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

138) Ontario passed a law in 1996 that the names of all public servants earning $100 000 or more will be published starting 1997. What is the equivalent of that amount in 2013 assuming Ontario's recommended discount rate is 2.5% per year? Answer: PV = $100 000 in 1997 i = 0.025 n = 2013 -1997 = 16 FV in 2013 = $100 000 × 1.02516 = $148 451 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

139) Ontario passed a law in 1996 that the names of all public servants earning $100 000 or more will be published starting 1997. This year (2013), Ontario is planning to revise the amount to an equivalent purchasing power amount in 2014. What is the amount in 2014, beyond which the names of the Ontario's public servants should be published, assuming the recommended discount rate is 2.5% per year? A) $65 720 B) $100 000 C) $102 500 D) $142 500 E) $152 162 Answer: E Diff: 1 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 140) Sagheer received two offers for his property that he is selling. Offer 1 consists of $700 000 immediately and offer 2 consists of $500 000 now and $110,000 per annum for the next 2 years. If money earns 6% compounded semi-annually, which offer has a better economic advantage and by how much. Assume no inflation or discount? A) Offer 1 is better by ~$1600 B) Offer 2 is better by ~$1600 C) Offer 1 is better by ~$5100 D) Offer 2 is better by ~$5100 E) Offer 2 is better by ~$20 000 Answer: B Diff: 1 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 141) For a nominal interest rate of 9%, what is the compounding frequency if the periodic interest is 0.75%? A) 8 B) 12 C) 6 D) 1 E) 24 Answer: B Diff: 1 Type: MC Page Ref: 337-343 Topic: 9.1 Basic Concepts and Computations Objective: 9-1: Calculate interest rates and the number of compounding periods.

142) A court ruled in favour of Professor Dumbledore. Professor Dumbledore successfully claimed that that Professor Umbridge defaulted on two payments of 1000 galleons each. One payment was due 18 months ago and the other 11 months ago. What is the appropriate amount for the court to order Umbridge to pay immediately if the court uses 6% compounded monthly for the interest rate money can earn? Answer: i =

= 0.005

PV = 1000 galleons The amount today that is equivalent to the payment 18 months ago is FV1 = PV(1 + i)n = 1000(1+ 0.005)18 = 1093.93 galleons The amount today that is equivalent to the payment 11 months ago is FV1 = PV(1 + i)n = 1000(1+ 0.005)11 = 1056.40 galleons The appropriate amount for Umbridge to pay = 1093.93 + 1056.40 = 2150.33 galleons Diff: 1 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 143) Two payments of $10 000 each must be made one year and four years from now. If money can earn 9% compounded monthly, what single payment two years from now would be equivalent to the two scheduled payments? A) $10 938 B) $8358 C) $11 964 D) $22 902 E) $19 296 Answer: E Diff: 1 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

144) Zara borrowed $4000 from Duca financial at an interest rate of 7% compounded semiannually. The loan is to be repaid in 3 equal payments in two years, three and five years, respectively. How much will be each payment? Answer: Sum of present value of the three payments = $4000 PV1 + PV2 + PV3 = $4000 Since the payments are equal, represent FV of each payment by x PV1 = x (1.035)-4 = 0.871442x PV2 = x (1.035)-6 = 0.813501x PV3 = x (1.035)-10 = 0.708919x 0.871442x + 0.813501x + 0.708919x = $4000 ⇒ 2.393862x = $4000 ⇒ x = $1670.94 Diff: 1 Type: SA Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 145) A five-year promissory note with a face value of $3500, bearing interest at 11% compounded semiannually, was sold 21 months after its issue date to yield the buyer 10% compounded quarterly. What amount was paid for the note? A) $3500 B) $5978.51 C) $4336.93 D) $6021.50 E) $4251.71 Answer: C Diff: 1 Type: MC Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 146) How much will Ahsan's registered retirement savings deposit of $30 000.00 be worth in 21 years at 5% compounded quarterly? A) $1 807 267.24 B) $83 578.88 C) $85 173.39 D) $38 941.88 E) $81 044.55 Answer: C Diff: 1 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

147) Chelsea's grand parents made a trust deposit of $5000.00 on November 30, 2002, at her birth. This trust deposit can be withdrawn on Chelsea's twenty-fifth birthday. To what will the deposit amount on that date at 19.11% compounded semi-annually? A) $531 830.45 B) $395 982.16 C) $31 360 374.23 D) $48 952.52 E) $479 269.91 Answer: E Diff: 1 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 148) Meridian Credit Union expects an average annual growth rate of 3% for the next four years. If the assets of the credit union currently amount to $11.4 billion, what will the forecasted assets be in four years? A) $12.83 billion B) $1.43 billion C) $12.85 billion D) $1.45 billion E) $32.56 billion Answer: A Diff: 1 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 149) Gabriel defaulted on her tax worth $4189.00. CRA audited her and after 31 months sent her the tax bill, which also included an interest rate compounded annually at 3.75% p.a. What is the total tax due, if Gabriel pays it immediately on receiving the notice? A) $13 114.19 B) $4614.12 C) $4222.90 D) $4606.94 E) $4592.83 Answer: D Diff: 1 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

150) Derik bought a promissory note this morning. Find the maturity value of the promissory note for $1200.00 dated March 31, 2016, and due on August 31, 2026, if interest is 3.64% compounded quarterly. A) $1741.51 B) $1318.75 C) $1750.28 D) $5323.00 E) $1724.05 Answer: C Diff: 1 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 151) A $25 641.00 investment matures in seven years, two months. Find the maturity value if interest is 6.11% p.a. compounded quarterly. Answer: PV = 25641.00; i = .015275; n = 4[7 + 2/12] = 28.666667 FV = 25461.00(1.015275)28.666667 = 25461.00(1.544303) = $39 319.50 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

152) An investment of $30 325 is accumulated at 9.45% compounded quarterly for nine and one-half years. At that time the interest rate is changed to 9.45% compounded monthly. How much is the investment worth two years after the change in interest rate? A) $73 646.89 B) $88 902.62 C) $49 319.03 D) $89 520.33 E) $88 773.12 Answer: B Diff: 2 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 153) RBC offers three-year term deposits at 3.125% compounded annually while Meridian credit union offers such deposits at 3.05% compounded quarterly. If you invest $23 000 at RBC and the same amount at Meridian, what is the maturity value of your deposit at each bank? Answer: At RBC, FV = 23000.00(1.03125)3 = 23000.00(1.0967102) = $25 224.33 At Meridian, FV = 23000.00(1.007625)12 = 23000.00(1.09543651) = $25 195.04 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

154) Two years after Michelle deposited $25 000 in a TD Bank savings account that earned interest at 2.25% compounded monthly, the rate of interest was changed to 5.75% compounded semi-annually. How much was in the account fifteen years after the deposit was made? A) $26 149.59 B) $54 641.30 C) $61 201.27 D) $58 510.72 E) $54 629.87 Answer: B Diff: 2 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 155) Suman deposited $95 000.00 in an RRSP on April 1, 2016, at 8.22% compounded quarterly. Subsequently the interest rate was changed to 8.5% compounded monthly on September 1, 2016, and to 8.8% compounded semi-annually on June 1, 2017. What was the value of the RRSP deposit on December 1, 2017, if no further changes in interest were made? A) $97 446.55 B) $103 837.72 C) $108 406.58 D) $101 734.20 E) $112 975.44 Answer: C Diff: 2 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 156) Romeo started an RRSP on March 1, 2016, with a deposit of $35 000.00. She added $19 000.00 on December 1, 2017, and $8500.00 on September 1, 2018. What is the accumulated value of her account on December 1, 2020, if interest is 6.6% compounded monthly? A) $39 272.78 B) $41 260.10 C) $47 846.07 D) $70 993.80 E) $80 850.58 Answer: E Diff: 3 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments.

157) Raman has a line of credit loan with the ICICI bank. The initial loan balance was $72 000.00. Payments of and $25 000.00 were made after four months and ten months respectively. At the end of one year, he borrowed an additional $42 500.00. Seven months later, the line of credit loan was converted into a collateral mortgage loan. What was the amount of the mortgage if the line of credit interest was at prime (3%) + 1.25% compounded monthly? A) $77 002.58 B) $45 368.70 C) $19 560.45 D) $63 125.35 E) $61 582.31 Answer: D Diff: 3 Type: MC Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 158) Aaron is planning to go on a vacation in 6 years. The vacation $7800.00. How much should he setaside today in his savings account if the current interest rate is 7.6% p.a., compounded quarterly? A) $4965 B) $7800 C) $5026 D) $12 254 E) $4951 Answer: A Diff: 1 Type: MC Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 159) What is the discounted value of $9900.00 due in five years, seven months if money is worth 2.2% compounded quarterly. A) $9031 B) $9118 C) $6855 D) $6876 E) $8757 Answer: A Diff: 1 Type: MC Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money.

160) A $95 600.00, non-interest-bearing note due August 1, 2017, discounted on March 1, 2015, at 1.49% compounded monthly. Find the proceeds. A) $92 220.90 B) $92 335.41 C) $92 106.54 D) $99 225.96 E) $92 243.34 Answer: A Diff: 1 Type: MC Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 161) A $50 000.00, 5-year note bearing interest at 3.25% compounded quarterly, discounted three and a half years after the date of issue at 2.5% compounded monthly. Find the proceeds. A) $48 161.60 B) $56 622.44 C) $58 783.81 D) $42 528.72 E) $40 965.02 Answer: B Diff: 2 Type: MC Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 162) A ten-year promissory note dated April 1, 2011, with a face value of $700.00 bearing interest at 7% compounded semi-annually, discounted six years later when money was worth 8.5% compounded monthly. What are the proceeds? A) $1392.85 B) $992.58 C) $998.38 D) $498.84 E) $250.70 Answer: B Diff: 2 Type: MC Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes.

163) Alflah Islamic bank issued $9300.00 promissory note issued without interest for nine years on September 30, 2001, is discounted on July 31, 2006, at 5.5% compounded quarterly. Find the compound discount. A) $15 205.27 B) $12 110.06 C) $9300.00 D) $7406.88 E) $1893.12 Answer: E Diff: 2 Type: MC Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 164) A twenty-year note for $1000.00 bearing interest at 9% compounded monthly is discounted at 5% compounded quarterly four years and six months before maturity. Find the proceeds of the note. A) $6009.15 B) $4805.10 C) $133.07 D) $4800.65 E) $4741.93 Answer: B Diff: 2 Type: MC Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 165) Calculate the proceeds of $5500.00 due in 20 years, nine months discounted at 8.0% compounded semi-annually. A) $1080.14 B) $1145.59 C) $1051.54 D) $1113.83 E) $1063.04 Answer: A Diff: 1 Type: MC Page Ref: 361-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes.

166) Arjun has taken a loan and debt can be repaid by payments of $425.00 today, $377.00 in two years and $560.00 in five years. What single payment would settle the debt three years from now if money is worth 6.25% p.a. compounded quarterly? A) $1362.00 B) $1407.70 C) $1546.98 D) $1387.92 E) $1341.13 Answer: B Diff: 2 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 167) Saeed has taken a loan such that debts of $4000.00, $4500.00 and $5000.00 are due in one year, eighteen months and two years from now respectively. He won a lottery and would like to make a single payment now that would settle the debts if interest is 5.99% p.a. compounded quarterly. A) $13 500.00 B) $12 324.59 C) $14 796.16 D) $13 079.58 E) $13 880.82 Answer: B Diff: 2 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 168) Ade has to settle a debt, for which scheduled debt payments of $3800.00 was due seven months ago, $4700.00 was due two months ago, and $8800.00 due in five months. Debt has to be settled by two equal payments now and three months from now respectively. Determine the size of the equal replacement payments at 4.9% p.a. compounded monthly. A) $8650.00 B) $8635.47 C) $8688.25 D) $17 270.93 E) $17 323.72 Answer: C Diff: 3 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

169) Debts of $2800.00 due three months from now, $3500 due 15 months and $4600.00 due twenty-one months from now are to be settled by two equal payments due in three months and nine months from now respectively. Determine the size of the equal replacement payments if interest is 5.5% p.a. compounded quarterly. A) $5450.00 B) $5176.03 C) $5246.71 D) $10 352.05 E) $10 900.00 Answer: C Diff: 3 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 170) Three debt payments, the first in the amount of $1600.00 due today, the second in the amount of $7200.00 due in 15 months and the third in the amount of $5000 due in 24 months, are to be settled by two equal payments nine months from now and a final payment in 21 months. Determine the size of the equal payments if the money is worth 12.12% p.a. compounded monthly. A) $6900.00 B) $6415.26 C) $6801.60 D) $12 830.51 E) $13 800.00 Answer: C Diff: 3 Type: MC Page Ref: 368-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 171) How many number of compounding periods are there when compounding quarterly for 876 days? A) 9.0 B) 7.2 C) 2.4 D) 9.6 Answer: D Diff: 1 Type: MC Page Ref: 337-343 Topic: 9.1 Basic Concepts and Computations Objective: 9-1: Calculate interest rates and the number of compounding periods. 172) For a sum of money borrowed at 4.50% p.a. compounded semi-annually for 2.5 years, state the numerical value of the compounding factor. Answer: i = 0.045/2 = 0.0225 = .25%; n = 2.5 * 2 = 5 Compounding factor: (1 + 0.0225)^5 = 1.0225^5 ≈ 1.117678 Diff: 1 Type: SA Page Ref: 337-343 Topic: 9.1 Basic Concepts and Computations Objective: 9-1: Calculate interest rates and the number of compounding periods.

173) For a sum of $2000 borrowed at 3.75% p.a. compounded monthly for 4.5 years, state the future value. Answer: PV = 2000; i = 0.0375/12 = 0.003125 = 0.3125%; n = 4.5 * 12 = 54 FV = 2000*(1 + 0.003125)^54 = 2000(1.183514) = $2367.03 Diff: 1 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 174) A sum of $10 000 earns 11.0% p.a. compounded quarterly for 3 years. After these three years, new policies force the interest rate to be changed to 6.5% p.a. compounded biweekly. What is the value of the deposit one year after the change in interest rate? Answer: First three years: PV = 10 000; i = 0.11/4 = 0.0275 = 2.75%; n = 3*4 = 12 FV = 10 000 * (1 + 0.0275)^12 = 10000 * (1.384784) = 13 847.8377547 Next year: PV = 13847.8377547; i = 0.065/26 = 0.0025 = 0.25%; n = 1 * 26 = 26 FV = 13847.837754*(1 + 0.0025)^26 = 13847.837754(1.067072) = $14 776.65 Diff: 2 Type: SA Page Ref: 344-354 Topic: 9.2 Using the Future Value Formula of a Compound Amount Objective: 9-2: Compute future (maturity) values of investments. 175) Find the numerical value of the discount factor if the interest rate is 8.0% compounded quarterly during a 3 year term. Answer: i = 0.08/4 = 0.02 = 2.0%; n = 3 * 4 = 12 Discount factor: (1 + 0.02)^-12 = 1/1.02^12 ≈ 0.788493 Diff: 1 Type: SA Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 176) What principal amount will approximately result in the future value of $9189.41, given a nine-year term and an interest rate of 1.2% compounding quarterly. A) $8250.00 B) $8250.45 C) $9189.41 D) $10 235.79 Answer: A Diff: 1 Type: MC Page Ref: 356-360 Topic: 9.3 Present Value and Compound Discount Objective: 9-3: Compute present values of future sums of money. 177) Determine the proceeds on a non-interest bearing promissory note for $4750, discounted 36 months before it's maturity date at 4% p.a. compounded semi-annually. Answer: FV = 4750; i = 0.04/2 = 0.02 = 2%; n = (36/12)(2) = 6 PV = 4750/(1 + 0.02)^6 = 4750/1.126162 = $4217.86 Diff: 1 Type: SA Page Ref: 362-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes.

178) Determine the proceeds for a $11 000 promissory note, compounding at 2% p.a. quarterly. The note is issued on October 1, 2017, due on April 1, 2020, and discounted on April 1, 2019 at 13% p.a. compounded biweekly. Answer: Calculating future value PV = 11000; i = 0.02/4 = 0.005 = 0.5%; n = (2.5)(4) = 10 FV= 11000 * (1 + 0.005)^10 = 11000(1.051140) = $11 562.54 Calculating proceeds FV = 11562.54; i = 0.13/26 = 0.005 = 0.5%; n = 1 * 2 = 2 PV = 11562.54/(1 + 0.005)^2 = 11562.54/1.010025 = $11 447.78 Diff: 2 Type: SA Page Ref: 362-366 Topic: 9.4 Application - Discount Negotiable Financial Instruments at Compound Interest Objective: 9-4: Discount long-term promissory notes. 179) Arjun wants to off some of his student debt through payments of $2200 six months from now, $1800 twelve months from now, $1400 eighteen months from now, and $1000 twenty-four months from now. Determine how much Arjun would have to pay if he chose to settle the debt now, if his money compounds 12% p.a. monthly. Answer: i = 0.12/12 = 0.01 = 1%; n1 = 6; n2 = 12; n3 = 18; n4 = 24 PV1 = 2200/(1.01^6) = 2,072.499517559… = $2072.50 PV2 = 1800/(1.01^12) = 1,597.408605477… = $1597.41 PV3 = 1400/(1.01^18) = 1,170.4242398675 = $1170.42 PV4 = 1000/(1.01^24) = 787.56612742372 = $787.57 Total = 2072.50 + 1597.41 + 1179.42 + 787.57 = $5636.90 Diff: 2 Type: SA Page Ref: 369-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values. 180) Zach has to make a tuition payment of $8200 today. However, he has the option to make three equal payments two months, four months, and eight months from now. Determine the size of the equal payments if his money is compounded at 3.6% p.a. monthly. Answer: Let now be the focal date. Let x represent the amount of the equal payment. Let E1, E2 and E3 represent the dated values of the two payments. i = 0.036/12 = 0.003 = 0.3%; n1 = 2; n2 = 4; n3 = 8 E1 + E2 + E3 = 8200 x(1.003^-2) + x(1.003^-4) + x(1.003^-8) = 8200 x(0.99402689… + 0.98808946… + 0.9763207865…) = 8200 x(2.95843714176…) = 8200 x = $2771.73 Diff: 2 Type: SA Page Ref: 369-381 Topic: 9.5 Equivalent Values Objective: 9-5: Solve problems involving equivalent values.

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 10 Compound Interest - Further Topics 1) A loan of $9000.00 was repaid together with interest of $3728.00. If interest was 9.4% compounded quarterly, how long was the loan taken out? (Give answer in years and months to the nearest tenth.) Answer: PV = 9000; FV = 9000 + 3728 = 12 728; i =

= 0.0235; P/Y = C/Y = 4

= 14.92069192 (quarters) ÷ 4 = 3.73017298 years = 3 years 8.8 months

Programmed solution:

Diff: 2 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 2) Calculate the number of years for money to triple at 3.6% p.a. compounded monthly. Answer: PV = 1; FV = 3; i =

= 0.003

= 366.7531281 (months) ÷ 12 = 30.56 years

Programmed solution:

Diff: 2 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

3) In how many days will $770.00 grow to $880.00 at 11.5% p.a. compounded monthly? Answer: PV = 770, FV = 880, i = 0.115 ÷ 12 = 0.0095833 n=

Time in days =

= 14.00037

= 1.166697509 × 365 = 425.85 = 426 days

Diff: 2 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 4) A loan of $4500.00 was repaid together with interest of $1164.00. If interest was 12 .4% compounded quarterly, for how many months was the loan taken out? Answer: PV = 4500, FV = 4500 + 1164 = 5664, i = 0.124 ÷ 4 = 0.031

= 7.535504

Number of months = 7.535507 × 3 = 22 .61 = 23 months Diff: 2 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 5) A 11-year $8000.00 promissory note, with interest at 8.4% compounded monthly, is discounted at 6.5% compounded semi-annually yielding proceeds of $14 631.15. How many months before the due date was the date of discount? Answer: Maturity value: PV = 8000.00; i = 0.084 ÷ 12 = 0.007, n = 11 × 12 = 132 FV = 8000.00(1 + 0.007)132 = 800.00(2 .5112509) = 20090.01 Discount:

PV = 14631.15 i = 0.065 ÷ 2 = 0.0325

= 9.913701

The number of months = 6(9.913701) = 59.48 months Diff: 3 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

6) In how many months will money double at 7.45% compounded semi-annually? Answer: PV = 1, FV = 2, i = 0.0745 ÷ 2 = 0.03725

= 18.95244

Number of months 18.95244 × 6 = 113.71 = 114 months Diff: 2 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 7) In how many years will money double at 6.62% compounded semi-annually? Answer: PV = 1, FV = 2, i = 0.0662 ÷ 2 = 0.0331

The number of years =

= 21.28569

= 10.64 years

Diff: 2 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 8) A five-year, $2000.00 note bearing interest at 10% compounded annually was discounted at 12% compounded semi-annually yielding proceeds of $1900.00. How many months before the due date was the discount date? Answer: PV = 2000, i = 0.10, n = 5 Maturity Value = 2000.00(1.10)5 = 2000.00(1.61051) = 3221.02 Discount: FV = 3221.02, PV = 1900, i = 0.12 ÷ 2 = 0.06

= 9.058762388

The discount date is 9.058762388 × 6 = 54.35 months before the due date. Diff: 3 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

9) In how many years will money triple at 12% compounded semi-annually? Answer: PV = 1, FV = 3, i = 0.12 ÷ 2 = 0.06

= 18.85417668 sa

The number of years =

= 9.43 years

Diff: 2 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 10) A promissory note for $3600.00 dated May 15, 2012, requires an interest payment of $370.00 at maturity. If interest is at 9.6% compounded monthly, determine the due date of the note. Answer: PV = 3600, FV = 3600 + 370 = 3970, i = 0.096 ÷ 12 = 0.008

= 12.27788231 = 1 year + 0.27788231 × 365 days

0.27788231 × 365 = 101.438 ≈ 101 days May 15 + 101 days = 135 + 101 = 236 = August 24, 2012. The due date is August 24, 2012. Diff: 3 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 11) A financial obligation requires the payment of $1000.00 in nine months, and $500.00 in twelve months. When can the obligation be discharged by a single payment of $1700.00 if interest is 12% compounded quarterly? Answer: Let the focal date be now; i = 0.12 ÷ 4 = 0.03; m = 4 Debts = 1000.00(1.03)-3 + 500.00(1.03)-4 = 1359.38 FV = 1700, PV = 1359.38

= 7.56456693 quarters = 1.891141733 years

The discharge date is 1 year 325 days from now. Diff: 3 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

12) A financial obligation requires the payment of $250.00 in eighteen months, $350.00 in thirty months, and $300.00 in fifty-four months. When can the obligation be discharged by a single payment of $800.00 if interest is 5% compounded semi-annually? Answer: Let the focal date be now; i = 0.05 ÷ 2 = 0.025; m = 2 For the obligation: 250.00(1.025)-3 + 350.00(1.025)-5 + 300.00(1.025)-9 250.00(.928599411) + 350.00(.883854288) + 300.00(.800728362) 232.15 + 309.35 + 240.22 = 781.72 For the payment: 800(1.025)-n

= 0.93611429 (semi-annual) ÷ 2 × 12 = 5.62 months Programmed solution:

Diff: 3 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

13) A six-year, $1650.00 note bearing interest at 9.51% compounded annually was discounted at 11.2% compounded semi-annually yielding proceeds of $1916.75. How many months before the due date was the discount date? Answer: PV = 1650, i = 0.0951, n = 6 Maturity Value = 1650.00(1.0951)6 = 1650.00(1.7247362) = 2845.81 Discount: FV = 2845.81, PV = 1916.75, i = 0.112 ÷ 2 = 0.056

= 7.253253

The discount date is 7.253253 × 6 = 43.52 months before the due date. Diff: 3 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 14) Janica owes two debt payments–a payment of $6700 due in twelve months and a payment of $8750 due in twenty-one months. If Janice makes a payment of $6000 now, when should she make a second payment of $7900 if money is worth 11.5% compounded semi-annually? Answer: Let the focal date be now. Today's balance 6700.00(1.0575)-2 + 8750.00(1.0575)-3.5 - 6000.00 6700.00(.8942094) + 8750.00(.822278) - 6000.00 5991.20 + 7194.93 - 6000.00 PV = 7186.13 owing today , FV = 7900 n = ln(7900/7186.13) / ln(1.0575) = ln(1.09933998)/ln(1.0575) n = 0.09470998/0.0559076 = 1.694045 The payment of $7900.00 should be made in 1.694045 × 6 = 10.164 months from now. Diff: 3 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

15) A financial obligation requires payments of $2000.00 today, $2500.00 in three years, and $3000.00 in five years. When can the obligation be discharged by a single payment equal to the sum of the required payments if interest is 6%p.a. compounded monthly? Answer: Let the focal date be now; i =

= 0.005; m = 12

For the obligation: 2000.00 + 2500.00(1.005)-36 + 3000.00(1.005)-60 2000.00 + 2500.00(.835644919) + 3000.00(.741372196) 2000.00 + 2089.11 + 2224.12 = 6313.23 For the payment: (2000 + 2500 + 3000) = 7500 FV = 7500, PV = 6313.23

= 34.53717401 months ÷ 12 = 2.878 years = 2 years 320 days

The obligation can be discharged 2 years 320 days from now. Programmed solution:

Diff: 3 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

16) A financial obligation requires the payment of $1500.00 in nine months, $700.00 in twenty-one months, and $1700.00 in 33 months. When can the obligation be discharged by a single payment of $3900.00 if interest is 8.44% compounded quarterly? Answer: Let the focal date be now; i = 0.0844 ÷ 4 = 0.0211; m = 4 Debts = 1500.00(1.0211)-3 + 700.00(1.0211)-7 + 1700.00(1.0211)-11 = 3364.86 FV = 3900, PV = 3364.86 N=

= 7.06833429 quarters = 1.76708 years

= 1.767 years The discharge date is 1 year 280 days from now. Diff: 3 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 17) Find the equated date at which two payments of $1600.00 due six months ago and $1850.00 due today could be settled by a payment of $4300.00 if interest is 9.48% compounded monthly. Answer: Let the focal date be today. Payments: 1600.00(1.0079)6 + 1850.00 = = 1600.00(1.0483461) + 1850.00 = = 1677.35 + 1850.00 = 3527.35 due today PV = 3527.35, FV = 4300, i = 0.0948 ÷ 12 = 0.0079

= 25.17 months from today

Diff: 3 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

18) Luciano sold a property and is to receive $14 200.00 in nine months, $14 000.00 in 42 months, and $15 500.00 in 57 months. The deal was renegotiated after six months at which time Luciano received a payment of ; he was to receive a further payment of $19 000.00 later. When should Luciano receive the second payment if money is worth 10% compounded quarterly? Answer: Let the focal date be at the time Luciano received $17000; i = 0.10 ÷ 4 = 0.025 He is still to receive: 14200.00(1.025)-1 + 14000.00(1.025)-12 + 15500.00(1.025)-17 - 17000.00 14200.00(.9756098) + 14000.00(.7435559) + 15500.00(.6571951) - 17000.00 13853.66 + 10409.78 + 10186.52 - 17000.00 34449.96 - 17000.00 = 17449.76 FV = 1900, PV = 17449.76

= 3.4464414 quarters

= 3.4464414 × 3 = 10.34 months from the Focal day The second payment is to be received in 10.34 months after the first payment. Diff: 3 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 19) A five-year, $4500.00 promissory note with interest at 7.5% compounded semi-annually was discounted at 8% compounded quarterly yielding proceeds of $6150.00. How many months before the due date was the discount date? Answer: PV = 4500, i = 0.075 ÷ 2 = 0.0375 , n = 5 × 2 = 10 Maturity Value = 4500.00(1.0375)10 = 4500.00(1.4450439) = 6502.70 PV = 6150

= 2.81606

It is due in 2.81606 × 3 = 8.448 = 8.448 months Diff: 3 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

20) At what nominal rate of interest compounded semi-annually will $11 800 earn $6800 interest in six years? Answer: PV = 11 800; FV = 11 800 + 6800 = 18 600; n = 6(2) = 12; P/Y = C/Y = 2 i=

- 1 = 0.038650046 Rate = 0.038650046(2) (100%) = 7.73%

Programmed solution:

Diff: 2 Type: SA Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 21) A principal of $4250.00 compounded monthly amounts to $4800.00 in 6.25 years. What is the nominal annual rate of interest? Answer: PV = 4250.00; FV = 4800.00; n = 6.25 × 12 = 75; m = 12 i=

- 1 = (1.1294118)1/75 - 1 = 1.0016239 - 1 = 0.001623942

The nominal annual rate is 0.001623943 × 12 × 100% = 1.94873% Diff: 2 Type: SA Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 22) A principal of $5000.00 compounded monthly amounts to $6000.00 in 7 years. What is the nominal annual rate of interest? Answer: PV = 5000.00; FV = 6000.00; n = 7 × 12 = 84; m = 12 i=

- 1 = (1.2)1/84 - 1 = 1.002172852 - 1 = 0.002172852

The nominal annual rate is 0.002172852 × 12 × 100% = 2.61% Diff: 2 Type: SA Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 23) At what nominal rate of interest compounded quarterly will $8100 earn $1700.00 interest in six years? Answer: PV = 8100, FV = 8100 + 1700 = 9800 n = 6 × 4 = 24 i=

- 1 = (1.2098765)1/24 - 1 = 1.0079699 - 1 = 0.0079698

The nominal annual rate = 0 .007969854 × 4 × 100% = 3.19% Diff: 2 Type: SA Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest.

24) Calculate the nominal rate of interest compounded semi-annually that will double money in eight years. Answer: PV = 1; FV = 2; n = 8(2) = 16; m = 2 2 = 1(1 + i)16 1 + i = 21/16 1 + i = 1.044273782 i = 0.044273782 i = 4.427% Nominal Rate = 4.427% × 2 = 8.85% p.a. compounded semi-annually Programmed solution:

Diff: 2 Type: SA Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 25) At what nominal rate of interest compounded monthly will $1700.00 earn $500.00 interest in three years? Answer: PV = 1700, FV = 1700 + 500 = 2200, N = 3 × 12 = 36 i=

- 1 = (1.2941176)1/36 - 1 = 1.0071876 - 1 = 0.0071876

The nominal annual rate is 0.71876 × 12 × 100% = 8.62512%. Diff: 2 Type: SA Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 26) At what nominal rate of interest compounded quarterly will $2000.00 earn $400.00 interest in three years? Answer: PV = 2000, FV = 2000 + 400 = 2400, n = 3 × 4 = 12 i=

- 1 = (1.2)1/12 - 1 = 1.01530947 - 1 = 0.01530947

The nominal annual rate is 0.01530947 × 4 × 100% = 6.12%. Diff: 2 Type: SA Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest.

27) What is the nominal rate of interest at which money will triple itself in eight years and six months if compounded semi-annually? Answer: PV = 1, FV = 3 n = 8 × 2 +

= 17

- 1 = (3)1/17 - 1 = 1.0667581 - 1 = 0.0667581

The nominal annual rate is 0.066758117 × 2 × 100% = 13.351623% Diff: 2 Type: SA Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 28) If $1400.00 accumulates to $2350.00 in five years, six months compounded semi-annually, what is the effective rate of interest? Answer: PV = 1400, FV = 2350 n = 5 × 2 +

= 11

- 1 = (1.6785714)1/11 - 1 = 1.0482118 - 1 = 0.0482118

f = (1+ .0482118)2 - 1 = 1.098748 - 1 = 0.098748 × 100% = 9.87% Diff: 2 Type: SA Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest. 29) If the effective rate of interest on an investment is 6.52%, what is the nominal rate of interest compounded monthly? Answer: i = (1+ 0.0652)1/12 - 1 = 1.005277 - 1 = 0.005277 The nominal annual rate is 0.005277 × 12 × 100% = 6.33% Diff: 3 Type: SA Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest.

30) If the effective rate of interest on an investment is 7.34%, what is the nominal rate of interest compounded quarterly? Answer: i = (1+ 0.0734)1/4 - 1 = 1.017865508 - 1 = 0.01786551 The nominal annual rate is 0. 017865508 × 4 × 100% = 7.15% Diff: 2 Type: SA Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest.

31) Calculate the nominal annual rate of interest compounded quarterly that is equivalent to 10% p.a. compounded semi-annually. Answer: i1 =

= 0.05

(1 + i)4 = (1 + 0.05)2 1 + i = (1.05)0.5

1 + i = 1.024695077 i = 0.024695077 × 4 × 100% = 9.878% i = 9.878% Programmed solution:

Diff: 2 Type: SA Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest. 32) What is the nominal rate of interest compounded semi-annually which is equivalent to an effective rate of 5.89%? Answer: i = (1 + 0.0589)(1/2) - 1 = 1.0290287 - 1 = 0.0290287 The nominal rate = 2(0.0290287)(100%) = 5.8057336% Diff: 3 Type: SA Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest.

33) The treasurer of Lynn Lake Credit Union proposes changing the method of compounding interest on premium savings accounts to daily compounding. If the current rate is 4% compounded quarterly, what nominal rate should the treasurer suggest to the Board of Directors to maintain the same effective rate of interest? Answer: Invest $1 for one year: /(1 + 0.01)4 = (1 + i)365 i = 1.01(4/365) - 1 = 1.0001091 - 1 = 0.0001091 Nominal rate = 0.0001091 × 365 = 0.0398215 = 3.98% compounded daily Diff: 3 Type: SA Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest.

34) The accountant of Crystal Credit Union proposes changing the method of compounding interest on premium savings accounts to yearly compounding. If the current rate is 8% compounded quarterly, what nominal rate should the treasurer suggest to the Board of Directors to maintain the same effective rate of interest? Answer: Invest $1 for one year: (1 + 0.02)4 = (1 + i)1 i = (1.02)4- 1 = 1.08243216 - 1 = 0.08243216 = 8.24% compounded yearly Diff: 3 Type: SA Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest.

35) Determine the effective rate of interest corresponding to 6% p.a. compounded monthly. Answer: i =

= 0.5% = 0.005; m = 12

f = (1 + i)m - 1 f = (1 + 0.005)12 - 1

f = 1.061677812 - 1 = 0.061677812 = 6.17% Programmed solution:

Diff: 1 Type: SA Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest. 36) Calculate the nominal rate of interest compounded semi-annually that is equivalent to 8.4% compounded monthly. Answer: For monthly; i =

= 0.007; n = 12

For semi-annually; n = 2 (1 + i)2 = (1 + 0.007)12

1 + i = (1.007)6 1 + i = 1.042741896 i = 0.042741896 = 4.2741%(2) = 8.55% Programmed solution:

Diff: 2 Type: SA Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest.

37) At what nominal rate of interest compounded semi-annually will $5900 earn $7400 interest in sixteen years? Answer: PV = 5900, FV = 5900 + 7400 = 13300, n = 16 × 2 = 32 i = (13300/5900)^1/32 - 1

-1

= (2.2542373)1/32 - 1 = 1.0257257 - 1 = 0.0257257 The nominal annual rate is 0. 0257257 × 2 × 100% = 5.15%. Diff: 2 Type: SA Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 38) At what nominal rate of interest compounded annually will $5000 earn $2500 interest in seven years? Answer: PV = 5000, FV = 5000 + 2500 = 7500, n = 7 i =

-1

= (1.5)1/7 - 1 = 1.059634023 - 1 = 0 .0596340227 The nominal annual rate is 0 .0596340227 × 100% = 5.693%. Diff: 2 Type: SA Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 39) What is the nominal rate of interest compounded semi-annually that is equivalent to an effective rate of 8.25%? Answer: i = (1+ 0.0825)0.5 - 1 = 1.0404326 - 1 = 0 .0404326 The nominal annual rate is 0 .0404326 × 2 × 100% = 8.08652% = 8.09%. Diff: 2 Type: SA Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest.

40) In how many years will money triple at 6.4% compounded monthly? A) 206.53 years B) 17.71 years C) 17.21 years D) 1.48 years E) 2.57 years Answer: C Diff: 2 Type: MC Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

41) In how many years will money double at 8% compounded yearly? A) 0.09 years B) 0.9 years C) 9.01 years D) 90.1 years E) 2 years Answer: C Diff: 2 Type: MC Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 42) How long, in compounding periods, will it take for $3327 to increase by $3799 if you are able to earn 6.2% compounded semi-annually? A) 42.95 B) 24.95 C) 4.56 D) 6.45 E) 4.95 Answer: B Diff: 1 Type: MC Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 43) How long did it take for $1635 to increase to $2310 if the investment earned interest at a rate of 8.08% compounded quarterly (give your final answer in years and months, e.g., 3 years and 7.23 months)? A) 7 years 7.69 months B) 4 years .64 months C) 3 years 4.79 months D) 4 years 3.84 months E) 8 years 1.69 months Answer: D Diff: 2 Type: MC Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 44) How many compounding periods does it take for a non-interest bearing bond of $420 150 to be reduced to $199 989 at a discount rate of 5.44% compounded monthly? A) 146.12 B) 121.12 C) 154.12 D) 184.12 E) 164.12 Answer: E Diff: 2 Type: MC Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

45) Calculate the nominal rate of interest p.a. compounded semi-annually if $5000.00 accumulates to $7002.64 in 78 months. A) 0.8656% B) 0.02625% C) 2.625% D) 0.0525% E) 5.25% Answer: E Diff: 1 Type: MC Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 46) At what nominal rate of interest compounded quarterly will $13 000.00 earn $8407.27 in interest in six years? (nearest hundredth) A) 1.83% B) 2.10% C) 7.33% D) 8.4% E) 18.4% Answer: D Diff: 1 Type: MC Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 47) What nominal rate of interest-compounded quarterly will have to be earned on a savings account for it to grow from $1525 to $1955 over a period of 21 months? A) 14.45% B) 15.45% C) 12.45% D) 3.61% E) 10.45% Answer: A Diff: 1 Type: MC Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 48) What is the quarterly interest rate that you will need to earn in order for an investment of $1635 to grow to be $1748.73 after 2.25 years? A) 12.93% B) 13.93% C) 10.93% D) 8.93% E) 3.00% Answer: E Diff: 2 Type: MC Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest.

49) What is the monthly discount rate expressed as an nominal annual rate for a future value of $182 500 to be reduced to $178 663.58 over a period of 4.25 years? A) 4.06% B) 8.04% C) 6.88% D) 4.88% E) 0.50% Answer: E Diff: 2 Type: MC Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 50) Calculate the effective annual rate for 5% p.a. compounded semi-annually. A) 0.050625 B) 5.0625% C) 0.0255% D) 2.55% E) 15.0625% Answer: B Diff: 1 Type: MC Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest. 51) Calculate the effective annual rate for 10% p.a. compounded quarterly. A) 0.1038% B) 1.038% C) 10.38% D) 10% E) 2.5% Answer: C Diff: 1 Type: MC Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest. 52) You have an interest rate of 8.44% compounded quarterly. What is the equivalent effective annual interest rate? A) 7.81% B) 7.71% C) 8.71% D) 8.91% E) 9.71% Answer: C Diff: 1 Type: MC Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest.

53) Deon has $4000.00 invested at 4.5% compounded semi-annually at her bank. In order to make a comparison with another financial institution she needs to know the effective rate of interest at her bank. What is the effective annual rate of interest? A) 1.12% B) 2.28% C) 9.10% D) 4.55% E) 4.5% Answer: D Diff: 1 Type: MC Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest. 54) According to the World Bank, the population in Georgia is decreasing by 0.83% per year. According to the 2011 consensus, the population of Georgia is 4.5 million. Assuming the population growth rate remains the same; calculate the year in which the population of Georgia will be half the population in 2011? Answer: FV =

i = -0.83%

PV = FV (1 + i)-n ⇒ 2 = 0.9917-n

⇒ ln2 = -n ln 0.9917 ⇒ -n = -83.16 ⇒ n = 83 Year = 2011 + 83 = 2094 Diff: 2 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 55) How long will it take Whitby's population to grow from 141 000 (in 2019) to 200 000, if the annual growth rate is 3%? Answer: PV = 141 000 FV = 200 000 i = 3% per year n = ln(200,000/141,000) / ln(1.03) = 0.3495575/0.029559 = 11.83 years Diff: 1 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

56) How long will it take an investment to double in value if it earns 6% compounded quarterly? Include accrued interest and round the answer to the nearest month A) 11 years 11 months B) 11 years 9 months C) 11 years 8 months D) 3 years 11 months E) 3 years 10 months Answer: C Diff: 2 Type: MC Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 57) BlackBerry took a loan contract which requires a payment of $40 million plus interest two years after the contract's date of issue. The interest rate on the $40 million face value is 9.6% compounded quarterly. Before the maturity date, the original lender sold the contract to a pension fund for $43 million. The sale price was based on a discount rate of 8.5% compounded semi-annually from the date of sale. How many months before the maturity date did the sale take place? Answer: The selling price represents the present value of the loan's maturity value on the date of sale. Therefore, the solution requires two steps: (1) calculate the maturity value of the debt; (2) determine the length of time over which the maturity value was discounted to give a present value of $43 million. n = 4 × 2 = 8 and i =

= 2.4%

FV = $40 (1.024)8 = $48.357 million For discounting the maturity value, i =

= 4.25%

= 2.82

Therefore, the date of sale was 2.82 × 6 = 17 months before the maturity date Diff: 2 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 58) Which is the most attractive of the following interest rates offered on a savings account? A) 2.7% compounded annually B) 2.68% compounded semiannually C) 2.66% compounded quarterly D) 2.64% compounded monthly E) All of them offer the same effective rate. Answer: A Diff: 2 Type: MC Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest.

59) The Brick store credit card quotes a rate of 1.75% per month on the unpaid balance. Calculate the effective rate of interest being charged. A) 21% B) 9.85% C) 10.97% D) 51.64% E) 23.14% Answer: E Diff: 2 Type: MC Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest. 60) Calculate the effective rate of interest, if $10 000 grew to $15 000 in 4 years with quarterly compounding. A) 12.28% B) 10.67% C) 4.06% D) 8.45% E) 0.85% Answer: B Diff: 2 Type: MC Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 61) The Brick store credit card quotes a rate of 1.75% per month on the unpaid balance. When Leons took over Brick's management, they reduced the card's effective rate by 4%. What will be the new periodic rate per month? Answer: Effective rate for 1.75% per month = (1.0175)12 - 1 = 23.14% New effective rate = 23.14% - 4% = 19.14%

⇒ ⇒

0.1914 = (1 + i)12 - 1 (1 + i)12 = 1.1914 1 + i = (1.1914)1/12 i = (1.1914)1/12 - 1

⇒ ⇒ i = 0.0147 = 1.47% Diff: 2 Type: SA Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest.

62) For a given interest rate of 10% compounded quarterly, what is the equivalent nominal rate of interest with monthly compounding? A) 10.381% B) 5.0625% C) 10.125% D) 0.8265% E) 9.918% Answer: E Diff: 2 Type: MC Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest. 63) Sajid invested $20 000 in strip bonds earning him $2536.50 in one year. What is the nominal annual rate of interest compounded monthly? A) 12.68% B) 2.05% C) 4.19% D) 12% E) 1% Answer: D Diff: 2 Type: MC Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 64) Over a 15-year period, Tariq's investment in Cameco stock grew in value from $6000 to $22 847. During the same period, the consumer price index rose from 88.31 to 118.91. What was his real compound annual rate of interest on the stock during this period? Answer: With the CPI up from 88.31 to 118.91, Tariq needed

times as many dollars at the end of

the period to purchase the same goods and services at the beginning. The $3000 value of the stock at the beginning had to grow to $6000 × i=

- 1 = (22,847/8079.04)^(1/15) - 1 = 0.0717610 = 7.18%

Diff: 3 Type: SA Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest.

= $8079.04

65) A loan of $15 000.00 was repaid together with interest of $1089.00. If interest was 5% compounded quarterly, how long was the loan taken out? (Answers are rounded off to years and months) A) 5 years 8 months B) 1 year 11 month C) 1 year 5 months D) 52 years 10 months E) 3 years 7 months Answer: C Diff: 2 Type: MC Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 66) Musa's parents deposited $20 000 in a long-term savings account as a wedding expenditure for their grand daughter at her birth, expecting it to triple by the time she gets married. Calculate the number of years for money to triple at 5% p.a. compounded monthly. A) 22 years B) 22 years 6 months C) 25 years D) 30 years E) 18 years 7 months Answer: A Diff: 2 Type: MC Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 67) An 11-year $5000.00 promissory note, with interest at 4.4% compounded monthly, is discounted at 3.5% compounded semi-annually yielding proceeds of $7651.15. How many months before the due date was the date of discount? A) 20 months B) 47 months C) 116 months D) 76 months E) 132 months Answer: A Diff: 3 Type: MC Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

68) A promissory note for $7200.00 dated May 15, 2016, requires an interest payment of $740.00 at maturity. If interest is at 9.6% compounded monthly, determine the due date of the note. A) 23 July 2016 B) 21 Sep 2016 C) 23 May 2017 D) 24 Aug 2017 E) 11 Oct 2018 Answer: C Diff: 3 Type: MC Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 69) A financial obligation requires the payment of $25 000.00 in 12 months, $35 000.00 in 18 months, and in 30 months. When can the obligation be discharged by a single payment of $83 000.00 if interest is 6.5% compounded semi-annually? A) 6 months B) 5 months C) 4 months D) 9 months E) 11 months Answer: B Diff: 3 Type: MC Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 70) Moriaty owes two debt payments–a payment of $6.7 million due in twelve months and a payment of $8.75 million due in twenty-one months. If Moriaty makes a payment of $7 million now, when should he make a second payment of $7.9 million if money is worth 11.5% compounded semi-annually? A) 26 months B) 35 months C) 14 months D) 19 months E) 21 months Answer: A Diff: 3 Type: MC Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

71) Musa's parents deposited $20 000 in a long-term savings account as a wedding expenditure for their grand daughter at her birth, expecting to triple by the time she gets married at the age of 22. Calculate the rate of return compounded monthly for the savings account. A) 4.17% B) 0.417% C) 0.5% D) 5% E) 0.05% Answer: D Diff: 2 Type: MC Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 72) At what nominal rate of interest compounded monthly will $40 000.00 earn $7200.00 interest in three years? A) 6% B) 0.461% C) 4.879% D) 59% E) 4.61% Answer: A Diff: 2 Type: MC Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 73) What is the nominal rate of interest at which money will be 5 times itself in 20 years and six months if compounded semi-annually? A) 0.04% B) 4% C) 48% D) 8% E) 4.61% Answer: D Diff: 2 Type: MC Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 74) What is the nominal rate of interest at which money will be 10 times itself in 30 years, if compounded monthly? A) 0.006% B) 0.642% C) 6.42% D) 7.7% E) 4.61% Answer: D Diff: 2 Type: MC Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest.

75) Calculate the nominal annual rate of interest compounded quarterly that is equivalent to 15% p.a. compounded semi-annually. A) 7.5% B) 3.682% C) 14.73% D) 7.7% E) 4.61% Answer: C Diff: 2 Type: MC Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest. 76) The PC financial is considering changing the method of compounding interest on line of credit accounts to daily compounding. If the current rate is 5.75% compounded quarterly, what nominal rate should PC Financial propose to maintain the same effective rate of interest? A) 0.576% B) 0.016% C) 1.6% D) 5.719% E) 4.61% Answer: D Diff: 3 Type: MC Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest. 77) PWC recently proposed to BMO to change the method of compounding interest on line of credit accounts to monthly compounding, as part of the marketing campaign. If the current rate is 2.5% compounded quarterly, what nominal rate should the PWC suggest to BMO executives to maintain the same effective rate of interest? A) 0.208% B) 2.495% C) 3.773% D) 2.5% E) 2.75% Answer: B Diff: 3 Type: MC Page Ref: 377-384 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest. 78) In how many days would $7250 accumulate to $11 000 at 5.50% p.a. compounded annually? Answer: Periodic rate of interest: 0.055/1 = 0.055 = 5.5% n = ln(11,000/7250) / ln(1.055) = 0.416893803931787 / 0.0535407669280298 n = 7.786474 years = 2842.06 days Diff: 2 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

79) How many months would it take to triple an amount of money at 9% p.a. monthly? Answer: Periodic rate of interest: 0.09/12 = 0.0075 = 0.75% n = ln(3) / ln(1.0075) = 1.0986122886681 / 0.0074720148387 = 147.03 months Diff: 1 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 80) High-profile movie star Rob Downer Sr. originally was contracted to receive four separate payments for his upcoming film: $1 million in 3 months, $2 million in 6 months, $4 million in a year and $8 million in two years. However, he was able to renegotiate terms of his contract and is now taking a single payment of $13.5 million. At a rate of 6% p.a. compounded monthly, when should Rob receive his payment? Answer: Let the focal date be now. Let n represent the number of compounding periods from the focal date to the equated date. Let E1, E2, E3, and E4 represent the equivalent values of the originally contracted payments. Let E5 represent the equivalent value of the single payment. Periodic rate of interest: 0.06/12 = 0.005 = 0.5% E5 = E1 + E2 + E3 + E4 13,500,000(1.005^-n) = 1,000,000(1.005^-3) + 2,000,000(1.005^-6) + 4,000,000(1.005^-12) + 8,000,000(1.005^-24) 13,500,000(1.005^-n) = 985,148.7593098 + 1,941,036.155939 + 3,767,621.358664 + 7,097,485.351129 13,500,000(1.005^-n) = 13,791,291.625042 n = ln(13,791,291.625042…/13,500,000) / ln(1.005) n = 0.02134766586 / 0.0049875415 = 4.28 months Diff: 3 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates. 81) Diana took out a loan of $600 today, and it must be repaid by payments of $200 in 3 months, $300 in 6 months, and then a payment of $125. If the interest rate is 7% p.a. compounded quarterly, when should the last payment be made? Answer: Let the focal point be now. $200 in first quarter, $300 in second quarter. Periodic interest rate: i = 0.07/4 = 0.0175 = 1.75% 600 = 200(1.0175^-1) + 300(1.0175^-2) + 125(1.0175^-n) 600 = 196.560197 + 289.769332 + 125(1.0175^-n) 113.670472 = 125(1.0175^-n) n = ln(125/113.670462) / ln(1.0175) n = 0.09501015922 / 0.01734863833 = 5.48 quarters Diff: 3 Type: SA Page Ref: 390-397 Topic: 10.1 Determining n and Related Problems Objective: 10-1: Determine the number of conversion periods and solve equated dates.

82) What is the annually compounded rate if $1379 accumulates to $1519 in six years? Answer: i = ((1519/1379)^(1/6)) - 1 i = (1.101522)^(1/6) - 1 i = 1.016246 - 1 i = 0.016246 = 1.62% Diff: 1 Type: SA Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 83) At what nominal interest rate will money quadruple itself over the course of 22 years and 3 months if the interest rate is compounded quarterly? Answer: Number of compounding terms: 22.25 years * 4 = 89 quarters i = 4^(1/89) - 1 i = 1.015698 - 1 i = 0.015698 = 1.569828% Nominal rate = 1.569828% * 4 = 6.279314 % Diff: 2 Type: SA Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 84) Shayan took a student loan of $3400 from OSAP and wanted to pay it off within two years. If he repaid $3800, what was the nominal interest rate if it was compounded semi-annually? Answer: Number of compounding terms: 2 years * 2 = 4 semi-annual terms. i = ((3800/3400)^(1/4)) - 1 i = 1.028196 - 1 i = 0.028196 = 2.819661% Nominal rate = 2.819661% * 2 = 5.639323% Diff: 2 Type: SA Page Ref: 398-401 Topic: 10.2 Calculating i and Related Problems Objective: 10-2: Compute periodic and nominal rates of interest. 85) Niyonella is trying to decide which local bank she should open a investment-linked savings account with. Uno Bank advertises a rate of 2.4% p.a. compounded monthly, while Bank de Deux promotes a rate of 2.6% compounded quarterly. Which bank would offer the higher rate of return on Niyonella's money? Answer: Uno Bank Periodic interest rate: i = 0.024/12 = 0.002 = 0.2%; m = 12 f = (1.002)^12 — 1 = 0.024266… = 2.426577…% Bank de Deux Periodic rate of interest: i = 0.026/4 = 0.0065 = 0.65%; m = 4 f = (1.0065)^4 — 1 = 0.026254… = 2.625460… % Therefore, Bank de Deux offers the better interest rate. Diff: 3 Type: SA Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest.

86) What nominal annual rate compounded daily is equivalent to 5% p.a. compounded semi-annually? Answer: Periodic rate of interest: i = 0.05/2 = 0.025 = 2.5% n=2 (1 + i)^365 = 1.025^2 i = 1.025^(2/365) - 1 i = 1.000135… - 1 i = 0.000135 = 0.013531… % Diff: 2 Type: SA Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest. 87) If the effective rate of interest on an investment is 5.25%, what is the nominal interest rate compounded biweekly? Answer: 0.0525 = (1 + i)^26 - 1 1.0525 = (1 + i)^26 1.001970… = 1 + i i = 0.001970… = 0.196995… % Diff: 2 Type: SA Page Ref: 402-410 Topic: 10.3 Effective and Equivalent Interest Rates Objective: 10-3: Compute effective and equivalent rates of interest.

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 11 Ordinary Simple Annuities 1) Jason deposited $100.00 at the end of each month for five years into an account paying 6% compounded monthly. What will be the balance in the account at the end of the five-year term? Answer: PMT = 100.00; n = 5(12) = 60; i = FV = 100

= 0.5% = 0.005; I/Y = 6; P/Y = C/Y = 12

= $6977.00

Programmed solution:

Diff: 1 Type: SA Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 2) What will deposits of $470.00 made at the end of each month amount to after seven years if interest is 10.8% compounded monthly? Answer: FVn = 470.00

= 470.00(124.7277964) = $58 622.06

Diff: 1 Type: SA Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 3) Find the amount to which semi-annual deposits of $200.00 will grow in four years at 6.6% p.a. compounded semi-annually. Answer: FVn = 200.00

= 200.00(8.9875671) = $1797.51

Diff: 1 Type: SA Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 4) Find the amount to which monthly deposits of $100.00 will grow in five years at 4.8% p.a. compounded monthly. Answer: FVn = 100.00

= 100.00(67.660180) = $6766.02

Diff: 1 Type: SA Page Ref: 394-403 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities.

5) Leanne Simon made ordinary annuity payments of $81.00 per month for fourteen years earning 9% compounded monthly. How much interest is included in the future value of the annuity? Answer: FVn = 81.00

= 81.00(334.5180794) = 27 095.96

Interest is 27095.96 - 81.00(168) = 27095.96 - 13608.00 = $13 487.96 Diff: 1 Type: SA Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 6) The Olfert Contractors, Inc., are saving $958.00 every month in order to purchase a new paving machine in twelve years. Their savings certificates pay 6% p.a. compounded monthly. How much of the maturity value will be interest? Answer: FVn = 958.00

= 958.00(210.1501631) = 201 323.86

Interest = 201323.86 - 958.00(144) = 201323.86 - 137952.00 = $63 371.86 Diff: 1 Type: SA Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 7) Mr. and Mrs. Fox have each contributed $1825.00 per year for the last eight years into RRSP accounts earning 6.93% compounded annually. Suppose they leave their accumulated contributions for another five years in the RRSP at the same rate of interest. a) How much will Mr. and Mrs. Fox have in total in their RRSP accounts? b) How much did the Fox's contribute? c) How much will be interest? Answer: a) FVn = 3650.00

= 3650.00(10.2339355) = $37 354.00

FV = 37354.00(1.0693)5 = 37354.00(1.3979699) = $52 219.77 b) Contribution = 8(3650.00) = $29 200.00 c) Interest = 52219.77 - 29200.00 = $23 019.77 Diff: 2 Type: SA Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities.

8) Mr. Hughes has contributed $4000.00 per year for the last ten years into a RRSP account earning 9.00% compounded annually. Suppose he leaves the accumulated contributions for another five years in the RRSP at the same rate of interest. a) How much will Mr. Hughes have in total in his RRSP account? b) How much did Mr. Hughes contribute? c) How much will be interest? Answer: a) FVn = 4000.00

= 4000.00(15.19292972) = 60 771.72

FV = 60771.72(1.09)5 = 60771.72(1.538623955) = 93 504.82 b) Contribution = 10(4000.00) = $40 000.00 c) Interest = 93504.82 -40000.00 = $53 504.82 Diff: 1 Type: SA Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 9) For the last four years Joe Thiesman has made deposits of $300.00 at the end of every six months earning interest at 5.7% compounded semi-annually. If he leaves the accumulated balance for another ten years at 3.2% compounded quarterly, what will the balance be in Joe's account? Answer: FVn = 300.00

= 300.00(8.8451439) = $2653.54

FV = 2653.54(1.008)40 = 2653.54(1.3753755) = $3649.61 Diff: 2 Type: SA Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 10) A man put aside $5710.00 at the end of every 3 months for seven years. How much will he have six years after the least deposit, if his account earned 5.6% p.a. compounded quarterly? Answer: FVn = 5710.00

= 5710.00(33.9942668) = $194 107.26

FV = 194107.26(1.014)24 = 194107.26(1.396082) = $270 989.65 Diff: 2 Type: SA Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 11) How much interest is included in the accumulated value of $4750 paid at the end of every six months for 4.5 years if the interest rate is 8.5% compounded semi-annually? Answer: FVn = 4750.00

= 4750.00(10.6918204) = $50 786.15

Total paid = 4750.00 × 9 = $42 750.00 Interest = 50786.15 - 42750.00 = $8036.15 Diff: 1 Type: SA Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities.

12) How much interest is included in the accumulated value of $250 paid at the end of every month for 5 years if the interest rate is 4.0% compounded monthly? Answer: FVn = 250.00

= 250.00(66.298971) = 16 574.74

Total paid = 250.00 × 60 = $15 000.00 Interest = 16574.74 - 15000 = $1574.74 Diff: 2 Type: SA Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 13) What is the discounted value of deposits of $150.00 made at the end of each month for fourteen years if interest is 4.5% compounded monthly? Answer: PVn = 150.00

= 150.00(124.4747396) = $18 671.21

Diff: 1 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 14) Find the present value for payments of $500.00 made at the end of each quarter for ten years, if interest is 8% compounded quarterly. Answer: PMT = 500.00; n = 10(4) = 40; I/Y = 8 P/Y = C/Y = 4; i = PV = PMT

= 500

= 2% = 0.02

= $13 677.74

Programmed solution:

Diff: 1 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 15) Find the present value of ordinary semi-annual payments of $810.00 for five years at 5.45% p.a., compounded semi-annually. Answer: PVn = 810.00 = 810.00(8.651179) = $7007.45 Diff: 1 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities.

16) An installment contract for the purchase of a car requires payments of $568.60 at the end of each month for the next three years. Suppose interest is 12.0% p.a. compounded monthly. a) What is the amount financed? b) How much is the interest cost? Answer: a) PVn = 568.60

= 568.60(30.10750504) = $17 119.13

b) Interest = 36(568.60) - 17228.86 = 20469.60 - 17119.13 = $3350.47 Diff: 2 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 17) An installment contract for the purchase of a computer requires payments of $30.00 at the end of each month for the next three years. Suppose interest is 18.00% p.a. compounded monthly. a) What is the amount financed? b) How much is the interest cost? Answer: a) PVn = 30.00

= 30.00(27.66068431) = 829.82

b) Interest = 36(30.00) - 829.82 = 1080.00 - 829.82 = $250.18 Diff: 2 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 18) Randy bought an annuity to pay him $2700.00 at the end of every six months for twenty years. How much of the total annuity payments is interest, if interest is 6.5% p.a. compounded semi-annually? Answer: PVn = 2700.00

= 2700.00(22.2084332) = $59 962.77

Interest = 2700.00(40) - 59962.71 = 108000.00 - 59962.77 = $48 037.23 Diff: 2 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 19) Annette bought a vacation property for $22 900.00 down and quarterly mortgage payments of $1224.51 at the end of each quarter for six years. Interest is 8.4% compounded quarterly. a) What was the purchase price of the property? b) How much interest will Annette pay? Answer: a) PVn = 1224.51

= 1224.51(18.7014044)

= 22 900.06 Purchase price = 22 900.06 + 22900.00 = $45800.06 b) Interest = 24(1224.51) - 22900.06 = 29388.24 - 22900.06 = $6488.18 Diff: 2 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities.

20) If a loan was repaid by ordinary monthly payments of $715.00 in seven years at 9.12%, compounded monthly, how much interest was paid? Answer: PVn = 715

= 715.00(61.9193329) = $44 272.32

Interest: 715.00(84) - 44272.32 = 60060.00 - 44272.32 = $15 787.68 Diff: 1 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 21) If a loan was repaid by ordinary yearly payments of $2000.00 in five years at 9.00%, compounded yearly, how much interest was paid? Answer: PVn = 2000.00

= 2000.00(3.88965126) = $7779.30

Interest: 2000.00(5) -7779.30 = 10000.00 -7779.30 = $2220.70 Diff: 1 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 22) Janice plans to retire in 10 years and would like to receive $3000.00 per month for fifteen years starting at the end of the first month after her retirement. Calculate the amount she must invest now if interest is 7.5% compounded monthly. Answer: Step 1: PMT = 3000.00; n = 15(12) = 180; I/Y = 7.5; P/Y = C/Y = 12; i=

= 0.625% = 0.00625

PV = PMT

= 3000

= $323 620.28

Step 2: FV = 323620.28; n = 10(12) = 120 PV = 323620.28(1.00625)-120 = $153 224.61 Programmed solution:

Diff: 2 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities.

23) You have inherited some money and you want to set some of that money aside for ten years. After ten years, you would like to receive $7600.00 at the end of each 6 months for nine years. If the interest is 6.5% compounded semi-annually, how much of your inheritance must you set aside? Answer: PVn = 7600.00

= 7600.00(13.4672608) = $102 351.18

PV = 102351.18(1.0325)-20 = 102351.18(.5274713) = $53 987.31 Diff: 2 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 24) Planning for their child's college education, David and Carol Roberts opened an account paying 6.36% compounded monthly. If ordinary annuity payments of $700.00 per month are to be paid out of the account for three years starting seven years from now, how much did the Roberts' deposit? Answer: Balance in account 7 years from now. PVn = 700.00

= 700.00(32.6954229) = $22 886.80

PV = 22886.80(1.0053)-84 = 22886.80(.6414498) = $14 680.73 Diff: 2 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 25) Planning for their child's college education, Sean and Jessica opened an account paying 8.00% compounded yearly. If ordinary annuity payments of $15 000.00 per year are to be paid out of the account for three years starting sixteen years from now, how much did the couple deposit? Answer: Balance in account 15 years from now. PVn = 15000.00

= 15000.00(2.577096987) = $38 656.45

PV = 38656.45(1.08)-15 = 38656.45(.315241705) = $12 186.13 Diff: 2 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 26) A father wants to provide an annuity of $3350.00 payable at the end of each 6 months during the three years his son attends college. It will be six years before his son goes to college. What single deposit must he make today that will finance the annuity, if he can invest his money at 7.84% p.a. compounded semiannually? Answer: PVn = 3350.00

= 3350.00(5.255817) = $17 606.99

PV = 17606.99(1.0392)-12 = 17606.99(.6303915) = $11 099.30 Diff: 2 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities.

27) Yankee Construction agreed to lease payments of $762.79 on construction equipment to be made at the end of each month for six years. Financing is at 15% compounded monthly. a) What is the value of the original lease contract? b) If, due to delays, the first eight payments were deferred, how much money would be needed after nine months to bring the lease payments up to date? c) How much money would be required to pay off the lease after nine months? d) If the lease were paid off after nine months, what would the total interest be? e) How much of the total interest would be due to deferring the first eight payments? Answer: a) PVn = 762.79 b) FVn = 762.79 c) PVn = 762.79

= 762.79(47.2924743) = $36 074.23 = 762.79(9.4633742) = $7218.57 = 762.79(43.4234299)

d) Interest: 40341.53 - 36074.23 = $4267.30 e) Payment needed after 9 months = $7218.57 minus Normal payments = 9 × 762.7 = 6865.11 Additional interest = $ 353.46 Diff: 3 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 28) A loan was repaid in 7.25 years by end-of month monthly payments of $472. If interest was 6.12% compounded monthly, how much interest was paid? Answer: PVn = 472.00

= 472.00(70.1212093) = $33 097.21

Amount paid = 472.00 × 87 = $41 064.00 Interest paid = 41064.00 - 33097.21 = $7966.79 Diff: 2 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 29) A car was purchased for $3500.00 down and payments of $575.00 at the end of each month for 5 years. Interest is 9.72% compounded monthly. a) What was the purchase price of the car? b) How much will be the amount of interest paid? Answer: a) PVn = 575.00

= 575.00(47.3719563) = $27 238.87

Purchase price = 27238.87+ 3500.00 = $30 738.87 b) Interest = 60(575.00) - 17764.48 = 34500.00 - 27238.87 = $7261.13 Diff: 2 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities.

30) A trust fund is set up to make payments of $1320.00 at the end of each month for seven years. Interest on the fund is 7.28% compounded monthly. a) How much money must be deposited into the fund? b) How much will be paid out of the fund? c) How much interest is earned by the fund? Answer: a) PVn = 1320.00 = 1320.00(65.6601227) = $86 671.36 b) Payout = 84(1320.00) = $110 880.00 c) Interest = 110880.00 - 86671.36 = $24 208.64 Diff: 2 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 31) If a loan was repaid by quarterly payments of $715.00 over 7 years at 8.36% compounded quarterly, how much money had been borrowed? Answer: PVn = 715.00

= 715.00(21.0352112) = $15 040.18

Diff: 1 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 32) Doris purchased a piano with $1300.00 down and monthly payments of $214.00 for two years at 9.72% compounded monthly. What was the purchase price of the piano? Answer: PVn = 214.00

= 214.00(21.7316663) = $4650.58

Purchase price = 4650.58 + 1300.00 = $5950.58 Diff: 1 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 33) Kam expects to retire in 22 years. Beginning one month after his retirement he would like to receive $590.00 per month for 25 years. How much must he deposit into a fund today to be able to do so if the rate of interest on the deposit is 6.12% compounded monthly? Answer: Amount needed at retirement PVn = 590.00 = 590.00(153.4550473) = $90 538.48 PV = 90538.48(1.0051)-264 = 90538.48(.261067) = $23 636.61

Diff: 2 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities.

34) What is the size of end-of-month monthly deposits which will accumulate to $122 200.00 after six years if interest is 5.7% compounded monthly? Answer: 122200.00 = PMT 122200.00 = PMT(85.6044029) $1427.50 = PMT The size of the monthly deposit is $1427.50 Diff: 1 Type: SA Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities. 35) What payment is required at the end of each month for 3 years to repay a loan of $8000.00 at 6.4% compounded monthly? Answer: 8000.00 = PMT 8000.00 = PMT(32.6760126) $244.83 = PMT Diff: 1 Type: SA Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities. 36) What payment is required at the end of each quarter for 8 years to repay a loan of $20 000.00 at 6.0% compounded quarterly? Answer: 20000.00 = PMT 20000.00 = PMT(25.26713874) $791.54 = PMT Diff: 1 Type: SA Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities. 37) What payment made at the end of each month for 13 years will amount to $17 520.00 at 9.66% compounded monthly? Answer: 17520.00 = PMT 17520.00 = PMT(309.6943889) $56.57 = PMT Diff: 1 Type: SA Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities.

38) The Chretiens bought a rental property valued at $125 000.00 by paying 25% down and mortgaging the balance over 21.5 years through equal payments at the end of each quarter at 8.6% compounded quarterly. What was the size of the quarterly payments? Answer: The mortgaged balance is 125 000 × (1 - 0.25) = 93 750 93750.00 = PMT 93750.00 = PMT(39.0460809) $2401.35 = PMT Diff: 2 Type: SA Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities. 39) The Savoias bought an investment property valued at $160 000.00 by paying 25% down and mortgaging the balance over 25 years through equal monthly payments at 6% compounded monthly. What was the size of the monthly payments? Answer: The mortgaged balance is 160 000 × (1-0.25) = 120 000 120000.00 = PMT 120000.00 = PMT(155.206864) $773.16 = PMT Diff: 2 Type: SA Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities. 40) What is the size of semi-annual deposits that will accumulate to $113 200.00 after 9.5 years at 6.5% compounded semi-annually? Answer: FV = 113 200, i = 0.065 ÷ 2 = 0.0325, n = 9.5 × 2 = 19 113200.00 = PMT 113200.00 = PMT(25.7280812 $4399.86 = PMT Diff: 1 Type: SA Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities.

41) A $5000.00 loan requires payments at the end of each quarter for five years. If the interest rate on the loan is 8% compounded quarterly, calculate the size of each payment. Answer: PV = 5000.00; n = 5(4) = 20; i =

= 2% = 0.02 ; I/Y = 8; P/Y = C/Y = 4

PV = PMT PV = PMT 5000 = PMT[16.35143334] PMT =

= $305.78

Programmed solution:

Diff: 1 Type: SA Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities. 42) A $15 000.00 loan requires payments at the end of each month for five years. If the interest rate on the loan is 12% compounded monthly, calculate the size of each payment. Answer: PV = 15000.00; n = 5(12) = 60; i =

= 1% = 0.01

PV = PMT 15000 = PMT 15000 = PMT[44.95503841] PMT =

= $333.67

Diff: 1 Type: SA Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities.

43) What deposit made at the end of each quarter will accumulate to $2510.00 in four years at 4% compounded quarterly? Answer: FV = 2510.00; n = 4(4) = 16; i =

= 1% = 0.01; I/Y = 4; P/Y = C/Y = 4

FV = PMT 2510 = PMT 2510 = PMT[17.25786449] PMT = $145.44 Diff: 1 Type: SA Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities. 44) What is the term of a mortgage of $235 000.00 repaid by monthly payments of $2475.00 if interest is 7.55% compounded monthly? Answer: PV = 235000, PMT = 2475, i = 0.0755 ÷ 12 = 0.0062917 235000 = 2475

n=-

= 145.056 = 145 ( months)

Diff: 2 Type: SA Page Ref: 448-452 Topic: 11.5 Determining the Term n of an Annuity Objective: 11-5: Compute the number of periods for ordinary simple annuities. 45) How many quarterly payments will it take for $400.00 deposited at the end of each quarter to amount to at 6% compounded quarterly? Answer: FV = 11453.40; PMT = 400.00; i =

= 1.5% = 0.015 ; I/Y = 6; P/Y = C/Y = 4

11453.40 = 400 0.4295025 = 1.015n - 1 1.4295025 = 1.015n

n ln 1.015 = ln 1.4295025 n = 24 (quarters) Programmed solution:

Diff: 2 Type: SA Page Ref: 448-452 Topic: 11.5 Determining the Term n of an Annuity Objective: 11-5: Compute the number of periods for ordinary simple annuities.

46) How many semi-annually payments will it take for $500.00 deposited at the end of each half year to amount to $10 000.00 at 6% compounded semi-annually? Answer: FV = 10000.00; PMT = 500.00; i =

= 3.0% = 0.03

10000.00 = 500 0.6 = 1.03n - 1 1.6 = 1.03n

n ln 1.03 = ln 1.6 n = 15.9 (payments) ≈ 16 payments Diff: 2 Type: SA Page Ref: 448-452 Topic: 11.5 Determining the Term n of an Annuity Objective: 11-5: Compute the number of periods for ordinary simple annuities. 47) How many monthly payments will it take for $1000.00 deposited at the end of each month to amount to at 9% compounded monthly? Answer: FV = 50000.00; PMT = 1000.00; i =

= 0.75% = 0.0075

50000.00 = 1000 0.375 = 1.0075n - 1 1.375 = 1.0075n

n ln 1.0075 = ln 1.375 n = 42.62 (months) ≈ 43 months Diff: 2 Type: SA Page Ref: 448-452 Topic: 11.5 Determining the Term n of an Annuity Objective: 11-5: Compute the number of periods for ordinary simple annuities. 48) Suppose $726.56 is deposited at the end of every six months into an account earning 6.45% compounded semi-annually. If the balance in the account four years after the last deposit is to be $31 300.00, how many deposits are needed? Answer: At the moment of last deposit: FV = 31300, n = 4 × 2 = 8, i = 0.0645 ÷ 2 = 0.03225 PV = 31300.00(1.03225)-8 = 31300.00(0.7757484) = $24280.92 = FVg Annuity: FVg = 24280.92, PMT = 726.56, i = 0.03225

= 23.04 ≈ 23 deposits

Diff: 3 Type: SA Page Ref: 448-452 Topic: 11.5 Determining the Term n of an Annuity Objective: 11-5: Compute the number of periods for ordinary simple annuities.

49) A loan of $19 850.00 is to be repaid in quarterly payments of $620.00. How many payments are required to repay the loan at 6.5% compounded quarterly? Answer: PV = 19 850, PMT 620, i = 0.065 ÷ 4 = 0.01625

n=-

= 45.56 ≈ 46 payments

The required number of payments is 41 Diff: 2 Type: SA Page Ref: 448-452 Topic: 11.5 Determining the Term n of an Annuity Objective: 11-5: Compute the number of periods for ordinary simple annuities. 50) How many yearly payments will it take for $25 000.00 deposited at the end of each year to amount to at 9% compounded yearly? Answer: FV = 1 000 000.00; PMT = 25000.00; i =

= 9% = 0.09

1 000 000.00 = 25000 3.6 = 1.09n - 1 4.6 = 1.09n

n ln 1.09 = ln 4.6 n = 17.71 (years) ≈ 18 years Diff: 2 Type: SA Page Ref: 448-452 Topic: 11.5 Determining the Term n of an Annuity Objective: 11-5: Compute the number of periods for ordinary simple annuities. 51) $1000.00 is deposited at the end of every month into an account earning 6.00% compounded monthly. If the balance in the account five years after the last deposit is to be $50 000.00, how many deposits are needed? Answer: The balance at the moment of last deposit: FV = 50 000, i = 0.06 ÷ 12 = 0.005, n = 5 × 12 = 60 PV = 50000.00(1.005)-60 = 50000.00(0.7413722) = $37 068.61 For annuity: FVg = 37068.61, PMT = 1000, i = 0.005

= 34.09 ... ≈ 34 deposits

Diff: 3 Type: SA Page Ref: 448-452 Topic: 11.5 Determining the Term n of an Annuity Objective: 11-5: Compute the number of periods for ordinary simple annuities.

52) In what period of time could you pay back a loan of $13 900.00 by making monthly payments of $296.00 if interest is 6.5% compounded monthly? Answer: n = -

= 54.33 months

Diff: 2 Type: SA Page Ref: 448-452 Topic: 11.5 Determining the Term n of an Annuity Objective: 11-5: Compute the number of periods for ordinary simple annuities. 53) What is the nominal rate of interest on a loan of $6800.00 repaid in semi-annual installments of $1175.00 in six years? Answer: PVn = 6800.00; PMT = 1175.00; n = 12

The nominal annual rate = 2(0.135)(100%) = 27% Diff: 2 Type: SA Page Ref: 452-453 Topic: 11.6 Finding the Periodic Rate of Interest i Using Preprogrammed Financial Calculators Objective: 11-6: Compute the interest rate for ordinary simple annuities. 54) Compute the nominal annual rate of interest (compounded monthly) at which $200.00 deposited at the end of each month for ten years will amount to $30 000.00. Answer: FV = 30 000.00; PMT = 200.00; n = 10(12) = 120; P/Y = C/Y = 12 Programmed solution:

Diff: 2 Type: SA Page Ref: 452-453 Topic: 11.6 Finding the Periodic Rate of Interest i Using Preprogrammed Financial Calculators Objective: 11-6: Compute the interest rate for ordinary simple annuities.

55) Note: The calculations for this question were done using Excel's RATE function. What nominal rate of interest is paid on quarterly RRSP contributions of $573.00 made for 17.75 years if the balance just after the last contribution is $106 000.00? Answer: Nper 71 Pmt -573.00 FV 106000.00 Result 0.0242781 The nominal rate is 2.42781% × 4 = 9.71124% Diff: 2 Type: SA Page Ref: 452-453 Topic: 11.6 Finding the Periodic Rate of Interest i Using Preprogrammed Financial Calculators Objective: 11-6: Compute the interest rate for ordinary simple annuities. 56) What is the nominal rate of interest if a six-year loan of $71 000.00 is repaid by monthly payments of $1701.58? Answer:

The nominal rate is 12(0.0166833)(100%) = 20.02% Diff: 2 Type: SA Page Ref: 452-453 Topic: 11.6 Finding the Periodic Rate of Interest i Using Preprogrammed Financial Calculators Objective: 11-6: Compute the interest rate for ordinary simple annuities. 57) Note: The calculations for this question were done using Excel's RATE function. A property worth $191 000.00 can be purchased for 15% down and quarterly mortgage payments of $1972.00 for 30 years. What effective rate of interest is charged? Answer: Nper 120 Pmt -1972.00 PV 191000.00 * (1 - .15) Result 0.0066885 f = (1 + 0.0066885)4 - 1 = 0.0270236 = 2.70236% Diff: 3 Type: SA Page Ref: 452-453 Topic: 11.6 Finding the Periodic Rate of Interest i Using Preprogrammed Financial Calculators Objective: 11-6: Compute the interest rate for ordinary simple annuities.

58) Bonny converted an RRSP balance of $176 875.67 into an RRIF that will pay her $2720.00 at the end of every month for 8 years. What is the nominal rate of interest? Answer: PVn = 176 875.67, PMT = 2 720, n = 8(12) = 96 Calculator answer: i = 0.0086532 The nominal rate = 12 (0.0086532)(100%) = 10.38384% Diff: 2 Type: SA Page Ref: 452-453 Topic: 11.6 Finding the Periodic Rate of Interest i Using Preprogrammed Financial Calculators Objective: 11-6: Compute the interest rate for ordinary simple annuities.

59) What nominal rate of interest is earned by monthly deposits of $560.00 made for 6 years if the balance just after the last deposit is $52 700.00? Answer: FVn = 52 700, PMT = 560, N = 6(12) = 72 Calculator answer: i = 0.007252 The nominal rate = 12(0.007252)(100%) = 8.7024% Diff: 2 Type: SA Page Ref: 452-453 Topic: 11.6 Finding the Periodic Rate of Interest i Using Preprogrammed Financial Calculators Objective: 11-6: Compute the interest rate for ordinary simple annuities.

60) Note: The calculations for this question were done using Excel's RATE function. What is the effective annual rate of interest on a loan of $9200.00 repaid in semi-annual installments of $810.00 in nine years? Answer: Nper 18 Pmt -810.00 PV 9200.00 Result 0.0537101 f = (1 + 0.0537101)2 - 1 = 11.0304975% Diff: 3 Type: SA Page Ref: 452-453 Topic: 11.6 Finding the Periodic Rate of Interest i Using Preprogrammed Financial Calculators Objective: 11-6: Compute the interest rate for ordinary simple annuities.

61) An annuity with periodic payments made at the end of each payment period is called: A) ordinary B) simple C) annuity due D) general E) none of the above Answer: A Diff: 1 Type: MC Page Ref: 418-420 Topic: 11.1 Introduction to Annuities Objective: 11-1: Distinguish between types of annuities based on term, payment date, and conversion period. 62) When the conversion period is different from the payment period, an annuity is called: A) ordinary B) certain C) deferred D) general E) different Answer: D Diff: 1 Type: MC Page Ref: 418-420 Topic: 11.1 Introduction to Annuities Objective: 11-1: Distinguish between types of annuities based on term, payment date, and conversion period.

63) An annuity due is an annuity for which: A) the payments are made at the beginning of each payment period B) the payments are made at the end of each payment period C) the payment period is not the same as the conversion period D) the payments are made to repay a loan E) the payments continue forever Answer: A Diff: 1 Type: MC Page Ref: 418-420 Topic: 11.1 Introduction to Annuities Objective: 11-1: Distinguish between types of annuities based on term, payment date, and conversion period. 64) The term of annuity is: A) the interval between the first and the last payments B) the length of the payment interval C) the length of time from the beginning of the first payment interval to the end of the last payment interval D) the length of the conversion interval E) the length of time between the successive payments Answer: C Diff: 1 Type: MC Page Ref: 418-420 Topic: 11.1 Introduction to Annuities Objective: 11-1: Distinguish between types of annuities based on term, payment date, and conversion period. 65) Regular contributions of $200 are made at the end of each month for five years into a savings account earning interest at 4% compounded quarterly. The annuity can be classified as: A) ordinary general annuity B) ordinary annuity due C) ordinary simple annuity D) perpetuity E) simple general annuity Answer: A Diff: 1 Type: MC Page Ref: 418-420 Topic: 11.1 Introduction to Annuities Objective: 11-1: Distinguish between types of annuities based on term, payment date, and conversion period.

66) Calculate the accumulated value of quarterly payments of $100.00 made at the end of each quarter for ten years just after the last payment has been made if interest is 8% p.a. compounded quarterly. A) $4003.91 B) $6040.20 C) $2040.00 D) $4415.88 E) $2015.88 Answer: B Diff: 1 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 67) Calculate the accumulated value of monthly payments of $100.00 made at the end of each month for ten years just after the last payment has been made if interest is 12% p.a. compounded monthly. A) $23 003.87 B) $33 003.87 C) $12 000 D) $1 046.22 E) $330 003.87 Answer: A Diff: 1 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 68) What is the future value of an annuity with monthly deposits of $277 for a period of 11.5 years at an interest rate of 7.44% compounded monthly? The deposits are made at the end of the month. A) $60 616.68 B) $60 161.68 C) $60 186.68 D) $60 611.68 E) $60 001.68 Answer: B Diff: 1 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 69) You want to save $710 per quarter for 15.5 years towards the purchase of a motor home. You think that you can earn 6.56% compounded quarterly for this period of time. If your first deposit is in 3 months, what is the most expensive motor home that you will be able to purchase? A) $44 020.00 B) $75 999.79 C) $75 399.79 D) $27 501.82 E) $79 999.79 Answer: C Diff: 1 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities.

70) You are offered payments of $475 at the end of each semi-annual period for 6.5 years. You think that the cost of money is 6.14% compounded semi-annually. What is the present cash value? A) $5092.06 B) $5029.06 C) $5209.06 D) $5292.06 E) $5290.06 Answer: B Diff: 1 Type: MC Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 71) Calculate the present value of five payments of $1500.00 made at the end of each of five consecutive years respectively if money is worth 6% compounded annually. A) $7950.00 B) $7486.52 C) $7050.00 D) $6318.55 E) $7486.00 Answer: D Diff: 1 Type: MC Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 72) You want to be able to travel and withdraw $3750 at the end of every month during your yearlong trip. During your trip your travel fund will earn 6.24% compounded monthly. What is the size of your travel fund if you want to be able to start your trip in 4 years and 7 months? A) $32 715.58 B) $45 315.20 C) $2849.74 D) $46 309.57 E) $45 555.20 Answer: A Diff: 2 Type: MC Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities.

73) Calculate the present value of 120 payments of $50.00 made at the end of each of 120 consecutive months respectively if money is worth 12% compounded monthly. A) $3485.03 B) $3485.02 C) $6000.00 D) $3000.00 E) $34 850.30 Answer: A Diff: 1 Type: MC Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 74) You currently have $4541 saved towards the purchase of a home theatre system. You want to be able to buy a system for $6225 plus GST in 4 years. Your money is earning 4.84% compounded monthly. What is the size of your monthly deposit? A) $12.80 B) $28.80 C) $22.80 D) $13.55 E) $20.80 Answer: D Diff: 2 Type: MC Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities. 75) You currently have $4000 saved towards the purchase of a new car. You want to be able to buy a car for $15000 in 4 years. Your money is earning 5% compounded yearly. What is the size of your yearly deposit? A) $928.05 B) $2320.12 C) $2552.13 D) $2352.13 E) $2000.99 Answer: D Diff: 2 Type: MC Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities.

76) How long will it take for quarterly deposits of $425 to accumulate to be $16 440 at an interest rate of 8.48% compounded quarterly? (Final answer in years and months e.g., 7 years and 4.24 months) A) 7 years 1.64 months B) 7 years .136 months C) 7 years 6.14 months D) 28 years .55 months E) 28 years .64 months Answer: A Diff: 2 Type: MC Page Ref: 448-452 Topic: 11.5 Determining the Term n of an Annuity Objective: 11-5: Compute the number of periods for ordinary simple annuities. 77) Sofia plans to save for next two years. She plans to put away $1000 every 6 months starting 6 months from now. What will be the value of her savings after two years if her savings accounts earns at a rate of 8% compounded semi-annually? A) $4000 B) $4246.46 C) $2080 D) $6414.51 E) $6632.98 Answer: B Diff: 2 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 78) Charter has been contributing $900 at the end of each month for the past 15 months to a savings plan that earns 4% compounded monthly. What amount will he have one year from now, if he continues with the plan? A) $25 382.84 B) $13 819.60 C) $6898.05 D) $12 669.42 E) $13 500 Answer: A Diff: 2 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities.

79) Salem plans to deposit $2200 every 6 months for 15 years to save for his son's higher education. The rate of return will be 4% compounded semi-annually for the first 5 years and 8% compounded semiannually for the subsequent 10 years. Calculate the future value of this simple annuity. A) $24 089.39 B) $52 782.81 C) $53 454.21 D) $106 237.02 E) $118 293.81 Answer: E Diff: 2 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 80) Salem plans to deposit $2200 every 6 months for 15 years to save for his son's higher education. The rate of return will be 4% compounded semi-annually for the first 5 years and 8% compounded semiannually for the subsequent 10 years. Calculate the future value of this simple annuity, if he stops payment after 5 years. A) $24 089.39 B) $52 782.81 C) $53 454.21 D) $106 237.02 E) $118 293.81 Answer: B Diff: 2 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 81) Don has just turned 43 and has accumulated $34 500 in his RRSP. He makes month-end contributions of $300 to the plan and intends to do so until he retires at the age of 60. The RRSP will be allowed to continue to accumulate until he reaches the age of 65. If the RRSP earn 8% compounded monthly for the next 22 years, what amount will his RRSP contain, when he reaches age 65? A) $129 539.17 B) $192 993.38 C) $227 493.81 D) $199 361.27 E) $392 354.65 Answer: E Diff: 2 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities.

82) Nuclear power plants are required to have full decommissioning costs accumulated during the useful life of the plant. The new nuclear plant at Darlington is designed for 60 years of operational life starting in 2025. OPG plans to set aside $1 million per month during the operation of the plant. What is the value of this annuity in 2025, if OPG uses Ontario's recommended discount rate of 2.5% compounded monthly? A) $373 million B) $1668 million C) $720 million D) $67 million E) $158 million Answer: A Diff: 2 Type: MC Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 83) Nuclear power plants are required to have full decommissioning costs accumulated during the useful life of the plant. The new nuclear plant at Darlington is designed for 60 years of operational life starting in 2025. OPG plans to set aside $1 million per month in a sinking fund for the operational of the plant. What is the value of the value of this sinking fund at the end of the plant life, if OPG expects an average rate of return of 2.5% compounded monthly? A) $373 million B) $1668 million C) $720 million D) $67 million E) $158 million Answer: B Diff: 2 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 84) How much will it cost Shaun to purchase a two-level retirement annuity that will pay $6000 at the end of each month for the first 10 years, and $9000 per month for the next 15 years? The payments represent a rate of return to Shaun of 6% compounded monthly. A) $1 066 532 B) $586 201 C) $540 441 D) $1 126 642 E) $2 193 174 Answer: D Diff: 2 Type: MC Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities.

85) Ali bought Kia Forte 5 from Durham Kia for $20 078.00 on finance agreement with BMO at 5.99% compounded monthly. Calculate the monthly payment on the loan, if the financing period is 60 months. A) $388.06 B) $334.63 C) $451.15 D) $336.30 E) $386.13 Answer: A Diff: 2 Type: MC Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities. 86) Pat won a lottery with two options: $1 million per year, at the beginning of the year, for 25 years or a single cash price of $18 million. If low-risk investments can earn 2.5% compounded annually, which option should Pat choose and what is the advantage in terms of current economic value? A) Lump sum has an advantage of $885 000 B) 25 year annuity has an advantage of $885 000 C) 25 year annuity has an advantage of $424 000 D) Lump sum has an advantage of $424 000 E) Both options are the same Answer: E Diff: 2 Type: MC Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities. 87) On his retirement, Louis received a lump sum of $800 000 from his employer. Taking advantage of existing tax legislation, he invested his money in an annuity that provides for payments of $60 000 at the end of every 3 months. If interest is 6.25% compounded quarterly, how long will the annuity exist? Answer: PV = $800 000; PMT = $60,000; i =

= 1.5625%

800000 = 60000 ⇒ 1 - 1.015625-n = 0.2083 ⇒ 1.015625-n = 0.7917

⇒ -n ln 1.015625 = ln 0.7917 ⇒ -n = ⇒ n = 15 quarters ⇒ n = 3 years 9 months Diff: 2 Type: SA Page Ref: 452-453 Topic: 11.6 Finding the Periodic Rate of Interest i Using Preprogrammed Financial Calculators Objective: 11-6: Compute the interest rate for ordinary simple annuities.

88) Jane borrowed $20 000 to help finance her BBA at UOIT with no interest until she finishes her education. She starts a job with a local bank immediately after finishing her BBA and contracts to pay her loan in monthly payments of $300 each. If the payments are due at the end of each month and the interest is compounded 4% monthly, how long will Jane have to make monthly payments? Answer: PV = $20,000; PMT = $300; i =

= 0.333%

20000 = 300 ⇒ 1 - 1.00333-n = 0.222 ⇒ 1.00333-n = 0.778

⇒ -n ln 1.00333 = ln 0.778 ⇒ -n = ⇒ n = 76 months ⇒ n = 6 years 4 months Diff: 2 Type: SA Page Ref: 448-452 Topic: 11.5 Determining the Term n of an Annuity Objective: 11-5: Compute the number of periods for ordinary simple annuities. 89) A loan of $20 000 is paid over 5 years by monthly payments of $388. What is the nominal rate of interest on the loan? A) 7.5% B) 0.625% C) 0.089% D) 6.14% E) 2.47% Answer: D Diff: 2 Type: MC Page Ref: 452-453 Topic: 11.6 Finding the Periodic Rate of Interest i Using Preprogrammed Financial Calculators Objective: 11-6: Compute the interest rate for ordinary simple annuities. 90) Sam got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded monthly. If he puts the money in his savings account at the end of each month, what will be the balance in the account at the end of the three-year term? A) $18 600.50 B) $18 408.04 C) $18 603.37 D) $27 284.81 E) $17 390.19 Answer: C Diff: 1 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities.

91) Sam got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded monthly. If he puts the money in his savings account at the end of each month, how much interest is included in the future value of the annuity? A) $18 000.00 B) $18 408.04 C) $18 603.37 D) $408.04 E) $603.37 Answer: E Diff: 1 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 92) Jordan is considering building a nuclear power plant, which will be ready for production in 2030. The country's governing body is also considering a decommissioning liability law for the operator to put aside $1 million every month towards decommissioning cost. If the production life of the plant is 60 years and the operator puts the money at the end of the month in a savings account, earning 7.25% compounded monthly, what will be the value of the decommissioning fund after 60 years of production? A) $12.49 billion B) $10.87 billion C) $720 million D) $11.88 billion E) $917.65 million Answer: A Diff: 2 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 93) Jordan is considering building a nuclear power plant, which will be ready for production in 2030. The country's governing body is also considering a decommissioning liability law for the operator to put aside $1 million every month towards decommissioning cost, when the plant is producing electricity. During the 60 years of production life of the plant, the operator will put $1 million at the end of the month in a savings account, earning 7.25% compounded monthly. At the end of the production life of the plant, there are no more contributions and the money is expected to grow at the rate of 6% compounded quarterly for the next 30 years. What will be the value of the decommissioning fund in 2120? A) $12.49 billion B) $74.57 billion C) $75.23 billion D) $70.94 billion E) $60.90 billion Answer: B Diff: 3 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities.

94) Jordan is considering building a nuclear power plant, which will be ready for production in 2030. The country's governing body is also considering a decommissioning liability law for the operator to put aside $1 million every month towards decommissioning cost, when the plant is producing electricity. During the 60 years of production life of the plant, the operator will put $1 million at the end of the month in a savings account, earning 7.25% compounded monthly. At the end of the production life of the plant, there are no more contributions and the money is expected to grow at the rate of 6% compounded quarterly for the next 30 years. How much interest is included in the future value in 2120? A) $12.49 billion B) $11.77 billion C) $74.57 billion D) $73.85 billion E) $70.27 billion Answer: C Diff: 3 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 95) Woods has contributed $3600.00 per year for the last ten years into a RRSP account earning 9% compounded annually. His wife Karen, on the other hand, contributed $300 at the end of every month for the same period into a RRSP account earning 9% compounded monthly. Who earns more interest at the end of 10 years and by how much? A) Wood by $3360 B) Karen by $3360 C) Wood by $54 695 D) Karen by $58 054 E) Karen by $22 054 Answer: B Diff: 2 Type: MC Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 96) Sam got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded monthly. If he puts the money in his savings account at the end of each month, what will be the present value of the balance in the account at the end of the three-year term? A) $18 000.00 B) $18 408.04 C) $18 603.37 D) $27 284.81 E) $17 390.19 Answer: E Diff: 1 Type: MC Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities.

97) Sam got a job at the Brick. He immediately finances a car paying $500 every month for 3 years. The interest is 2.25% compounded monthly. How much is the interest cost of the financing? A) $17 390.19 B) $408.04 C) $603.37 D) $9284.81 E) $609.81 Answer: E Diff: 2 Type: MC Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 98) Keith bought a lakeside cottage for 30% down and monthly mortgage payments of $1224.51 at the end of each month for 20 years. Interest is 4.4% compounded monthly. What is the purchase price of the property? A) $194 378 B) $195 214 C) $278 878 D) $469 885 E) $671 265 Answer: C Diff: 2 Type: MC Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 99) Siri plans to retire now and would like to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. Calculate the amount he must invest now if interest is 4.9% compounded monthly. A) $1 528 015 B) $594 425 C) $633 665 D) $574 626 E) $539 042 Answer: A Diff: 2 Type: MC Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities.

100) Knowing that the education costs for her kids would be huge, Roberta puts lump sum money from her inheritance in a savings account paying an interest of 4.69% compounded monthly. If ordinary annuity payments of $1500.00 per month are to be paid out of the account at the end of each month for four years, how much did Roberta deposit? A) $65 533 B) $32 474 C) $65 592 D) $32 723 E) $54 343 Answer: A Diff: 2 Type: MC Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 101) Siri plans to retire in 18 years and would like to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. Calculate the amount he must invest now every month (at the end of the month) if interest is 4.9% compounded monthly. A) $1 528 015 B) $4420.74 C) $633 665 D) $10 660 E) $11 755 Answer: B Diff: 2 Type: MC Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities. 102) Jamal plans to retire in 17 years. He is saving $2000 every month in a retirement savings account paying him a long-term interest of 9% compounded monthly. What will be the size of his payments per month from the ordinary simple annuity for 20 years following his retirement? A) $957 836 B) $1434 C) $8618 D) $333 E) $2000 Answer: C Diff: 1 Type: MC Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities.

103) Siri plans to retire when her simple annuity savings account has enough money to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. She starts saving $4420 per month. Calculate when should Siri retire from today if her savings account pays 4.9% compounded monthly. A) 18 years B) 216 years C) 2.41 year D) 1.41 years E) 36 years Answer: A Diff: 2 Type: MC Page Ref: 448-452 Topic: 11.5 Determining the Term n of an Annuity Objective: 11-5: Compute the number of periods for ordinary simple annuities. 104) Jamal plans to retire in 17 years. He is saving $2000 every month in a retirement savings account paying him a long-term interest of 9% compounded monthly until retirement. The rate changes following the retirement. He wants $7000 per month paid to him for 20 years after the retirement. At what rate should his savings account pay him to fulfill his dream? (Use the Rate function in Excel) A) 9% B) 10% C) 0.86% D) 10.27% E) 0% Answer: D Diff: 2 Type: MC Page Ref: 452-453 Topic: 11.6 Finding the Periodic Rate of Interest i Using Preprogrammed Financial Calculators Objective: 11-6: Compute the interest rate for ordinary simple annuities. 105) What's the difference between a simple annuity and a general annuity? A) payments are made at the end of each payment period for simple annuities B) simple annuities are a type of general annuity C) the compounding frequency and payment period are the same for simple annuities D) payments are made at the beginning of each payment period for simple annuities Answer: C Diff: 1 Type: MC Page Ref: 418-420 Topic: 11.1 Introduction to Annuities Objective: 11-1: Distinguish between types of annuities based on term, payment date, and conversion period.

106) What type of annuity does not require a payment in the first payment period? A) general annuity B) annuity certain C) contingent annuity D) deferred annuity Answer: D Diff: 1 Type: MC Page Ref: 418-420 Topic: 11.1 Introduction to Annuities Objective: 11-1: Distinguish between types of annuities based on term, payment date, and conversion period. 107) Lease payments on equipment, installment payments on loans, and interest payments on bonds are all examples of what type of annuity? A) Annuity certain B) Simple annuity C) Annuity due D) Perpetuity Answer: A Diff: 1 Type: MC Page Ref: 418-420 Topic: 11.1 Introduction to Annuities Objective: 11-1: Distinguish between types of annuities based on term, payment date, and conversion period. 108) Ali deposits $25 of his biweekly paycheck into an RRSP that has an interest rate of 1.95% p.a. compounded biweekly. If he continues to do this for the next 47 years, what will his total contribution be and what will be the balance of the RRSP? Answer: $25 * 26 bi-weeks/year * 47 years = $30 550 Periodic interest rate: i = 0.0195/26 = 0.00075 = 0.075% n = 26(47) = 1222 FV = 25

= 25

= $49 988.81

Diff: 2 Type: SA Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities.

109) Asifa invested $6000 in her TFSA at the end of every year for 4 years. Her TFSA is linked to an index mutual fund that returns around 4.8% p.a. compounded annually. After 4 years, she gave the deposit to her son to aid in the purchase of a house in the future. Her son left the deposit in a savings account that earned interest at 3.5% p.a. compounded annually. What is the balance of the savings account after 3 years? Answer: TFSA Return Periodic rate of interest: i = 0.048 n=4 FV1 = 6000

= 6000

= $25 783.96

Savings Account PV = FV1 = $25 783.96 Periodic rate of interest: i = 0.035 n=3 FV2 = 25783.96(1.035^3) = $28 587.14 Diff: 2 Type: SA Page Ref: 421-429 Topic: 11.2 Ordinary Simple Annuity - Calculating Future Value FV Objective: 11-2: Compute the future value for ordinary simple annuities. 110) Find the present value for an annuity with payments of $100 paid at the end of every three months, with an interest rate of 5.2% p.a. compounded quarterly and a term of 8.75 years. Answer: Periodic rate of interest: 0.052/4 = 0.013 = 1.3%; n = 8.75 * 4 = 35 PV = 100

= $2797.61

Diff: 1 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities. 111) Buster and Dave purchase a cottage by making a down payment of $22 500 and paying $745 on a monthly basis for three years. If the interest rate is 4.2% p.a compounded monthly, what's the cash value of the cottage? Answer: Periodic rate of interest: 0.042/12 = 0.0035 = 0.35%; n = 3 * 12 = 36 PV = 745

= 745(33.768910) = $25 157.84

Cash Value = 22500 + 25157.84 = $47 567.84 Diff: 2 Type: SA Page Ref: 431-440 Topic: 11.3 Ordinary Simple Annuity - Calculating Present Value PV Objective: 11-3: Compute the present value for ordinary simple annuities.

112) What payment is required at the end of every six months for 78 months to accumulate to $13 650 at 9.4% p.a. compounded semi-annually? Answer: Periodic rate of interest: 0.094/2 = 0.047 = 4.7%; n = (78/12) * 2 = 13 PMT =

= $785.44

Diff: 1 Type: SA Page Ref: 442-446 Topic: 11.4 Ordinary Simple Annuities - Determining the Periodic Payment PMT Objective: 11-4: Compute the payment for ordinary simple annuities. 113) Jordan wants to buy a car selling for $31 500 by paying a down payment of $11 500 and then the remaining balance in equal monthly payments for 5 years. Determine the monthly payments if the interest rate is 1.2% p.a. compounded monthly. Answer: Periodic rate of interest: 0.012/12 = 0.001 = 0.1%; PV = 31500 - 11500 = 20000; n = 5 * 12 = 60 PMT =

= $343.60

= 13.815395 years = 14 years

Diff: 1 Type: SA Page Ref: 448-452 Topic: 11.5 Determining the Term n of an Annuity Objective: 11-5: Compute the number of periods for ordinary simple annuities.

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 12 Ordinary General Annuities 1) Calculate the accumulated value after ten years of payments of $1000.00 made at the end of each quarter if interest is 4% compounded semi-annually. Answer: PMT = 1000.00; n = 10(4) = 40; i =

= 2% = 0.02; c = 2/4 = 0.5

p = (1 + 0.02)^0.5 - 1 = 0.00995049 FV = 1000

= $48 836.51

Diff: 2 Type: SA Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 2) What is the accumulated value of deposits of $1120.00 made at the end of every six months for three years if interest is 8.48% compounded quarterly? Answer: PMT = 1120, i = 0.0848 ÷ 4 = 0.0212, n = 3 × 2 = 6, c =

p = (1.0212)2 - 1 = 0.0428494 FVg = 1120.00

= 1120.00(6.6804902) = $7482.34

Diff: 2 Type: SA Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 3) What is the accumulated value of deposits of $2000.00 made at the end of every year for three years if interest is 6.00% compounded quarterly? Answer: PMT = 2000, i = 0.06 ÷ 4 = 0.015, n = 3, c =

p = (1.015)4 - 1 = 0.0613636 FVg = 2000.00

= 2000.00(3.1878563) = $6375.71

4) What is the future value of payments of $230.00 made at the end of every three months in twenty-one years if interest is 7.65% compounded annually? Answer: PMT = 230, i = 0.0765 , n = 21 × 4 = 84, c = p= FVg = 230.00

- 1 = 0.0185996 = 230.00(199.0432233) = $45 779.94

Diff: 2 Type: SA Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 5) Conchita Martinez saves $17.25 at the end of each month and deposits the money in an account, paying 4.96% compounded quarterly. a) How much will she accumulate in 22 years? b) How much of the accumulated amount is interest? Answer: PMT = 17.25, i = 0.0496 ÷ 4 = 0.0124, n = 22 × 12 = 264, c =

a) p = (1 + 0.0124)1/3 - 1 = 0.00411637 FVg = 17.25

= 17.25(475.6458279) = $8204.89

b) Interest = 8204.89 - 264(17.25) = 8204.89 - 4554.00 = $3650.89 Diff: 2 Type: SA Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 6) Ali saves $25.00 at the end of each month and deposits the money in a savings account, paying 3.00% compounded semi-annually. a) How much will she accumulate in 20 years? b) How much of the accumulated amount is interest? Answer: PMT = 25, i = 0.03 ÷ 2 = 0.015, n = 20 × 12 = 240, c =

a) p = (1 + 0.015)1/6 - 1 = 0.00248452 FVg = 25.00

= 25.00(327.636659) = $8190.92

b) Interest = 8190.92 - 240(25.00) = 8190.92 - 6000.00 = $2190.92 Diff: 2 Type: SA Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities.

7) Mrs. Engleder has made deposits of $750.00 at the end of every six months for twenty-three years. If interest is 3.72% compounded monthly, how much will Mrs. Engleder have accumulated six years after the last deposit? Answer: PMT = 750, i = 0.0372 ÷ 12 = 0.0031, n = 23 × 2 = 46, c =

Amount after 23 years. p = (1 + 0.0031)6 - 1 = 0.0187447 FVg = 750.00

= 750.00(72.0026939) = $54 002.02

Amount after another 6 years FV = 54002.02(1.0031)72 = $67 483.03

Diff: 2 Type: SA Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 8) What is the accumulated value after 11.25 years of monthly deposits of $415.00 earning interest at 5.56% compounded semi-annually if the deposits are made at the end of each month? Answer: PMT = 415, i = 0.0556 ÷ 2 = 0.0278, n = 11.25 × 12 = 135, c =

p = (1 + 0.0278)1/6 - 1 = 0.0045806 FVg = 415.00

= 415.00(186.2861243) = $77 308.74

Diff: 2 Type: SA Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 9) What is the accumulated value after 15.50 years of deposits of $500.00 earning interest at 4.00% compounded semi-annually if the deposits are made at the end of each quarter? Answer: PMT = 500, i = 0.04 ÷ 2 = 0.02, n = 15.5 × 4 = 62, c =

= 0.5

p = (1 + 0.02)0.5 - 1 = 0.0099505 FVg = 500.00

=500.00(85.18052518) = 42 590.30

10) A loan is repaid by making payments of $6000.00 at the end of every six months for twelve years. If interest on the loan is 8% compounded quarterly, what was the principal of the loan? Answer: PMT = 6000.00; n = 12(2) = 24; i =

= 2% = 0.02; c =

= 2; I/Y = 8; P/Y = 2; C/Y = 4

p = (1 + 0.02)2 - 1 = 1.0404 - 1 = 0.0404 PVg = PMT

= 6000

= $91 108.28

Diff: 2 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 11) What is the discounted value of $4760.00 paid at the end of each year for 7 years if interest is 6.52% compounded quarterly? Answer: PMT = 4760, i = 0.0652 ÷ 4 = 0.0163, n = 7, c =

p = (1 + 0.0163)4 - 1 = 0.0668115 PVg = 4760.00

= 4760.00(5.4497186) = $25 940.66

Diff: 2 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 12) What is the discounted value of $3000.00 paid at the end of every six months for 5 years if interest is 8.00% compounded yearly? Answer: PMT = 3000, i = 0.08, n = 5 × 2 = 10, c = p = (1 + 0.08)1/2 - 1 = 0.0392305 PVg = 3000.00

= 3000.00(8.142055396) = 24 426.17

Diff: 2 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 13) What is the principal from which $2279.00 can be withdrawn at the end of each month for 17.5 years if interest is 5.44% compounded quarterly? Answer: PMT = 2279, i = 0.0544 ÷ 4 = 0.0136, n = 17.5 × 12 = 210, c =

p = (1 + 0.0136)1/3 - 1 = 0.0045129 PVg = 2279.00

= 2279.00(135.5103403) = $308 828.07

14) A real estate developer bought land for $170 000.00 down and monthly payments of $10 450.00 for 5 years. What is the equivalent cash price if money is worth 7.75% compounded semi-annually? Answer: PMT = 10450, i = 0.0775 ÷ 2 = 0.03875, n = 5 × 12 = 60, c =

p = (1 + 0.03875)1/6 - 1 = 0.0063565 PVg = 10450.00

= 10450.00(49.7543772) = $519 933.24

Cash price = 170000.00 + 519933.24 = $689 933.24 Diff: 2 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 15) A student bought a rental property for $30 000.00 down and monthly payments of $1100.00 for 5 years. What is the equivalent cash price if money is worth 5.25% compounded semi-annually? Answer: PMT = 1100, i = 0.0525 ÷ 2 = 0.02625, n = 5 × 12 = 60, c =

p = (1 + 0.02625)^(1/6) - 1 = 0.004327902 PVg = 1100.00

= 1100.00(52.74261) = $58 016.87

Cash price = 30000 + 58016.87 = $88 016.87 Diff: 2 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 16) If a loan was repaid by monthly payments of $9230.00 in 7.5 years at 6.15% compounded annually, how much interest was paid? Answer: PMT = 9 230, i = 0.0615, n = 7.5 × 12 = 90, c = p = (1 + 0.0615)1/12 - 1 = 0.004986 PVg = 9230.00

= 9230.00(72.3737824) = $668 010.79

Interest paid = 90(9230.00) - 66801.01 = 830700 - 668010.01 = $162 689.99 Diff: 2 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities.

17) If a loan was repaid by semi-annual payments of $4720.00 in 6.5 years at 8.24% compounded quarterly, how much interest was paid? Answer: PMT = 4720, i = 0.0824 ÷ 4 = 0.0206, n = 6.5 × 2 = 13, c =

p = (1 + 0.0206)2 - 1 = 0.0416244 PVg = 4720.00

= 4720.00(9.8857467) = $46 660.74

Interest paid = 13(4720.00) - 46660.72 = 61360.00 - 46660.72 = $14 699.28 Diff: 2 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 18) If a student loan was repaid by monthly payments of $800.00 in 5.0 years at 8.00% compounded yearly, how much interest was paid? Answer: PMT = 800, i = 0.08, n = 5 × 12 = 60, c = p = (1 + 0.08)1/12 - 1 = 0.0064340 PVg = 800.00

= 800.00(49.6449458) = $39 715.92

Interest paid = 60(800.00) - 39715.96 = 48000.00 -39715.96 = $8284.04 Diff: 2 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 19) A 21-year mortgage is amortized by making payments of $3052.61 at the end of every three months. If interest is 8.45% compounded annually, what was the original mortgage balance? Answer: PMT = 3052 .61, i = 0.0845, n = 21 × 4 = 84, c = p = (1 + 0.0845)1/4 - 1 = 0.0204868 PVg = 3052.61

= 3052.61(39.925945) = $121 178.36

20) For her daughter's university education, Carla Hackl has invested an inheritance in a fund paying 9.2% compounded quarterly. If ordinary annuity payments of $4750.00 per month are to be made out of the fund for 5 years and the annuity begins 7.75 years from now, how much was the inheritance? Answer: PMT = 4750, i = 0.092 ÷ 4 = 0.023, n = 5 × 12 = 60, c =

p = (1 + 0.023)1/3 - 1 = 0.0076083 The amount in the fund 7.75 years from now (annuity begins): PVg = 4750.00

= 4750.00(48.0273124) = $228 129.73

Today's balance:

PV = 228129.73 (1+ 0.023)-7.75 × 4 = $112 729.41 Diff: 3 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 21) You won $150 000.00 in a lottery and you want to set some of that sum of money aside for 6 years. After 6 years you would like to receive $2710.00 at the end of every 6 months for 8 years. If interest is 6% compounded quarterly, how much of your winnings must you set aside? Answer: PMT = 2710, i = 0.06 ÷ 4 = 0.015, n = 8 × 2 = 16, c =

Balance at 6 years (annuity starts): p = (1 + 0.015)2 - 1 = 0.030225 PVg = 2710.00

= 2710.00(12.539523) = $33 982.11

Balance now:

PV = 33982.11(1 + 0.015)-6 × 4 = $23 771.98 Diff: 3 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 22) You won $250 000.00 in a lottery and you want to set some of that sum of money aside for 11.5 years. After 11.5 years, you would like to receive $4500.00 at the end of each month for 4.75 years. If the interest is 6.79% compounded annually, how much of your winnings must you set aside? Answer: PMT = 4500, i = 0.0679, n = 4.75 × 12 = 57, c = Balance after 11.5 years (annuity starts) p = (1 + 0.0679)1/12 - 1 = 0.0054895 PVg = 4500.00

= 4500.00(48.8299465) = $219 734.76

Balance today

PV = 219734.76(1.0679)-11.5 = $103 227.88 Diff: 3 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities.

23) What is the principal invested at 6.76% compounded semi-annually from which monthly withdrawals of $450.00 can be made at the end of each month for 22 .75 years? Answer: PMT = 450, i = 0.0676 ÷ 2 = 0.0338, n = 22 .75 × 12 = 273, c =

p = (1 + 0.0338)1/6 - 1 = 0.0055556 PVg = 450.00

= 450.00(140.3336252) = $63 150.13

Diff: 2 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 24) A 22-year mortgage is amortized by payments of $1761.50 made twice a year. If interest is 9.65% compounded semi-annually, what is the mortgage principal? Answer: PMT = 1761.50, i = 0.0965 ÷ 2 = 0.04825, n = 22 × 12 = 264, c = 2/2 = 1 p = 1.04825 - 1 = 0.04825 PVg = 1761.50

= 1761.50(20.725307) = $36 507.63

Diff: 2 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 25) A loan was repaid in 6.5 years by quarterly payments of $515.00 at 9.85% compounded semi-annually. What is the value of the loan? Answer: PMT = 515, i = 0.0985 ÷ 2 = 0.04925, n = 6.5 × 4 = 26, c =

= 0.5

p = (1 + 0.04925)0.5 - 1 = 0.024329 PVg = 515.00

= 515.00(19.108482) = $9837.45

26) Calculate the amount of money that must be deposited at the end of every three months into an account paying 6% compounded monthly to accumulate to $12 500.00 in ten years? Answer: FV = 12 500.00; n = 10(4) = 40; i =

= 0.5% = 0.005; I/Y = 6; P/Y = 4; C/Y = 12; c =

p = (1 + i)c - 1 = (1 + 0.005)3 - 1 = 0.015075125 FV = PMT 12500 = PMT 12500 = PMT[54.35422486] PMT = $229.97 Programmed solution:

Diff: 1 Type: SA Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities. 27) A $ 120 000.00 mortgage is amortized over 25 years. If interest on the mortgage is 8.5% compounded semi-annually, calculate the size of monthly payments made at the end of each month. Answer: PV = 120 000.00; n = 25(12) = 300; i =

= 4.25% = 0.0425; I/Y = 8.5; P/Y = 12; C/Y = 2; c =

p = (1 + i)c - 1 = (1 + 0.0425)1/6 - 1 = 0.006961062 120000 = PMT 120000 = PMT[125.7286733] PMT =

= $954.44

Programmed solution:

Diff: 1 Type: SA Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities.

28) How much must be deposited at the end of each quarter for 5 years to accumulate to $27 000.00 at 6.84% compounded biweekly? Answer: FVg = 27000, i = 0.0684 ÷ 26 = 0.002631, n = 5 × 4 = 20, c = 26/4 = 6.5 p = (1 + 0.002631)6.5 - 1 = 0.0171977 27000.00 = PMT 27000.00 = PMT(23.637339) $1142.26 = PMT Diff: 1 Type: SA Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities. 29) How much must be deposited at the end of each month for 10 years to accumulate to $100 000.00 at 6.00% compounded yearly? Answer: FVg = 100 000, i = 0.06, n = 10 × 12 = 120, c = p = (1 + 0.06)1/12 - 1 = 0.0048675506 100 000 = PMT 100 000.00 = PMT(162.4734413) $615.49 = PMT Diff: 1 Type: SA Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities. 30) A loan of $91 450.00 is to be repaid by equal semi-annual payments for 11 years. What is the size of each semi-annual payment if the interest is 9.24% compounded annually? Answer: PVg = 91 450, i = 0.0924, n = 11 × 2 = 22, c = p = (1 + 0.0924)1/2 - 1 = 0.0451794 91450.00 = PMT 91450.00 = PMT(13.7613682) $6645.41 = PMT Diff: 2 Type: SA Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities.

31) A loan of $40 000.00 is to be repaid by equal quarterly payments for 5 years. What is the size of each semi-annual payment if the interest is 5.00% compounded annually? Answer: PVg = 40 000, i = 0.05, n = 5 × 4 = 20, c = p = (1 + 0.05)1/4 - 1 = 0.0122722344 40000.00 = PMT 40000.00 = PMT(17.63931702) $2267.66 = PMT Diff: 2 Type: SA Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities. 32) Kelsey bought a car priced at $19 700.00 for 10% down and equal monthly payments for 4.5 years. If interest is 8.22% compounded semi-annually, what is the size of the monthly payment? Answer: The mortgaged amount is 19700(1-0.1) = 17730 = PVg i = 0.0822 ÷ 2 = 0.0411, n = 4.5 × 12 = 54, c =

p = (1 + 0.0411)1/6 - 1 = 0.0067356 17730.00 = PMT 17730.00 = PMT(45.1433762) $392.75 = PMT Diff: 2 Type: SA Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities. 33) Edward Martin borrowed $15 810.00 from his uncle to finance his graduate studies. The loan agreement calls for equal payments at the end of each month for 6 years. The payments are deferred for 3 years and interest is 8.12% compounded semi-annually. What is the size of the monthly payments? Answer: Balance in 3 years: i = 0.0812 ÷ 2 = 0.0406, N = 3 × 2 = 6 FV = 15810.00(1.0406)6 = 15810.00(1.2694802) = $20 074.04 For annuity: PVg = 20074.04, i = 0.0406, n = 6 × 12 = 72, c =

p = (1 + 0.0406)1/6 - 1 = 0.006655 20074.04 = PMT 20074.04 = PMT(57.05678) $351.83 = PMT Diff: 3 Type: SA Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities.

34) A mortgage of $198 000.00 is to be amortized by payments over 22.5 years. If the payments are made at the end of each month and interest is 8.75% compounded semi-annually, what is the size of the monthly payments? Answer: PVg = 198000, n = 22 .5 × 12 = 270, i = 0.0875 ÷ 2 = 0.04375, c =

p = (1 + 0.04375)1/6 - 1 = 0.0071622 198000.00 = PMT 198000.00 = PMT(119.2931506) $1659.78 = PMT Diff: 2 Type: SA Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities. 35) What sum of money can be withdrawn from a fund of $46 950.00 invested at 6.5% compounded semiannually at the end of every three months for twelve years? Answer: PVg = 46 950, i = 0.065 ÷ 2 = 0.0325, n = 12 × 4 = 48, c =

= 0.5

p = (1 + 0.0325)0.5 - 1 = 0.0161201 46950.00 = PMT 46950.00 = PMT(33.24246) $1412.35 = PMT Diff: 2 Type: SA Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities. 36) Sean Paul borrowed $20 000.00 from his father to finance his new business. The loan agreement calls for equal payments at the end of each month for 10 years. The payments are deferred for 4 years and interest is 4.00% compounded semi-annually. What is the size of the monthly payments? Answer: Balance in 4 years: i = 0.04 ÷ 2 = 0.02, n = 4 × 2 = 8 FV = 20000(1.02)8 = $23433.19 = PVg For annuity: i = 0.02, n = 10 × 12 = 120, c =

p = (1 + 0.02)1/6 - 1 = 0.0033058903 23433.19 = PMT 23433.19 = PMT(98.9230334) $236.88 = PMT Diff: 3 Type: SA Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities.

37) How much must be contributed into an RRSP at the end of each year for 37 years to accumulate to if interest is 8.04% compounded quarterly? Answer: FVg = 213000, i = 0.0804 ÷ 4 = 0.0201, c =

p = (1 + 0.0201)4 - 1 = 0.0828567 213000.00 = PMT 213000.00 = PMT(217.4395489) $979.58 = PMT Diff: 2 Type: SA Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities. 38) How many deposits of $6100.00 made at the end of every 6 months are needed to accumulate to $172 000.00 at 5.24% compounded quarterly? Answer: FVg = 172 000, PMT = 6100, i = 0.0524 ÷ 4 = 0.0131, c =

p = (1 + 0.0131)2 - 1 = 0.0263716

= 21.36

Twenty-one deposits of $6100.00 each and a final deposit smaller than $6100.00 are needed. Diff: 2 Type: SA Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities. 39) For how long will Harold have to make payments of $550.00 at the end of every three months to repay a loan of $16 900.00 if interest is 9% compounded monthly? Answer: PVg = 16900, PMT = 550, i = 0.09 ÷ 12 = 0.0075, c =

p = (1.0075)3 - 1 = 0.022669172

n=-

= 53.2 ≈ 53 quarters

Diff: 2 Type: SA Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities.

40) For how long will Jessica have to make payments of $1200.00 at the end of every three months to repay a loan of $10 000.00 if interest is 12% compounded semi-annually? Answer: PVg = 10 000, PMT = 1200, i = 0.12 ÷ 2 = 0.06, c =

p = (1.06)1/2 -1 = 0.029563014

= 9.708 ≈ 10 periods.

Diff: 2 Type: SA Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities. 41) Mr. Lavergne accumulated $320 000.00 in an RRSP. He converted the RRSP into an RRIF and started to withdraw $7500.00 at the end of every three months from the fund. If interest is 4.68% compounded monthly, how many withdrawals can Mr. Lavergne make? Answer: PVg = 320 000, PMT = 7500, i = 0.0468 ÷ 12 = 0.0039, c =

p = (1 + 0.0039)3 - 1 = 0.01174569

= 59.56 (withdrawals)

Diff: 2 Type: SA Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities. 42) Mr. Amer accumulated $102 800.00 in an RRSP. He converted the RRSP into an RRIF and started to withdraw $2510.00 at the end of every month from the fund. If interest is 4.5% compounded yearly, for how long can Mr. Sepaba make withdrawals? Answer: PVg = 102 800, PMT = 2510, i = 0.045, c = p = (1 + 0.045)1/12 - 1 = 0.0036748094

= 44.47 months = 3.71 years

Diff: 2 Type: SA Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities.

43) Deon makes monthly contributions of $250.00 at the end of every month into an RRSP. Calculate, to the nearest month, how long it will take to grow to $100 000.00 if interest is 8% compounded annually. Answer: FV = 100 000.00; PMT = 250.00; i =

= 8% = 0.08; I/Y = 8; P/Y = 12; C/Y = 1; c =

p = (1 + 0.08)1/12 -1 = 0.00643403 100000.00 = 250 2.573612044 = (1.00643403)n - 1 3.573612044 = (1.00643403)n

n ln 1.00643403 = ln 3.573612044 n(0.00641342) = 1.273576861 n = 198.58 ≈ 199 months Programmed solution:

Diff: 1 Type: SA Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities. 44) A loan of $54 000.00 is repaid by making payments of $3000.00 at the end of every three months. If the interest is 9% compounded monthly, calculate how long the payments have to be made. Answer: PV = 54 000.00; PMT = 3000.00; i = !

= 0.0075; I/Y = 9; P/Y = 4; C/Y = 12; c =

p = (1 + 0.075)3 -1 = 0.022669172 54000.00 = 3000 0.408045094 = 1 - (1.022669172)-n 0.591954906 = (1.022669172)-n

-n ln 1.022669172 = ln 0.591954906 -n(0.022416045) = 0.524324819 n = 23.39060393 quarters ÷ 4 ≈ 6 years Programmed solution:

Diff: 1 Type: SA Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities.

45) For how long must $179.23 be deposited at the end of each month to accumulate to $9700.00 at 6.18% compounded quarterly? Answer: FVg = 9700, PMT = 179.23, i = 0.0618 ÷ 4 = 0.01545, c =

p = (1 + 0.01545)1/3 - 1 = 0.0051237

= 47.88968 ≈ 48 months

Diff: 2 Type: SA Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities. 46) A $360 000.00 mortgage is amortized by making monthly payments of $2600. If interest is 7.5% compounded semi-annually, what is the term of the mortgage? Answer: PVg = 360 000, PMT = 2564.19, i = 0.075 ÷ 2 = 0.0375, c =

p = (1 + 0.0375)1/6 - 1 = 0.0061545

= 311.5625381 ≈ 312 months

Diff: 2 Type: SA Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities. 47) Sabrina deposits $300.00 into a savings account at the end of each month for five years. If the accumulated value of the deposits is $20 000.00 and interest is compounded semi-annually, calculate the nominal rate of interest Answer: FV = 20 000.00; PMT = 300; n = 5(12) = 60; P/Y = 12; C/Y = 2; c =

Programmed solution:

The rate is 4.26% p.a. compounded semi-annually. Diff: 1 Type: SA Page Ref: 479-481 Topic: 12.5 Ordinary General Annuities - Finding the Periodic Interest Rate i Objective: 12-5: Compute the interest rate for ordinary general annuities.

48) Calculate the nominal interest rate of interest compounded quarterly if a loan of 43 000.00 is repaid in seven years by payments of $4000.00 made at the end of every six months. Answer: PV = 43 000.00; PMT = 4000.00; n = 7(2) = 14; P/Y = 2; C/Y = 4 Programmed solution:

The rate is 7.40% p.a. compounded quarterly. Diff: 1 Type: SA Page Ref: 479-481 Topic: 12.5 Ordinary General Annuities - Finding the Periodic Interest Rate i Objective: 12-5: Compute the interest rate for ordinary general annuities. 49) Note: The calculations for this question were done using Excel's RATE function. Compute the nominal annual rate of interest on a loan of $67 000.00 repaid in semi-annual installments of $6540.00 in 11.5 years. Answer: Nper 23 Pmt -6540.00 PV 67000.00 Result 0.0815144 The nominal annual rate is 8.15144% × 2 = 16.30288% Diff: 2 Type: SA Page Ref: 479-481 Topic: 12.5 Ordinary General Annuities - Finding the Periodic Interest Rate i Objective: 12-5: Compute the interest rate for ordinary general annuities. 50) Note: The calculations for this question were done using Excel's RATE function. What nominal annual rate of interest compounded semi-annually is earned by quarterly deposits of $7327.00 made for six years if the balance just after the last deposit is $289 000.00? Answer: Nper 24 Pmt -7327.00 FV 289000.00 Result 0.0407147 p = (1 + 0.0407147)2 - 1 = 0.0830871 2(0.0830871) = 16.61742% per year comp. semi-annually Diff: 2 Type: SA Page Ref: 479-481 Topic: 12.5 Ordinary General Annuities - Finding the Periodic Interest Rate i Objective: 12-5: Compute the interest rate for ordinary general annuities.

51) Note: The calculations for this question were done using Excel's RATE function. What is the rate of interest compounded quarterly if a loan of $31 500.00 is repaid in seven years by payments of $2712.00 made at the end of every six months? Answer: Nper 14 Pmt -2712.00 PV 31500.00 Result 0.0259414 f = (1 + 0.0259414).5 - 1 = .0128877 The nominal annual rate is 1.28877 * 4 = 5.15506% Diff: 3 Type: SA Page Ref: 479-481 Topic: 12.5 Ordinary General Annuities - Finding the Periodic Interest Rate i Objective: 12-5: Compute the interest rate for ordinary general annuities.

52) Note: The calculations for this question were done using Excel's RATE function. A $345 000.00 mortgage is repaid in 19 years by making monthly payments of $2486.44. What is the nominal annual rate of interest compounded semi-annually? Answer: Nper 228 Pmt -2486.44 PV 345000.00 Result 0.0047739 1.0047739 = (1 + i)1/6 1 + i = 1.00477396 i = 0.0289874 The nominal annual interest rate is 0.0289874 × 2 × 100% = 5.79748% Diff: 2 Type: SA Page Ref: 479-481 Topic: 12.5 Ordinary General Annuities - Finding the Periodic Interest Rate i Objective: 12-5: Compute the interest rate for ordinary general annuities. 53) Note: The calculations for this question were done using Excel's RATE function. A property worth $15 000.00 is purchased for 10% down and semi-annual payments of $1750.00 for twelve years. What is the effective rate of interest if interest is compounded semi-annually? Answer: Nper 24 Pmt -1750.00 PV 15000.00 * (1 - 0.10) Result 13500 Effective annual rate f = 1.1213292 - 1 = 25.5737873% Diff: 3 Type: SA Page Ref: 479-481 Topic: 12.5 Ordinary General Annuities - Finding the Periodic Interest Rate i Objective: 12-5: Compute the interest rate for ordinary general annuities.

54) Note: The calculations for this question were done using Excel's RATE function. A $160 000.00 mortgage with a 20-year term is repaid by making monthly payments of $1361.00. What is the rate of interest compounded semi-annually on the mortgage? Answer: Nper 240 Pmt -1361.00 PV 160000.00 Result 0.0068559 f = (1 + 0.0068559)12/2 - 1 = 0.0418469 The interest rate is 4.18469% × 2 = 8.36938% Diff: 3 Type: SA Page Ref: 479-481 Topic: 12.5 Ordinary General Annuities - Finding the Periodic Interest Rate i Objective: 12-5: Compute the interest rate for ordinary general annuities.

55) For their daughter's future education David and Bonnie set a fund in which they invest at the end of every six months. The first investment is $215 and they plan to increase their investments at a constant rate of 0.5%. If the fund earns 2.4% compounded semi-annually, how much money will be in the fund in 18 years when their daughter goes to college? Answer: PMT = 215, k = 0.005, i = 0.024 ÷ 2 = 0.012, n = 18 × 2 = 36 FV = 215

= $10433.61

Diff: 1 Type: SA Page Ref: 482-486 Topic: 12.6 Constant-Growth Annuities Objective: 12-6: Compute future value and present value for constant-growth annuities. 56) How much should you invest today into account paying 3.2% compounded quarterly if you want to be able to withdraw at the end of every 3 month the amounts of money growing at a constant rate of 3% for the period of 5 years , if the first withdrawal is to be $100 five years three months from now? Answer: The amount needed to be in the account five years from now: PMT = 100, i = 0.032 ÷ 4 = 0.008, k = 0.03, n = 5 × 4 = 20 PV = 100

= $2454.76

The amount to be invested today: PV = 2454.76(1 + 0.008)-20 = $2093.14

Diff: 2 Type: SA Page Ref: 482-486 Topic: 12.6 Constant-Growth Annuities Objective: 12-6: Compute future value and present value for constant-growth annuities.

57) Calculate the accumulated value after ten years of payments of $1000.00 made at the end of each year if interest is 6% compounded monthly. A) $10 228.03 B) $13 285.11 C) $10 228.56 D) $10 600.00 E) $13 228.03 Answer: B Diff: 2 Type: MC Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 58) Calculate the accumulated value after twenty years of payments of $12 000.00 made at the end of each year if interest is 6% compounded semi-annually. A) $445 721.73 B) $322 444.73 C) $642 766.07 D) $240 000.00 E) $441 427.09 Answer: A Diff: 2 Type: MC Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 59) What is the final amount of an investment where you are making quarterly deposits of $672.31 at the end of the payment period for 7 years and 9 months? The interest rate is 6.16% compounded semiannually. A) $26 811.16 B) $28 611.16 C) $26 611.57 D) $28 811.16 E) $26 407.57 Answer: E Diff: 2 Type: MC Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities.

60) You miss the 11th to 13th payments of a loan. The loan payments are $519.27 each month and the interest rate is 8.45% compounded annually. How much are you behind in your payments on the day that you miss the 13th payment? A) $1658.40 B) $1586.40 C) $1568.40 D) $1508.40 E) $1458.40 Answer: C Diff: 3 Type: MC Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities.

61) You are to receive $16 700 from your trust fund in 7.5 years. You negotiate a deal where you can withdraw $350 per month starting 2.75 years before you can withdraw the remainder of the funds. You are able earn 4.18% compounded annually on your money. How much is your trust fund reduced to? A) $4195.40 B) $4159.40 C) $4459.40 D) $4495.40 E) $4155.40 Answer: D Diff: 3 Type: MC Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 62) A mortgage requires payments of $4000.00 at the end of every six months for twelve years. If interest is 8% compounded quarterly, calculate the principal of the loan. A) $75 655.70 B) $84 948.26 C) $60 738.85 D) $88 320.00 E) none of the above Answer: C Diff: 2 Type: MC Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities.

63) A mortgage requires payments of $1000.00 at the end of every month for twenty-five years. If interest is 6% compounded semi-annually, calculate the principal of the loan. A) $156 297.23 B) $155 206.86 C) $46 188.41 D) $300 000.00 E) $33 328.64 Answer: A Diff: 2 Type: MC Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 64) The loan will have a term of 54 months and monthly payments of $385.13. The interest rate on the loan is 8.84% compounded quarterly. What is the amount of money that was borrowed? A) $17 310.71 B) $17 430.71 C) $17 030.71 D) $17 130.71 E) $20 368.27 Answer: D Diff: 2 Type: MC Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 65) You lease a car for monthly payments of $610.23 at the end of each month for 54 months. There is a no security deposit sale going on. The interest rate on the lease is 3.9% compounded annually. The buy-out is $19 552 at the end of the lease. How expensive is the vehicle? A) $48 683.26 B) $48 863.26 C) $46 683.26 D) $46 863.26 E) $48 633.26 Answer: C Diff: 2 Type: MC Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 66) What is the payment size of a loan that has semi-annual payments and an interest rate of 5.52% compounded monthly? The loan principal is $1641.12 and the loan is for 48 months. A) $321.74 B) $231.74 C) $123.74 D) $274.31 E) $221.31 Answer: B Diff: 2 Type: MC Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities.

67) Calculate the amount of money that must be deposited at the end of every three months into an account paying 6% compounded monthly to accumulate to $12 500.00 in ten years. A) $293.75 B) $80.77 C) $304.94 D) $229.97 E) none of the above Answer: D Diff: 2 Type: MC Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities. 68) Calculate the amount of money that must be deposited at the end of every month into an account paying 12% compounded annually to accumulate to $100 000.00 in ten years. A) $434.71 B) $450.59 C) $833.33 D) $45.06 E) $416.67 Answer: B Diff: 2 Type: MC Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities. 69) You buy a motorcycle for $52 523 plus freight of $724 and other delivery charges of $561. You have saved 25% of the total purchase price towards a down payment. The bank is willing to finance the purchase of the motorcycle at 9.86% compounded quarterly. What is the size of your monthly payment if the loan is for 4.5 years? A) $296.80 B) $926.80 C) $962.80 D) $996.80 E) $262.80 Answer: B Diff: 3 Type: MC Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities.

70) You can buy a piece of equipment for $8600. The equipment is expected to last for 9 years and you want to have it completely paid off 3.5 years before it becomes unusable. Current loan rates are 9.5% compounded annually. What is the size of your semi-annual payment? A) $1515.98 B) $1105.98 C) $1055.98 D) $1015.98 E) $1115.98 Answer: D Diff: 3 Type: MC Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities. 71) Mark deposits $900 at the end of each month in a savings account earning 4.5% quarterly. How soon can he buy his first car assuming that he needs $8 000 for a decent car. A) 8.76 months B) 8.52 months C) 9.06 months D) 9 years E) 8.86 months Answer: A Diff: 2 Type: MC Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities. 72) A mortgage of $326 000 is to be repaid by making payments of $2500 at the end of each month. If interest is 4.75% compounded semi-annually, what is the term of the mortgage? A) 183 months B) 106 months C) 182 months D) 184 months E) 105 months Answer: A Diff: 2 Type: MC Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities. 73) You make 6 quarterly deposits starting at $135 and increasing at a constant rate of 4.5%. The interest rate is 8.4% compounded quarterly. What is the size of the last deposit? A) $186.23 B) $168.23 C) $128.63 D) $135.00 E) $668.23 Answer: B Diff: 1 Type: MC Page Ref: 482-486 Topic: 12.6 Constant-Growth Annuities Objective: 12-6: Compute future value and present value for constant-growth annuities.

74) You make 4 semi-annual deposits starting at $350 and increasing at a constant rate of 2.14%. The interest rate is 5.12% compounded annually. How much has been deposited in total? A) $1445.59 B) $1454.59 C) $1400.00 D) $1459.59 E) $1444.59 Answer: A Diff: 2 Type: MC Page Ref: 482-486 Topic: 12.6 Constant-Growth Annuities Objective: 12-6: Compute future value and present value for constant-growth annuities. 75) Saba gets a year-end bonus in December every year before the Christmas. She plans to invest $1000 from the bonus at the end of every year into mutual funds RRSP, which are expected to grow at the rate of 8% compounded semi-annually. What will be the total value of her periodic investments after 25 years? Answer: PMT = $1000; n = 1 × 25 = 25; i = c=

= 4%

p = (1 + i)c - 1 = (1+ 0.04)2 - 1 = 0.0816 per year FV = PMT

= $1000

= $74 836.81

Diff: 2 Type: SA Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 76) Bhamani is contributing to a RESP they have set up for their grandchildren. What amount will they have in the RESP after 13 years of contributing $185 at the end of every month if the plan earns 5.5% compounded quarterly? Answer: PMT = $185;

n = 12 × 13 = 156;

= 1.375%

= 0.3333

p = (1 + i)c - 1 = (1+ 0.01375)0.3333 - 1 = 0.0045625 FV = PMT

= $185

= $41 937

77) A series of payments of $900 are being made at the end of each year for the next 8 years. What is the future value 8 years from now, if money can earn 9% compounded semi-annually? A) $9999 B) $30 220 C) $77 200 D) $14 346 E) $7200 Answer: A Diff: 2 Type: MC Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 78) Don turned 60 years old today and has retired from Hydro One. Current economic value of his pension is his significant asset. He will be receiving pension payments of $7000 at the end of each month. Based on a 22-year life expectancy and money worth 8% compounded semi-annually, calculate the current economic value of Don's pension. Answer: PMT = $7000; n = 12 × 22 = 264; i = c=

= 4%

= 0.1667

p = (1 + i)c - 1 = (1+ 0.04)0.1667 - 1 = 0.0065582 PV = PMT

= $7000

= $877 325.83

Diff: 2 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 79) Silva is contemplating to buy a 20 year annuity paying $2500 at the end of each month. What amount will be required to purchase the annuity, if the annuity provides a return of 6.75% compounded annually? A) $334 000 B) $47 246 C) $600 000 D) $37 037 E) $567 436 Answer: A Diff: 2 Type: MC Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities.

80) A dealer sold a car to Kristy for $2000 down and monthly payments of $259.50 for 3.5 years, including interest at 7.5% compounded annually. What was the selling price of the car? A) $9600 B) $11 600 C) $6012.50 D) $12 899 E) $10 899 Answer: B Diff: 2 Type: MC Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 81) Sam took a $5000 loan to build a deck in the backyard. The loan requires payments at the end of each quarter or 4 years. If the interest on the loan is 9% compounded monthly, what is the size of each payment Sam is required to make? A) $447.83 B) $353.98 C) $332.80 D) $1321.64 E) $376.08 Answer: E Diff: 2 Type: MC Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities. 82) In order to purchase a truck Shandex Transport (ST) obtained a $50 000 loan for 5 years at 7.8% compounded semi-annually. The loan is structured to reduce the balance owing to $10 000 at the end of the 5 year period. How much are ST's payments at the end of each month? A) $1006.08 B) $804.87 C) $868.84 D) $834.79 E) $1070.11 Answer: C Diff: 3 Type: MC Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities.

83) Dion is planning to buy a brand new BMW. He plans to keep the car for 3 years and then trade it in. He borrows $37 000 at 7.2% compounded semi-annually to purchase the car. What monthly payment will reduce the balance on the loan after 3 years to the expected trade-in value of $16 500? Answer: $37000 = (PV of the loan payment annuity) + (PV of the $16500 balance) For PVannuity n = 36; i = 3.6% c= p = 1.0362/12 - 1 = 0.0059 PVannuity = PMT

= 32.34PMT

For PVbalance n = 6; i = 3.6%

Vbalance = FV(1 + i)-n = $16500 (1+ 0.036)-6 = $13 345.21 32.34PMT = $37000 - $13,345.21 = $23 654.79 ⇒ PMT = $731.44 Diff: 3 Type: SA Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities. 84) One month from now, Shiva will make his first monthly contribution of $250 to a Tax-Free Savings account (TFSA). Over the long run, he expects to earn 8% compounded annually. How long will it take for the contributions and accrued earnings to reach $100 000? Answer: FV = $100 000; PMT = $250; i = 8% c= p = 1.081/12 -1 = 0.006434

= 198.58

⇒ n = 199 months = 16 years 7 months Diff: 2 Type: SA Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities.

85) Momin plans to buy a house on cash instead of paying mortgage. He targets spending $400 000 on his dream house. He plans to set aside $1200 every month to reach his target. He puts his savings in a Tax Free Savings Account (TFSA) and invests them in a high risk mutual fund, which has traditionally earned 12.2% annually. How long will it take Momin to reach his target? A) 12 years 6 months B) 27 years 10 months C) 24 years 9 months D) 15 years 3 months E) 13 years 9 months Answer: A Diff: 2 Type: MC Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities. 86) Note: This problem is solved using CalcPro for iPad Lahore Gym allows members to pay the members to pay the annual membership fee by a single payment of $240 at the beginning of the year, or by payments of $22 at the beginning of each month. What effective rate of interest is paid by members, who choose to pay by month? Answer: The first monthly payment is due on the same date as the full annual fee. In effect, Lahore Gym lends $240 - $22 = $218 to a member choosing the monthly payment option. The member then repays this loan in 11 month-end payments of $22. PV = $218; PMT = $22; P/Y = 12; C/Y = 1; n = 11 Using above values in CalcPro, i = 23.62% compounded annually Diff: 3 Type: SA Page Ref: 479-481 Topic: 12.5 Ordinary General Annuities - Finding the Periodic Interest Rate i Objective: 12-5: Compute the interest rate for ordinary general annuities. 87) What annually compounded rate of return must Thiva earn in her TFSA account in order for month end contributions of $950 to accumulate $1 million after 25 years? A) 0% B) 8.97% C) 8.62% D) 0.716% E) 7.16% Answer: B Diff: 3 Type: MC Page Ref: 479-481 Topic: 12.5 Ordinary General Annuities - Finding the Periodic Interest Rate i Objective: 12-5: Compute the interest rate for ordinary general annuities.

88) Momin plans to buy a house on cash instead of paying mortgage. He plans to set aside $14 400 at the end of each year for 15 years. He puts his savings in a Tax Free Savings Account (TFSA) and invests them in a high risk mutual fund, which has traditionally earned 12.2% annually. Money decreases in value by 2.5% per annum as well. How long will it take Momin to reach his target? Answer: PMT = $14 400; k = -2.5%; i = 12.2%; n = 15 years FV = PMT

= $14400

= $14400 × 33.59

FV = $483 723 Diff: 3 Type: SA Page Ref: 482-486 Topic: 12.6 Constant-Growth Annuities Objective: 12-6: Compute future value and present value for constant-growth annuities. 89) Don turned 60 years old today and has retired from Hydro One. Current economic value of his pension is his significant asset. He will be receiving pension payments of $84 000 every year. Based on a 22-year life expectancy and money worth 8% compounded annually and money depreciating at 2% annually, calculate the current economic value of Don's pension. A) $740 932 B) $877 326 C) $1 001 884 D) $1 234 888 E) $832 256 Answer: A Diff: 3 Type: MC Page Ref: 482-486 Topic: 12.6 Constant-Growth Annuities Objective: 12-6: Compute future value and present value for constant-growth annuities. 90) Sam got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded semi-annually. If he puts the money in his savings account at the end of each month, what will be the balance in the account at the end of the three-year term? A) $18 408.04 B) $18 600.50 C) $18 603.37 D) $27 284.81 E) $17 390.19 Answer: B Diff: 2 Type: MC Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities.

91) Sam got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded semi-annually. If he puts the money in his savings account at the end of each month, how much interest is included in the future value of the annuity? A) $600.50 B) $18 600.50 C) $18 603.37 D) $408.04 E) $603.37 Answer: A Diff: 2 Type: MC Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 92) Jordan is considering building a nuclear power plant, which will be ready for production in 2030. The country's governing body is also considering a decommissioning liability law for the operator to put aside $1 million every month towards decommissioning cost. If the production life of the plant is 60 years and the operator puts the money at the end of the month in a savings account, earning 7.25% compounded semi-annually, what will be the value of the decommissioning fund after 60 years of production? A) $12.49 billion B) $10.87 billion C) $720 million D) $11.88 billion E) $917.65 million Answer: D Diff: 2 Type: MC Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 93) Jordan is considering building a nuclear power plant, which will be ready for production in 2030. The country's governing body is also considering a decommissioning liability law for the operator to put aside $1 million every month towards decommissioning cost, when the plant is producing electricity. During the 60 years of production life of the plant, the operator will put $1 million at the end of the month in a savings account, earning 7.25% compounded semi-annually. At the end of the production life of the plant, there are no more contributions and the money is expected to grow at the rate of 6% compounded quarterly for the next 30 years. What will be the value of the decommissioning fund in 2120? A) $12.49 billion B) $74.57 billion C) $11.88 billion D) $70.94 billion E) $60.90 billion Answer: D Diff: 3 Type: MC Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities.

94) Silva saves $100 at the end of each month and deposits the money in an account, paying 7.49% compounded semi-annually. How much of the accumulated amount is interest after 40 years? A) $291 872 B) $243 872 C) $299 531 D) $251 531 E) $253 521 Answer: B Diff: 2 Type: MC Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 95) Woods has contributed $3600.00 per year for the last ten years into a RRSP account earning 9% compounded semi-annually. His wife Karen, on the other hand, contributed $300 at the end of every month for the same period into a RRSP account earning 9% compounded quarterly. Who has bigger saving at the end of 10 years and by how much? A) Wood by $3360 B) Karen by $3360 C) Wood by $2580 D) Karen by $2580 E) Karen by $2262 Answer: D Diff: 2 Type: MC Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 96) Sam got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded semi-annually. If he puts the money in his savings account at the end of each month, what will be the present value of the balance in the account at the end of the three-year term? A) $18 600.50 B) $18 408.04 C) $18 603.37 D) $17 392.96 E) $17 390.19 Answer: D Diff: 2 Type: MC Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities.

97) Sam got a job at the Brick. He immediately finances a car paying $500 every month for 3 years. The interest is 2.25% compounded semi-annually. How much is the interest cost of the financing? A) $17 392.96 B) $408.04 C) $603.37 D) $607.04 E) $609.81 Answer: D Diff: 2 Type: MC Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 98) Keith bought a lakeside cottage for 30% down and monthly mortgage payments of $1224.51 at the end of each month for 20 years. Interest is 4.4% compounded semi-annually. What is the purchase price of the property? A) $194 378 B) $195 881 C) $278 878 D) $279 831 E) $254 646 Answer: D Diff: 2 Type: MC Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 99) Siri plans to retire now and would like to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. Calculate the amount he must invest now if interest is 4.9% compounded quarterly. A) $1 528 015 B) $1 530 569 C) $633 665 D) $793 072 E) $539 042 Answer: B Diff: 3 Type: MC Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities.

100) Knowing that the education costs for her kids would be huge, Roberta puts lump sum money from her inheritance in a savings account paying an interest of 4.69% compounded semi-annually. If ordinary annuity payments of $1500.00 per month are to be paid out of the account at the end of each month for four years, how much did Roberta deposit? A) $65 533 B) $32 474 C) $65 592 D) $32 723 E) $54 343 Answer: C Diff: 2 Type: MC Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 101) Siri plans to retire in 18 years and would like to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. Calculate the amount he must invest now every month (at the end of the month) if interest is 4.9% compounded semi-annually. A) $1 528 015 B) $4420.74 C) $1 534 362 D) $4461.50 E) $11 755 Answer: D Diff: 3 Type: MC Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities. 102) Jamal plans to retire in 17 years. He is saving $2000 every month in a retirement savings account paying him a long-term interest of 9% compounded semi-annually. What will be the size of his payments per month from the ordinary annuity for 20 years following his retirement? A) $957 836 B) $8721 C) $8618 D) $941 547 E) $8372 Answer: E Diff: 1 Type: MC Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities.

103) Siri plans to retire when her simple annuity savings account has enough money to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. She starts saving $4420 per month. Calculate when should Siri retire from today if her savings account pays 4.9% compounded semi-annually. A) 18 years B) 216 years C) 217.35 years D) 18.11 years E) 36 years Answer: D Diff: 3 Type: MC Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities. 104) Jamal plans to retire in 17 years. He is saving $2000 every month in a retirement savings account paying him a long-term interest of 9% compounded semi-annually until retirement. The rate changes following the retirement. He wants $7000 per month paid to him for 20 years after the retirement. At what rate (semi-annually) should his savings account pay him to fulfill his dream? (Use the Rate function in Excel.) A) 0.54% B) 3.28% C) 0.86% D) 10.27% E) 6.55% Answer: E Diff: 3 Type: MC Page Ref: 479-481 Topic: 12.5 Ordinary General Annuities - Finding the Periodic Interest Rate i Objective: 12-5: Compute the interest rate for ordinary general annuities. 105) Yu Yi receives annuity payments on a quarterly basis and then puts the money into an account earning 3.0% p.a. monthly. What's the equivalent rate of interest per payment period? Answer: c = 12/4 = 3; i = 0.03/12 = 0.0025; p = 1.0025^3 - 1 = 0.751877% Diff: 1 Type: SA Page Ref: 463-469 Topic: 12.1 Ordinary General Annuities - Calculating the Future Value FV Objective: 12-1: Compute the future value (or accumulated value) for ordinary general annuities. 106) Every 3 months, Carolyn gets an allowance of $80 that she puts into a savings account with a 1.2% p.a. return compounded semi-annually. After 6 years, what is the balance of her savings account? Answer: c = 2/4 = 0.5; i = 0.012/2 = 0.006 p = 1.006^0.5 - 1 = 0.002996 n = 6 * 2 = 12 FV = 80

= 80(12.199691) = $975.98

107) While planning for his retirement, Alishan purchased an annuity that paid him $3200 every year for 13 years at a rate of 3.8% compounding semi-annually. What was Alishan's initial investment into the annuity? Answer: c = 6/12 = 0.5; i = 0.038/2 = 0.019 = 1.9% p = 1.019^0.5 — 1 = 0.009455 n = 13 * 2 = 26 PV = 3200

= 3200

= 3200(22.955082) = $73.456.26

Diff: 2 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 108) What is the loan principal for a loan in which payments of $460 must be made at the end of every two weeks for 1.5 years, and the interest rate is 4.4% p.a. compounded quarterly? Answer: c = 4/26 = 2/13; i = 0.044/4 = 0.011 = 1.1% p = 1.011^(2/13) — 1 = 0.001684 n = 1.5 * 26 = 39 PV = 460

= 460

= 460(37.715822) = $17349.28

Diff: 2 Type: SA Page Ref: 471-472 Topic: 12.2 Ordinary General Annuities - Calculating the Present Value PV Objective: 12-2: Compute the present value (or discounted value) for ordinary general annuities. 109) Raahim, a new business owner, puts a certain amount of money aside into a mutual fund for future use. The fund returns 4.6% compounding interest quarterly. He hopes that after 5.5 years, he'll have $500 000 in the account. How much money on a monthly basis should Raahim put into the fund? Answer: c = 4/12 = 1/3; i = 0.046/4 = 0.0115 = 1.15% p = 1.0115^(1/3) — 1 = 0.003819 n = 12 * 5.5 = 66 PMT =

= $6675.51

Diff: 1 Type: SA Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities. 110) Antonio bought a big boat for $17 400 and paid $1100 as a down payment. The remainder of the balance must be paid in equal installments at the end of each year for six years. If interest compounds on a semi-annual basis at 8.2% p.a., what is the size of the yearly payment? Answer: c = 2/1 = 2 i = 0.082/2 = 0.041 = 4.1% p = 1.041^2 — 1 = 0.083681 n=6 PMT =

= $3565.41

Diff: 1 Type: SA Page Ref: 473-475 Topic: 12.3 Ordinary General Annuities - Determining the Periodic Payment PMT Objective: 12-3: Compute the payment for ordinary general annuities.

111) A start-up known as Spec-lights recently decided to lease new equipment valued at $33 500. The company has to make payments of $6500 at the end of each year, with an interest rate of 4.5% p.a. compounded quarterly. How long will payments have to be made? Answer: c = 4/1 = 4; i = 0.045/4 = 0.01125 = 1.125% p = 1.01125^4 — 1 = 0.045765

= 6.011617 years

Diff: 2 Type: SA Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities. 112) How many monthly payments of $100 would it take for an RRSP to have a balance of $12,000 at an interest rate of 3.4% p.a. compounding semi-annually? Answer: c = 2/12 = 1/6; i = 0.034/2 = 0.017 = 1.7% p = 1.017^(1/6) - 1 = 0.002813…

= 103.53700 payments

Therefore it would take 103 monthly payments of $100 and a smaller final payment to result in a $12 000 balance in the RRSP. Diff: 2 Type: SA Page Ref: 476-479 Topic: 12.4 Ordinary General Annuities - Determining the Term n Objective: 12-4: Compute the number of periods for ordinary general annuities. 113) Withdrawals from Mo's RRSP are made semi-annually, starting with $3000 and increasing by 3% per term for 8 years. The interest rate on the money is 6.8% p.a. compounding semi-annually. a) What is the size of the 13th withdrawal? b) Compute the compounding factor for the constant-growth annuity. Answer: a) 13th payment = 3000(1.03)^12 = 4227.28 b) i = 0.068/2 = 0.034 = 3.4%; n = 8 * 2 = 16 compounding factor = =

= 25.666847

Diff: 2 Type: SA Page Ref: 482-486 Topic: 12.6 Constant-Growth Annuities Objective: 12-6: Compute future value and present value for constant-growth annuities.

114) Petra has started to make contributions towards her child's RESP, starting with a contribution of $4500 and decreasing by 2% at the end of every year when she makes the next contribution. If she plans on doing this for 18 years with an interest rate of 3.0% p.a. compounded annually, what will be the balance of the RESP at the end of the 18 years? Answer: k = -2% = -0.02, i = 3% = 0.03, n = 18 FV = 4500

= 4500

= 4500 (20.145955) = $90 656.80

Diff: 1 Type: SA Page Ref: 482-486 Topic: 12.6 Constant-Growth Annuities Objective: 12-6: Compute future value and present value for constant-growth annuities.

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 13 Annuities Due, Deferred Annuities, and Perpetuities 1) Hector makes payments of $2000.00 into an RRSP at the beginning of every six months starting today for 33 years. If the interest rate is 8% compounded semi-annually, calculate the accumulated value. Answer: PMT = 2000.00; n = 33(2) = 66; i = FV(due) = 2000

= 0.04; I/Y = 8; P/Y = C/Y = 2

= $640 155.60

Programmed solution:

Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 2) A contract valued at 60 000.00 requires payments of $3000.00 at the beginning of each quarter. If interest is 6% compounded quarterly, calculate the term of the contract. Answer: PMT = 3000.00; PV = 60 000; i =

= 0.015; I/Y = 6; P/Y = C/Y = 4

60 000 = 3000(1.015) 0.295566502 = 1 - 1.015-n -n ln 1.015 = ln 0.704433498 -n(0.014888612) = -0.350361349 n = 23.53216923 quarters ÷ 4 = 5.88 ≈ 6 years Programmed solution:

Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

3) Calculate the present value of payments of $100.00 made at the beginning of each quarter for ten years if the interest is 6% compounded quarterly. Answer: PMT = 100.00; n = 10(4) = 40; i = PV(due) = 100

= 0.015; I/Y = 6; P/Y = C/Y = 4

= $3036.46

Programmed solution:

Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 4) A lease agreement valued at $15 000.00 requires payment of $900.00 at the beginning of every quarter for five years. What is the nominal rate of interest charged (compounded quarterly)? Answer: PV = 15 000.00; n = 5(4) = 20; PMT = 900.00; P/Y = C/Y = 4 Programmed solution:

The nominal rate of interest is 8.03% (approx.) p.a. compounded quarterly. Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

5) What semi-annual payment must be made into a fund at the beginning of every six months to accumulate to $4800.00 in ten years at 7% compounded semi-annually? Answer: FV(due) = 4800; n = 10(2) = 20; i =

= 0.035; I/Y = 7; P/Y = C/Y = 2

FV(due) = PMT(1 + i) 4800 = PMT(1 + 0.035) 4800 = PMT(1 + 0.035)[28.27968181] 4800 = PMT(29.26947068) PMT = PMT = $163.99 Programmed solution:

Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 6) Ashlyn just graduated and plans to buy a house in 15 years worth $1 million. How much does she have to save at the beginning of every three months for 15 years to accumulate $1 million dollars if interest is 4% compounded quarterly? Answer: FVn (due) = 1000000 1000000.00 = PMT

(1.01)

1000000.00 = PMT(81.66966986)(1.01) $12 123.22 = PMT Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

7) Payments of $500.00 are made at the beginning of each month for four years. The interest rate is 4.5% compounded monthly. If no further deposits are made a) Calculate the accumulated value twelve years after the first deposit. b) Calculate the amount deposited. c) Calculate the interest. Answer: a) Step 1: PMT = 500; n = 4(12) = 48; i =

= 0.00375; I/Y = 4.5; P/Y = C/Y = 12

FV(due) = 500

= $26 340.32

Step 2: PV = 26340.32; n = 8(12) = 96 FV = 26340.32(1.00375)96 = $37 728.94 Programmed solution:

b) Amount deposited is 48(500.00) = $24 000.00 c) Interest is 37 728.94 - 24 000.00 = $13 728.94 Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 8) Bala bought mutual funds within TFSA as part of his retirement planning in 2019. Payments of $500.00 will be made into the fund at the beginning of every month for 20 years. If the fund is expected to earn interest at 8.4% compounded monthly, what will be the balance in the fund after 20 years? Answer: FV (due)

= 500.00

(1.007)

= 500.00(619.1778116)(1.007) = $311 756.03 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

9) Determine the accumulated value after 13.5 years of deposits of $470.00 made at the beginning of every three months and earning interest at 7.2% compounded quarterly. Answer: FV (due)

= 470.00

(1.018)

= 470.00(90.026386)(1.018) = $43 074.02 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 10) Payments of $1000.00 are made into a fund at the beginning of every month for 10 years. If the fund earns interest at 12% compounded monthly, what will the balance in the fund at the end of the term? Answer: FV (due)

= 1000.00

(1.01)

= 1000.00(230.0386895)(1.01) = $232 339.08 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 11) Determine the accumulated value after 15 years of deposits of $9000.00 made at the beginning of every three months and earning interest at 8% compounded quarterly. Answer: FV (due)

= 9000.00

(1.02)

= 9000.00(114.05153394)(1.02) = $1 046 993.13 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 12) What is the discounted value of deposits of $570.00 made at the beginning of every three months for 8.25 years if money is worth 8.4% compounded quarterly? Answer: PV (due)

= 570

(1.021)

= 570(23.6345174)(1.021) = $13 754.58 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

13) What is the discounted value of deposits of $250.00 made at the beginning of every three months for 10.5 years if money is worth 6% compounded quarterly? Answer: PV (due)

= 250

(1.015)

= 250(30.99405004)(1.015) = $7864.74 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 14) Linda contributes $3000.00 at the beginning of every six months into an RRSP paying interest at 6% compounded semi-annually. a) How much will her RRSP deposits amount to in 15 years? b) How much of the amount will be interest? Answer: a) FV (due)= 3000

(1.03)

= 3000(47.57541571)(1.03) = $147008.03 b) Total deposits = (3000) × 30 = $90 000.00 Interest = 147008.03 - 90000 = $57 008.03 Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 15) Steve deposited $600.00 in a trust account on the day of his son's birth and every three months thereafter in order to plan and pay for his son's higher education. If interest paid is 7.2% compounded quarterly, what will the balance in the trust account be before the deposit is made on the son's 18th birthday?

Answer: FVn (due) = 600

(1.018)

= 600(145.1537512)(1.018) = $88 659.91 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

16) The monthly rent payment on a storage warehouse is $3971.00 payable in advance. What yearly payment in advance would satisfy the lease if interest is 6.84% compounded monthly? Answer: PV (due)

= 3971

(1.0057)

= 3971(11.5669781)(1.0057) = $46 194.29 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 17) Judy paid for a living room suite over time, signing an installment contract that requires semi-annual payments of $2560.00 for 8.5 years. The first payment is made on the signing date and interest is 14.5% compounded semi-annually. What was the cash price? Answer: PV (due)

= 2560

(1.0725)

= 2560(9.5964039)(1.0725) = $26 347.89 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 18) Jessica bought a bedroom suite on credit, signing an installment contract with a finance company that requires monthly payments of $100.00 for 5 years. The first payment is made on the date of signing and interest is 24% compounded monthly. a) What was the cash price? b) How much will Tricia pay in total? c) How much of what she pays will be interest? Answer: a) PV (due) = 100

(1.02)

= 100(34.76088668)(1.02) = $3545.61 b) Total paid = 60(100) = $6000 c) Interest = 6000 - 3545.61 = $2454.39 Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

19) Roxanne contributes $167.00 at the beginning of every six months into an RRSP paying interest at 8.9% compounded semi-annually. a) How much will her RRSP deposits amount to in 31.5 years? b) How much of the amount will be interest? Answer: a) FV (due) = 167

(1.0445)

= 167(326.5655028)(1.0445) = $56963.31 b) Total deposits = (167) × 63 = $10 521 Interest = 56963.31 - 10521 = $46 442.31 Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 20) The monthly rent payment on computer equipment is $30.00 payable in advance. What yearly payment in advance would satisfy the lease if interest is 12% compounded monthly? Answer: PV (due)

= 30

(1.01)

= 30(11.25507747)(1.01) = $341.03 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 21) Vicki paid for a kitchen set over time, signing an installment contract that requires semi-annual payments of $500.00 for 5 years. The first payment is made on the signing date and interest is 18% compounded semi-annually. What was the cash price? Answer: PV (due)

= 500

(1.09)

= 500(6.4176577)(1.09) = $3497.62 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

22) Gibran bought a boat valued at $110 194.00 on the installment plan. He made equal semi-annual payments for 6 years. If the first payment is due on the date of purchase and interest is 7.5% compounded semi-annually, what is the size of the semi-annual payments? Answer: PVn (due) = 110 194 110 194 = PMT

(1.0375)

110 194.00 = PMT(9.5226939)(1.0375) $11 153.47 = PMT Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 23) Determine the future value and the present value of monthly payments of $200.00 each for five years at 12% compounded monthly if a) the payments form an annuity due; b) the payments form an ordinary annuity. Answer: i = 0.12 ÷ 12 = 0.01 a) FVn (due) = 200

(1.01 ) = 200(81.66966986)(1.01) = $16 497.27

PVn (due) = 200

(1.01)

= 200(44.95503841)(1.01) = $9080.92 b) FVn = 200(81.66966986) = $16 333.93 PVn = 200(44.95503841) = $8991.01 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

24) Tricia bought a living-room suite on credit, signing an installment contract with a finance company that requires monthly payments of $167.25 for 4.5 years. The first payment is made on the date of signing and interest is 28.8% compounded monthly. a) What was the cash price? b) How much will Tricia pay in total? c) How much of what she pays will be interest? Answer: a) PV (due) = 167.25

(1.024)

= 167.25(30.0897982)(1.024) = $5153.30 b) Total paid = 54(167.25) = $9031.50 c) Interest = 9031.50 - 5153.30 = $3878.20 Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 25) Determine the future value and the present value of monthly payments of $173.00 each for six years at 6.48% compounded monthly if a) the payments form an annuity due; b) the payments form an ordinary annuity. Answer: i = 0.0648 ÷ 12 = 0.0054 a) FVn (due) = 173.00

(1.0054 ) = 173.00(87.7161525)(1.0054) = $15 256.84

PVn (due) = 173.00

(1.0054)

= 173.00(59.522361)(1.0054) = $10352.97 b) FVn = 173.00(87.7161525) = $15 174.89 PVn = 173.00(59.522361) = $10 297.37 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

26) Jack Chen makes deposits of $415.00 at the beginning of every three months. Interest earned by the deposits is 4.12% compounded quarterly. a) What will the balance in Jack's account be after eight years? b) How much of the balance will Jack have contributed? c) How much of the balance is interest? Answer: a) FVn (due) = 415.00

(1.0103)

= 415.00(37.676677)(1.0103) = $15 796.87 b) Total deposit = 32(415.00) = $13 280.00 c) Interest = 15796.87 - 13280.00 = $2516.87 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 27) Sean Paul makes deposits of $500.00 at the beginning of every three months. Interest earned by the deposits is 8% compounded quarterly. a) What will the balance in Sean's account be after ten years? b) How much of the balance will Sean have contributed? c) How much of the balance is interest? Answer: a) FVn (due) = 500

(1.02)

= 500(60.40198318)(1.02) = $30 805.01 b) Total deposit = 40(500.00) = $20 000.00 c) Interest = 30805.01 - 20000.00 = $10 805.01 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

28) How many months will it take to accumulate $17 897.20 at 5.87% compounded semi-annually, if $1140.00 is deposited in an account at the beginning of every 6 months? Answer: FVn (due) = 17897.20 = 1140.00

(1.02935)

15.251662 = 0.4476363 = 1.02935n - 1 ln 1.447363 = n ln 1.02935 =n n = 12 .78 ≈ 13 semi-annual periods Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 29) How much does a depositor have to save at the beginning of every 6 months for twenty years to accumulate $1 000 000.00 if interest is 6.00% compounded semi-annually? Answer: FVn (due) = 1 000 000 1 000 000.00 = PMT

(1.03)

1 000 000.00 = PMT(75.40125973)(1.03) $12876.10 = PMT Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 30) A sum of money is deposited at the beginning of each year for 3 years at 12% compounded annually. After the last deposit interest for the account is to be 8.24% compounded quarterly and the account is to be paid out by payments of $370.00, made at the beginning of each quarter for nine years. What is the size of the annual deposit? Answer: PVn (due) = 370.00

(1.0206)

= 370.00(25.2449471)(1.0206) = $9533.05 = FVn (due) 9533.05 = PMT

(1.12)

9533.05 = PMT(3.3744)(1.12) $2522.42 = PMT Diff: 3 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

31) Kelly is saving $1600.00 at the beginning of each month. How soon can he retire if he wants to have a retirement fund of $345 000.00 and interest is 11.04% compounded monthly? Answer: 345 000.00 = 1600.00

(1.0092)

213.6593341 = 1.9656659 = 1.0092n - 1 2.9656659 = 1.0092n

ln 2 .9656659 = n ln 1.0092 1.0871016 = n(.0091579) n = 118.71 ≈ 119 months Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 32) How much does a depositor have to save at the beginning of every three months for 13.25 years to accumulate $41 600.00 if interest is 8% compounded quarterly? Answer: FVn (due) = 41 600 41 600.00 = PMT

(1.02)

41 600.00 = PMT(92.816737)(1.02) $439.41 = PMT Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 33) A contract valued at $81 500.00 requires payment of $3350.00 at the beginning of every six months. If interest is 6.5% compounded semi-annually, what is the term of the contract? Answer: PVn (due) = 81 500, PMT = 3350 81 500 = 3350

(1.0325)

23.5625745 = 0.7657837 = 1 - 1.0325-n -n ln 1.0325 = ln 0.2342163 -n(0.031983) = -1.4515102 n = 45.383804 Answer is: 45 full payments and one less than $3350 Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

34) Robert Bernard deposited $1000.00 in a trust account on the day of his son's birth and every three months thereafter. If interest paid is 8% compounded quarterly, what will the balance in the trust account be before the deposit is made on the son's twenty-first birthday? Answer: Sn (due)

= 1000

(1.02) = 1000(213.8666068)(1.02) = $218 143.94

Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 35) How much does a depositor have to save at the beginning of every 6 months for eleven years to accumulate $17 448.00 if interest is 6.25% compounded semi-annually? Answer: FVn (due) = 17 448 17 448 = PMT

(1.03125)

17 448.00 = PMT(30.9733986)(1.03125) $546.25 = PMT Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 36) Equal sums of money are withdrawn monthly from a fund of $33 000.00 for 15.25 years. If interest is 9% compounded monthly, what is the size of each withdrawal if the withdrawal is made at the beginning of each month? Answer: PVn (due) = 33 000 33 000 = PMT

(1.0075)

33 000 = PMT(99.3634773)(1.0075) $329.64 = PMT Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 37) Equal sums of money are withdrawn monthly from a fund of $60 000.00 for 15 years. If interest is 12% compounded monthly, what is the size of each withdrawal if the withdrawal is made at the beginning of each month? Answer: PVn (due) = 60 000 60 000 = PMT

(1.01)

60 000 = PMT(83.32166399)(1.01) $712.97 = PMT Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for

simple annuities due. 38) What is the accumulated value after 15 years of monthly deposits of $500.00 earning interest at 8% compounded semi-annually if the deposits are made a) at the end of each month? b) at the beginning of each month? Answer: a)

- 1 = 0.006558197

FVg (due) = 500

= 500(342.0753463) = $171 037.67

FVg = 500(342.0753463)(1.006558197) = $172 159.37 Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 39) A lease requires payments of $1200.00 at the beginning of every quarter for ten years. If interest is 6% compounded monthly, calculate the value of the lease. Answer: PMT = 1200; n = 10(4) = 40; i =

= 0.005; I/Y = 6; P/Y = 4; C/Y = 12; c =

p = (1.005)3 -1 = 0.015075125 PV(due) = 1200(1.015075125)

= $36 390.27

Programmed solution:

Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due.

40) What deposit made at the beginning of each month will accumulate to $36 000.00 at 5% compounded quarterly at the end of eight years? Answer: FV(due) = 36 000.00; n = 8(12) = 96; i = c=

= 0.0125; I/Y = 5; P/Y = 12; C/Y = 4

p = (1 + i)c - 1 = (1 + 0.0125)1/3 - 1 = 0.004149425 FV(due) = PMT(1 + p) 36 000 = PMT(1 + 0.004149425) 36 000 = PMT(1.004149425)[117.6381043] 36 000 = PMT(118.1262348) PMT =

= $304.76

Programmed solution:

Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 41) A vacation property was bought by making semi-annual payments of $7500.00 for seven years. If the first payment is due on the date of purchase and interest is 6% p.a. compounded quarterly, calculate the purchase price of the property. Answer: PMT = 7500.00; n = 7(2) = 14; i =

= 0.015; c =

= 2; I/Y = 6; P/Y = 2; C/Y = 4

p = (1.015)2 - 1 = 0.030225 PV(due) = 7500

= $87 147.51

Programmed solution:

Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due.

42) Ellora wants to accumulate $150 000.00 in an RRSP by making annual contributions of $5000.00 at the beginning of each year. If interest is 5.5% compounded quarterly, calculate how long she has to make contributions. Answer: PMT = 5000.00; FV(due) = 150 000; i =

= 0.01375; c =

= 4; I/Y = 5.5; P/Y = 1; C/Y = 4

p = (1.01375)4 - 1 = 0.056144809 150 000 = 5000 1.594804293 = 1.056144809n - 1 n ln 1.056144809 = ln 2.594804293 n(0.054625306) = 0.953511097 n = 17.455483 Programmed solution:

n = 18 years (approx.).The last payment will be less than $5000.00. Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 43) Compute the nominal annual rate of interest compounded monthly at which $1500.00 deposited at the beginning of every three months for ten years will amount to $90 000.00. Answer: FV(due) = 90 000.00; n = 10(4) = 40; PMT = 1500.00; P/Y = 4; C/Y = 12 Programmed solution:

The nominal rate of interest is 7.48% (approx.) p.a. compounded monthly. Diff: 1 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due.

44) Determine the accumulated value after 14.5 years of deposits of $172.00 made at the beginning of every 6 months and earning interest at 4.45% compounded monthly? Answer: p = FVg (due) = 172.00

- 1 = 0.0224573 (1.0224573)

= 172.00(40.26262)(1.0224573) = $7080.69 Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 45) Determine the accumulated value after 15 years of deposits of $500.00 made at the beginning of every 6 months and earning interest at 12% compounded monthly? Answer: p = FVg (due) = 500

- 1 = 0.061520151 (1.061520151)

= 500(81.20594577)(1.061520151) = $43 100.87 Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 46) Sean contributes $545.00 at the beginning of each month into TFSA paying interest at 6.25% compounded semi-annually. What will be the accumulated balance in the TFSA at the end of 10.25 years? Answer: p = FVg (due) = 545

- 1 = 0.0051418 (1.0051418)

= 545(170.9815136)(1.0051418) = $93 664.06 Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due.

47) A property was purchased for quarterly payments of $1070.00 for 9 years. If the first payment was made on the date of purchase and interest is 5.57% compounded annually, what was the purchase price of the property? Answer: p = PVg (due) = 1070

- 1 = 0.013643245 (1.013643245)

= 1070(28.2958162)(1.013643245) = $30 689.59 Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 48) A property was purchased for quarterly payments of $3000.00 for 10 years. If the first payment was made on the date of purchase and interest is 6% compounded annually, what was the purchase price of the property? Answer: p = PVg (due) = 3000

- 1 = 0.014673846 (1.014673846)

= 3000(30.09471567)(1.014673846) = $91 608.96 Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 49) Slick Sean contributes $500.00 at the beginning of each month into RRSP paying interest at 6.00% compounded semi-annually. What will be the accumulated balance in the RRSP at the end of 15 years? Answer: p = FVg (due) = 500

- 1 = 0.004938622 (1.004938622)

= 500(289.0001444)(1.004938622) = $145 213.70 Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due.

50) Aubrey made quarterly deposits of $750.00 at the beginning of every 3 months for 11.25 years in a non-registered savings account. If the interest is prime+0.5 compounded monthly, what amount can be withdrawn 6 years after the last deposit? Assume a fixed prime rate of 2.25%. Answer: p =

- 1 = 0.006891

FVg (due) = 750

(1.006891)

= 750(52.5455)(1.006891) = $39 680.69 FV = 39680.69(1.006891)24 = 39680.69(1.179177) = $46 790.56

Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 51) What is the accumulated value after 14 years of monthly deposits of $415.00 earning interest at 5.8% compounded semi-annually if the deposits are made a) at the end of each month? b) at the beginning of each month? Answer: a)

p= FVg = 415

- 1 = 0.004775944 = 415(256.8161929) = $106 578.72

FVg (due) = 415(256.8161929)(1.004775944) = $107 087.73 Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 52) Quarterly deposits of $2000.00 were made at the beginning of every 3 months for 10 years. If the interest is 6% compounded monthly, what amount can be withdrawn 7 years after the last deposit? Answer: p = FVg (due) = 2000

- 1 = 0.015075125 (1.015075125)

= 2000(54.35422486)(1.015075125) = $110 347.24 FV = 110347.24 (1.015075125)28 = 110347.24(1.520369636) = $167 768.59

Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due.

53) Gary and Sandra Duncan bought a property by making semi-annual payments of $4210.00 for 8.5 years. If the first payment is due on the date of purchase and interest is 8.53% compounded quarterly, what is the purchase price of the property? Answer: p =

- 1 = 0.0431048

PVg (due) = 4210

(1.0431048)

= 421(11.8778937)(1.0431048) = $52 161.43 Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 54) Bow Valley Credit Union entered a lease contract valued at $3100.00. The contract provides for payments at the beginning of each month for three years. If interest is 8.8% compounded quarterly, what is the size of the monthly payment? Answer: PVn (due) = 3100 p= 3100.00 = PMT

- 1 = 0.0072802 (1.0072802)

3100.00 = PMT(31.5685288)(1.0072802) $97.49 = PMT Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 55) A lease requires monthly payments of $1150.00 due in advance. If interest is 16% compounded quarterly and the term of the lease is 4 years, what is the cash value of the lease? Answer: p = PVn (due) = 1150

- 1 = 0.013159404 (1.013159404)

= 1150(35.41891644)(1.013159404) = $41 267.76 Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due.

56) At the beginning of every 6 months, $1280.00 is deposited for 8 years at 6.6% compounded monthly. After 8 years interest for the account is to be 9.66% compounded quarterly and the account is to be paid out by equal end-of-quarter quarterly payments over 6 years. What is the size of the quarterly payment? Answer: p =

- 1 = 0.03345709

FVg(due) = 1280.00

(1.03345709) = 27403.23

27403.23 = PMT 27403.23 = PMT(18.0540866) 1517.84 = PMT The size of the quarterly payment is $1517.84 Diff: 3 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 57) Amy contributed $1011.00 at the end of every six months for 21.5 years into an RRSP earning interest at 6.65% compounded semi-annually. Seven years after the last contribution, Amy converted the RRSP into an RRIF that is to pay her equal quarterly amounts for 17.25 years. If the first payment is due three months after the conversion into the RRIF and interest on the RRIF is 7.66% compounded quarterly, how much will Amy receive every three months? Answer: FVn = 1011 = 1011(92.680366) = $93 699.85 FV = 93699.85(1.03325)14 = 93699.85(1.5807959) = $148 120.34

148120.34 = PMT 148120.34 = PMT(38.1133915) $3886.31 = PMT Diff: 3 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities.

58) Starting three months after her grandchild Crystal's birth, Mrs. Robinson made deposits of $63.25 into a trust fund every three months until Crystal was 19 years old. The trust fund provides for equal withdrawals at the end of each quarter for 5 years beginning three months after the last deposit. If interest is 8.24% compounded quarterly, how much will Crystal receive every three months? Answer: FVn = 63.25 = 63.25(180.0981175) = $11391.21 PVn = 11391.21 = PMT 11391.21 = PMT(16.2571498) $700.69 = PMT Diff: 3 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 59) Karen Peters bought their neighbor's farm for $450 000.00 down and payments of $67100.00 at the end of every six months for 7 years. What is the purchase price of the farm if the semi-annual payments are deferred for four years and interest is 5.5% compounded semi-annually? Answer: PVn = 67100.00 = 67100.00(11.4910081) = $771046.64 PV = 771046.64(1.0275)-8 = 771046.64(0.8049064) = 620620.34

Purchase price = 620620.34 + 450000 = $1 070 620.34 Diff: 2 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 60) Mike acquired his sister's share of their business by agreeing to make payments of $4711.00 at the end of each year for 8 years. If the payments are deferred for three years and money is worth 5.96% compounded quarterly, what is the cash value of the sister's share of the business? Answer: p =

- 1 = 0.0609453

PVn = 4711.00 = 4711.00(6.1866279) = $29 145.20 PV = 29145.20(1.0609453)-3 = 29145.20(0.837377) = $24 405.52

Diff: 2 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities.

61) The Saskatchewan Junior Achievers need to borrow to finance a project. Repayment of the loan involves payments of $8711.00 at the end of every three months for six years. No payments are to be made during the development period of three years. Interest is 9.24% compounded quarterly. a) How much did the Achievers borrow? b) What amount will be repaid? c) How much of that amount will be interest? Answer: a) PVn = 8711.00 = 8711.00(18.2662014) = $159116.88 PV = 159116.88(1.0231)-12 = 159116.88(.7602965) = $120 976.01

b) Total repaid = 24(8711.00) = $209 064.00 c) Interest = 209064.00 - 120976.01 = $88 087.99 Diff: 2 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 62) Equal payments are to be made at the end of each month for 12 years with interest at 9.66% compounded monthly. After the last payment, the fund is to be invested for 10 years at 6.82% compounded quarterly and have a maturity value of $27 600.00. What is the size of the monthly payment? Answer: Balance after 15 years: PV = 27600.00(1.01705)-40 = 27600.00(.50851976) = $14 035.15 Deposits:

FVn = 14035.15 = PMT 14035.15 = PMT(269.891221) $52.00 = PMT Diff: 3 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 63) Flin Flon Aqua Club bought swimming equipment on a contract requiring monthly payments of $815.50 for 4 years beginning 19 months after the date of purchase. What was the cash value of the equipment if interest is 7.8% compounded monthly? Answer: PVn = 815.50

= 815.50(41.119857) = $33 533.24

PV = 33533.24(1.0065)-18 = $29 841.97 Diff: 2 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities.

64) A sum of $15 700.00 is invested at 6.48% compounded semi-annually for eight years. After the eight years, the balance in the fund is converted into an annuity paying equal payments at the end of every 6 months for 5.5 years. If interest on the annuity is 7.95% compounded monthly, what is the size of the equal payments? Answer: FV = 15700.00(1.0324)16 = 15700.00(1.6655893) = $26 149.75 p=

- 1 = 0.0404142

26149.75 = PMT 26149.75 = PMT(8.7409393) $2991.64 = PMT Diff: 3 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 65) Oakville Aqua Club bought swimming equipment on a contract requiring monthly payments of $1000 for 5 years beginning 6 months after the date of purchase. What was the cash value of the equipment if interest is 12% compounded monthly? Answer: PVn = 1000

= 1000(44.95503841) = $44 955.04

PV = 44955.04 (1.01)-5 = 44955.04(0.951465687) = $42 773.18 Diff: 2 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 66) A sum of money is deposited at the end of every month for 9 years at 8.5% compounded monthly. After the last deposit, interest for the account is to be 8.1% compounded quarterly and the account is to be paid out by end-of-quarter quarterly payments of $1800.00 over 7 years. What is the size of the monthly deposit? Answer: PVn = 1800 = 1800(21.2125066) = $38182.51 = FVn 38182.51 = PMT 38182.51 = PMT(161.3936213) $236.58 = PMT The monthly deposit is $236.58. Diff: 3 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities.

67) The local high school soccer club need to borrow to finance a new soccer field. Repayment of the loan involves payments of $10 000.00 at the end of every three months for eight years. No payments are to be made during the development period of five years. Interest is 8% compounded quarterly. a) How much did the Achievers borrow? b) What amount will be repaid? c) How much of that amount will be interest? Answer: a) PVn = 1000.00 = 10000.00(23.46833482) = $234683.35 PV = 234683.35(1.02)-20 = 234683.35(.6729713331) = $157 935.17

b) Total repaid = 32(10000) = $320 000 c) Interest = 320000 - 157935.17 = $162 064.83 Diff: 2 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 68) Paul received an inheritance payment of $38 000.00 at age 50. He invested that sum of money at 8.5% compounded semi-annually until he was 65. At that time he converted the existing balance into an ordinary annuity paying $9150.00 every 6 months, with interest at 5% compounded semi-annually. How many payments will be made? Answer: FV = 38000.00(1.0425)30 = 38000.00(3.485635) = $132 254.13 = PVn n= n=

= 18.16

Diff: 3 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities.

69) A sum of money is deposited at the end of every 6 months for 9.5 years at 8% compounded semiannually. After the last deposit, interest for the account is to be 4.5% compounded quarterly and the account is to be paid out by making payments of $4360.00 at the end of each month over 7.75 years. What is the size of the semi-annual deposit? Answer: p =

- 1 = 0.003736025

PVn = 4360.00 = 4360.00(78.44041381) = $342000.20 = FVn 342000.20 = PMT 342 000.20 = PMT(27.6712294) $12 359.42 = PMT Diff: 3 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 70) The amount of $75 400.00 is invested at 9.65% compounded quarterly for 3 years. After 3 years the balance in the fund is converted into an annuity. If interest on the annuity is 6.75% compounded semiannually and payments are made at the end of every six months for 8 years, what is the size of the payments? Answer: FV = 75400.00(1.024125)12 = 75400.00(1.3311764) = $100 370.70 = PVn

100370.70 = PMT 100370.70 = PMT(12.2084378) $8221.42 = PMT Diff: 3 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities.

71) Monica would like to receive $3760.00 at the end of every six months for 8 years after her retirement. If she retires ten years from now and interest is 6.5% compounded semi-annually, how much must she deposit into an account every six months starting now? Answer: PVn = 3760.00 = 3760.00(12.3243576) = $46339.58 = FVn (due) 46339.58 = PMT

(1.0325)

46339.58 = PMT(27.5642438)(1.0325) $1628.23 = PMT Diff: 2 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 72) Jack took a loan of $140 000.00 to rebuild his cottage. The loan is to be repaid in installments due at the end of each month for 10 years. If the payments are deferred for a year and interest is 7.28% compounded quarterly, what is the size of the monthly payments? Answer: FV = 140000.00(1.0182)4 = 140000.00(1.074812) = $150 473.63 = PVn

- 1 = 0.0060302

150473.63 = PMT 150473.63 = PMT(85.2299222) $1765.50 = PMT Diff: 3 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities.

73) Mark contributed $270.00 at the end of every 3 months for 5.75 years into an RRSP earning 12.6% interest compounded quarterly. Four years after the last contribution he converted the RRSP into a RRIF that is to pay him quarterly payments for 29.5 years. If the first payment is due three months after the conversion into the RRIF and interest on the RRIF is 9.5% compounded quarterly, how much will Mark receive every 3 months? Answer: FVn = 270.00 270.00(33.0400611) = $8920.82 FV = 8920.82(1.0315)16 = 8920.82(1.6425088) = $14 652.53 = PVn 14652.53 = PMT 14652.53 = PMT(39.4661717) $371.27 = PMT Diff: 3 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 74) $6420.00 was invested at a fixed rate of 6.5% compounded semi-annually for 5 years. After 5 years, the fund was converted into an ordinary annuity paying $550.00 per month. If interest on the annuity was 6% compounded monthly, what was the term of the annuity? Answer: FV = 6420(1.0325)10 = 6420(1.3768943) = $8839.66 = PVn

n= n=

= 16.7966

The annuity term is 17 months. Diff: 3 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities.

75) Mr. Caldwell received a retirement gratuity of $481 700.00, which he deposited in an RRSP. He intends to leave the money for 6 years, then transfer the balance into an RRIF and make equal withdrawals at the end of every six months for 25 years. If interest is 8.5% compounded semi-annually, what will be the size of each withdrawal? Answer: FV = 481700.00(1.0425)12 = 481700.00(1.6478314) = $793 760.39 = PVn

793760.39 = PMT 793760.39 = PMT(20.5930613) $38 545.04 = PMT Diff: 2 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 76) Mrs. Lavergne deposits $562.71 at the beginning of every three months. Starting three months after the last deposit, she intends to withdraw $587.00 every three months for 16 years. If interest is 8.14% compounded quarterly, for how long must Mrs. Lavergne make deposits? Answer: PVn = 587.00 = 587.00(35.6040979) = $20899.61 = FVn (due) 20899.61 = 562.71

(1.02035)

36.4002511 = 0.7407451 = 1.02035n - 1 1.7407451 = 1.02035n

ln 1.7407451 = n ln 1.02035 0.5543132 = n(0.0201457) n = 27.51521 ≈ 28 quarters Diff: 3 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities.

77) TJ invested a retirement gratuity of $43 000.00 in an income averaging annuity paying 7% compounded monthly. He withdraws the money in equal monthly amounts over five years. If the first withdrawal is made nine months after the deposit, what is the size of each withdrawal? Answer: FV = 43000(1.0058333)8 = 43000(1.0476304) = $45048.12 = PVn

45048.12 = PMT 45048.12 = PMT(50.5020416) $892.01 = PMT Diff: 2 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 78) Anne-Marie received $445 000.00 from her mother's estate. She wants to set aside part of her inheritance for her retirement 12 years from now. At that time she would like to receive a pension supplement of $2600.00 at the end of each month for 15 years. If the first payment is due one month after her retirement and interest is 3.6% compounded monthly, how much must Anne-Marie set aside? Answer: PVn = 2600 = 2600(138.9268243) = $361209.74 =FV PV = 361209.74(1.003)-144

= 361209.74(.6496294) = $234 652.45 Diff: 3 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities.

79) Mrs. Martin contributes $1830.00 at the beginning of every three months to an RRSP. Interest on the account is 7.44% compounded quarterly. a) What will the balance in the account be after 6 years? b) How much of the balance will be interest? c) If Mrs. Martin converts the balance after 6 years into an RRIF paying 5.41% compounded quarterly and makes equal quarterly withdrawals for 11 years starting three months after the conversion into the RRIF, what is the size of the quarterly withdrawal? d) What is the combined interest earned by the RRSP and the RRIF? Answer: a) FVn (due) = 1830

(1.0186)

= 1830.00(29.9076016)(1.0186) = $55 748.91 b) 55748.91 - 24(1830.00) = 55748.91 - 43920.00 = $11 828.91 c) 55748.91 = PMT 55748.91 = PMT(32.9973043) $1689.50 = PMT d) Total interest = 44(1689.50) - 24(1830.00) = 74338.00 - 43920.00 = $30 418.00 Diff: 3 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 80) Beverley intends to retire in 7.25 years and would like to receive $2700.00 every month for 12.25 years starting on the date of her retirement. How much must she deposit in an account today if interest is 11.36% compounded annually? Answer: p = PVg (due) = 2700

- 1 = 0.00900682 (1.0090068)

= 2700(81.3107476)(1.00900682) = $221516.02 PV = 221516.02(1.00900682)-87 = 221516.27(.4583676) = $101 535.59

Diff: 2 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

81) Jennifer intends to retire in 15 years and would like to receive $5000.00 every month for 20 years starting on the date of her retirement. How much must she deposit in an account today if interest is 9% compounded annually? Answer: p =

- 1 = 0.0072073233

PVg (due) = 5000

× 1.0072073233

= 5000.00(114.362113)(1.0072073233) = $575 931.79 PV = 575931.79 (1.0072073233 )-180 = 575931.79(0.274536) = $158 115.19

Diff: 2 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 82) Sheila intends to retire in 14 years and would like to receive $2472.00 every six months for 9 years starting on the date of her retirement. How much must Sheila deposit in an account today if interest is 6.65% compounded semi-annually? Answer: PV (due) = 2472

(1.03325)

= 2472(13.383091)(1.03325) = $34183.01 PV = 34183.01(1.03325)-28 = 34183.01(.4001736) = $13 679.14

Diff: 2 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 83) Sandra intends to retire in 15 years and would like to receive $3000.00 every six months for 19 years starting on the date of her retirement. How much must Sandra deposit in an account today if interest is 8% compounded semi-annually? Answer: PV(due) = 3000

(1.04)

= 3000(12.65929697)(1.04) = $39497.01 PV = 39497.01(1.04)-30 = 39497.01(.308318668) = $12 177.67

Diff: 2 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

84) Sundeep wants to withdraw $1000.00 at the beginning of each quarter for twelve years. If the withdrawals are deferred to begin ten years from now and interest is 4.5% compounded monthly calculate the amount that must be invested today to be able to make the withdrawals. Answer: Step 1: PMT = 1000.00; n = 12(4) = 48; i =

= 0.00375; c =

= 3; I/Y = 4.5; P/Y = 4; C/Y = 12

p = (1.00375)3 - 1 = 0.01129224 PV(due) = 1000

= $37 314.82

Step 2: FV = 37314.82; n = 10(4) = 40

PV = 37314.82(1 - 1.01129224)-40 = $23 813.01 Programmed solution:

Diff: 2 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 85) The sale of property provides for payments of $4500.00 due at the beginning of every three months for five years. If the payments are deferred for two years and interest is 6% compounded monthly, what is the cash value of the property? Answer: p = PVg (due)= 4500

- 1 = 0.015075125 (1.015075125)

= 4500(17.15593096)(1.0150751) = $78 365.51 PV = 78365.51(1.015075125)-8 = 78365.51(0.887185668) = $69 524.76

Diff: 3 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

86) The sale of property provides for payments of $10 000.00 due at the beginning of every three months for ten years. If the payments are deferred for five years and interest is 12% compounded monthly, what is the cash value of the property? Answer: p = PVg (due)= 10000

- 1 = 0.030301 (1.030301)

= 10000(23.00271345)(1.030301) = $236997.19 PV = 236997.19(1.030301)-20 = 236997.19(0.5504496159) = $130 455.01

Diff: 3 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 87) Erica intends to retire in 37.5 years and would like to receive $6600.00 every 6 months for 14 years, starting on the date of her retirement. How much must she deposit in an account today if interest is 6.25% compounded semi-annually? Answer: PV (due) = 6600

(1.03125)

= 6600(18.480549)(1.03125) = $125783.24 PV537-543 = 125783.24(1.03125)-75 = 125783.24(.0994725) = $12 511.97

Diff: 2 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 88) Paul intends to retire in 15 years and would like to receive $8000.00 every 6 months for 20 years, starting on the date of his retirement. How much must he deposit in an account today if interest is 5% compounded semi-annually? Answer: PV (due) = 8000

(1.025)

= 8000(25.10277505)(1.025) = $205 824.76 PV = 205824.76(1.025)-30 = 205824.76(0.4767426852) = $98 125.45

Diff: 2 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

89) An investment in a lease offers returns of $3713 per month due at the beginning of each month for 4.5 years. What investment is justified if the returns are deferred for two years and the interest required is 6.12% compounded monthly? Answer: i = 0.0612 ÷ 12 = 0.0051 PVn (due) = 3713

(1.0051)

= 3713(47.0980497)(1.0051) = $175 766.92 PV = 175766.92(1.0051)-24 = 175766.92(.8850696) = $155 565.96

Diff: 2 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 90) An investment in a lease offers returns of $4000 per month due at the beginning of each month for 5 years. What investment is justified if the returns are deferred for three years and the interest required is 12% compounded monthly? Answer: i = 0.12 ÷ 12 = 0.01 PVn (due) = 4000

(1.01)

= 4000(44.95503841)(1.01) = $181 618.36 PV = 181618.36 (1.01)-36 = 181618.36 (.6989249496) = $126 937.60

Diff: 2 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 91) What is the principal invested at 6.5% compounded semi-annually from which monthly withdrawals of $840.00 can be made at the beginning of each month for 15.5 years but deferred for 2 years? Answer: p = PVg (due) = 840

- 1 = 0.00534474 (1.00534474)

= 840(117.6803282)(1.00534474) = $99 379.81 PV = 99379.81(1.00534474)-24 = 99380.12(0.879913049) = $87 445.59

Diff: 3 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

92) What is the principal invested at 8% compounded semi-annually from which monthly withdrawals of $1000.00 can be made at the beginning of each month for 10 years but deferred for 3 years? Answer: p =

- 1 = 0.006558197

PVg (due) = 1000

(1.006558197)

= 1000(82.89062668)(1.006558197) = $83 434.24 PV = 83,434.24 (1.006558197)-36 = 83333.65 (.790314523) = $65 939.29

Diff: 3 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 93) A lease requires semi-annual payments of $6100.00 for 7.5 years. If the first payment is due in 3 years and interest is 9.72% compounded monthly, what is the cash value of the lease? Answer: p =

- 1 = 0.04959484

PVg (due) = 6100

(1.04959484) = 6100(10.40814249)(1.04959484) = $66 638.43

PV = 66638.43(1.04959484)-6 = 66638.43(0.747945354) = $49 841.90 Diff: 3 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 94) Payments on a six-year lease valued at $12 975.00 are to be made at the beginning of each month during the last 3 years of the lease. If interest is 9% compounded monthly, what is the size of the monthly payments? Answer: FV = 12975.00(1.0075)36 = 12975.00(1.3086454) = $16979.67 = PVn (due)

16979.67 = PMT

(1.0075)

16979.67 = PMT(31.4468053)(1.0075) $535.93 = PMT Diff: 2 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

95) Joshua deposited $3560.00 at the beginning of every 6 months for 11 years into a fund paying 3.5% compounded semi-annually. Fourteen years after the first deposit he converted the then existing balance into an annuity due paying him equal annual payments for 22 years. If interest for the annuity due is 5% compounded annually, what is the size of the annual payment? Answer: FVn(due) = 3560

(1.0175) = 3560.00(26.5559262)(1.0175)

= $96193.53 FV = 96193.53(1.0175)6 = 96193.53(1.1097024) = $106 746.19 FVn (due) = 106746.19 = PMT

(1.05)

106746.19 = PMT(13.1630026)(1.05) $7723.39 = PMT Diff: 3 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 96) Mr. Dawson wants to receive payments of $1230.00 at the beginning of every three months for 19 years starting on the date of his retirement. If he retires in 21 years, how much must he deposit in an account at the beginning of every three months if interest on the account is 6.84% compounded quarterly? Answer: PVn (due) = 1230

(1.0171)

= 1230(42.3594904)(1.0171) = $52993.12 = FVn (due) 52993.12 = PMT

(1.0171)

52993.12 = PMT(184.4894598)(1.0171) $282.41 = PMT Diff: 3 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

97) An annuity provides payments of $4580.00 at the end of every three months. The annuity is bought for $53 500.00 and payments are deferred for 10 years. If interest is 9.72% compounded monthly, for how long will payments be received? Answer: p =

- 1 = 0.0244974

FV = 53500.00(1.0081)120 = 53500.00(2.6328964) = $140859.96 = PVn

n =

= 57.8502657 ≈ 58 quarters

98) If you save $150.00 at the beginning of each quarter for 20 years, for how long can you withdraw $800.00 at the beginning of each quarter, starting 20 years from now, and assuming that interest is 9% compounded quarterly? Answer: PMT = 150.00; n = 20(4) = 80; i = FV = 150.00(1 + 0.0225)

= 2.25% = 0.0225; I/Y = 9; P/Y = 4; C/Y = 4 = 150(1.0225)(219.1175688) = 33607.16

PV = 33607.16; PMT = 800.00; 33607.16 = 800(1.0225) 0.924402323-1 = -(1.0225)-n -n ln 1.0225 = -ln 0.075597677 -n = n = 116.0565865 (quarters) ÷ 4 ≈ 29 years Programmed solution:

Diff: 3 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

99) Fraser deposits $790.00 at the beginning of every three months. He wants to build up his account so that he can withdraw $1550.00 every three months starting three months after the last deposit. If he wants to make the withdrawals for 16 years and interest is 6.64% compounded quarterly, for how long must Fraser make the quarterly deposits? Answer: PVn (due) = 1550.00

(1.0166)

= 1550.00(39.237778)(1.0166) = $61 828.14 = FVn (due) 61828.14 = 790.00

(1.0166)

76.986256 = 1.2779719 = 1.0166n - 1 ln 2.2779719 = n ln 1.0166 0.8232855 = n(0.0164637) 50.01 deposits = n Approximately 12.5 years. Diff: 3 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 100) Mrs. Muirhead paid $27 700.00 into a retirement fund paying interest at 9.32% compounded semiannually. If she retires in 13.5 years, for how long can Mrs. Muirhead withdraw $5000.00 from the fund every six months? Assume that the first withdrawal is on the date of retirement. Answer: FV = 27700(1.0466)27 = 27700.00(3.4204222) = $94 745.70 = PVn (due) 94745.70 = 5000

(1.0466)

18.1054271 = 0.843712902 = 1 - 1.0466-n -n ln 1.0466 = ln.156287097 -n(0.045546814) = -1.856060596 n = 40.75 semi-annual periods = 20.4 years Diff: 2 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

101) The sum of $59 760.00 is invested at 9.5% compounded monthly for four years. After the four years, the balance in the fund is converted into an annuity due paying annually $15 240.00. If interest on the annuity is 9.57% compounded annually, what is the term of the annuity? Answer: FV = 59760.00(1.0079167)48 = 59760.00(1.4601006) = $87255.47 = PVn (due)

87255.47 = 15240.00

(1.0957)

5.225357799 = 0.500066741 = 1 - 1.0957-n -n ln 1.0957 = ln .499933258 -n(0.0913934) = -0.693280672 n ≈ 7.6 years Diff: 3 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 102) A lease valued at $47 600.00 requires payments of $4500.00 every three months. If the first payment is due 2 years after the lease was signed and interest is 6.88% compounded quarterly, what is the term of the lease? Answer: FV = 47600.00(1.0172)8 = 47600.00(1.1461747) = $54 557.91 = PVn (due) 54557.91 = 4500

(1.0172)

11.91897365 = .205006346 = 1 - 1.0172-n -n ln 1.0172 = ln 0.794993653 -n(0.01705375) = -0.229421147 n = 13.4528 ≈ 13 quarters Diff: 3 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

103) A retirement gratuity of $123 900.00 is invested in annuity deferred for 10 years. The annuity provides payments of $9600.00 due at the beginning of every six months. If interest is 7.41% compounded annually, for how long will annuity payments be made? Answer: p =

- 1 = 0.03638796

FV = 123900.00(1.03638796)20 = 123900.00(2.043841277) = $253 231.93 = PVg (due) 253231.93 = 9600

(1.03638796)

25.45217342 = 0.926152623 = 1 - 1.03638796-n -n ln 1.03638796 = ln 0.073847376 -n(0.0357416) = -2.605754795 n = 72.9054766 semi-annual periods ≈ 36.5 years Diff: 3 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 104) A debt of $23 900.00 is repaid by making payments of $3617.00 at the beginning of every six months deferred for three years. If interest is 9.6% compounded monthly, for how long will payments have to be made? Answer: p =

- 1 = 0.0489703

FV = 23900.00(1 + .008)36 = 23900(1.3322298) = $31840.29 = PVg (due) 31840.29 = 3617.00

(1.0489703)

8.3919969 = .4109586 = 1 - 1.0489703-n -n ln 1.0489703 = 0.5890414 -n(0.047809) = -0.5292588 n = 11.0702755 ≈ 11 semi-annual periods Diff: 3 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

105) Carla Craig invested a retirement gratuity of $17 570.00 in an RRSP paying 6.5% compounded semiannually for 11.5 years. At the end of 11.5 years, she rolled the RRSP balance over into an RRIF paying $1500.00 at the beginning of each month starting with the date of rollover. If interest on the RRIF is 9.5% compounded monthly, for how long will Carla receive monthly payments? Answer: FV = 17570(1 + .0325)23 = 17570(2.0867546) = $36664.28 = PVn (due)

36664.28 = 1500

(1.007916667)

24.2508673 = 0.191986032 = 1 - 1.0079167-n -n ln 1.007916667 = ln 0.808013967 -n(0.007885494) = -0.213175934 n = 27.0333934 ≈ 27 months Diff: 3 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 106) Note: The calculations for this question were done using Excel's RATE function. A vacation property valued at $62 000.00 was bought for 40 payments of $2150.00 due at the beginning of every six months. What nominal rate of interest was charged? Answer: Nper 40 Pmt -2150.00 PV 62000.00 Type 1 Result 0.0180267 The nominal rate is 1.80267% * 2 = 3.60534% Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

107) You bought a camper priced at $15 660.00 on February 1. You agreed to make monthly payments of $575.00 beginning December 1 of the same year. For how long will you have to make these payments if interest is 8.5% compounded monthly? Answer: FV = 15660.00(1.0070833)9 = 15660.00(1.0655861) = $16687.08 = PVn

= 32.60312148

Payments are required for 33 months. Diff: 2 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 108) Shaun's maternal grandmother Abbey gifted a handsome amount of money to Shaun on his birthday. The money is held in a trust fund providing a scholarship of $1000.00, payable at the beginning of each 3 months, if interest is 8.4% compounded quarterly? Answer: PV (due) = 1000.00 +

= 1000.00 + 47619.05 = $48 619.05

Diff: 1 Type: SA Page Ref: 525-531 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities. 109) A rental property provides a monthly income of $1472.00 due at the beginning of every month. What is the cash value of the property if money is worth 7.32% compounded monthly? Answer: PV(due) = 1472.00 +

= 1472.00 + 241311.48 = $242 783.48

Diff: 2 Type: SA Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities. 110) Triad is considering a medical process that is expected to reduce annual costs by $43 000.00. What is the maximum amount of money that could be invested in the process to be economically feasible if interest is 7.44% compounded quarterly? Answer: p = PV =

- 1 = 0.0765016 = $562 079.75

111) Enermax offers to acquire a right-of-way from a property owner who receives annual lease payments of $3725.00 due in advance. What is a fair offer if money is worth 6.5% compounded quarterly? Answer: p = PV (due) = 3725.00 +

- 1 = 0.0666016 = 3725.00 + 55929.59 = $59 654.59

Diff: 2 Type: SA Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities. 112) Aaron Perry bought an income property for $41 321.00 three years ago. He has held the property for the three years without renting it. If he rents the property now, what should be the size of the monthly payment due in advance be if money is worth 6.24% compounded monthly? Answer: FV = 41321.00(1.0052)36 = 41321.00(1.2052837) = $49803.53 = PV(due) 49803.53 = PMT + 49803.53 = PMT + 192.3076923 PMT 49803.53 = 193.3076923 PMT $257.64 = PMT Diff: 3 Type: SA Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities. 113) What monthly lease payment due in advance should be charged for a tract of land valued at $135 000 if the agreed interest is 8.15% compounded semi-annually? Answer: p =

- 1 = 0.006679141

135000 = PMT + 135000 = PMT + 149.7198483 PMT 135000 = (150.7198483)PMT $895.70 = PMT Diff: 2 Type: SA Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities.

114) Geomax pays $8710.00 at the beginning of each year for using a tract of land. What should the company offer the property owner as a purchase price if interest is 10.18% compounded annually? Answer: PV(due) = 8710.00 + = 8710.00 + 85559.92 = $94 269.92 Diff: 1 Type: SA Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities. 115) A fund to provide an annual scholarship of $5700.00 is to be set up. If the first payment is due in three years and interest is 6.4% compounded quarterly, what sum of money must be deposited in the scholarship fund today? Answer: p = PV1 =

- 1 = 0.06555245 = $86 953.27

PV2 = 86953.27(1.06555245)-3 = 86953.27(.82656213) = $71 872.28 Diff: 2 Type: SA Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities. 116) What sum of money must be deposited in a trust fund to provide a scholarship of $630.00 payable at the end of each month if interest is 7.8% compounded monthly? Answer: PV =

= $96 923.08

117) Calculate the amount of money that needs to be invested today at 4.8% compounded semi-annually to provide monthly payments of $500.00 in perpetuity starting: a) one month from today b) one year from today Answer: a)

PMT = 500; i =

= 0.024; c =

p = (1.024)1/6 - 1 = 0.003960577 PV = b)

PV = 500 +

= $126244.24 = 500 + 126244.24 = 126744.24

FV = 126244.24; i = 0.024; n = 1(12) = 12 PV = 126744.24(1.003960577)-12 = $12 0872.73

Diff: 3 Type: SA Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities. 118) Calculate the amount of money that needs to be invested today at 7% compounded semi-annually to provide monthly payments of $2000.00 in perpetuity starting: a) one month from today b) one year from today Answer: a) PMT = 2000; i =

= 0.035; c =

p = (1.035)1/6 - 1 = 0.005750039 PV = b) PV = 2000 +

= $347823.70 = 2000 + 347823.70 = 349823.70

FV = 349823.70; i = 0.035; n = 1(12) = 12 PV = 349823.70(1.005750039)-12 = $326 564.17

119) Carla plans to invest in a property that after 6.25 years will yield $1457.00 at the end of each month indefinitely. How much should Carla be willing to pay if an alternative investment yields 12.96% compounded monthly? Answer: PV =

= $134907.41

PV2 = 134907.41(1.0108)-75 = 134907.41(0.4467942) = $60 275.85 Diff: 2 Type: SA Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities. 120) Note: The calculations for this question were done using Excel's RATE function. What is the effective rate of interest charged on a lease valued at $17 040.00 if payments of $1180.00 are made at the beginning of each 3 months for four years? Answer: Nper 16 Pmt -1180.00 PV 17040.00 Type 1 Result 0.0140451 f = (1 + 0.0140451)4 - 1 = 0.057375 Effective rate = 5.7375% Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

121) Note: The calculations for this question were done using Excel's RATE function. At what nominal rate of interest compounded semi-annually will $2200.00 deposited at the beginning of every six months accumulate to $68 000.00 in 11 years? Answer: Nper 22 PMT -2200.00 FV 68000.00 Type 1 Result .0285814 The nominal rate is 2.85814% × 2 = 5.711628% Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

122) Note: The calculations for this question were done using Excel's RATE function. An insurance policy provides for a lump sum benefit of $450 000.00 fifteen years from now. Alternatively, payments of $25700.00 may be received at the beginning of each of the next fifteen years. What is the effective rate of interest if interest is compounded quarterly? Answer: Nper 15 Pmt = -25700.00 FV = 450000.00 Type 1 Result 0.019101047 The effective annual rate is 1.9101047% Diff: 3 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 123) Note: The calculations for this question were done using Excel's RATE function. What is the nominal annual rate of interest charged quarterly on a lease valued at $2340.00 if payments of $240.00 are made at the beginning of every three months for 3.25 years? Answer: Nper 13 Pmt -240.00 PV 2340.00 Type 1 Result 0 .0522876 Nominal annual rate = 5.22876% × 4 = 20.91503% Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 124) You make six monthly deposits of $125 starting now. The interest rate is 6.36% compounded monthly. How much interest has been earned in total? A) $11.01 B) $9.01 C) $12.10 D) $13.10 E) $14.04 Answer: E Diff: 1 Type: MC Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

125) You start to save for a major purchase. You can invest $329 every three months starting today for 5 years and 9 months. You are able to earn 5.87% compounded semi-annually. What is the amount of interest that you earn during the entire term? A) $2092.49 B) $2029.49 C) $2292.49 D) $2229.49 E) $1475.72 Answer: E Diff: 2 Type: MC Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 126) You start to save for a major purchase. You can invest $500 every three months for 10 years. You are able to earn 8% compounded semi-annually. What is the amount of interest that you earn during the entire term? A) $10 072.94 B) $20 072.94 C) $30 072.94 D) $15 072.94 E) $25 072.94 Answer: A Diff: 2 Type: MC Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 127) Starting in 5 years you want to be able to withdraw $6220 every six months for 7.5 years. You want to deposit a single amount immediately and then let it grow at a rate of 4.88% compounded quarterly. How much do you have deposit? A) $61 216.62 B) $62 216.62 C) $62 116.62 D) $61 126.62 E) $62 226.62 Answer: C Diff: 3 Type: MC Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

128) Starting in 5 years and 6 months you want to be able to withdraw $950 every three months for 3 years and 3 months. You want to deposit a single amount immediately and then let it grow at a rate of 7.24% compounded quarterly. How much interest did you earn during the entire period of time? A) $2090.47 B) $2099.47 C) $4589.47 D) $4859.47 E) $4888.47 Answer: D Diff: 2 Type: MC Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 129) How much extra interest do you gain by starting today rather than waiting one year if you want to deposit $1710 per year for 28 years? You are able to earn 7.07% compounded annually. A) $9689.63 B) $9896.63 C) $9869.63 D) $9889.63 E) $9999.63 Answer: C Diff: 1 Type: MC Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 130) Your child is currently six years old. When he turns 19 you want him to be able to withdraw $14700 per year in order to either travel or go to school for four years. If you are able to earn 6.62% compounded semi-annually for the entire time period, what is the single amount of the deposit three years from now that will ensure that your child is able to do this? A) $28 777.49 B) $28 877.49 C) $27 777.49 D) $27 877.49 E) $29 877.49 Answer: D Diff: 2 Type: MC Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

131) A trust fund will start to pay you $15 200 per year at the beginning of each year once you turn 18. The prevailing interest rate is 8.79% compounded annually. For how long can you receive payments if you are celebrating your 14th birthday today and the current value of your trust fund is $111 345? A) 20.97 years B) 29.07 years C) 20.07 years D) 29.77 years E) 24.97 years Answer: A Diff: 3 Type: MC Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 132) If you want to be able to set up a scholarship fund that will last forever and you feel that you can earn 4.8% compounded annually. How much money do you have to deposit today in order to be able to provide 6 scholarships of $3000 each? A) $735 000.00 B) $375 000.00 C) $37 500.00 D) $357 000.00 E) $333 000.00 Answer: B Diff: 1 Type: MC Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities. 133) Jawad makes semi-annual contributions of $2000 starting on his 27th birthday. How much will he accumulate in his RRSP by age 60, if RRSP earns 8% compounded semi-annually and no contribution is made on his sixtieth birthday? A) $615 535 B) $640 156 C) $565 324 D) $587 937 E) $669 842 Answer: B Diff: 2 Type: MC Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

134) Jawad makes semi-annual contributions of $2000 starting on his 27th birthday. How much will he accumulate in his RRSP by age 60, if RRSP earns 8% compounded annually and no contribution is made on his sixtieth birthday? A) $615 535 B) $640 156 C) $565 324 D) $595 254 E) $618 606 Answer: E Diff: 2 Type: MC Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 135) Jenna intends to contribute $2500 to her RRSP at the beginning of every six months, starting today. The financial adviser has advised Jenna that if she continues her contributions for 20 years, the RRSP would earn 8% compounded semi-annually for the first 7 years and 7% compounded semi-annually, thereafter. What amount will Jenna have in the plan after 20 years? Answer: For the first 7 years, i =

= 4% and n = 2 × 7 = 14

FV after first 7 years of contribution = ⇒

FV1 = $2500

= PMT

(1 + i)

(1.04) = $47 559

The future value of $47 559 after an additional 13 years at i =

= 3.5% and n = 2 × 13 = 26 is:

FV = $47559 × 1.03526 = $116 327 The FV for the last 26 payments will be FV2 = $2500

= $106 898

The total amount in the RRSP after 20 years will be $116327 + $106898 = $223 225 Diff: 3 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

136) Heavy Mechanical Complex (HMC) acquired a giant lathe machine under a capital lease agreement. HMC pays the lessor $2400 at the beginning of every 3 months for 5 years. If HMC can obtain 5-year financing at 10% compounding quarterly, what long-term lease liability will HMC initially record? A) $37 414 B) $48 000 C) $61 307 D) $62 840 E) $38 349 Answer: E Diff: 2 Type: MC Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 137) Heavy Mechanical Complex (HMC) acquired a giant lathe machine under a capital lease agreement. HMC pays the lessor $2400 at the beginning of every 3 months for 5 years. If HMC can obtain 5-year financing at 10% compounding quarterly, what liability will be reported 2 years later? A) $37 414 B) $38 349 C) $33 109 D) $33 937 E) $25 234 Answer: E Diff: 2 Type: MC Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 138) Dell Canada advertises a computer bundle including printer, router and home network attached storage for $2995. The same system bundle may be leased for 24 months at $130 per month, at the beginning of each month. At the end of the lease, the bundle may be purchased for 10% of its retail price. What is the economic cost of the lease or purchase in present value, if a two year loan can be obtained at 12% compounded annually to purchase the computer? A) Lease is better by $48.64 B) Purchase is better by $48.64 C) Lease is better by $109.38 D) Purchase is better by $109.38 E) Both options are the same. Answer: D Diff: 3 Type: MC Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due.

139) Sue won a "Millionaire Life" lottery. As a winner of the Grand Prize, she can choose either $1 million per year for 25 years or a single cash payment of $17 million. What option should be chosen if the payments are made at the beginning of each year and on low risk investments, money can earn 3.2% compounded annually? Answer: PMT = $1 000 000; n = 25; i = 3.2% PV(due)

= PMT = $1 000 000

(1 + i) 1.032

= $17 576 379 ∴ 25 year annuity has $576 379 more economic value than the lump-sum payment. Diff: 3 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 140) Ford Oshawa sells Ford Fusion special edition for a price of $27 900. During a month long promotion, the manufacturer offers an interest rate of 1.8% compounded monthly on a 3-year lease. If the residual value is $14 500, what will be the lease payment at the start of each month with a $2500 down payment? A) $403.82 B) $724.30 C) $350 D) $332.50 E) $311 Answer: D Diff: 3 Type: MC Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 141) Olivia plans to have $1 000 000 in her RRSP at her retirement at the age of 60. How long will it take her to accumulate $1 000 000 in an RRSP if the first quarterly contribution of $2000 is made today and RRSP is expected to earn 8% compounded quarterly? A) 30 years and 3 months B) 30 years 6 months C) 47 years and 3 months D) 29 years E) 28 years 9 months Answer: A Diff: 2 Type: MC Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

142) Jeff is retired and has an investment fund worth $210 000 in an RRSP, which earns 9% compounded semi-annually. He plans to withdraw only $2000 at the beginning of each month starting today. When will the fund become depleted? A) 16 years and 6 months B) 16 years 10 months C) 8 years and 9 months D) 9 years E) 17 years Answer: A Diff: 2 Type: MC Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 143) Cameron is 35 years old and intends to retire at the age of 60. He is planning to contribute $3000 at the beginning of each six-month period to an RRSP. What semi-annually compounded rate of interest must his RRSP earn in order to reach $600 000 after 25 years? A) 9.72% B) 9.43% C) 4.86% D) 4.72% E) 9.52% Answer: B Diff: 2 Type: MC Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 144) Li buys a $100 000 life insurance policy, which requires an annual premium of $420 or a monthly premium of $37. In either case, the premium is payable at the beginning of the period of coverage. What is the effective rate of interest Li pays, if he chooses monthly payment plan? A) 10.89% B) 13.04% C) 10.39% D) 1.09% E) 5.7% Answer: B Diff: 2 Type: MC Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due.

145) Whitby Honda advertised Honda Pilot for sale at $33 610. The same vehicle could also be leased for 5 years at $379 per month, based on $4310 down payment and a residual value of $10 711. What monthly compounded interest rate was built into the lease? A) 4.05% B) 4.13% C) 0.38% D) 7.45% E) 0.9% Answer: B Diff: 2 Type: MC Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 146) Sagheer is setting up a fund to help his grandson's university education. He wants his grandson to be able to withdraw $3000 every 3 months for 3 years after he starts university. His first withdrawal will be 5.5 years from now. If the fund can earn 7.2% compounded quarterly, what single amount contributed today will provide for the withdrawal? A) $32 119 B) $22 083 C) $52 077 D) $42 041 E) $36 000 Answer: B Diff: 2 Type: MC Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 147) Basim purchased a deferred annuity from an insurance company for $10 971. The money used to purchase the annuity will earn 6% quarterly. The annuity will provide 16 quarterly payments of $1000. If the first payment is to be received on October 1, 2017, when did Basim purchase the deferred annuity? A) April 1, 2011 B) April 1, 2013 C) October 1, 2017 D) October 1, 2013 E) June 1, 2014 Answer: B Diff: 2 Type: MC Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities.

148) Leons has a promotion on a refrigerator selling for $1750. Buyers will pay "no money down and no payments for six months." The first of 12 equal monthly payments is required 6 months from the purchase date. What should the monthly payments be if Leons is to earn 15% compounded monthly on its account receivable during both the deferred period and repayment period? A) $1862.14 B) $1885.42 C) $170.17 D) $182.24 E) $168.07 Answer: E Diff: 2 Type: MC Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 149) Asif received a bonus of $10 000, which he immediately invested in a fund earning 9.25% compounded semi-annually through his Tax Free Savings Account. Five years later, Asif plans to make the first semi-annual withdrawal of $1000. After how many withdrawals, the fund will be completely depleted? A) 29 B) 28 C) 27 D) 20 E) 10 Answer: C Diff: 2 Type: MC Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 150) National Research Council is considering the establishment in perpetuity of a Visiting Professor chair at UOIT. The ongoing cost will be $12 500 at the end of each month. If the money can earn 8% compounded monthly in perpetuity, what endowment is required to fund the position? A) $156 250 B) $312 500 C) $937 500 D) $1 875 000 E) $2 000 000 Answer: D Diff: 1 Type: MC Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities.

151) National Research Council is considering the establishment of $2 million endowment held in perpetuity of a Visiting Professor chair at UOIT. The ongoing cost will be $12 500 at the end of each month. What monthly compounded nominal rate of return must be earned to fully fund the position? A) 0.625% B) 6.25% C) 0.075% D) 7.5% E) 6% Answer: B Diff: 1 Type: MC Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities. 152) Goggle Inc. is planning to pay a quarterly dividend of $0.50 on its perpetual preferred share. The market requires a dividend yield of 5% compounded annually on preferred shares of similar risks. What is the fair market value of its perpetual preferred share just after payment of a dividend at the end of the quarter? A) $10 B) $40 C) $40.74 D) $20.25 E) $20 Answer: C Diff: 1 Type: MC Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities. 153) After retirement, Andy plans to buy bonds in a perpetual fund. What amount must Andy place in a perpetual fund today, if it earns 4.8% compounded semi-annually and the first monthly payment of $500 in perpetuity will be made one year from today? A) $126 244 B) $120 873 C) $120 396 D) $119 629 E) $125 000 Answer: B Diff: 2 Type: MC Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities.

154) At the age of 75, Mrs. Walker decided to set up a trust fund for her grandson, which can take care of some of his expenses for the rest of his life. She made an initial contribution of $150 000 to set up the trust fund, which would make the first semi-annual payment of perpetuity to her grandson, 5 years from now. If the funds in trust earn 5% compounded annually, what is the maximum payment the trust can make to Mrs. Walker's grandson in perpetuity? A) $4613.74 B) $4727.68 C) $6033.85 D) $5746.53 E) $4844.43 Answer: A Diff: 2 Type: MC Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities. 155) Sam received a lump sum of $135 000 as an inheritance from his great grandmother. He plans to invest them for 20 years in a long-term investment fund with an expected rate of return of 9% compounded monthly. After 20 years, what payment can he withdraw at the beginning of each month for 10 years? A) $10 200 B) $7245 C) $2955 D) $4161 E) $66 985 Answer: A Diff: 2 Type: MC Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 156) When Nancy died, she left $20 000 for her 13 year old grandson, Max, to be paid to him as an annual scholarship of $8000 at the beginning of each year, if he ends up in the university. The inheritance was invested in a scholarship fund for 5 years at 9% compounded annually. How long will the scholarship be paid, once Max enters the university? A) 3 years B) 4 years C) 5 years D) 6 years E) 7 years Answer: C Diff: 2 Type: MC Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

157) OPG has set up a fund to help pay the expenses for the monitoring of Pickering nuclear power plant once it is in safe storage. The balance in the fund today is $700 million, and it will earn interest at 7% compounded annually. Starting 10 years from now, $100 million will be withdrawn from the fund at the beginning of every year for 30 years. What is the annually compounded interest rate for the fund during the final 30 years? A) 6.61% B) 4.33% C) 0.72% D) 0.67% E) 7% Answer: A Diff: 2 Type: MC Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 158) Sam got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded monthly. If he puts the money in his savings account at the beginning of each month, what will be the balance in the account at the end of the three-year term? A) $18 408.04 B) $18 600.50 C) $18 603.37 D) $18 638.25 E) $17 390.19 Answer: B Diff: 1 Type: MC Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 159) Sam got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded monthly. If he puts the money in his savings account at the beginning of each month, what will be the present value of the balance in the account at the end of the three-year term? A) $17 390.19 B) $18 408.04 C) $18 603.37 D) $18 638.25 E) $17 422.79 Answer: E Diff: 2 Type: MC Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

160) Keith bought a lakeside cottage for 30% down and monthly mortgage payments of $1224.51 at the beginning of each month for 20 years. Interest is 4.4% compounded monthly. What is the purchase price of the property? A) $194 378 B) $195 214 C) $278 878 D) $195 930 E) $279 900 Answer: E Diff: 2 Type: MC Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 161) Jamal plans to retire in 17 years. He is saving $2000 every start of the month in a retirement savings account paying him a long-term interest of 9% compounded monthly. What will be the size of his payments at the start of each month from the ordinary simple annuity for 20 years following his retirement? A) $957 836 B) $965 020 C) $8618 D) $8682 E) $2000 Answer: D Diff: 2 Type: MC Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 162) Siri plans to retire when her simple annuity savings account has enough money to receive $10 000.00 per month for twenty years starting at the start of the first month after her retirement. She starts saving $4420 per month. Calculate when should Siri retire from today if her savings account pays 4.9% compounded monthly. A) 18 years B) 216 years C) 2.41 year D) 1.41 years E) 36 years Answer: A Diff: 2 Type: MC Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

163) Jamal plans to retire in 17 years. He is saving $2000 every start of the month in a retirement savings account paying him a long-term interest of 9% compounded monthly until retirement. The rate changes following the retirement. He wants $7000 per month paid to him at the start of the month for 20 years after the retirement. At what rate should his savings account pay him to fulfill his dream? (Use the Rate function in Excel) A) 9% B) 10% C) 6.22% D) 10.27% E) 0% Answer: C Diff: 2 Type: MC Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 164) Sam got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded semi-annually. If he puts the money in his savings account at the beginning of each month, what will be the balance in the account at the end of the three-year term? A) $18 635.22 B) $18 600.50 C) $18 603.37 D) $18 638.25 E) $17 390.19 Answer: A Diff: 2 Type: MC Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 165) Sam got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded monthly. If he puts the money in his savings account at the beginning of each month, what will be the present value of the balance in the account at the end of the three-year term? A) $17 390.19 B) $17,425.42 C) $18 603.37 D) $18 638.25 E) $17 422.79 Answer: B Diff: 2 Type: MC Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due.

166) Keith bought a lakeside cottage for 30% down and monthly mortgage payments of $1224.51 at the beginning of each month for 20 years. Interest is 4.4% compounded semi-annually. What is the purchase price of the property? A) $194 378 B) $196 593 C) $280 847 D) $195 930 E) $279 900 Answer: C Diff: 2 Type: MC Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 167) Jamal plans to retire in 17 years. He is saving $2000 every start of the month in a retirement savings account paying him a long-term interest of 9% compounded quarterly. What will be the size of his payments at the start of each month from the annuity for 20 years following his retirement? A) $958 266 B) $965 020 C) $8618 D) $8682 E) $8517 Answer: E Diff: 2 Type: MC Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 168) Siri plans to retire when her simple annuity savings account has enough money to receive $10 000.00 per month for twenty years starting at the start of the first month after her retirement. She starts saving $4420 per month. Calculate when should Siri retire from today if her savings account pays 4.9% compounded quarterly. A) 18 years B) 216 years C) 18 years 6 months D) 1.41 years E) 36 years Answer: C Diff: 2 Type: MC Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due.

169) Jamal plans to retire in 17 years. He is saving $2000 every start of the month in a retirement savings account paying him a long-term interest of 9% compounded quarterly until retirement. The rate changes following the retirement. He wants $7000 per month paid to him at the start of the month for 20 years after the retirement. At what rate should his savings account pay him to fulfill his dream? (Use the Rate function in Excel.) A) 9% B) 10% C) 6.22% D) 10.27% E) 6.34% Answer: E Diff: 2 Type: MC Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 170) Knowing that the education costs for her kids would be huge, Roberta puts a lump sum of money from her inheritance in a savings account paying an interest of 4.69% compounded monthly. If ordinary annuity payments of $1500.00 per month are to be paid out of the account at the end of the month for four years starting fifteen years from now, how much did Roberta deposit? A) $65 533 B) $32 474 C) $65 592 D) $32 723 E) $54 343 Answer: B Diff: 3 Type: MC Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 171) Knowing that the education costs for her kids would be huge, Roberta puts lump sum money from her inheritance in a savings account paying an interest of 4.69% compounded semi-annually. If ordinary annuity payments of $1500.00 per month are to be paid out of the account at the end of the month for four years starting fifteen years from now, how much did Roberta deposit? A) $65 533 B) $32 474 C) $65 592 D) $32 723 E) $54 343 Answer: D Diff: 3 Type: MC Page Ref: 485-492 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities.

172) Siri plans to retire in 18 years and would like to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. Calculate the amount he must invest now if interest is 4.9% compounded monthly. A) $1 528 015 B) $594 425 C) $633 665 D) $574 626 E) $539 042 Answer: C Diff: 2 Type: MC Page Ref: 513-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 173) Siri plans to retire in 18 years and would like to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. Calculate the amount he must invest now if interest is 4.9% compounded quarterly. A) $1 528 015 B) $1 530 569 C) $633 665 D) $636 990 E) $539 042 Answer: D Diff: 2 Type: MC Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 174) Rob bought a yacht on a contract requiring monthly payments of $2500 for 5 years beginning on the 6th month after the date of purchase. What was the cash value of the yacht, if interest is 6.9% compounded monthly? A) $126 556 B) $122 980 C) $122 277 D) $135 267 E) $136 979 Answer: B Diff: 3 Type: MC Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities.

175) Abe deposited a sum of money at the end of every month for 15 years at 8.5% compounded monthly. After the last deposit, interest for the account is to be 3.5% compounded quarterly and the account is to be paid out by end-of-quarter quarterly payments of $5905.00 over 17 years. What is the size of the monthly deposit? A) $560 735 B) $1550 C) $756 D) $1160 E) $1466 Answer: B Diff: 3 Type: MC Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 176) Witzke retired from Ontario Hydro and received a payment of $980 000.00 at age 56. He invested that sum of money at 5.5% compounded semi-annually until he was 65. At that time he converted the existing balance into an ordinary annuity paying $10 250.00 every month, with interest at 5% compounded monthly. How many payments will be made? A) 252 B) 21 C) 271 D) 183 E) 23 Answer: A Diff: 3 Type: MC Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 177) Rousan contributed $1900.00 at the beginning of every month to an RRSP. Interest on the account is 4.95% compounded monthly. If she converted the balance after 13 years into an RRIF paying 7.1% compounded quarterly and makes equal quarterly withdrawals for 20 years starting three months after the conversion into the RRIF, what is the size of the quarterly withdrawal? A) $9789.48 B) $8232.21 C) $11 463.51 D) 3831.66 E) 1071.47 Answer: A Diff: 3 Type: MC Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

178) Babar wants to withdraw $27 000.00 at the beginning of each quarter for twelve years. If the withdrawals are deferred to begin ten years from now and interest is 5.6% compounded monthly calculate the amount that must be invested today to be able to make the withdrawals. A) $950 943 B) $1 859 180 C) $543 895 D) $1 063 365 E) $546 034 Answer: C Diff: 2 Type: MC Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 179) Tom started his career with Oracle with an intention to retire in 25 years. When he retired, he started receiving $66 000.00 every 6 months for 15 years, starting on the date of his retirement. How much did Tom deposit deposit in an account at the start of his career, if interest is 9.25% compounded semiannually? A) $1 108 432 B) $115 594 C) $1 449 334 D) $147 332 E) $79 360 Answer: B Diff: 2 Type: MC Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 180) Payments on a six-year lease value of a car at $27 175.00 are to be made at the beginning of each month during the last 3 years of the lease. If interest is 7.25% compounded monthly, what is the size of the monthly payments? A) $33 755 B) $1039.85 C) $937.65 D) $754.86 E) $1029.42 Answer: B Diff: 2 Type: MC Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due.

181) What sum of money must be deposited in a savings scholarship account to provide a scholarship of $3320.00, payable at the beginning of each 3 months, if interest is 3.59% compounded quarterly? A) $369 916 B) $373 236 C) $92 479 D) $92 799 E) $558 195 Answer: B Diff: 1 Type: MC Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities. 182) What monthly lease payment due in advance should be charged for a house valued at $835 000 if the agreed interest is 4.4% compounded semi-annually? A) $3022.99 B) $17 974.56 C) $3050.48 D) $18 136.54 E) $3039.40 Answer: A Diff: 2 Type: MC Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities. 183) Cote heard about the Tesla truck being launched in 3 years and plans to buy the truck in cash for $78 000. How much does he have to save at the beginning of every three months for 3 years to accumulate $78 000 dollars if interest is 4% compounded quarterly? Answer: FVn (due) = 78000 78000.00 = PMT

(1.01)

78000.00 = PMT(16.09689554)(1.01) $4797.68 = PMT Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

184) Rick wants to celebrate his 25th wedding anniversary in style and has a budget of $50 000. How much does he have to save at the beginning of every three months for 3 years to accumulate $50 000 dollars if interest is 6% compounded monthly? Answer: FVn (due) = 50000 50000.00 = PMT

(1.005)

50000.00 = PMT(39.33610496)(1.005) $1264.77 = PMT Diff: 2 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 185) Lauren bought mutual funds within TFSA as part of her retirement planning in 2019, immediately after graduation at the age of 22. Payments of $500.00 will be made into the fund at the beginning of every month for 33 years. If the fund is expected to earn interest at 8.4% compounded monthly, what will be the balance in the fund after 33 years? How much interest did she earn in 33 years? Answer: FV (due)

= 500.00

(1.007)

= 500.00(2119.558001)(1.007) = $1 067 197.45 Interest = 1067197.45 — (500 × 12 × 33) = $869 197.45 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 186) In the previous question, what is the discounted value of deposits of $500.00 made at the beginning of every month for 33 years if money is worth 8.4% compounded quarterly? Answer: PV (due)

= 500

(1.007)

= 500 × 134.7734782 = $67 386.34 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due.

187) The monthly rent payment on a domestic water heater is $16.99 payable in advance. What yearly payment in advance would satisfy the lease if interest is 6.84% compounded monthly? Answer: PV (due)

= 16.99

(1.0057)

= 16.99(11.5669781)(1.0057) = $197.64 Diff: 1 Type: SA Page Ref: 494-506 Topic: 13.1 Simple Annuities Due Objective: 13-1: Compute the future value, present value, periodic payment, term, and interest rate for simple annuities due. 188) Steve wants to save for a dream vacation with his girlfriend. What deposit made at the beginning of each month will accumulate to $16 000.00 at 5% compounded quarterly at the end of eighteen months? Answer: FV(due) = 16000.00; n = 18; i = c=

= 0.0125; I/Y = 5; P/Y = 12; C/Y = 4

p = (1 + i)c - 1 = (1 + 0.0125)1/3 - 1 = 0.004149425 FV(due) = PMT(1 + p) 16000 = PMT(1 + 0.004149425) 16000 = PMT(1.004149425)[18.64913287] 16000 = PMT(18.72651605) PMT = $854.40 Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 189) Ryan has bought a vacation property in Florida by making semi-annual payments of $9000.00 for 10 years. If the first payment is due on the date of purchase in 2019 and interest is 6% p.a. compounded quarterly, calculate the total amount Ryan paid for the property. Answer: PMT = 9000.00; n = 10(2) = 20; i =

= 0.015; I/Y = 6; P/Y = 2; C/Y = 4; c =

p = (1.015)2 - 1 = 0.030225 PV(due) = 9000

= $137 657.80

Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due.

190) Arjun is 25 years old and planning his retirement. The financial advisor has advised him to accumulate a minimum of $2 150 000.00 in an RRSP to live comfortably after retirement. He can make annual contributions of $15 000.00 at the beginning of each year. If interest is 5.5% compounded quarterly, at what age can he retire. Answer: PMT = 15000.00; FV(due) = 2150000; i =

= 0.01375= 4; I/Y = 5.5; P/Y = 1; C/Y = 4 ; c =

p = (1.01375)4 - 1 = 0.056144809 2150000 = 15000 7.619620486 = 1.056144809n - 1 n ln 1.056144809 = ln 8.619620486 n(0.054625306) = 2.154041057 n = 39.433

n = 40 years (approx.).The last payment will be less than $15000.00.n Arjun can retire at the age of 65 Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 191) Asim bought a cottage property near lake Huron for quarterly payments of $2470.00 for 18 years. If the first payment was made on the date of purchase and interest is 3.57% compounded annually, what was the purchase price of the property? Answer: p = PVg (due) = 2470

- 1 = 0.00880795 (1.00880795)

= 2470(52.71499667)(1.00880795) = $131 352.89 Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due.

192) What is the difference in accumulated value after 15 years of monthly deposits of $995.00 earning interest at 5.8% compounded semi-annually, between the deposits made at the end of each month vs. at the beginning of each month? Answer: p =

- 1 = 0.004775944

FVg = 995

= 995(284.2477768) = $282 826.54

FVg (due) = 995(284.2477768)(1.004775944) = $284 177.30 Difference = 284177.30 — 282826.54 = $1350.76 Diff: 2 Type: SA Page Ref: 508-513 Topic: 13.2 General Annuities Due Objective: 13-2: Compute the future value, present value, periodic payment, term, and interest rate for general annuities due. 193) Mohsin contributed $9100.00 at the end of every six months for 25 years into an RRSP earning interest at 6.65% compounded semi-annually. Two years after the last contribution, Mohsin converted the RRSP into an RRIF that is to pay her equal monthly amounts for 20 years. If the first payment is due a month after the conversion into the RRIF and interest on the RRIF is 7.66% compounded monthly, how much will Mohsin receive every month? Answer: FVn = 9100 FV

= 9100(124.265) = $1 130 811.57 = 1130811.57(1.03325)4 = 1130811.57(1.139782) = $1 288 878.26

0.00638333 240 1288878.26 = PMT 1288878.26 = PMT(122.6385) $10 509.57 = PMT Diff: 3 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities.

194) A month after his grandchild Mobin's birth, Mr. Akbar started a savings accounts to help in Mobin's higher education when he reaches 18 years of age. Mr. Akbar made deposits of $150 into the savings account every month until Mobin was 18 years old. At the age of 18, the savings account was converted into a trust fund providing for equal withdrawals at the end of each month for 4 years beginning one month after the last deposit. If interest is 7.2% compounded quarterly, how much will Mobin receive every month? Answer: FVn = 150 = 150(440.0872454) = $66 013.09 PVn = 66013.09 = PMT 66013.09 = PMT(41.598819) $1586.90 = PMT Diff: 3 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 195) Suncor bought an oil field in Libya for $4.5 billion down and payments of $671 million at the end of every six months for 7 years. What is the purchase price of the oil field if the semi-annual payments are deferred for four years and interest is 5.5% compounded semi-annually? Answer: PVn = 671 = 671(11.4910081) = $7710.47 million -8 PV = 7710.47(1.0275) = 7710.47(0.8049064) = $ 6206.20 million Purchase price = 6206.20 + 4500 = $10 706.20 million or $10.71 billion Diff: 2 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 196) Delawri acquired his partner's share in Dehli Hydrogenated Oil by agreeing to make payments of $471 100.00 at the end of each year for 5 years. If the payments are deferred for three years and money is worth 5.96% compounded quarterly, what is the cash value of the partner's share of the business? Answer: p =

- 1 = 0.0609453

PVn = 471100.00 = 471100.00(4.2015533) = $1 979 351.77 PV = 1979351.77(1.0609453)-3 = 1979351.77(0.837377) = $1 657 463.65

197) The Sask Power needs to borrow to finance the carbon capture project for their coal power plants. Repayment of the loan involves payments of $871 100.00 at the end of every three months for six years. No payments are to be made during the design, development and installation period of three years. Interest is 9.24% compounded quarterly. a) How much did the Sask Power borrow? b) What amount will be repaid? c) How much of that amount will be interest? Answer: a) PVn = 871100.00 = 871100.00(18.2662014) = $15 911 688.04 PV = 15911688.04(1.0231)-12 = 15911688.04(.7602965) = $12 097 600.73

b) Total repaid = 24(871100.00) = $20 906 400 c) Interest = 20906400 - 12097600.73 = $8 808 799.27 Diff: 2 Type: SA Page Ref: 516-523 Topic: 13.3 Ordinary Deferred Annuities Objective: 13-3: Compute the future value, present value, periodic payment, term, and interest rate for ordinary deferred annuities. 198) Dave would like to receive $22 500.00 at the end of every six months for 15 years after his retirement. If he retires 12 years from now and interest is 7% compounded semi-annually, how much must he deposit into an account every six months starting now? Answer: PVn = 22500.00 = 22500 (18.392) = $413 821.02= FVn (due) 413821.02 = PMT

(1.035)

413821.02 = PMT(32.9878333)(1.035) $12 120.44 = PMT Diff: 2 Type: SA Page Ref: 525-531 Topic: 13.4 Deferred Annuities Due Objective: 13-4: Compute the future value, present value, periodic payment, term, and interest rate for deferred annuities due. 199) A farm provides a monthly income of $7972.00 due at the beginning of every month. What is the cash value of the farm if money is worth 5.32% compounded monthly? Answer: PV(due) = 7972.00 +

= 7972.00 + 1798195.49 = $1 806 167.49

200) Gopal invested $40 000 two years ago to pay for his son's tuition fee at the Ivey School of Business. His son dropped out and started his own business. Gopal converted that amount into a trust fund, such that it provides a monthly scholarship to his son. What will be the amount of scholarship if money is worth 6.24% compounded monthly? Answer: FV = 40000.00(1.0052)24 = 40000.00(1.13256) = $45302.22 = PV(due) 45302.22 = PMT + 45302.22 = PMT + 192.3076923 PMT 45302.22 = 193.3076923 PMT $234.35 = PMT Diff: 3 Type: SA Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities. 201) Knowing he has cancer, Dier has created a fund to provide an annual scholarship to his children of $9500.00. If the first payment is due in three years and interest is 5.0% compounded quarterly, what sum of money must be deposited in the scholarship fund today? Answer: p = PV =

- 1 = 0.050945 = $186 474.38 = FV

PV2 = 186474.38(1.050945)-3 = $160 649.44 Diff: 2 Type: SA Page Ref: 533-538 Topic: 13.5 Perpetuities Objective: 13-5: Compute the present value, periodic payment, and interest rate for ordinary perpetuities, perpetuities due, and deferred perpetuities.

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 14 Amortization of Loans, Residential Mortgages, and Sinking Funds 1) A debt of $10 000.00 with interest at 8% compounded quarterly is to be repaid by equal payments at the end of every three months for two years. a) Calculate the size of the monthly payments. b) Construct an amortization table. c) Calculate the outstanding balance after three payments. Answer: a) PV = 10 000.00; n = 2(4) = 8; i =

= 2% = 0.02; I/Y = 8; P/Y = C/Y = 4;

10000 = PMT 10000 = PMT[7.32548144] PMT = PMT = $1365.10 Programmed solution:

b) Amortization Table (done using Excel):

c) PMT = 1365.10; n = 5(number of payments left); i = 0.02 PV = 1365.10

= $6434.34

Programmed solution:

The difference from the chart is due to rounding. Diff: 2 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 2) A contractor's price for a new building was $85 100.00. Stampede Inc., the buyers of the building, paid down and financed the balance by making equal payments at the end of every six months for 14 years. Interest is 9.8% compounded semi-annually. a) What is the size of the semi-annual payment? b) How much will Stampede Inc. owe after 6 years? c) What is the total cost of the building for Stampede Inc.? d) What is the total interest included in the payments? Answer: a) 68850.00 = PMT 68850.00 = PMT(15.06140866) $4571.29 = PMT b) PVn = 4571.29

= 4571.29(8.913452135) = $40745.97

Total paid = 16250.00 + 28(4571.29) = 16250.00 + 127996.12 = $144246.12 d) Total interest = 144246.12 - 85100.00 = $59 146.12 Diff: 2 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

3) Vicki receives payments of $4170.00 at the beginning of each month from a pension fund of $208 500.00. Interest earned by the fund is 6% compounded monthly. What is the number of payments Vicki will receive? Answer: 208500 = 4170

(1.005)

49.75124378 = 0.2487562189 = 1 - 1.005-n -n ln 1.005 = ln 0.7512437811 -n(0.0049875415) = -0.2860250712 n = 57.3479079 payments Diff: 2 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 4) A investor's price for a townhouse was $160 000.00. Sepaba Investments., the buyers of the rental unit, paid $40 000.00 down and financed the balance by making equal payments at the end of every six months for 25 years. Interest is 6% compounded semi-annually. a) What is the size of the semi-annual payment? b) How much will Sepaba Investments. owe after 20 years? c) What is the total cost of the building for Sepaba Investments? d) What is the total interest included in the payments? Answer: a) 120000.00 = PMT 120000.00 = PMT(25.72976401) $4663.86 = PMT b) An = 4663.86

= 4663.86(8.530202837) = $39 783.67

Total paid = 40000.00 + 50(4663.86) = 40000.00 + 233193.00 = $273 193.00 d) Total interest = 273193.00 - 160000.00 = $113 193.00 Diff: 2 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

5) A loan of $14 100.00 is amortized over 11 years by equal monthly payments at 5.4% compounded monthly. Construct an amortization schedule showing details of the first three payments, the fortieth payment, the last three payments, and totals. Answer: 14100 = PMT 14100 = PMT(99.3665357) $141.90 = PMT Partial Amortization Table (rounded to the nearest cent):

PVn = 141.90 = 141.90(75.8552917) = 10763.87 PVn = 141.90 = 141.90(2 .97320114) = $421.90 Diff: 2 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

6) A loan of $8380.00 is repaid by equal payments made at the end of every three months for 3 years. If interest is 7% compounded quarterly, find the size of the quarterly payments and construct an amortization schedule showing the total paid and the total cost of the loan. Answer: 8380.00 = PMT 8380.00 = PMT(10.7395497) $780.29 = PMT Amortization Schedule (done using Excel):

Diff: 2 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

7) Rola Inc. borrowed $42 000.00 at 7% compounded semi-annually. The loan is repaid by payments of $4700.00 due at the end of every six months. a) How many payments are needed? b) How much of the principal will be repaid by the 6th payment? c) Prepare a partial amortization schedule showing the details of the last three payments and totals. Answer: a)

= 10.90304697

The number of payments is 11. b)

PVn = 4700 = 4700.00(5.2509011) = $24 679.24 Interest = 24679.24(.035) = $863.77 Principal repaid = 4700.00 - 863.77 = $3836.23

c) Partial Amortization (rounded the nearest cent):

PVn = 4700 = 4700(2.71554292) = $12 763.05 Diff: 2 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

8) A debt of $12 500.00 with interest at 5.5% compounded semi-annually is amortized by making payments of $1000.00 at the end of every six months. Calculate the outstanding balance after seven years. Answer: PV = 12 500.00; n = 7(2) = 14; i =

% = 0.0275; I/Y = 5.5; P/Y = C/Y = 2

The accumulated value of the original principal after five years FV = 12500(1.0275)14 = $18 274.93 The accumulated value of the first ten payments FV = 1000

= $16 799.79

The outstanding balance after five years is $18 274.93 - $16 799.79 = $1475.14 Diff: 1 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 9) Olfert Inc. is repaying a loan of $52 500.00 by making payments of $4700.00 at the end of every six months. If interest is 7.5% compounded semi-annually, construct an amortization schedule showing the total paid and the total cost of the loan. Answer: Amortization Schedule (done using Excel)

10) Barbara borrowed $12 000.00 from the bank at 9% compounded monthly. The loan is amortized with end-of-month payments over five years. a) Calculate the interest included in the 20th payment. b) Calculate the principal repaid in the 36th payment. c) Construct a partial amortization schedule showing the details of the first two payments, the 20th payment, the 36th payment, and the last two payments. d) Calculate the totals of amount paid, interest paid, and the principal repaid. Answer: a) PV = 12 000.00; n = 5(12) = 60; i =

= 0.75% = 0.0075; I/Y = 9; P/Y = C/Y = 12;

12000 = PMT 12000 = PMT[48.17337352] PMT = PMT = $249.10 Programmed solution:

The outstanding balance after the 19th payment: PV = 249.10

= $8764.10

Programmed solution:

The interest paid in the 20th payment is 8764.10(0.0075) = $65.73 b) The outstanding balance after the 35th payment: PV = 249.10

= $5659.24

Programmed solution:

The interest included in Payment 36 is 5659.24(0.0075) = $42.44 The principal repaid by Payment 36 is 249.10 - 42.44 = $206.66

The outstanding balance after Payment 58: PV = 249.10

= $492.65

Programmed solution:

The principal repaid is $12 000.00. The total amount paid is 249.10(59) + 249.09 = $14 945.99 The total interest paid is 14945.99 - 12000.00 = $2945.99 Partial Amortization Schedule: (rounded to the nearest cent):

Diff: 3 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

11) Mr. Lamb borrowed $8321.00 at 11.12% compounded monthly. He agreed to repay the loan in equal monthly payments over five years. a) What is the size of the monthly payment? b) How much of the 24th payment is interest? c) What is the principal repaid in the 37th payment? d) Prepare a partial amortization schedule showing details of the first three payments, the last three payments, and totals. Answer: a) 8321.00 = PMT 8321.00 = PMT(45.86668688) $181.42 = PMT b) PVn = 181.42 = 181.42(31.2027738) = $5660.81 5660.81(0.009266667) = $52.46 c) PVn = 181.42 = 181.42(21.42999316) = $3887.83 181.42 - 3887.83(.009266667) = 181.42 - 36.03 = $145.39 d) Partial amortization table (done by hand, rounded to the nearest cent).

12) A loan of $19 000.00 is repaid by quarterly payments of $900.00 each at 8% compounded quarterly. What is the principal repaid by the 21st payment? Answer: n =

= 27.7016753

PVn = 900 = 900(7.07263077) = $6365.37 Interest = 6365.37(0.02) = $127.31 Principal repaid = 900.00 - 127.31 = $772.69 Diff: 2 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 13) A loan of $12 000.00 is repaid by quarterly payments of $1200.00 each at 12% compounded quarterly. What is the principal repaid by the 11th payment? Answer: n =

= 12.06662371

PVn = 1200 = 1200(10) = $12000 Interest = 12000(0.03) = $360 Principal repaid = 12000 - 360 = $11 640 Diff: 2 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

14) Cody borrowed $9100.00 from his credit union. He agreed to repay the loan by making equal monthly payments for 4 years. Interest is 9% compounded monthly. a) What is the size of the monthly payments? b) How much will the loan cost him? c) How much will Cody owe after 16 months? d) How much interest will he pay in his 26th payment? e) How much of the principal will be repaid in the 18th payment? f) Prepare a partial amortization schedule showing details of the first three payments, Payments 24, 25, 26, the last three payments, and totals. Answer: a) 9100.00 = PMT 9100.00 = PMT(40.1847819) $226.45 = PMT b) Total paid = 226.45(48) = $10 869.60 Amount borrowed = 9100.00 Cost of financing = $ 1769.60 c)

Balance after 16 months: n = 48 - 16 = 32 An = 226.45

= 226.45(28.3556505) = $6421.14

d) Balance after 25th payment: n = 48 - 25 = 23 An = 226.45

= 226.45(21.0533147) = $4767.52

Interest in 26th payment = 4767.52(.0075) = $35.76 e) Balance after 17th payment: n = 48 - 17 = 31 An = 226.45

= 226.45(27.5683178) = $6242.85

Interest in 18th payment = 6242.85(.0075) = $46.82 Principal repaid = 226.45 - 46.82 = $179.63 f)

Balance after 23rd payment: n = 48 - 23 = 25 An = 226.45

= 226.45(22.7187555) = $5144.66

An = 226.45

= 226.45(2.9555562) = $669.29

Partial Amortization Table (each step rounded to the nearest cent).

15) Duguid and Partners bought a property valued at $87 300.00 for $17 000.00 down and a mortgage amortized over 17 years. The firm makes equal payments due at the end of every three months. Interest on the mortgage is 6.85% compounded annually and the mortgage is renewable after five years. a) What is the size of each quarterly payment? b) What is the outstanding principal at the end of the five-year term? c) What is the cost of the mortgage for the first five years? d) If the mortgage is renewed for a further five years at 7.17% compounded semi-annually, what will be the size of each quarterly payment? Answer: a) p =

- 1 = 0.01670189

70300 = PMT 70300 = PMT(40.4615867) $1737.45 = PMT b) PVg = 1737.45 = 1737.45(32.8376079) = $57053.70 c)

Total paid = 1737.45(20) = $34 749.00 Principal repaid = 70300.00 - 57053.70 = $13 246.30 Interest cost = 34749.00 - 13246.30 = $21 502.70

d) p =

- 1 = 0.01776716

57053.70 = PMT 57053.70 = PMT(32.1146991) $1776.56 = PMT Diff: 3 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

16) Sean and friends bought a property valued at $50 000 for $5000.00 down and a mortgage amortized over 10 years. The group makes equal payments due at the end of every three months. Interest on the mortgage is 6.00% compounded annually and the mortgage is renewable after five years. a) What is the size of each quarterly payment? b) What is the outstanding principal at the end of the five-year term? c) What is the cost of the mortgage for the first five years? d) If the mortgage is renewed for a further five years at 9% compounded semi-annually, what will be the size of each quarterly payment? Answer: a) p =

- 1 = 0.01467385

45000 = PMT 45000 = PMT(30.0947153) $1495.28 = PMT b) PVg = 1,495.28 = 1519.67(17.2239653) = $25754.65 c)

Total paid = 1495.28(20) = $29 905.60 Principal repaid = 45000.00 - 25754.65 = $19 245.35 Interest cost = 29905.60 -19245.35 = $10 660.25

d) p =

- 1 = 0.022252415

25754.65 = PMT 25754.65 = PMT(16.00151134) $1609.51 = PMT Diff: 3 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

17) The owner of the Pink Flamingo Motel borrowed $19 500.00 at 8.11% compounded semi-annually and agreed to repay the loan by making payments of $1110.00 at the end of every three months. a) How many payments will be needed to repay the loan? b) How much will be owed at the end of five years? c) How much of the payments made at the end of five years will be interest? Answer: a) p =

- 1 = 0.02007353

= 21.879938

The number of payments is 22. b) PVg = 1110

= 1110(1.826974901) = $2027.94

Total paid = 1110(20) = $22 200 Principal repaid = 19500.00 - 2027.94 = $17 472.06 Interest = 22200.00 - 17472.06 = $4727.94 Diff: 3 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

18) A debt of $12 970.00 with interest at 7.23% compounded semi-annually is repaid by payments of $1880.00 made at the end of every three months. Construct an amortization schedule showing the total paid and the total cost of the debt. Answer: p =

- 1 = 0.01791454

Loan Amortization Schedule (done in Excel):

Diff: 3 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

19) The Taylors agreed to monthly payments rounded up to the nearest $100.00 on a mortgage of $136 000.00 amortized over 15 years. Interest for the first five years was 8.5% compounded semi-annually. After 30 months, as permitted by the mortgage agreement, the Taylors increased the rounded monthly payment by 10%. a) Determine the mortgage balance at the end of the five-year term. b) If the interest rate remains unchanged over the remaining term, how many more of the increased payments will amortize the mortgage balance? c) How much did the Taylors save by exercising the increase-in-payment option? Answer: a) p =

- 1 = 0.00696106

136000 = PMT 136000 = PMT(102 .442161) $1327.58 = PMT Payment rounded to $1400.00 FV = 136000(1.00696106)60 = 136000.00(1.51621447) = $206 205.17 FVn = 1400

= 1400(74.1574287) = $103 820.40

Balance = 206205.17 - 103820.40 = $102 384.77

b) n =

= 89.5752224 payments

c) n =

= 102.561916 payments

PVn = 1400.00 = 1400.00(0.58865597) = $782.43 782.43(1.00696106) = $787.88 PVn = 1540.00 = 1540.00(0.57208741) = $881.01 881.01(1.00696106) = $887.15 Rounded payments = 102(1400.00) + 787.88 = 142800.00 + 787.88 = $143 587.88 Increased rounded payments = 89(1540.00) + 887.15 = 137,060.00 + 887.15 = $137 947.15 Savings = 143 587.88 - 137 947.15 = $5640.73 Diff: 3 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

20) A $30 000.00 mortgage is amortized by monthly payments over twenty years and is renewable after five years. a) If the interest rate is 8.5% compounded semi-annually, calculate the outstanding balance at the end of the five-year term. b) If the mortgage is renewed for a further three-year term at 8% compounded semi-annually, calculate the size of the new monthly payment. c) Calculate the payout figure at the end of the three-year term. Answer: a) PV = 30 000.00; n = 20(12) = 240; i =

= 0.0425; I/Y = 8.5; P/Y = 12; C/Y = 2; c =

p = (1 + i)c = (1 + 0.0425)1/6 - 1 = 0.006961062 30000 = PMT 30000 = PMT[116.4742048] PMT = $257.57 The number of outstanding payments after five years is 15(12) = 180 PV = 257.57 PV = $26 386.10 Programmed solution:

b) The outstanding balance of $26 386.10 is to be amortized the remaining 15 years. PV = 26 386.10; n = 15(12) = 180; i = p=

= 0.04; I/Y = 8; P/Y = 12; C/Y = 2; c =

- 1 = 0.006558197

26386.10 = PMT 26386.10 = PMT[105.4682161] PMT = $250.18 Programmed solution:

c) At the end of the 3 year term the outstanding payments is 144. PV = 250.18 PV = $23 265.45 Programmed solution:

Diff: 2 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

21) A debt of $43 100.00 is repaid by payments of $3350.00 made at the end of every six months. Interest is 8.4% compounded quarterly. a) What is the number of payments needed to retire the debt? b) What is the cost of the debt for the first five years? c) What is the interest paid in the 11th payment? d) Construct a partial amortization schedule showing details of the first three payments, the last three payments, and totals. Answer: a) p =

- 1 = 0.042441

= 18.9998078

18 full payments are needed and one smaller payment. b) PVg = 3350.00

= 3350.00(7.3531857) = $24 633.17

Total paid = 10(3350.00) Principal repaid = 43100.00 - 24633.17 Cost = 33500.00 - 18466.83 c)

= $33 500.00 = $18 466.83 = $15 033.17

$24633.17(.042441) = $1045.46

d) Partial Amortization Schedule (rounded to 2 decimal places):

22) A $180 000.00 mortgage is to be amortized by making monthly payments for 22.5 years. Interest is 7.2% compounded semi-annually for a four-year term. a) Compute the size of the monthly payment. b) Determine the balance at the end of the four-year term. c) If the mortgage is renewed for a five-year term at 8.66% compounded semi-annually, what is the size of the monthly payment for the renewal term? Answer: a) p =

- 1 = 0.00591193

180000.00 = PMT 180000.00 = PMT(134.708006) $1336.22 = PMT b) PVn = 1336.22 = 1336.22(123.444897) = $164 949.54 c)

- 1 = 0.00708981

164949.54 = PMT 164949.54 = PMT(111.6558184) $1477.30 = PMT Diff: 2 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

23) A $248 000.00 mortgage amortized by monthly payments over 35 years is renewable after five years. Interest is 8.12% compounded semi-annually. a) What is the size of the monthly payments? b) How much interest is paid during the first year? c) How much of the principal is repaid during the first five-year term? d) If the mortgage is renewed for a further five-year term at 7.16% compounded semi-annually, what will be the size of the monthly payments? Answer: p =

- 1 = 0.00665496

a) 248000 = PMT 248000 = PMT(140.995881) $1758.92 = PMT b) PVn = 1758.92 = 1758.92(140.2280408) = $246,649.91 Total paid = 12(1758.92) = $ 21 107.04 Principal repaid = 248000.00 - 246,649.91 = 1350.09 Interest paid = $ 19756.95 c)

PVn = 1758.92 = 1758.92(136.465612) = $240 032.09 248 000.00 - 240 032.09 = $7967.91 principal repaid

d) p =

- 1 = 0.0058796

240032.09 = PMT 240032.09 = PMT(149.469597) $1605.89 = PMT Diff: 3 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

24) A debt of $42 500.00 is repaid by payments of $4850.00 made at the end of each year. Interest is 7% compounded semi-annually. a) How many payments are needed to repay the debt? b) What is the cost of the debt for the first three years? c) What is the principal repaid in the 3rd year? d) Construct an amortization schedule showing details of the first three payments, the last three payments, and totals. Answer: a) p =

- 1 = 0.071225

n= =

= 14.22222256 payments

b) PVn = 4850.00

= 4850.00(7.5530728) = $36 632.42

Total paid = 3(4850.00) Principal repaid 42500.00 - 36632.42 Total cost c)

= $14 550.00 = 5867.58 $ 8682.42

PVn = 4850.00 = 4850.00(7.9843889) = $38 724.29 38724.29 - 36632.42 = $2091.87 repaid in year 3.

d) (Rounded to the nearest cent)

PVn = 4850 = 4850(1.99059863) = $9654.40 Diff: 3 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules. 25) Karen receives pension payments of $2604.90 at the end of every six months from a retirement fund of . The fund earns 8.06% compounded semi-annually. a) How many payments will Karen receive? b) What is the size of the final pension payment? Answer: a) n =

b) PVn = 2604.90

= 32.5019768

= 2604.90(0.48727778) = $1269.31

Final payment = 1269.31(1.0403) = $1320.46 Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

26) Tom receives pension payments of $6000.00 at the end of every six months from a retirement fund of . The fund earns 8.00% compounded semi-annually. a) How many payments will Tom receive? b) What is the size of the final pension payment? Answer: a) n =

b) PVn = 6000.00

= 23.36241894

= 6000(0.352844567) = $2117.07

Final payment = 2117.07 (1.04) = $2201.75 Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 27) The owner of the Guelph Inc. borrowed $20 000.00 at 1.2% compounded semi-annually and agreed to repay the loan by making payments of $1000.00 at the end of every three months. a) How many payments will be needed to repay the loan? b) How much will be owed at the end of five years? Answer: a) P =

- 1 = 0.0029955134

= 20.6550349

The number of payments is 21. b) PVg = 1000

= 1000(0.653415465) = 653.42

Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

28) A debt of $41 000.00 is repaid in monthly installments of $660.00. If interest is 8.4% compounded quarterly, what is the size of the final payment? Answer: p =

- 1 = 0.006951564

= 81.6095827

Present value of final payment: PVg = 660

= 660(0594-601.6069114) = $400.09

Final payment = 400.09(1.006951564) = $402.87 Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

29) Far East Imports Inc. owes $64 000.00 to be repaid by monthly payments of $2475.00. Interest is 6.12% compounded monthly. a) How many payments will Far East Imports Inc. have to make? b) How much interest is included in the 15th payment? c) How much of the principal will be repaid by the 17th payment? d) Construct a partial amortization schedule showing details of the first three payments, the last three payments, and totals. Answer: a) n=

= 27.8008341

28 payments are required. b) PVn = 2475.00 = 2475.00(13.29365829) = $32 901.80 32901.80(.0051) = $167.80 c)

PVn = 2475.00 = 2475.00(11.4244994) = $28 275.64 2475.00 - 28,275.64(0.0051) = 2475.00 - 144.21 = $2330.79

d) Partial Amortization Table (rounded to the nearest cent)

Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

30) A debt of $42 000.00 is repaid in monthly installments of $525.00. If interest is 8.0% compounded quarterly, what is the size of the final payment? Answer: p =

- 1 = 0.00662271

= 114.323133

Present value of final payment: PVg = 525

= 525(0.321724458) = $168.91

Final payment = 168.91(1.00662271) = $170.02 Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 31) A factory valued at $50 000.00 is purchased for a down payment of 28% and payments of $2000.00 at the end of every three months. If interest is 9% compounded monthly, calculate the size of the final payment. Answer: PV = 50 000(0.72) = 36 000.00;PMT = 4000.00; i =

= 0.0075; I/Y = 9; P/Y = 4;C/Y = 12; c =

p = (1 + 0.0075)3 - 1 = 0.022669172 36000 = 2000 0.408045094 = 1 - 1.022669172-n -n ln 1.022669172 = ln 0.591954906 n = 23.39060393 quarters Present Value of final payment: PMT = 2000.00; p = 0.022669172; n = 0.39060393 PV = 2000 PV = $769.12 Interest is 769.12(0.022669172) = $17.44 Final Payment is 769.12 + 17.44 = $786.56 Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

32) Payments of $1000.00 deferred for nine years are received at the end of each month from a fund of $20 000.00 deposited today at 6% compounded monthly. Calculate the size of the final payment. Answer: PV = 20 000.00; n = 9(12) = 108; i =

= 0.005;I/Y = 6; P/Y = C/Y = 12

FV = 20000 (1 + 0.005)108 = $34 273.99 34273.99 = 1000 0.17136995 = 1 - 1.005-n 1.005-n = 0.82863005

-n ln 1.005 = ln 0.82863005 n = 37.6902094 months PV = 1000 PV = $687.31 Programmed solution:

The size of the final payment is 687.31(1.005) = $690.75 Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

33) Andre has saved $152 000.00. If he withdraws $1750.00 at the beginning of every month and interest is 7.5% compounded monthly, what is the size of the last withdrawal? Answer: 152000.00 = 1750.00

(1.00625)

86.3176575 = 0.53948536 = 1 - 1.00625-n -n ln 1.00625 = ln 0.46051464 -n(0.00623055) = -0.77541063 n = 124.4530037 PVn(due) = 1750.00

(1.00625) = 1750.00(.45095723)(1.00625) = $794.11

The final payment is $794.11. Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 34) Lisa saved $750 000.00. If she withdraws $12 000.00 at the beginning of every month and interest is 10.8% compounded monthly, what is the size of the last withdrawal? Answer: 750000 = 12000

(1.009)

61.9425 = 0.55748 = 1 - 1.01-n -n = ln(1 - 0.55748)/ln(1.01) n = 81.934488 PVn(due) = 12000

(1.009) = $11 217.14

The final payment is $11 217.14 Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

35) Angelina has agreed to purchase her partner's share in the business by making payments of $1720.00 every three months. The agreed transfer value is $19 500.00 and interest is 9.11% compounded annually. If the first payment is due at the date of the agreement, what is the size of the final payment? Answer: p =

- 1 = 0.02203587

19500 = 1720

(1.02203587)

11.0927704 = 0.24443886 = 1 - 1.0220359-n -n ln 1.0220359 = ln 0.75556114 -n(0.021796590) = -0.280294578 n = 12.85956074 PVn(due) = 1720

(1.02203587)

= 1720(0.84231184)(1.02203587) = $1480.70 The final payment is $1480.70. Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 36) John has agreed to purchase his partner's share in real estate by making payments of $5650.00 every three months. The agreed transfer value is $45 200.00 and interest is 10% compounded annually. If the first payment is due at the date of the agreement, what is the size of the final payment? Answer: p =

- 1 = 0.0241136891

45200 = 5650

(1.0241136891)

7.811632717 = 0.18836728271 = 1 - 1.0241136891-n -n ln 1.0241136891 = ln 0.8116327173 -n(0.023827545) = -0.2087073597 n = 8.75907946 PVn(due) = 5650

(1.0241136891) = 4301.08

The final payment is $4301.08 Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

37) A contract valued at $47 500.00 requires payments of $6951.00 every six months. The first payment is due in four years and interest is 6.6% compounded semi-annually. a) How many payments are required? b) What is the size of the last payment? c) How much will be paid in total? d) How much of what is paid is interest? Answer: a) FV = 47500.00(1.033)8 = 47500.00(1.2965897) = $61 588.01 61588.01 = 6951.00

(1.033)

8.577259937 = 0.283049578 = 1 - 1.033-n -n ln 1.033 = ln 0.716950422 -n(0.032467190) = -0.332748578 n = 10.24876426 The number of payments is 11. b) Present value of last payment: PVn (due) = 6951.00

(1.033) = 6951.00(0.237620547)(1.033)

= $1750.30 The final payment is $1750.30. c)

Total paid = 6951.00(10) + 1750.30 = 69510.00 + 1750.30 = $71 260.30 d) Interest = 71260.30 - 47500.00 = $23 760.30 Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

38) A mortgage balance of $137 960.70 is to be repaid over a eleven-year term by equal monthly payments at 8.1% compounded semi-annually. At the request of the mortgagor, the monthly payments were set at $1140.00. a) How many payments will the mortgagor have to make? b) What is the size of last payment? c) Determine the difference between the total actual amount paid and the total amount required to amortize the mortgage by the contractual monthly payments. Answer: a) p =

- 1 = 0.006638834

= 245.8378879

246 payments b) PVn = 1140.00

= 1140.00(.8293897) = $949.40

949.40(1.006638834) = $955.70 c)

137960.70 = PMT

137960.70 = PMT(87.73874749) $1572.40 = PMT Contractual = 1572.40(132) = $207 556.80 Actual = 1140.00(245) + 955.70 = 280 255.70 Difference in payments = $72 698.90 Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

39) Catrina receives payments of $1120.00 at the beginning of each month from a pension fund of $79 500.00. Interest earned by the fund is 11.36% compounded monthly. a) What is the number of payments Catrina will receive? b) What is the size of the final payment? Answer: a) 79500.00 = 1120

(1.0094667)

70.31648017 = 0.665662679 = 1 - 1.0094667-n -n ln 1.0094667 = ln 0.334337321 -n(0.009422139) = -1.095604853 n = 116.2798491 payments b) PVn = 1120

(1.009466667)

= 1120(0.2789165895)(1.009466667) = $314.50 Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 40) A contract worth $59 000.00 provides benefits of $21 000.00 at the end of each year. The benefits are deferred for six years and interest is 8.95% compounded quarterly. a) How many payments are to be made under the contract? b) What is the size of the last benefit payment? Answer: a) p =

- 1 = 0.092548902

FV = 59000.00(1.092548902)6 = 59000.00(1.700768902) = $100 345.37

n= =

= 6.595730861 payments

b) PVn = 21000 = 21000(0.554993627) = $11 654.87 Final payment = 11654.87(1.092548902) = $12 733.52 Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

41) A $176 000.00 mortgage is to be repaid over a 21-year period by monthly payments rounded up to the next higher $100.00. Interest is 9.44% compounded semi-annually. a) Determine the number of rounded payments required to repay the mortgage. b) Determine the size of the last payment. Answer: a)

- 1 = 0.007716274

176000.00 = PMT 176000.00 = PMT(110.914783) $1586.77 = PMT The rounded payment is $1600.00.

n= =

= 245.7618375 payments.

PVn = 1600

= 1600.00(0.756695498) = $1210.71 Final payment = 1210.05(1.007716274) = $1220.05 Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

42) The Acme Parts Company is repaying a debt of $15 000.00 by payments of $1410.00 made at the end of every three months. Interest is 7.2% compounded monthly. a) How many payments are needed to repay the debt? b) What is the size of the final payment? Answer: a) p =

- 1 = 0.018108216

= 11.92375947

The number of payments is 12. b) PVg = 1410

= 1410.00(0.907947708) = $1280.21

1280.21(1.018108216) = $1303.39 Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 43) A debt of $17 000.00 is repaid by quarterly payments of $1430.00. If interest is 6.12% compounded quarterly, what is the size of the final payment? Answer: n=

n = 13.22144230 PVn = 1430 = 1430.00(.219396293) = $313.74 = Balance Final Payment = 313.74(1.0153) = $318.54 Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

44) A $151 000.00 mortgage is to be repaid over a 17-year period by monthly payments rounded up to the next higher $50.00. Interest is 8.2% compounded semi-annually. a) Determine the number of rounded payments required to repay the mortgage. b) Determine the size of the last payment. c) Calculate the amount of interest saved by rounding the payments up to the next higher $50.00. Answer: a) p =

- 1 = 0.00671944

151000.00 = PMT 151000.00 = PMT(110.8605388) $1362.07 = PMT Rounded payments $1400.00

= 192.6301891

192 equal payments plus one less than $1400 b) PVg = 1400

= 1400.00(0.626757778) = $877.46

Last Payment = 877.46(1.00671944) = $883.36 c)

Unrounded 204(1362.07) = $277 862.28 Rounded 192(1400.00) + 883.36 = 269 683.36 Savings = $ 8178.92 Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

45) A demand (variable rate) mortgage of $137 000.00 is amortized over 20 years by equal monthly payments. After 21 months the original interest rate of 6% p.a was raised to 6.6% p.a. Three years after the mortgage was taken out, it was renewed at the request for the mortgagor for a five-year term at a fixed rate of 7.25% p.a. a) Calculate the mortgage balance after 21 months. b) Compute the size of the new monthly payment at the 6.6% rate of interest. c) Determine the mortgage balance at the end of the five-year term. Answer: a) p =

- 1 = 0.00486755

137000 = PMT 137000 = PMT(141.3843090) $968.99 = PMT PVn = 968.99 = 968.99(134.5076431) = $130 336.56 b) p =

- 1 = 0.005340319

130335.56 = PMT 130335.56 = PMT(128.9285035) $1010.91 = PMT c)

PVn = 1010.91 1010.91(124.0775295) = $125431.22 p=

- 1 = 0.005849741

125431.22 = PMT 125431.22 = PMT(118.9350584) $1054.62 = PMT PVn = 1054.62

= 1054.62(97.14104632) = $102 446.89

Diff: 3 Type: SA Page Ref: 585-594 Topic: 14.4 Residential Mortgages in Canada Objective: 14-5: Compute the periodic payments for fixed-rate mortgages and for demand mortgages.

46) A bank has an interest rate of 5% compounded semi annually for a five year closed mortgage. If the mortgage has monthly payments, calculate the effective rate of interest per each term. Answer: i =

= 0.025; c =

p = 1.0251/6 - 1 = 0.004123915 = 0.4124% Programmed solution:

p = 0.4124% Diff: 2 Type: SA Page Ref: 585-594 Topic: 14.4 Residential Mortgages in Canada Objective: 14-4: Compute the effective interest rate for fixed-rate residential mortgages. 47) How much interest is paid in total on a 3-year loan for $28 600 (rounded to the nearest dollar)? The interest rate is 8.7% compounded monthly and the payments are monthly. A) $3997 B) $3779 C) $3977 D) $3999 E) $4000 Answer: A Diff: 2 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 48) How much interest is paid in total on a 1-year loan for $5000? The interest rate is 12 % compounded monthly and the payments are monthly. A) $330.00 B) $333.88 C) $333.00 D) $300.88 E) $330.88 Answer: E Diff: 2 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

49) How much principal is repaid in the 74th payment interval on a $142 300 mortgage? The mortgage is amortized over 25 years and the payments are monthly. The interest rate is 7.44% compounded monthly. A) $257.16 B) $275.16 C) $527.16 D) $572.16 E) $574.16 Answer: A Diff: 1 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 50) How much total principal is repaid between the 1st and 7th payment interval of a 4.5-year loan for $4887 at an interest rate of 7.4% compounded monthly and the payments are also monthly. Round your answer to the nearest dollar. A) $556 B) $546 C) $456 D) $445 E) $423 Answer: B Diff: 2 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 51) What is the outstanding balance after 23 payments on a 20-year mortgage that has monthly payments of $1062.32 and an interest rate of 6.6% compounded monthly? A) $143 402.54 B) $144 302.54 C) $134 402.54 D) $134 302.54 E) $134 322.54 Answer: C Diff: 2 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

52) What is the monthly payment size of a 21-year mortgage for $169 600 and an interest rate of 7.2% compounded semi-annually? A) $1169.11 B) $1296.11 C) $1926.11 D) $1211.96 E) $1222.96 Answer: B Diff: 2 Type: MC Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules. 53) What is the monthly payment size of a 25-year mortgage for $100 000 and an interest rate of 6% compounded semi-annually? A) $639.81 B) $1639.81 C) $630.81 D) $600.81 E) 633.81 Answer: A Diff: 2 Type: MC Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules. 54) How much interest is paid in the 53rd monthly payment interval of a loan for $43 200? The loan is amortized at a rate of 9.17% compounded annually over 7 years. A) $411.11 B) $111.11 C) $144.11 D) $444.11 E) $400.11 Answer: C Diff: 2 Type: MC Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

55) How much total interest is paid between the 6th and 13th payments on a loan that has monthly payments for 3 years and an original principal of $14 750? The loan rate is 6.6% compounded quarterly. A) $540 B) $405 C) $450 D) $504 E) $545 Answer: D Diff: 2 Type: MC Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules. 56) How much interest is paid after the first payment interval on a $100 000 mortgage? The mortgage is amortized over 25 years and the payments are monthly. The interest rate is 6% compounded semiannually. Round your answer to the nearest dollar. A) $146 B) $640 C) $527 D) $494 E) $401 Answer: D Diff: 1 Type: MC Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules. 57) What is the outstanding balance after the 23rd payment interval of a 16-year loan for $14 734 with semi-annual payments and an interest rate of 5.95% compounded quarterly? A) $5625.21 B) $5652.21 C) $5562.21 D) $5526.21 E) $5566.21 Answer: A Diff: 1 Type: MC Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

58) What is the size of the final unequal payment of a loan that has 69.47 payments and the equal payment size is $215.64? The interest rate is 3.1% per compounding period. A) $120.17 B) $210.17 C) $102.17 D) $99.10 E) $99.17 Answer: C Diff: 3 Type: MC Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 59) What is the size of the final unequal payment of a loan for $2723.44 if it has quarterly payments of $500? The interest rate is 8.4% compounded quarterly. A) $432.14 B) $423.14 C) $441.14 D) $414.14 E) $424.14 Answer: B Diff: 2 Type: MC Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 60) Munawar is a first time buyer and bought a brand new townhouse in Oshawa for $212 000. He has saved enough to make a 5% down payment and will have to pay mortgage loan insurance at 2.75% of the mortgage balance. Munawar wants the insurance premium added to the maximum allowable loan balance. Determine his maximum initial mortgage balance. A) $5538.50 B) $201 400 C) $206 938.50 D) $212 000 E) $212 477 Answer: C Diff: 1 Type: MC Page Ref: 585-594 Topic: 14.4 Residential Mortgages in Canada Objective: 14-6: Create statements for various types of residential mortgages.

61) Li bought a house in Whitby for $305 000 5 years ago. He made a down payment of $46 000 and financed the house through Scotia Bank with a mortgage amortized over 25 years. Interest for the first 5 years was 4.4% compounded semi-annually and payments were made monthly. At the end of the 5 year term, Li has renewed the mortgage for another 5 years, at 3.7% compounded semi-annually. What is the monthly payment for the second term? Answer: First calculate the balance at the end of the first 5 year term. This requires calculating the FV of the payments made in first 5 years and the future value of the original loan. PV = 305000 - 46000 = $259 000;

∴p=

n = 12 × 25 = 300;

= 2.2%; c =

- 1 = 0.0036335

PMT for original term =

= $1419.13

∴FV after 1st term = 1419.13

= $94 950

FV of the original loan at the end of 5 year term = 259000 × 1.003633560 = $321 965 Balance = $321 965 - $94 950 = $227 015 Balance is the PV for the second term For second term PV = $227,015;

∴p=

n = 12 × 20 = 240;

= 1.85%;

- 1 = 0.00306

PMT for second term =

= $1336.72

Diff: 3 Type: SA Page Ref: 585-594 Topic: 14.4 Residential Mortgages in Canada Objective: 14-5: Compute the periodic payments for fixed-rate mortgages and for demand mortgages. 62) Shiva transferred his mortgage of $145 981 from CIBC to BMO for the remaining amortization period at 3.59% compounded semi-annually. He has agreed to pay rounded up payment of $2000 every month. How long will it take Shiva to pay-off his mortgage? A) 6 years 11 months B) 6 years 1 months C) 8 years D) 6 years 6 months E) 10 years Answer: A Diff: 3 Type: MC Page Ref: 585-594 Topic: 14.4 Residential Mortgages in Canada Objective: 14-5: Compute the periodic payments for fixed-rate mortgages and for demand mortgages.

63) John makes a monthly mortgage payment of $1940 at the end of every month. On October 31, his loan balance was $145 000 after making a payment of $1940. His financial institute allows him to make one additional payment equal to or less than his monthly payment, once a month. Taking advantage of that, John made a payment of $1500 on November 15 and $1900 on December 20. What is his year-end balance after making monthly payment on December 31? The effective monthly fixed rate is 0.306%. Answer: For p = 0.306%, equivalent simple annual rate is 3.67% Using MS-Excel,, mortgage statement is created as follows Payment Date 31-Oct 15-Nov 30-Nov 20-Dec 31-Dec

Number of Amount Days Paid

Interest Paid Principal

15 15 20 11

$218.69 $216.76 $285.55 $155.27

$1500 $1940 $1900 $1940

$1281.31 $1723.24 $1614.45 $1784.73

Balance Repaid $145 000.00 $143 718.69 $141 995.45 $140 381.00 $138 596.26

Year-end balance = $138 596.26 Diff: 3 Type: SA Page Ref: 585-594 Topic: 14.4 Residential Mortgages in Canada Objective: 14-6: Create statements for various types of residential mortgages. 64) Peter worked for GM for 31 years. On his retirement, he opted a pension buy-out and received $1 467 921. He invested his money in an annuity that provided for payments of $9000 at the end of every month. If interest is 4.4% compounded monthly, determine the size of the final payment. Answer: PV = $1 467 921; PMT = $9000; P/Y = 12 C/Y = 12; I/Y = 4.4% i=

= 0.367%

1467921 = 9000 ⇒ -n ln 1.00367 = ln 0.402 ⇒ n = 248.7953157 months For last month's payment: PMT = $9000; i = 0.367%; PV = $9000 × 1.00367 ×

n = 0.7953157 = $7160.52

Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

65) The Korean community is planning to buy a place for their community in Markham valued at $1 697 000. They made a down payment of 10% and payments of $400 000 are required at the end of every 3 months. If interest is 9% compounded monthly, what is the size of the final payment A) $15 685 B) $205 512 C) $131 373 D) $355 547 E) $48 381 Answer: A Diff: 2 Type: MC Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 66) Kirk had accumulated $975 000 in his RRSP at the age of 55. He converted the RRSP into an RRIF at that time and started withdrawing $5000 at the beginning of each month. If interest is 4.5% compounded monthly, what is the size of his final withdrawal? A) $1739.27 B) $1409.28 C) $4947.31 D) $412.07 E) $33.82 Answer: A Diff: 2 Type: MC Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 67) Honda is selling a 2014 Accord Coupe Ex M for $30 056.90 including freight, PDI and all applicable fees. The lease payment of $382 is due at the beginning of each month. If the interest rate compounded annually is 3.99% and the residual value is $13 200, determine the size of the final lease payment. A) $240.51 B) $285.95 C) $150.28 D) $214.13 E) $5.92 Answer: D Diff: 2 Type: MC Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

68) Zara bought a 2013 Kia Forte by taking a $19 500 loan from BMO with interest at 5.99% compounded annually and amortized by payments of $378 at the end of every month. She receives an inheritance after making 36 payments and plans to pay the balance in full. What is the payout figure just after the last regular payment made? Answer: PV = $19 500; n = 3; i = 5.99% FV = 19500 × 1.05993 = $23 218.24 Accumulate the value of 36 payments: PMT = $378; n = 36; P/Y = 12; C/Y = 1; I/Y = 5.99%; c =

p= FV = 378

; i = 5.99%

- 1 = 0.00485965 = $14 831.65

Outstanding balance after 36 months = $23 218.24 - $14 831.65 = $8386.59 Diff: 2 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

69) Rebecca bought a house in Oshawa and took a $259 000 mortgage amortized by monthly payments over 25 years that is renewable after 5 years. If interest is 4.4% compounded semi-annually, what is the outstanding balance at the end of the five year term? If the mortgage is renewed for a further 5 year term at 3.7% compounded semi-annually, what is the balance after the second 5 year term? Answer: PV = $259 000; n = 25 × 12 = 300; P/Y = 12; C/Y = 2; I/Y = 4.4%; c =

;i=

= 2.2%

- 1 = 0.0036335

259000 = PMT ⇒ PMT = $1419.13 FV = 259000 × 1.003633560 = $321 965.04 FV60 = 1419.13

= $94 950.38

Balance at the end of 5 year term = $321 965.04 - $94 950.38 = $227 014.66 For second 5 year term: PV = $227 014.66; n = 20 × 12 = 240; P/Y = 12; C/Y = 2; I/Y = 4.4%; c =

;i=

= 1.85%

-1 = 0.003059831

227014.66 = PMT ⇒ PMT = $1336.72 FV = 227014.66 × 1.00305983160 = $272 686.88 FV60 = 1336.72

= $87 890.34

Balance at the end of second 5 year term = $272686.88 - $87890.34 = $184 796.54 Diff: 3 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

70) Amanda borrowed $57 800 from BMO to buy a 2014 E-Class Mercedes Sedan at 3.9% compounded monthly. Payments of $2056 are required at the end of every month. Construct a complete amortization schedule in MS Excel showing the details of all payments and the total. Answer: Payment Principal Outstanding Number Amount Paid Interest Paid Repaid Principal balance 0 $57,800 1 $2,056 $188 $1,868 $55,932 2 $2,056 $182 $1,874 $54,058 3 $2,056 $176 $1,880 $52,177 4 $2,056 $170 $1,886 $50,291 5 $2,056 $163 $1,893 $48,398 6 $2,056 $157 $1,899 $46,500 7 $2,056 $151 $1,905 $44,595 8 $2,056 $145 $1,911 $42,684 9 $2,056 $139 $1,917 $40,766 10 $2,056 $132 $1,924 $38,843 11 $2,056 $126 $1,930 $36,913 12 $2,056 $120 $1,936 $34,977 13 $2,056 $114 $1,942 $33,035 14 $2,056 $107 $1,949 $31,086 15 $2,056 $101 $1,955 $29,131 16 $2,056 $95 $1,961 $27,170 17 $2,056 $88 $1,968 $25,202 18 $2,056 $82 $1,974 $23,228 19 $2,056 $75 $1,981 $21,248 20 $2,056 $69 $1,987 $19,261 21 $2,056 $63 $1,993 $17,267 22 $2,056 $56 $2,000 $15,267 23 $2,056 $50 $2,006 $13,261 24 $2,056 $43 $2,013 $11,248 25 $2,056 $37 $2,019 $9,229 26 $2,056 $30 $2,026 $7,203 27 $2,056 $23 $2,033 $5,170 28 $2,056 $17 $2,039 $3,131 29 $2,056 $10 $2,046 $1,085 30 $1,089 $4 $1,085 $0 Totals $60,713 $2,913 $57,800 Diff: 3 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

71) Rebecca borrowed $20 400 from her local credit union to pay for the upgrades in her kitchen. She wants to pay the loan with monthly payments over the period of 3 years. What is the interest included in the tenth payment, if the rate charged by the credit union is 6% monthly? A) $78.19 B) $5698.49 C) $21 336.56 D) $15 638.09 E) $75.48 Answer: A Diff: 2 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 72) Samuel paid $850 000.00 for a new house. He paid $140 000.00 down and financed the balance by making equal payments at the end of every month for 25 years. Interest is 5.1% compounded monthly. What is the total cost of the building for Samuel? A) $1 397 618 B) $1 412 058 C) $1 400 254 D) $1 404 198 E) $1 257 618 Answer: A Diff: 2 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 73) Keith receives payments of $14 500.00 at the beginning of each month from a pension fund of $1 500 000.00. Interest earned by the fund is 7.6% compounded monthly. What is the number of payments Keith will receive? A) 167 B) 169 C) 156 D) 140 E) 104 Answer: A Diff: 2 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

74) A car loan of $46 700.00 is amortized over 8 years by equal monthly payments at 2.9% compounded monthly. What is the monthly payment and outstanding principal balance after 40 payments. A) $545.66 and $28 546.88 B) $545.66 and $18 153.12 C) $544.34 and $28,478.06 D) $486.46 and $27 241.67 Answer: A Diff: 2 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 75) Zam Zam Inc. borrowed $142 000.00 at 4.4% compounded semi-annually. The loan is repaid by payments of $9700.00 due at the end of every six months. How many payments are needed and what will be the outstanding balance after 8 payments? A) 18 and $85 158 B) 18 and $81 520 C) 18 and $60 480 D) 18 and $56 842 E) 15 and $64 800 Answer: A Diff: 2 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

76) Big Printing Inc. is repaying a loan of $37 300.00 by making payments of $2300.00 at the end of every three months. If interest is 3.5% compounded quarterly, construct an amortization schedule showing the total paid and the total cost of the loan. What is the total interest paid and cost of the loan? Answer: Amortization Schedule (done using Excel)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Amount Paid

Interest Paid

$2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $1,305.11

$326.38 $309.11 $291.69 $274.11 $256.39 $238.50 $220.47 $202.27 $183.92 $165.40 $146.72 $127.88 $108.87 $89.70 $70.36 $50.85 $31.17 $11.32

Principal Repaid $1,973.63 $1,990.89 $2,008.31 $2,025.89 $2,043.61 $2,061.50 $2,079.53 $2,097.73 $2,116.08 $2,134.60 $2,153.28 $2,172.12 $2,191.13 $2,210.30 $2,229.64 $2,249.15 $2,268.83 $1,293.79

Outstanding Principal Balance $37,300.00 $35,326.38 $33,335.48 $31,327.17 $29,301.28 $27,257.67 $25,196.17 $23,116.64 $21,018.91 $18,902.82 $16,768.22 $14,614.94 $12,442.82 $10,251.70 $8,041.40 $5,811.76 $3,562.62 $1,293.79 $0.00

Total Interest paid = $3105.11 Total cost of the loan = $37300 + $3105.11 = $40 405.11 Diff: 2 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

77) Big Printing Inc. is repaying a loan of $37 300.00 by making payments of $2300.00 at the end of every three months. If interest is 3.5% compounded monthly, construct an amortization schedule showing the total paid and the total cost of the loan. What is the total interest paid and cost of the loan? Answer: p = 0.88% Loan Amortization Schedule (done in Excel):

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Amount Paid

Interest Paid

$2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $1,315.13

$327.33 $310.02 $292.55 $274.94 $257.17 $239.24 $221.15 $202.91 $184.51 $165.94 $147.22 $128.32 $109.27 $90.04 $70.65 $51.08 $31.35 $11.44

Principal Repaid $1,972.67 $1,989.98 $2,007.45 $2,025.06 $2,042.83 $2,060.76 $2,078.85 $2,097.09 $2,115.49 $2,134.06 $2,152.78 $2,171.68 $2,190.73 $2,209.96 $2,229.35 $2,248.92 $2,268.65 $1,303.69

Outstanding Principal Balance $37,300.00 $35,327.33 $33,337.34 $31,329.90 $29,304.83 $27,262.00 $25,201.24 $23,122.39 $21,025.31 $18,909.81 $16,775.76 $14,622.97 $12,451.30 $10,260.57 $8,050.61 $5,821.26 $3,572.34 $1,303.69 $0.00

Total Interest paid = $3115.13 Total cost of the loan = $37300 + $3115.13 = $40 415.13 Diff: 3 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

78) Tariq bought a commercial property valued at $71 000 for $26 000.00 down and a mortgage amortized over 10 years. He makes equal payments due at the end of every quarter. Interest on the mortgage is 6% compounded annually. a) What will be the payment amount needed to repay the loan? b) How much will be owed at the end of five years? c) How much of the payments made at the end of five years will be interest? Answer: a) p =

- 1 = 0.01467385

45000 = PMT 45000 = PMT(30.0947153) $1495.28 = PMT b) PVg = 1,495.28 = 1519.67(17.2239653) = $25754.65 c)

Total paid = 1495.28(20) = $29 905.60 Principal repaid = 45000.00 - 25754.65 = $19 245.35 Interest cost = 29905.60 -19245.35 = $10 660.25 Diff: 3 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

79) The owner of the Pink Flamingo Motel borrowed $19 500.00 at 8.11% compounded semi-annually and agreed to repay the loan by making payments of $1110.00 at the end of every three months. a) How many payments will be needed to repay the loan? b) How much will be owed at the end of five years? c) How much of the payments made at the end of five years will be interest? Answer: a) p =

- 1 = 0.02007353

= 21.879938

The number of payments is 22. b) PVg = 1110

= 1110(1.826974901) = $2027.94

80) The Taylors agreed to monthly payments rounded up to the nearest $100.00 on a mortgage of $136 000.00 amortized over 15 years. Interest for the first five years was 8.5% compounded semi-annually. After 30 months, as permitted by the mortgage agreement, the Taylors increased the rounded monthly payment by 10%. a) Determine the mortgage balance at the end of the five-year term. b) If the interest rate remains unchanged over the remaining term, how many more of the increased payments will amortize the mortgage balance? c) How much did the Taylors save by exercising the increase-in-payment option? Answer: a) p =

- 1 = 0.00696106

136000 = PMT 136000 = PMT(102 .442161) $1327.58 = PMT Payment rounded to $1400.00 FV = 136000(1.00696106)60 = 136000.00(1.51621447) = $206 205.17 FVn = 1400

= 1400(74.1574287) = $103 820.40

Balance = 206205.17 - 103820.40 = $102 384.77

b) n =

= 89.5752224 payments

c) n =

= 102.561916 payments

PVn = 1400.00 = 1400.00(0.58865597) = $782.43 782.43(1.00696106) = $787.88 PVn = 1540.00 = 1540.00(0.57208741) = $881.01 881.01(1.00696106) = $887.15 Rounded payments = 102(1400.00) + 787.88 = 142800.00 + 787.88 = $143 587.88 Increased rounded payments = 89(1540.00) + 887.15 = 137 060.00 + 887.15 = $137 947.15 Savings = 143 587.88 - 137 947.15 = $5640.73 Diff: 3 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

81) A $30 000.00 mortgage is amortized by monthly payments over twenty years and is renewable after five years. a) If the interest rate is 8.5% compounded semi-annually, calculate the outstanding balance at the end of the five-year term. b) If the mortgage is renewed for a further three-year term at 8% compounded semi-annually, calculate the size of the new monthly payment. c) Calculate the payout figure at the end of the three-year term. Answer: a) PV = 30 000.00; n = 20(12) = 240; i =

= 0.0425; I/Y = 8.5; P/Y = 12; C/Y = 2; c =

b) The outstanding balance of $26 386.10 is to be amortized the remaining 15 years. PV = 26 386.10; n = 15(12) = 180; i =

= 0.04; I/Y = 8; P/Y = 12; C/Y = 2; c =

p = (1 + i)c - 1 = 0.006558197 26386.10 = PMT 26386.10 = PMT[105.4682161] PMT = $250.18 Programmed solution:

c) At the end of the 3 year term the outstanding payments is 144. PV = 250.18 PV = $23 265.45 Programmed solution:

82) A debt of $43 100.00 is repaid by payments of $3350.00 made at the end of every six months. Interest is 8.4% compounded quarterly. a) What is the number of payments needed to retire the debt? b) What is the cost of the debt for the first five years? c) What is the interest paid in the 11th payment? d) Construct a partial amortization schedule showing details of the first three payments, the last three payments, and totals. Answer: a) p =

- 1 = 0.042441

= 18.9998078

18 full payments are needed and one smaller payment. b) PVg = 3350.00

= 3350.00(7.3531857) = $24 633.17

Total paid = 10(3350.00) Principal repaid = 43100.00 - 24633.17 Cost = 33500.00 - 18466.83

= $33 500.00 = $18 466.83 = $15 033.17

c) $24633.17(.042441) = $1045.46 d) Partial Amortization Schedule (rounded to 2 decimal places):

83) A $180 000.00 mortgage is to be amortized by making monthly payments for 22.5 years. Interest is 7.2% compounded semi-annually for a four-year term. a) Compute the size of the monthly payment. b) Determine the balance at the end of the four-year term. c) If the mortgage is renewed for a five-year term at 8.66% compounded semi-annually, what is the size of the monthly payment for the renewal term? Answer: a)

- 1 = 0.00591193

180000.00 = PMT 180000.00 = PMT(134.708006) $1336.22 = PMT b)

PVn = 1336.22 = 1336.22(123.444897) = $164 949.54

- 1 = 0.00708981

84) A $248 000.00 mortgage amortized by monthly payments over 35 years is renewable after five years. Interest is 8.12% compounded semi-annually. a) What is the size of the monthly payments? b) How much interest is paid during the first year? c) How much of the principal is repaid during the first five-year term? d) If the mortgage is renewed for a further five-year term at 7.16% compounded semi-annually, what will be the size of the monthly payments? Answer: p =

- 1 = 0.00665496

a) 248000 = PMT 248000 = PMT(140.995881) $1758.92 = PMT b) PVn = 1758.92 = 1758.92(140.2280408) = $246,649.91 Total paid = 12(1758.92) Principal repaid = 248000.00 - 246,649.91 Interest paid

= $ 21 107.04 = 1350.09 = $ 19756.95

c) PVn = 1758.92 = 1758.92(136.465612) = $240 032.09 248 000.00 - 240 032.09 = $7967.91 principal repaid d) p =

- 1 = 0.0058796

85) A debt of $42 500.00 is repaid by payments of $4850.00 made at the end of each year. Interest is 7% compounded semi-annually. a) How many payments are needed to repay the debt? b) What is the cost of the debt for the first three years? c) What is the principal repaid in the 3rd year? d) Construct an amortization schedule showing details of the first three payments, the last three payments, and totals. Answer: a) p =

- 1 = 0.071225

= =

= 14.22222256 payments

b) PVn = 4850.00

= 4850.00(7.5530728) = $36 632.42

Total paid = 3(4850.00) Principal repaid 42500.00 - 36632.42 Total cost

= $14 550.00 = 5867.58 $ 8682.42

c) PVn = 4850.00 = 4850.00(7.9843889) = $38 724.29 38724.29 - 36632.42 = $2091.87 repaid in year 3.

d) (Rounded to the nearest cent)

PVn = 4850 = 4850(1.99059863) = $9654.40 Diff: 3 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules. 86) Karen receives pension payments of $4570.00 at the end of every six months from a retirement fund of . The fund earns 8.06% compounded semi-annually. a) How many payments will Karen receive? b) What is the size of the final pension payment? Answer: a)

PVn = 4570.00

= 32.5019768

= 4570.00(0.48727778) = $2226.86

Final payment = 2226.86(1.0403) = $2316 .60 Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

87) Tom receives pension payments of $6000.00 at the end of every six months from a retirement fund of . The fund earns 8.00% compounded semi-annually. a) How many payments will Tom receive? b) What is the size of the final pension payment? Answer: a)

PVn = 6000.00

= 23.36241894 = 6000(0.352844567) = $2117.07

Final payment = 2117.07 (1.04) = $2201.75 Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 88) The owner of the Guelph Inc. borrowed $20 000.00 at 1.2% compounded semi-annually and agreed to repay the loan by making payments of $1000.00 at the end of every three months. a) How many payments will be needed to repay the loan? b) How much will be owed at the end of five years? Answer: a)

- 1 = 0.0029955134

= 20.6550349

The number of payments is 21. b)

PVg = 1000

= 1000(0.653415465) = 653.42

Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

89) A debt of $41 000.00 is repaid in monthly installments of $660.00. If interest is 8.4% compounded quarterly, what is the size of the final payment? Answer: p =

- 1 = 0.006951564

= 81.6095827

Present value of final payment: PVg = 660

= 660(0594 - 601.6069114) = $400.09

90) Far East Imports Inc. owes $64 000.00 to be repaid by monthly payments of $2475.00. Interest is 6.12% compounded monthly. a) How many payments will Far East Imports Inc. have to make? b) How much interest is included in the 15th payment? c) How much of the principal will be repaid by the 17th payment? d) Construct a partial amortization schedule showing details of the first three payments, the last three payments, and totals. Answer: a) n =

= 27.8008341

28 payments are required. b) PVn = 2475.00 = 2475.00(13.29365829) = $32 901.80 32901.80(.0051) = $167.80 c) PVn = 2475.00 = 2475.00(11.4244994) = $28 275.64 2475.00 - 28275.64(0.0051) = 2475.00 - 144.21 = $2330.79 d) Partial Amortization Table (rounded to the nearest cent)

Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

91) A debt of $40 000.00 is repaid in monthly installments of $500.00. If interest is 8.0% compounded quarterly, what is the size of the final payment? Answer: p =

- 1 = 0.00662271

= 114.323133

Present value of final payment: PVg = 500

= 500(0.321724458) = $160.86

Final payment = 160.86(1.00662271) = $161.93 Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

92) A factory valued at $100 000.00 is purchased for a down payment of 28% and payments of $4000.00 at the end of every three months. If interest is 9% compounded monthly, calculate the size of the final payment. Answer: PV = 100 000(0.72) = 72 000.00;PMT = 4000.00; i =

= 0.0075; I/Y = 9; P/Y = 4;C/Y = 12; c =

p = (1 + 0.0075)3 - 1 = 0.022669172 72000 = 4000 0.408045094 = 1 - 1.022669172-n -n ln 1.022669172 = ln 0.591954906 n = 23.39060393 quarters Programmed solution:

Present Value of final payment: PMT = 4000.00; p = 0.022669172; n = 0.39060393 PV = 4000 PV = $1538.23 Programmed solution:

Interest is 1538.23(0.022669172) = $34.87 Final Payment is 1538.23 + 34.87 = $1573.10 Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

93) Payments of $1000.00 deferred for nine years are received at the end of each month from a fund of $20 000.00 deposited today at 6% compounded monthly. Calculate the size of the final payment. Answer: PV = 20 000.00; n = 9(12) = 108; i = FV = 20000

= 0.005; I/Y = 6; P/Y = C/Y = 12

= $34 273.99

34273.99 = 1000 0.17136995 = 1 - 1.005-n 1.005-n = 0.82863005

-n ln 1.005 = ln 0.82863005 n = 37.6902094 months PV = 1000 PV = $687.31 Programmed solution:

94) Andre has saved $152 000.00. If he withdraws $1750.00 at the beginning of every month and interest is 7.5% compounded monthly, what is the size of the last withdrawal? Answer: 152000.00 = 1750.00

(1.00625)

86.3176575 = 0.53948536 = 1 - 1.00625-n -n ln 1.00625 = ln 0.46051464 -n(0.00623055) = -0.77541063 n = 124.4530037 PVn(due) = 1750.00

(1.00625) = 1750.00(.45095723)(1.00625) = $794.11

The final payment is $794.11. Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 95) Lisa saved $750 000.00. If she withdraws $12 000.00 at the beginning of every month and interest is 12% compounded monthly, what is the size of the last withdrawal? Answer: 750000 = 12000

(1.01)

61.88 = 0.6188 = 1 - 1.01-n ln 0.38118811 = -nln 1.01 (0.9644622765) = -n(-0.0099503309) n = 96.927659 PVn(due) = 12000

(1.01)

= 12000(0.9303234212) = $11 135.91 The final payment is $11 135.91 Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

96) Angelina has agreed to purchase her partner's share in the business by making payments of $1720.00 every three months. The agreed transfer value is $19 500.00 and interest is 9.11% compounded annually. If the first payment is due at the date of the agreement, what is the size of the final payment? Answer: p =

- 1 = 0.02203587

19500 = 1720

(1.02203587)

11.0927704 = 0.24443886 = 1 - 1.0220359-n -n ln 1.0220359 = ln 0.75556114 -n(0.021796590) = -0.280294578 n = 12.85956074 PVn(due) = 1720

(1.02203587)

= 1720(0.84231184)(1.02203587) = $1480.70 The final payment is $1480.70. Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 97) John has agreed to purchase his partner's share in real estate by making payments of $5000.00 every three months. The agreed transfer value is $40 000.00 and interest is 10% compounded annually. If the first payment is due at the date of the agreement, what is the size of the final payment? Answer: p = 40000 = 5000

- 1 = 0.0241136891 (1.0241136891)

7.811632717 = 0.18836728271 = 1 - 1.0241136891-n -n ln 1.0241136891 = ln 0.8116327173 -n(0.023827545) = -0.2087073597 n = 8.75907946 PVn(due) = 5000

(1.0241136891) = 3806.27

The final payment is $3806.27 Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

98) A contract valued at $47 500.00 requires payments of $6951.00 every six months. The first payment is due in four years and interest is 6.6% compounded semi-annually. a) How many payments are required? b) What is the size of the last payment? c) How much will be paid in total? d) How much of what is paid is interest? Answer: a) FV = 47500.00(1.033)8 = 47500.00(1.2965897) = $61 588.01 61588.01 = 6951.00

(1.033)

8.577259937 = 0.283049578 = 1 - 1.033-n -n ln 1.033 = ln 0.716950422 -n(0.032467190) = -0.332748578 n = 10.24876426 The number of payments is 11. b) Present value of last payment: PVn (due) = 6951.00

(1.033) = 6951.00(0.237620547)(1.033)

= $1750.30 The final payment is $1750.30. c)

Total paid = 6951.00(10) + 1750.30 = 69510.00 + 1750.30 = $71 260.30 d) Interest = 71260.30 - 47500.00 = $23 760.30 Diff: 3 Type: SA Page Ref: 544-5576-58351 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

99) A mortgage balance of $137 960.70 is to be repaid over an eleven-year term by equal monthly payments at 8.1% compounded semi-annually. At the request of the mortgagor, the monthly payments were set at $1140.00. a) How many payments will the mortgagor have to make? b) What is the size of last payment? c) Determine the difference between the total actual amount paid and the total amount required to amortize the mortgage by the contractual monthly payments. Answer: a) p =

- 1 = 0.0066638834

= 245.8378879

246 payments b) PVn = 1140.00

= 1140.00(.8293897) = $949.40

949.40(1.006638834) = $955.70 c) 137960.70 = PMT 137960.70 = PMT(87.73874749) $1572.40 = PMT Contractual = 1572.40(132) = $207 556.80 Actual = 1140.00(245) + 955.70 = 280 255.70 Difference in payments = $72 698.90 Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

100) Catrina receives payments of $1120.00 at the beginning of each month from a pension fund of $79 500.00. Interest earned by the fund is 11.36% compounded monthly. a) What is the number of payments Catrina will receive? b) What is the size of the final payment? Answer: a) 79500.00 = 1120

(1.0094667)

70.31648017 = 0.665662679 = 1 - 1.0094667-n -n ln 1.0094667 = ln 0.334337321 -n(0.009422139) = -1.095604853 n = 116.2798491 payments b) PVn = 1120

(1.009466667)

= 1120(0.2789165895)(1.009466667) = $314.50 Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 101) A contract worth $59 000.00 provides benefits of $21 000.00 at the end of each year. The benefits are deferred for six years and interest is 8.95% compounded quarterly. a) How many payments are to be made under the contract? b) What is the size of the last benefit payment? Answer: a) p =

- 1 = 0.092548902

FV = 59000.00(1.092548902)6 = 59000.00(1.700768902) = $100 345.37

= 6.595730861 payments

102) A $176 000.00 mortgage is to be repaid over a 21-year period by monthly payments rounded up to the next higher $100.00. Interest is 9.44% compounded semi-annually. a) Determine the number of rounded payments required to repay the mortgage. b) Determine the size of the last payment. Answer: a)

- 1 = 0.007716274

176000.00 = PMT 176000.00 = PMT(110.914783) $1586.77 = PMT The rounded payment is $1600.00. n= = 245.7618375 payments. b)

PVn = 1600

103) The Acme Parts Company is repaying a debt of $15 000.00 by payments of $1410.00 made at the end of every three months. Interest is 7.2% compounded monthly. a) How many payments are needed to repay the debt? b) What is the size of the final payment? Answer: a)

- 1 = 0.018108216

= 11.92375947

The number of payments is 12. b)

PVg = 1410

= 1410.00(0.907947708) = $1280.21

1280.21(1.018108216) = $1303.39 Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 104) A debt of $17 000.00 is repaid by quarterly payments of $1430.00. If interest is 6.12% compounded quarterly, what is the size of the final payment? Answer: n =

105) A $151 000.00 mortgage is to be repaid over a 17-year period by monthly payments rounded up to the next higher $50.00. Interest is 8.2% compounded semi-annually. a) Determine the number of rounded payments required to repay the mortgage. b) Determine the size of the last payment. c) Calculate the amount of interest saved by rounding the payments up to the next higher $50.00. Answer: a) p =

- 1 = 0.00671944

151000.00 = PMT 151000.00 = PMT(110.8605388) $1362.07 = PMT Rounded payments $1400.00

= 192.6301891

192 equal payments plus one less than $1400 b) PVg = 1400

= 1400.00(0.626757778) = $877.46

Last Payment = 877.46(1.00671944) = $883.36 c) Unrounded 204(1362.07) = $277 862.28 Rounded 192(1400.00) + 883.36 = 269 683.36 Savings = $ 8178.92 Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

106) A demand (variable rate) mortgage of $137 000.00 is amortized over 20 years by equal monthly payments. After 21 months the original interest rate of 6% p.a was raised to 6.6% p.a. Three years after the mortgage was taken out, it was renewed at the request for the mortgagor for a five-year term at a fixed rate of 7.25% p.a. a) Calculate the mortgage balance after 21 months. b) Compute the size of the new monthly payment at the 6.6% rate of interest. c) Determine the mortgage balance at the end of the five-year term. Answer: a) p =

- 1 = 0.00486755

137000 = PMT 137000 = PMT(141.3843090) $968.99 = PMT PVn = 968.99 = 968.99(134.5076431) = $130 336.56 b) p =

- 1 = 0.005340319

130335.56 = PMT 130335.56 = PMT(128.9285035) $1010.91 = PMT c) PVn = 1010.91 1010.91(124.0775295) = $125431.22 p=

- 1 = 0.005849741

125431.22 = PMT 125431.22 = PMT(118.9350584) $1054.62 = PMT PVn = 1054.62

= 1054.62(97.14104632) = $102 446.89

Diff: 3 Type: SA Page Ref: 585-594 Topic: 14.4 Residential Mortgages in Canada Objective: 14-5: Compute the periodic payments for fixed-rate mortgages and for demand mortgages.

107) A bank has an interest rate of 5% compounded semi annually for a five year closed mortgage. If the mortgage has monthly payments, calculate the effective rate of interest per each term. Answer: i =

= 0.025; c =

p = 1.0251/6 - 1 = 0.004123915 = 0.4124% Programmed solution:

p = 0.4124% Diff: 2 Type: SA Page Ref: 585-594 Topic: 14.4 Residential Mortgages in Canada Objective: 14-4: Compute the effective interest rate for fixed-rate residential mortgages. 108) How much interest is paid in total on a 3-year loan for $28 600 (rounded to the nearest dollar)? The interest rate is 8.7% compounded monthly and the payments are monthly. A) $3997 B) $3779 C) $3977 D) $3999 E) $4000 Answer: A Diff: 2 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 109) How much interest is paid in total on a 1-year loan for $5000? The interest rate is 12 % compounded monthly and the payments are monthly. A) $330.00 B) $333.88 C) $333.00 D) $300.88 E) $330.88 Answer: E Diff: 2 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

110) How much principal is repaid in the 74th payment interval on a $142 300 mortgage? The mortgage is amortized over 25 years and the payments are monthly. The interest rate is 7.44% compounded monthly. A) $257.16 B) $275.16 C) $527.16 D) $572.16 E) $574.16 Answer: A Diff: 1 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 111) How much total principal is repaid between the 1st and 7th payment interval of a 4.5-year loan for $4887 at an interest rate of 7.4% compounded monthly and the payments are also monthly. Round your answer to the nearest dollar. A) $556 B) $546 C) $456 D) $445 E) $423 Answer: B Diff: 2 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 112) What is the outstanding balance after 23 payments on a 20-year mortgage that has monthly payments of $1062.32 and an interest rate of 6.6% compounded monthly? A) $143 402.54 B) $144 302.54 C) $134 402.54 D) $134 302.54 E) $134 322.54 Answer: C Diff: 2 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

113) What is the monthly payment size of a 21-year mortgage for $169 600 and an interest rate of 7.2% compounded semi-annually? A) $1169.11 B) $1296.11 C) $1926.11 D) $1211.96 E) $1222.96 Answer: B Diff: 2 Type: MC Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules. 114) What is the monthly payment size of a 25-year mortgage for $100 000 and an interest rate of 6% compounded semi-annually? A) $639.81 B) $1639.81 C) $630.81 D) $600.81 E) 633.81 Answer: A Diff: 2 Type: MC Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules. 115) How much interest is paid in the 53rd monthly payment interval of a loan for $43 200? The loan is amortized at a rate of 9.17% compounded annually over 7 years. A) $411.11 B) $111.11 C) $144.11 D) $444.11 E) $400.11 Answer: C Diff: 2 Type: MC Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

116) How much total interest is paid between the 6th and 13th payments on a loan that has monthly payments for 3 years and an original principal of $14 750? The loan rate is 6.6% compounded quarterly. A) $540 B) $405 C) $450 D) $504 E) $545 Answer: D Diff: 2 Type: MC Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules. 117) How much interest is paid after the first payment interval on a $100 000 mortgage? The mortgage is amortized over 25 years and the payments are monthly. The interest rate is 6% compounded semiannually. Round your answer to the nearest dollar. A) $146 B) $640 C) $527 D) $494 E) $401 Answer: D Diff: 1 Type: MC Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules. 118) What is the outstanding balance after the 23rd payment interval of a 16-year loan for $14 734 with semi-annual payments and an interest rate of 5.95% compounded quarterly? A) $5625.21 B) $5652.21 C) $5562.21 D) $5526.21 E) $5566.21 Answer: A Diff: 1 Type: MC Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules.

119) What is the size of the final unequal payment of a loan that has 69.47 payments and the equal payment size is $215.64? The interest rate is 3.1% per compounding period. A) $120.17 B) $210.17 C) $102.17 D) $99.10 E) $99.17 Answer: C Diff: 3 Type: MC Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 120) What is the size of the final unequal payment of a loan for $2723.44 if it has quarterly payments of $500? The interest rate is 8.4% compounded quarterly. A) $432.14 B) $423.14 C) $441.14 D) $414.14 E) $424.14 Answer: B Diff: 2 Type: MC Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 121) Munawar is a first time buyer and bought a brand new townhouse in Oshawa for $212 000. He has saved enough to make a 5% down payment and will have to pay mortgage loan insurance at 2.75% of the mortgage balance. Munawar wants the insurance premium added to the maximum allowable loan balance. Determine his maximum initial mortgage balance. A) $5538.50 B) $201 400 C) $206 938.50 D) $212 000 E) $212 477 Answer: C Diff: 1 Type: MC Page Ref: 585-594 Topic: 14.4 Residential Mortgages in Canada Objective: 14-6: Create statements for various types of residential mortgages.

122) Li bought a house in Whitby for $305 000 5 years ago. He financed the house through Scotia Bank with a mortgage of $259 000 amortized over 25 years. Interest for the first 5 years was 4.4% compounded semi-annually and payments were made monthly. At the end of the 5 year term, Li has renewed the mortgage for another 5 years, at 3.7% compounded semi-annually. What is the monthly payment for the second term? Answer: First calculate the balance at the end of the first 5 year term. This requires calculating the FV of the payments made in first 5 years and the future value of the original loan. PV = $259 000; n = 12 × 25 = 300;

∴p=

= 2.2%;

- 1 = 0.0036335

PMT for original term =

= $1419.13

∴FV after 1st term = 1419.13

= $94 950

FV of the original loan at the end of 5 year term = 259000 × 1.003633560 = $321 965 Balance = $321 965 - $94 950 = $227 015 Balance is the PV for the second term For second term PV = $227015;

∴p=

n = 12 × 20 = 240;

= 1.85%;

- 1 = 0.00306

PMT for second term =

= $1336.72

Diff: 3 Type: SA Page Ref: 585-594 Topic: 14.4 Residential Mortgages in Canada Objective: 14-5: Compute the periodic payments for fixed-rate mortgages and for demand mortgages. 123) Shiva transferred his mortgage of $145 981 from CIBC to BMO for the remaining amortization period at 3.59% compounded semi-annually. He has agreed to pay rounded up payment of $2000 every month. How long will it take Shiva to pay-off his mortgage? A) 6 years 11 months B) 6 years 1 months C) 8 years D) 6 years 6 months E) 10 years Answer: A Diff: 3 Type: MC Page Ref: 585-594 Topic: 14.4 Residential Mortgages in Canada Objective: 14-5: Compute the periodic payments for fixed-rate mortgages and for demand mortgages.

124) John makes a monthly mortgage payment of $1940 at the end of every month. On October 31, his loan balance was $145 000 after making a payment of $1940. His financial institute allows him to make one additional payment equal to or less than his monthly payment, once a month. Taking advantage of that, John made a payment of $1500 on November 15 and $1900 on December 20. What is his year-end balance after making monthly payment on December 31? The effective monthly fixed rate is 0.306%. Answer: For p = 0.306%, equivalent simple annual rate is 3.67% Using MS-Excel,, mortgage statement is created as follows Payment Date 31-Oct 15-Nov 30-Nov 20-Dec 31-Dec

Number of Amount Days Paid

Interest Paid

Principal

15 15 20 11

$218.69 $216.76 $285.55 $155.27

$1281.31 $1723.24 $1614.45 $1784.73

$1500 $1940 $1900 $1940

Balance Repaid $145,000.00 $143,718.69 $141,995.45 $140,381.00 $138,596.26

Year-end balance = $138 596.26 Diff: 3 Type: SA Page Ref: 585-594 Topic: 14.4 Residential Mortgages in Canada Objective: 14-6: Create statements for various types of residential mortgages. 125) Peter worked for GM for 31 years. On his retirement, he opted for a pension buy-out and received $1 467 921. He invested his money in an annuity that provided for payments of $9000 at the end of every month. If interest is 4.4% compounded monthly, determine the size of the final payment. Answer: PV = $1 467 921; PMT = $9000; P/Y = 12 C/Y = 12; I/Y = 4.4% i=

= 0.367%

1467921 = 9000 ⇒ -n ln 1.00367 = ln 0.402 ⇒ n = 248.7953157 months For last month's payment: PMT = $9000; i = 0.367%; n = 0.7953157 PV = $9000 × 1.00367 ×

= $7160.52

Diff: 3 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

126) The Korean community is planning to buy a place for their community in Markham valued at $1 697 000. They made a down payment of 10% and payments of $400 000 are required at the end of every 3 months. If interest is 9% compounded monthly, what is the size of the final payment? A) $15 685 B) $205 512 C) $131 373 D) $355 547 E) $48 381 Answer: A Diff: 2 Type: MC Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 127) Kirk had accumulated $975 000 in his RRSP at the age of 55. He converted the RRSP into an RRIF at that time and started withdrawing $5000 at the beginning of each month. If interest is 4.5% compounded monthly, what is the size of his final withdrawal? A) $1739.27 B) $1409.28 C) $4947.31 D) $412.07 E) $33.82 Answer: A Diff: 2 Type: MC Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 128) Honda is selling a 2014 Accord Coupe Ex M for $30 056.90 including freight, PDI and all applicable fees. The lease payment of $382 is due at the beginning of each month. If the interest rate compounded annually is 3.99% and the residual value is $13 200, determine the size of the final lease payment. A) $240.51 B) $285.95 C) $150.28 D) $214.13 E) $5.92 Answer: D Diff: 2 Type: MC Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

129) Zara bought a 2013 Kia Forte by taking a $19 500 loan from BMO with interest at 5.99% compounded annually and amortized by payments of $378 at the end of every month. She receives an inheritance after making 36 payments and plans to pay the balance in full. What is the payout figure just after the last regular payment made? Answer: PV = $19 500; n = 3; i = 5.99% FV = 19500 × 1.05993 = $23 218.24 Accumulate the value of 36 payments:

PMT = $378; n = 36; P/Y = 12; C/Y = 1; I/Y = 5.99%; c =

p= FV = 378

; i = 5.99%

-1 = 0.00485965 = $14 831.65

130) Rebecca bought a house in Oshawa and took a $259 000 mortgage amortized by monthly payments over 25 years that is renewable after 5 years. If interest is 4.4% compounded semi-annually, what is the outstanding balance at the end of the five year term? If the mortgage is renewed for a further 5 year term at 3.7% compounded semi-annually, what is the balance after the second 5 year term? Answer: PV = $259 000; n = 25 × 12 = 300; P/Y = 12; C/Y = 2; I/Y = 4.4%; c =

;i=

= 2.2%

-1 = 0.0036335

259000 = PMT ⇒ PMT = $1419.13 FV = 259000 × 1.003633560 = $321 965.04 FV60 = 1419.13

= $94 950.38

Balance at the end of 5 year term = $321965.04 - $94950.38 = $227 014.66 For second 5 year term: PV = $227 014.66; n = 20 × 12 = 240; P/Y = 12; C/Y = 2;

I/Y = 4.4%; c =

;i=

= 1.85%

-1 = 0.003059831

227014.66 = PMT ⇒ PMT = $1336.72 FV = 227014.66 × 1.00305983160 = $272 686.88 FV60 = 1336.72

= $87 890.34

131) Amanda borrowed $57 800 from BMO to buy a 2014 E-Class Mercedes Sedan at 3.9% compounded monthly. Payments of $2056 are required at the end of every month. Construct a complete amortization schedule in MS Excel showing the details of all payments and the total. Answer: Outstanding Payment Principal Principal Number Amount Paid Interest Paid Repaid balance 0 $57,800 1 $2,056 $188 $1,868 $55,932 2 $2,056 $182 $1,874 $54,058 3 $2,056 $176 $1,880 $52,177 4 $2,056 $170 $1,886 $50,291 5 $2,056 $163 $1,893 $48,398 6 $2,056 $157 $1,899 $46,500 7 $2,056 $151 $1,905 $44,595 8 $2,056 $145 $1,911 $42,684 9 $2,056 $139 $1,917 $40,766 10 $2,056 $132 $1,924 $38,843 11 $2,056 $126 $1,930 $36,913 12 $2,056 $120 $1,936 $34,977 13 $2,056 $114 $1,942 $33,035 14 $2,056 $107 $1,949 $31,086 15 $2,056 $101 $1,955 $29,131 16 $2,056 $95 $1,961 $27,170 17 $2,056 $88 $1,968 $25,202 18 $2,056 $82 $1,974 $23,228 19 $2,056 $75 $1,981 $21,248 20 $2,056 $69 $1,987 $19,261 21 $2,056 $63 $1,993 $17,267 22 $2,056 $56 $2,000 $15,267 23 $2,056 $50 $2,006 $13,261 24 $2,056 $43 $2,013 $11,248 25 $2,056 $37 $2,019 $9,229 26 $2,056 $30 $2,026 $7,203 27 $2,056 $23 $2,033 $5,170 28 $2,056 $17 $2,039 $3,131 29 $2,056 $10 $2,046 $1,085 30 $1,089 $4 $1,085 $0 Totals $60,713 $2,913 $57,800 Diff: 3 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

132) Rebecca borrowed $20 400 from her local credit union to pay for the upgrades in her kitchen. She wants to pay the loan with monthly payments over a period of 3 years. What is the interest included in the tenth payment, if the rate charged by the credit union is 6% monthly? A) $78.19 B) $5698.49 C) $21 336.56 D) $15 638.09 E) $75.48 Answer: A Diff: 2 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 133) Howden is interested in purchasing a jet ski for $15 834. He chose to finance this purchase across a 5year term at an interest rate of 3.6% compounded quarterly. For the first, second and third payment, how much of the payment is interest? A) $726.13, $728.67, $735.23 B) $142.51, $142.51, $142.51 C) $864.64, $864.64, $864.64 D) $153.98, $147.42, $145.88 E) $142.51, $135.97, $129.41 Answer: E Diff: 3 Type: MC Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 134) Jacob took out a loan of $11 256 to furnish his new studio apartment that charged an interest rate of 10.8% compounded monthly over 3 and a half years. After the third payment, how much of the loan remains to be paid? Answer: PV = 11 256; n = 3.5(12) = 42; i = 0.108/12 = 0.009; 11 256 = PMT PMT = 323.03 First period interest = 11 256(0.009) = 101.30 First period principal payment = 323.03 - 101.30 = 221.73 Owed after first period = 11 256 - 221.73 = 11034.27 Second period interest = 11034.27(0.009) = 99.31 First period principal payment = 323.03 - 99.31 = 223.72 Owed after second period = 11 034.27 - 223.72 = 10810.55 Third period interest = 10810.55(0.009) = 97.29 Third period principal payment = 323.03 - 97.29 = 225.74 Owed after third period: 10810.55 - 225.74 = 10584.81 Diff: 2 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

135) Huzaifa received a loan of $43 610 from the bank at 9.1% compounded semi-annually to help start his business. As per the terms of the loan, Huzaifa must pay off the loan in 8 years and 6 months. What's the loan balance after 4 years? Answer: PV = 43 610; n = 8.5(2) = 17; i = 0.091/2 = 0.0455; 43 610 = PMT 3739.25 = PMT Original Debt:

FV = 43610(1.0455)8 = $62255.59 4 years of payments: FV = 3739.25

= $35136.88

Balance:

Outstanding balance = $62255.59 - $35136.88 = $27 118.71 Diff: 2 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules. 136) The Allenoyin Association needed to cover some of their off-season costs and took a loan of $24 755 from the bank at 6.24% compounded bi-weekly. Payment of $680 were expected to be made every other week. How many payments are necessary? Answer: PV = $24 755; PMT = $680; i = 0.0624/26 = 0.0024 24 755 = 680 0.087370588 = 1 - 1.0024-n 1.0024-n = 0.9126294118 -n = n = 38.13960377 Therefore, 38 payments of $680 and one smaller payment will be necessary to cover the loan. Diff: 2 Type: SA Page Ref: 549-566 Topic: 14.1 Amortization Involving Simple Annuities Objective: 14-1: Perform computations associated with amortization of debts involving simple annuities, and construct complete or partial amortization schedules.

137) Mirsu Materials acquired a loan of $68 900 at 5.4% compounded quarterly to begin operations. They must make equal monthly instalments for the next 11 years to pay off this loan. After the first payment, how much does Mirsu Materials still owe? Answer: PV = $68 900; n = 11(12) = 132; c = 4/12 = 1/3; i = 0.054/4 = 0.0135 p=

- 1 = 0.004479901

68 900 = PMT 692.56 = PMT First period interest = 68900(0. 004479901) = 308.67 First period principal payment = 692.56 - 308.67 = 383.89 Owed after first period = 68900 - 383.89 = 68 516.11 Diff: 2 Type: SA Page Ref: 569-574 Topic: 14.2 Amortization Involving General Annuities Objective: 14-2: Perform computations associated with amortization of debts involving general annuities, and construct complete or partial amortization schedules. 138) In the contract for actor Dwayne "The Large Pebble" Johnson's new movie, he has agreed to receive payments of $15 000 every month for the next 6.5 years. At the time the contract was established, he interest rate was 8.6% compounded semi-annually. Using the prospective method, determine the present value of all these payments. Answer: PMT = 15 000; n = 6.5(12) = 78; c = 2/12 = 1/6; i = 0.086/2 = 0.043 p= PV = 15000

- 1 = 0.007041539 = $897 884.38

139) At the end of every year, The Train Trust must make a payment of $675 to the bank towards their loan of . The loan agreement has an interest rate of 2.3% compounded annually. What's the size of the final payment? Answer: PV = $8400; PMT = $675; i = 0.073 8400 = 675 0.28622222 = 1 - 1.023-n 1.023-n = 0.71377777 -n = n = 14.828109 Fractional value of n for last payment period: 14.828109 - 14 = 0.828109 PV = 675 PV = $547.47 Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size. 140) Ivey Inc has decided to lease a new building for operations. The lease obligation is $10 900 with lease payments of $315 made every three months. If the interest rate is 4.4% compounded quarterly, what's the size of the last payment? Answer: PV = $10 900; PMT = $315; i = 0.044/4 = 0.011; 10 900 = 315 0.389635 = 1.011-n 1.011-n = 0.619365 -n = n = 43.7900498 Fraction value of n for last payment period: 0.7900498 PV = 315

= $246.44

Diff: 2 Type: SA Page Ref: 576-583 Topic: 14.3 Finding the Size of the Final Payment Objective: 14-3: Find the size of the final payment when all payments except the final payment are equal in size.

141) ________ is an affordability measure used to determine qualification for mortgage default insurance. It takes the monthly housing costs such as property tax as a percentage of the gross household monthly income, and to qualify, this measure must be no more than ________. A) Total debt service; 35% B) Gross debt service; 25% C) Gross debt service; 35% D) Variable equity mortgage ratio; 25% E) Total debt service; 25% Answer: C Diff: 1 Type: MC Page Ref: 585-594 Topic: 14.4 Residential Mortgages in Canada Objective: 14-5: Compute the periodic payments for fixed-rate mortgages and for demand mortgages.

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 15 Bond Valuation 1) A $5000 bond that pays 6% semi-annually is redeemable at par in 14 years. Calculate the purchase price if it is sold to yield 8% compounded semi-annually. Answer: PMT =

= $150

PP = 5000(1 + 0.04)-28 + 150 PP = 1667.39 + 2499.46 PP = $4166.85 Programmed solution:

The purchase price is 1667.39 + 2499.46 = $4166.85 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 2) A $25 000, 6% bond redeemable at par is purchased 11 years before maturity to yield 6.9% compounded semi-annually. If the bond interest is payable semi-annually, what is the purchase price of the bond? Answer: FV = $5000; n = 11(2) = 22; P/Y = C/Y = 2; I/Y = 6.9; i =

= 0.0345; PMT = 25000(0.06)

$750.00 PP = 25000(1.0345)-22 + 750 = 25000.00(.4741646) + 750.00(15.2416064) = 11854.11 + 11431.20 = $23 285.32 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

3) A $50 000 bond bearing interest at 6.5% bond payable semi-annually matures in 10 years. If it is bought to yield 5.7% compounded semi-annually, what is the purchase price of the bond? Answer: FV = $50 000; n = 10(2) = 20; P/Y = C/Y = 2; PMT = 50000(0.065) PP = 50000(1.0285)-20 + 1625

= $1625; i =

= 0.0285

= 28502.57 + 24514.61 = $53 017.18

Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 4) A $200 000 bond is redeemable at 108 in 14 years. If interest on the bond is 5.5% payable semi-annually, what is the purchase price to yield 4% compounded semi-annually? Answer: PP = 200000.00(1.08)(1.02)-28 + 5500.00 = 200000.00(1.08)(.5743746) + 5500.00(21.2812724) = 124064.90 + 117047.00 = $241 111.90 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 5) What is the purchase price of a $10 000, 3.5% bond with semi-annual coupons redeemable at 104 in seven years if the bond is bought to yield 2.5% compounded semi-annually? Answer: FV = $10000(1.04) = $10400; P/Y = C/Y = 2; PMT = 10000(0.035/2) = $175.00; n = 7(2) = 14. PP = 10400.00(1.0125)-14 + 175.00 = 10400.00(.8403681) + 175.00(12.7705528) = 87398.28143 + 2234.84673 = $89 633.13 Diff: 1 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

6) A $100 000 bond is redeemable at 105 in 20 years. If interest on the bond is 5% payable semi-annually, what is the purchase price to yield 6% compounded semi-annually? Answer: FV = 100000(1.05) = $105000; PMT = 100000(0.05)

= $2500; n = 20(2) = 40; P/Y = C/Y = 2; i =

0.03 PP = 105000(1.03)-40 + 2500 = 105000(.3065568408) + 2500(23.11477197) = 32188.47 + 57786.93 = $89 975.40 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 7) A $100 000, 4% bond with semi-annual coupons is redeemable at 108. What is the purchase price to yield 5.5% compounded semi-annually seven years before maturity? Answer: FV = $100000(1.08) = $108 000; PMT = 100000(0.04/2) = $2000; P/Y = C/Y = 2; i =

= 0.0275; n =

7(2) = 14. PP = 108000(1.0275)-14 + 2000 = 73871.70584 + 22982.01627 = $96 853.72 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 8) A $5000, 6.5% bond with semi-annual coupons redeemable at 108 is bought two years before maturity to yield 6% compounded semi-annually. What is the purchase price? Answer: FV = $5000(1.08) = $5400; P/Y = C/Y = 2; PMT = $5000

= $162.50; i =

= 0.03

PP = 5400(1.03)-4 + 162.50 = 5400.00(.888487) + 162.50(3.7170984) = 4797.83 + 604.03 = $5401.86 Diff: 1 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

9) A $200 000.00, 6% bond with semi-annual coupons is redeemable at par. What is the purchase price of the bond six years before maturity to yield 8% compounded semi-annually? Answer: FV = $200 000; P/Y = C/Y = 2; PMT = 200000(0.06/2) = $6000; i =

= 0.04; n = 6(2) = 12.

PP = 200000(1.04)-12 + 6000 = 124919.4099 + 56310.44256 = $181 229.85 Diff: 1 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 10) A $100 000, 7.5% bond with semi-annual coupons redeemable at 106 is bought three years before maturity to yield 8% compounded semi-annually. What is the purchase price? Answer: FV = $100000(1.06) = $106000; P/Y = C/Y = 2; PMT = $100000

= $3 750; n = 3(2) = 6

PP = 106000.00(1.04)-6 + 3750.00 = 106000.00(.7903145257) + 3750.00(5.242136857) = 83773.34 + 19658.01 = $103 431.35 Diff: 1 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 11) A $150 000 bond bearing interest at 6% payable semi-annually is bought eight years before maturity to yield 4.5% compounded annually. If the bond is redeemable at par, what is the purchase price? Answer: FV = $150 000; PMT = 150000(0.06)0.5 = $4500; n = 8(2) = 16; P/Y = 2; C/Y = 1; c =

= 0.5; i =

= 0.045

p = (1 + 0.054)1/2 - 1 = 0.0222524 PP

= 150000(1.0222524)-16 + 4500

= 150000(.7031853) + 4500(13.3385481) = 105477.77 + 60023.46 = $165 501.23 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

12) A $100 000 bond bearing interest at 8% payable semi-annually is bought five years before maturity to yield 6% compounded annually. If the bond is redeemable at par, what is the purchase price? Answer: FV = $100000; PMT = $100000(0.08)0.5 = $4000; P/Y = 2; C/Y = 1; c = i=

= 0.06; p =

; n = 5(2) = 10;

- 1 = .029563014

PP = 100000(1.029563014)-10 + 4000 = 100000(.7472581729) + 4000(8.549257745) = 74725.82 + 34197.03 = $108 922.85 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 13) Six $1500 bonds with 4.5% coupons payable semi-annually are bought to yield 5% compounded monthly. If the bonds are redeemable at par in eight years, what is the purchase price? Answer: FV = $1500(6) = $9 000.00; PMT = 9000(0.05)0.5 = $225.00; n = 8(2) = 16; P/Y = 2; C/Y = 12; c=

= 6; p =

- 1 = 0.02526187

= 9000 (1.02526187)-16 + 225

= 6037.90 + 2931.40 = $8969.30 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

14) A $150 000 bond redeemable at par on October 1, 2026, is purchased on January 15, 2014. Interest is 6% payable semi-annually and the yield is 7.5% compounded semi-annually. a) What is the market price of the bond? b) How much interest has accrued? c) What is the cash price? Answer: FV = $150 000; PMT = $4500; P/Y = C/Y = 2; The interest date preceding the date of purchase is October 1, 2013. The time period from October 1, 2013 to the date of maturity is 13 years. n = 13(2) = 26; i =

= 0.0375

a) PP on October 1, 2013 = 150000(1.0375)-26 + 4500 = 57597.09 + 73922.33 = $131 519.42 Market price = $131 519.42 b) Days in the interest period Oct 1, 2013-Apr 1, 2014: 182 days; Days between Oct 1, 2013 and Jan 15, 2014: 106 Accrued interest = 150000(.03)

= $2620.88

c) Cash price = 131519.42 + 2620.88 = $134 140.30 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 15) Bonds in denominations of $10 000 redeemable at par in six years and four months are offered for sale. If the coupon rate is 6.5% payable quarterly and the expected yield is 8% compounded quarterly, determine a) the market price; b) the accrued interest; c) the cash price. Answer: FV = $10 000; PMT = 10 000(0.065)(0.25) = $162.50; a) The interest date preceding the purchase date is 6.5 years before maturity. n = 6.5(4) = 26; P/Y = 4; C/Y = 4; I/Y = 8; i =

= 0.02

PP = 10000(1.02)-26 + 162.50 = 5975.79 + 3269.67 = $9245.46 Market price = $9245.46 b) The accrued interest two months later PV = $10 000; r =

= 0.01625; t =

= 0.5

Interest = 10 000(0.01625)0.5 = $81.25 c) Cash price = $9245.46 + 81.25 = $9326.71 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

16) Bonds in denominations of $100 000 redeemable at 104 are offered for sale. If the bonds mature in ten years and six months and the coupon rate is 5.5% payable quarterly, what is the market price of the bonds to yield 7.2% compounded quarterly? Answer: FV = 100 000(1.04) = $104 000; PMT = 100000(0.055)0.25 = $1375.00; n = 10.5(4) = 42; P/Y = 4; C/Y = 4; I/Y = 7.2; i =

= 0.018;

The purchase price on the interest date preceding the date of purchase PP = 104000(1.018)-42 + 1375 = 49161.67 + 40279.22 = $89 440.89 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 17) A $100 000, 8.75% bond with interest payable annually is redeemable at 103 in four years. What is the purchase price to yield 7% compounded quarterly? Answer: FV = $100 000; P/Y = 1; C/Y = 4; PMT = 100000(0.0875/1) = $8750.00; c = p=

= 4; n = 4(1) = 4.

- 1 = 0.071859031

PP = 103000.00(1.071859031)-4 + 8750 = 103000.00(.7576164) + 8750.00(3.373044266) = 78034.49 + 29514.14 = $107 548.63 Diff: 1 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 18) A $1 000 000, 9% bond with interest payable annually is redeemable at 104 in 5 years. What is the purchase price to yield 8% compounded quarterly? Answer: FV = $1 000 000(1.04) = $1040 000; P/Y = 1; C/Y = 4; PMT = 1 000 000(0.09/1) = $90 000; c = p=

= 4;

- 1 = .08243216

PP = 1040000(1.08243216)-5 + 90000 = 1040000.00(.6729713331) + 90000.00(3.96724612) = 699890.19 + 357052.15 = $1 056 942.34 Diff: 1 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

19) A $50 000 bond bearing interest at 5.5% payable semi-annually is redeemable at par on August 10, 2033. The bond is sold on the primary market on December 10, 2013, to yield 5% compounded semiannually. Determine a) calculate the market price; b) the accrued interest; c) the cash price. Answer: FV = $50 000; P/Y = C/Y = 2; PMT = 50000

= $1 375.00;

a) The interest date preceding the purchase is August 10, 2013. The time period until maturity is 20 years. n = 20(2) = 40; i =

= 0.025;

PP = 50000(1 + 0.025)-40 + 1375 PP = 18621.53118 + 34516.3157 = $53 137.85 The purchase price is $53 137.85 b) The number of days from August 10, 2013 to February 10, 2014 is 184 days. The number of days from August 10, 2013 to December 10, 2013 is 122 days. PV = $53 137.85; r =

= 0.0275; t =

Interest = 53137.85(0.0275)

= $968.90

The accrued interest is $968.90 c) The cash price is $53 137.85 + 968.90 = $5106.75 Diff: 3 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

20) A $50 000 bond bears interest at 4.5% payable semi-annually and is redeemable at par on November 4, 2022. The bond is sold on March 20, 2013, to yield 5.5% compounded semi-annually. What is the cash price? Answer: FV = 50 000; PMT = 5000

= $1125.00; P/Y = C/Y = 2; i =

= 0.0275

Since the maturity date is Nov.4, the semi-annual interest dates are May 4 and Nov. 4. The interest date preceding the date of purchase is November 4, 2012. The time period from Nov. 4, 2012 to November 4, 2022 is 10 years. n = 10(2) = 20 The purchase price on the date preceding the date of purchase PP(Nov. 4, 2012) = 50000(1.0275)-20 + 1125 = 29062.52832 + 17130.65865 = $46 193.19 The purchase price is $46 193.19 The number of days from Nov.4, 2012 to May 4, 2013 is 181 days. The number of days from Nov.4, 2012 to March 20, 2013 is 136 days. PV = $46 193.19; r = i = 0.0225; t = FV = $46193.19

= $46 974.14

The cash price on March 20, 2013 is $46 974.14 Diff: 3 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

21) A $250 000, 7% bond redeemable at par with interest payable semi-annually is bought 4 years before maturity. Determine the premium or discount and the purchase price if the bond is purchased to yield: a) 6.5% compounded semi-annually b) 8% compounded semi-annually Answer: FV = $250 000; P/Y = 2; C/Y = 2; b = a) i =

= 0.035; PMT = 250 000(0.035) = $8750.00

= 0.0325; n = 4(2) = 8.

Since b > i, the bond will sell at a premium. PP = 250 000(1.0325)-8 + 8750

= 193 561.744 + 60 779.661 = $254 341.41 The premium is $254 341.41-$250 000 = $4341.41 b) i =

= 0.04; n = 4(2) = 8.

Since b < i, the bond will sell at a discount. PP = 250 000(1.04)-8 + 8750

= 182 675.551 + 58 911.518 = $241 587.07 The discount is $250 000 - $241 587.07 = $8412.93 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 22) A $10 000, 7.2% bond with semi-annual coupons is purchased 3 years before maturity. Calculate the discount or premium if the bond is sold to yield 6% compounded semi-annually. Answer: FV = $10 000; P/Y = 2; C/Y = 2; n = 3(2) = 6; b =

= 0.036; i =

= 0.03.

Since the bond rate is greater than the yield rate (b > i), the bond is sold at a premium. PMT = 10 000

= $360.00

PP = 10000(1.03)-6 + 360 = 8374.84 + 1950.189 = $10 325.03 The purchase price is $10 325.03 The premium is 10 325.03 - 10 000 = $325.03 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

23) A $50 000, 4% bond with semi-annual coupons is purchased three years before maturity. Calculate the discount or premium if the bond is sold to yield 6% compounded semi-annually. Answer: FV = $50 000; P/Y = C/Y = 2; PMT = 50 000

= $1 000; n = 3(2) = 6; i =

= 0.03.

Since the bond rate is less than the yield rate (b < 1), the bond is sold at a discount. PP = 50000(1 + 0.03)-6 + 1000 PP = 41 874.21283 + 5 417.19144 = $47 291.40 The purchase price is $47 291.40 The discount is 50 000 - 47 291.40 = $2708.60 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 24) A $1000, 7% bond redeemable at par in eight years bears coupons payable annually. Compute the premium or discount and the purchase price if the yield, compounded annually, is: a) 6.5% b) 7.5% c) 8.5% Answer: FV = $1000; P/Y = C/Y = 1; b = a) i =

= 0.07; n = 8(1) = 8

= 0.065; Since b > i the bond will sell at a premium.

Premium = [1000(0.07) - 1000(0.065)] = 5(6.088751) = $30.44 PP = 1000.00 + 30.44 = $1034.44 b) i =

= 0.075; Since b < i the bond will sell at a discount.

Discount = [1000(0.07) - 1000(0.075)] = -5(5.8573036) = -$29.29 PP = 1000 - 29.29 = $970.71 c) i =

= 0.085; Since b < i the bond will sell at a discount.

Discount = [1000(0.07) - 1000(0.085)] = -15.00(5.639183) = -$84.59 PP = 1000 - 84.59 = $915.41 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

25) A $1000, 6.5% bond redeemable at par in four years bears coupons payable annually. Compute the premium or discount and the purchase price if the yield, compounded annually, is 7.5%. Answer: FV = $1000; P/Y = C/Y = 1; n = 4(1) = 4; b = 0.065; i = 0.075; Since b < i, the bond will sell at a discount Discount = [1000(0.065) - 1000(0.075] = -10(3.349326) = $33.49 PP = 1000 - 33.49 = $966.51 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 26) Bonds with a face value of $130 000 redeemable at par bearing 4.5% coupons payable quarterly are sold six years before maturity to yield 5.25% quarterly. Determine a) the premium or discount; b) the purchase price? Answer: FV = $130 000; P/Y = C/Y = 4; n = 6(4) = 24; b =

= 0.01125; i =

= 0.013125

a) Since b < i , the bond will sell at a discount Discount = [130 000(0.01125) - 130 000(0 .013125)] = -243.75(20.473452) = - $4990.40 b) PP = 130 000 - 4990.40 = $125 009.60 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 27) Nine $10 000, 4% bonds with interest payable semi-annually and redeemable at par are purchased 8.5 years before maturity. Find the premium or discount and the purchase price if the bonds are bought to yield 6%. Answer: FV = 9($10 000) = $90 000; P/Y = C/Y = 2; b =

= 0.02; i =

= 0.03; n = 8.5(2) = 17.

Since b < i, the bonds are sold at a discount. Discount = [90000(0.02) - 90000(0.03)] = -900(13.1661185) = $11 849.51 PP = 90000 - 11849.51 = $78 150.49 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

28) A $50 000.00, 8% bond with quarterly coupons redeemable at par is purchased five years before maturity to yield 4% compounded semi-annually. Determine the premium or discount. Answer: FV = $50 000; P/Y = 4; C/Y = 2; n = 5(4) = 20; c =

= 0.5; p ==

- 1 = .00995049

Since b > i adjusted, the bond sells at a premium Premium = [50000(0.02) - 50 000(0.00995049)] 502.48(18.054551) = $9072.05 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 29) Ten $10 000, 4% bonds with interest payable semi-annually and redeemable at par are purchased 18.5 years before maturity. Find the premium or discount and the purchase price if the bonds are bought to yield 7%. Answer: FV = 10($10 000) = $100 000; P/Y = C/Y = 2; b =

= 0.02; i =

= 0.035; n = 18.5(2) = 37.

Since b < i, the bonds are sold at a discount. Discount = [100000(0.02) - 100000(0.035)] = -1500(20.570525) = -$30 855.79 PP = 100000.00 - 30 855.79 = $69 144.21 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 30) Twenty $5 000 bonds redeemable at par bearing 6% coupons payable quarterly are sold eight years before maturity to yield 5.5% compounded annually. What is the purchase price of the bonds? Answer: FV = 20(5 000) = $100 000; P/Y = 4; C/Y = 1; c = p=

= 0.25; n = 8(4) = 32; b =

= 0.015;

- 1 = 0.01347517. Since b > i, the bonds are sold at a premium.

Premium = [100000(0.015 ) - 100 000(0.01347517] = 152.482556(25.855038) = $3942.44 PP = 100 000.00 + 3 942.44 = $103 942.44 Diff: 3 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

31) A $125 000 bond, redeemable at par in three years with 7.5% coupons payable quarterly, is bought to yield 6% compounded quarterly. (i) Compute the premium or discount and the purchase price. (ii) Construct a schedule for amortization of premium. Answer: FV = $125 000; P/Y = C/Y = 4, b =

= 0.01875; i =

= 0.015; n = 3(4) = 12;

PMT = 125 000(0.01875) = $2 343.75. Since b > i, bond sells at a premium. (i) Premium = [125000(0.01875) - 125000(0.015)] = 468.75(10.907505) = $5 112.89 PP = 125 000.00 + 5112.89 = $130 112.89 (ii)

Diff: 3 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts.

32) A $100 000 bond, redeemable at 110 in seven years with 6.75% coupons payable annually, is bought to yield 7.25% compounded annually. (i) Determine the discount and the purchase price. (ii) Construct a schedule of accumulation of discount. Answer: FV = $100 000; P/Y = C/Y = 1; n = 7(1) = 7; b =

= 0.0675; i =

= 0.0725;

PMT = 100 000(0.0675) = $6750.00. Since b < i, the bond sells at a discount. (i) Discount = [100000(0.0675) -110000(0.0725)] = -1225(5.342633) = -$6544.73 PP = 110000.00 - 6544.73 = $103 455.27 (ii)

Diff: 3 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts.

33) A $5000.00 6.5% bond redeemable at 109 with interest payable annually is purchased six years before maturity to yield 7% compounded annually. Construct the appropriate bond schedule. Answer: FV = $5 000(1.09) = $5450.00; P/Y = C/Y = 1; b =

= 0.065; i =

= 0.07;

PMT = $5000(0.065) = $325 Discount = [5000(0.065) - 5450(0.07)] = -56.50(4.7665397) = -$269.31 PP = 5450.00 - 269.31 = $5180.69

Diff: 3 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts.

34) A $5000, 6.25% bond with interest payable annually redeemable at par in eight years is purchased to yield 7.5% compounded annually. Find the premium or discount and the purchase price and construct the appropriate bond schedule. Answer: FV = $5000; P/Y = C/Y = 1; b = 0.0625; i = 0.075; n = 8(1) = 8 Discount = [5000.00 * .0625 - 5000.00 * .075] = -$62.50(5.8573036) = -$366.08 PP = 5000.00 - 366.08 = $4633.92

Diff: 3 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts. 35) A $1000, 6% bond redeemable at par with semi-annual coupons was purchased 10 years before maturity to yield 5% compounded semi-annually. The bond was sold 3 years later at 102. Calculate the gain or loss on the sale of the bond. Answer: FV = 1000; P/Y = 2; C/Y = 2; n = (10 - 3)(2) = 14; b = 3%; i = 2.5% Since b > i, the bond sells at a premium Premium = [0.03(1000) - 0.025(1000)] = 5(11.690912) = $58.45 Programmed solution: PMT = (0.03 × 1000 - 0.025 × 1000) = 5

The purchase price is 1000 + 58.45 = $1058.45 The proceeds from the bond are 1000(1.02) = $1020.00 The loss on the sale is 1058.45 - 1020 = $38.45 Diff: 2 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts.

36) Find the gain or loss on the sale without constructing a bond schedule for six $5000, 9.5% bonds with interest payable semi-annually redeemable at par bought twenty-one years before maturity to yield 10.5% compounded semi-annually. The bonds were sold three years later at 103.625. Answer: FV = 6(5000) = $30 000; P/Y = C/Y = 2; n = (21-3)2 = 36; b =

= 0.0475; i =

= 0.0525.

Since b < i, the bonds are sold at a discount. Discount = [30 000(0.0475) - 30 000(0.0525)] = -150(16.0287454) = -$2404.31 Book value 30000.00 - 2404.31= $27 595.69 Proceeds = 30000(1.03625 ) = $31 087.50 Gain $ 3491.81 Diff: 2 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts. 37) Find the gain or loss on the sale without constructing a bond schedule for a $100 000, 8% bond with semi-annual coupons redeemable at par purchased eleven-and-a-half years before maturity to yield 9% compounded semi-annually. The bond was sold five years later at 99.125. Answer: FV = $100 000; P/Y = C/Y = 2; b =

= 0.035; i =

= 0.04; n = (11.5 - 5)2 = 13

Since b < i, the bonds are sold at a discount. Discount = [100000(0.035) - 100 000(0.04)] = -500(9.985648) = -$4992.83 Book value = 100 000 - 4992.83 = $95 007.17 Proceeds = 100000(0.99125 = $99 125.00 Gain $ 4117.83 Diff: 2 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts.

38) Find the gain or loss on the sale without constructing a bond schedule for a $50 000, 6% bond with semi-annual coupons redeemable at par purchased twelve years before maturity to yield 10% compounded semi-annually. The bond was sold five years later at 99.375. Answer: FV = $50 000; P/Y = C/Y = 2; b =

= 0.03; i =

= 0.05; n = (12 - 5)2 = 14;

Since b < i, the bonds are sold at a discount. Discount = [50000(0.03) - 50 000(0.05)] = -1000(9.898641) = -$9898.64 Book value = 50 000 - 9898.64 = $40 101.36 Proceeds = 50000(0.99375) = $49 687.50 Gain = 49 687.50 - 40 101.36 = $9586.14 Diff: 2 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts. 39) Three $25 000, 6% bonds with semi-annual coupons redeemable at par were bought eight years before maturity to yield 7% compounded semi-annually. Determine the gain or loss if the bonds are sold at 89.625 four years later. Answer: FV = 3($25 000) = $75 000; P/Y = C/Y = 2; b Discount = [75000(.03 - .035)]

= 0.03; i =

= 0.035; n = 4(2) = 8.

= -375.00(6.873956) = -$2577.73

Book value = 75 000.00 - 2577.73 = $72 422.27 Proceeds = 75000.00 * .89625 = $67 218.75 Loss on sale = $5203.52 Diff: 3 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts. 40) A $40 000.00, 5% bond with semi-annual coupons redeemable at par in 14 years is purchased to yield 7% compounded semi-annually. What is the gain or loss if the bond is sold three years before maturity at 99.75? Answer: Discount = [40000.00 * .025 - 40000.00 * .035] = -400.00(5.328553) = -$2131.42 Book value = 40000.00 - 2131.42 = $37868.58 Proceeds = 40000.00 * .9975 = 39900.00 Gain = $2031.42 Diff: 3 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts.

41) A $50 000.00, 6% bond with semi-annual coupons redeemable at par in 14 years is purchased to yield 8% compounded semi-annually. What is the gain or loss if the bond is sold four years before maturity at 99.25? Answer: FV = $50 000; P/Y = C/Y = 2; n = 4(2) = 8; Discount = [50000.00 * .03 - 50000.00 * .04] = -500.00(6.732744875) = -$3366.37 Book value = 50000.00 - 3366.37 = $46633.63 Proceeds = 50000.00 * .9925 = 49625.00 Gain = $2991.37 Diff: 3 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts. 42) Find the gain or loss on the sale without constructing a bond schedule for three $100 000, 5.5% bonds with quarterly coupons redeemable at par on September 1, 2017, bought on May 1, 2011, to yield 6% compounded quarterly. The bonds were sold on January 21, 2014, at 93.5. Answer: FV = $3(100 000) = $300 000; P/Y = C/Y = 4; b =

= 0.01375; i =

= 0.015;

The interest date preceding the purchase date is December 1, 2013, 3.75 years before maturity n = (3.75)4 = 15 Since b < i, the bonds are sold at a discount. Discount = [300000(0.01375) - 300 000(0.015)] = -375(13.343233) = -$5003.71 Book value = 300 000 -5003.71 = $294 996.29 on December 1, 2013 The number of days from Dec.1, 2013 to Jan.21,2014 is 51 days. The number of days from Dec.1,2013 to March 1, 2014 is 90 Accumulated price on Jan. 21, 2014 294 996.29

= $297 503.76

Accrued interest to Jan. 21, 2004 Interest = 300000(0.01375)

= $2337.50

Proceeds = 300000(0.935)+ 2337.50 = $282 837.50 Loss on sale = 297 503.76 - 282 837.50 = $14 666.26 Diff: 2 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts.

43) A 4.5% annuity bond of $500 000 with interest payable quarterly is to be redeemable at par in seven years. a) What is the purchase price to yield 6% compounded quarterly? b) What is the book value after 6 years? c) What is the gain or loss if the bond is sold six years after the date of purchase at Answer: a) Discount = [500000.00 * .01125 - 500000.00 * .015] = -1875.00(22.7267167) = -$42612.59 PP = 500000.00 - 42612.59 = $457 387.41 b) Discount = 500000.00(-.00375) -1875.00(3.8543847) = -$7226.97 Book value = 500000.00 - 7226.91 = $492 773.09 c) Proceeds: 500000.00 * .99625 = $498 125.00 Gain $5351.91 Diff: 1 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts. 44) A $25 000, 8% bond with semi-annual coupons, redeemable at par in 12 years, is purchased to yield 6% compounded semi-annually. Determine the gain or loss if the bond is sold two years later at 107.25. Answer: FV = $25 000; P/Y = C/Y = 2; b =

= 0.04; i =

= 0.03; n = 10(2) = 20

Since b > i, the bond sells at a premium. Premium = [25 000(0.04 ) - 25 000(0.03)] = 250(14.877475) = $3719.37 Book value: 25 000 + 3719.37 = $28 719.37 Proceeds: 25 000(1.0725) = $26 812.50 Loss = $1906.87 Diff: 1 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts.

45) A $25 000, 7% bond with semi-annual coupons redeemable at par in twenty-two years is purchased to yield 6% compounded semi-annually. Determine the gain or loss if the bond is sold seven years after the date of purchase at 98.25. Answer: FV = $25 000; P/Y = C/Y = 2; b =

= 0.035; i =

= 0.03; n = (22 - 7)2 = 30;

Premium = [25000(0.035) - 25 000(0.03)] = 125(19.6004414) = $1274.03 Book value: 27000.00 + 1274.03 = $28 274.03 Proceeds: 25000(0.9825) = $24 562.50 Loss = $3711.53 Diff: 1 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts. 46) A $50 000, 6% bond with semi-annual coupons redeemable at par on April 25, 2018, was purchased on June 25, 2009, at 94.378. What was the approximate yield rate? Answer: Quoted price = 50000(0.94375) = $47 187.50 Redemption price = $50 000.00 Average book value = .5(50 000 + 47 187.50) = $48 593.75 the number of interest payments to maturity is 9(2) = 18 Total interest = 18(50000)(.03) = $27 000.00 Bond discount = 50 000.00 - 47 187.50 = $2 812.50 Average income per payment interval = Approximate value of i =

(27 000 + 2 812.50) = $1656.25

= .0340836

Approximate yield = .0340836 * 2 = 6.81672% Diff: 2 Type: SA Page Ref: 627-629 Topic: 15.4 Finding the Yield Rate Objective: 15-4: Calculate the yield rate for bonds bought on the market by the method of averages. 47) A $100 000, 8% bond with semi-annual coupons redeemable at par on April 25, 2018, was purchased on June 25, 2009, at 94.125. What was the approximate yield rate? Answer: Quoted price = 100000(0.94125) = $94 125.00 Redemption price = $100000.00 Average book value = .5(100000.00 + 94125.00) = $97 062.5 Total interest = 18(100000.00)(.04) = $72 000.00 Bond discount = 100000.00 - 94125.00 = $5 875.00 Average income per payment interval = Approximate value of i =

(72000.00 + 5875.00 ) = $4326.39

= .044573239

Approximate yield = .044573239 * 2 = 8.1946% Diff: 2 Type: SA Page Ref: 627-629 Topic: 15.4 Finding the Yield Rate Objective: 15-4: Calculate the yield rate for bonds bought on the market by the method of averages.

48) A $40 000 bond with semi-annual coupon payments at 5.5% compounded semi-annually is redeemable at par in 12 years. Calculate the yield rate if the bond is purchased at 102.5. Answer: The quoted price (initial book value) is 40 000(1.025) = $41 000 Redemption Value is 40 000 Average book value =

(41000 + 40000) = $40 500

The semi-annual interest payment is 40 000

= $1100.00

The number of interest payments to maturity is 12(2) = 24 The total interest payments 24(1100) = $26 400 The premium is 41000 - 40 000 = $1000 Average income per interest payment interval is Approximate value of i =

= 1058.33

= 0.026131605 = 2.61%

The yield rate is 2(2.61) = 5.22% Diff: 2 Type: SA Page Ref: 627-629 Topic: 15.4 Finding the Yield Rate Objective: 15-4: Calculate the yield rate for bonds bought on the market by the method of averages. 49) A $100 000, 6.0% bond with semi-annual coupons redeemable at par is bought 17.5 years before maturity at 74.25. What was the approximate yield rate? Answer: Quoted price: 100000.00 ∗ .7425 = $74 250.00 Average book value: .5(100000.00 + 74250.00) = $87 125.00 Semi-annual interest: 100000.00 * .03 = $3000.00 Total interest: 35(3000.00) = $105000.00 Discount: 100000.00 - 74250.00 = $25750.00 Average income:

(3000.00 + 105000.00) = $3085.71

Approximate value of i is

= .0354170445

Annual rate: 2(.0354170445) = 7.0834% Diff: 2 Type: SA Page Ref: 627-629 Topic: 15.4 Finding the Yield Rate Objective: 15-4: Calculate the yield rate for bonds bought on the market by the method of averages.

50) A $10 000.00, 5% bond with semi-annual coupons redeemable at 105 in 23 years is purchased at 102.5. What is the approximate yield rate? Answer: Initial book value: 100000.00 * 1.025 = $102 500.00 Redemption price: 100000.00 * 1.05 = $105 000.00 Average book value: .5(102500.00 + 105000.00) = $103 750.00 Total interest: 46(100000(0.025) = $115 000.00 Discount = 102500.00 - 105000.00 = $2500.00 Average income per interest payment interval = Approximate value of i:

= $2554.35

= .0246202

Annual rate: 2(.0246202) = 4.92405% Diff: 2 Type: SA Page Ref: 627-629 Topic: 15.4 Finding the Yield Rate Objective: 15-4: Calculate the yield rate for bonds bought on the market by the method of averages. 51) A $50 000 bond that pays 5% semi-annually is redeemable at par on July 15, 2025.It is quoted at 97.5 on December 2, 2013. Determine the yield rate. Answer: The interest dates are January 15 and July 15. The nearest interest date is January 15, 2004 (11.5 years to maturity) The quoted price is 50 000(0.975) = $48 750 The redemption value is 50 000 The average book value is

(48750 + 50000) = $49375.00

The semi-annual interest is 50000

= $1250

The number of interest payments to maturity is 11.5(2) = 23 The total interest payments are 23(1250) = $28 750.00 Discount is 50 000 - 48 750 = $1250 Average income per interest interval is Approximate value of i is

= $1304.35

= 0.026417215 = 2.64%

The yield rate is 2(2.64) = 5.28% Diff: 2 Type: SA Page Ref: 627-629 Topic: 15.4 Finding the Yield Rate Objective: 15-4: Calculate the yield rate for bonds bought on the market by the method of averages.

52) A $250 000, 6.5% bond with semi-annual coupons redeemable at par is bought 17.5 years before maturity at 78.25. What was the approximate yield rate? Answer: Quoted price: 250000(0.7825) = $195 625.00 Average book value: .5(250 000 + 195 625) = $222 812.50 Semi-annual interest: 250 000(0.0325) = $8125.00 Total interest: 35(8125.00) = $284 375.00 Discount: 250000.00 - 195625.00 = $54 375.00 Average income:

(8125.00 + 284375.00) = $8357.14

Approximate value:

= .0375075

Annual rate: 2(.0375075) = 7.5015% Diff: 2 Type: SA Page Ref: 627-629 Topic: 15.4 Finding the Yield Rate Objective: 15-4: Calculate the yield rate for bonds bought on the market by the method of averages. 53) A $100 000, 9.0% bond with semi-annual coupons redeemable at par on March 1, 2015, was purchased on September 22, 2007, to yield 10.00% compounded semi-annually. What was the purchase price? Answer: 100000.00(1.05)-15 + 4500.00 100000.00(.4810170981) + 4500.00(10.37965804) 48101.71 + 46708.46 = $94 810.17 PP = 94810.17 = 94810.17(1.005769231) = $95 357.15 Diff: 3 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 54) A $10 000, 8.5% bond with semi-annual coupons redeemable at par on March 1, 2005, was purchased on September 22, 1997, to yield 10.06% compounded semi-annually. What was the purchase price? Answer: 10000.00(1.0503)-15 + 425.00 10000.00(.4789603) + 425.00(10.3586421) 4789.60 + 4402.42 = $9192.02 PP = 9192.02 = 9192.02(1.005803846) = $9245.37 Diff: 3 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

55) A $10 000 bond bearing interest at 6% payable semi-annually redeemable at par on March 1, 2002, was purchased on September 30, 1996, to yield 5% compounded semi-annually. Determine the purchase price. Answer: [10000.00 * .03 - 10000.00 * .025] 50.00(9.5142087) = $475.71 Premium PP1 = 10000.00 + 475.71 = $10 475.71 PP2 = 10475.71 = 10475.71(1.0040055) = $10 517.67 Diff: 3 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 56) A $200 000, 6% bond with semi-annual coupons redeemable at par March 1, 2009, was purchased on November 15, 1999, to yield 5% compounded semi-annually. What was the purchase price of the bond? Answer: PP1 = 200000.00(1.025)-19 + 6000.00 = 200000.00(.6255277) + 6000.00(14.9788913) = 125105.54 + 89873.35 = $214 978.89 PP2 = 214978.89 = 214978.89(1.0103591) = $217 205.88 Diff: 1 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 57) A $100 000.00, 7% bond with semi-annual coupons redeemable at 105 on November 1, 2001, is purchased on July 23, 1999, to yield 8% compounded semi-annually. Determine the quoted price. Answer: 105000.00(1.04)-5 + 3500.00 105000.00(.8219271) + 3500.00(4.4518223) 86302.35 + 15581.38 = $101883.73 PP = 101883.73 = 101883.73(1.0180435) = $103 722.07 Accrued interest = 100000.00 * .035 *

= $1578.80

Quoted price = 103722.07 - 1578.80 = $102 143.27 Diff: 3 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

58) What is the purchase price of a bond that has 4 years and 6 months until it matures? The face value of the bond is $3000 and the coupon rate is 5.2% compounded semi-annually. The yield rate is 7.5% compounded semi-annually. A) $2740.53 B) $2440.53 C) $2770.53 D) $2447.53 E) $2547.53 Answer: A Diff: 2 Type: MC Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 59) What is the purchase price of a bond that is issued on May 1, 2002 for 22 years at a coupon rate of 9.4% compounded semi-annually? The purchase date of the bond is May 11, 2007 and the yield rate is 8.5% compounded semi-annually. The bond has a face value of $5000 and is redeemable at par. A) $4513.30 B) $5413.30 C) $5431.30 D) $4531.30 E) $4533.30 Answer: B Diff: 1 Type: MC Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 60) What is the amount of premium/discount for a bond that has 4 years till it matures? The face value of the bond is $100 000 and it has a bond rate of 6.4% compounded semi-annually. The bond is sold to yield 7.4% compounded semi-annually. A) $3084.47 B) $3840.47 C) $3048.47 D) $3408.47 E) $3184.47 Answer: D Diff: 2 Type: MC Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

61) What is the amount of premium/discount amortized or accumulated in the first payment interval for a bond that has a face value of $6000 and it sold for $5700? The coupon rate is 9% compounded semiannually and the market rate is 10.1% compounded semi-annually. A) $18.75 B) $17.85 C) $11.85 D) $15.75 E) $14.75 Answer: B Diff: 2 Type: MC Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts. 62) What is the amount of premium/discount amortized or accumulated for the second payment interval of a $1500 bond that sold for $1627.32? The bond rate is 6.95% compounded semi-annually and the yield rate is 6% compounded semi-annually. A) $4.30 B) $5.30 C) $1.40 D) $3.40 E) $2.30 Answer: D Diff: 2 Type: MC Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts. 63) What is the quoted price of a bond that has 4.5 years until maturity? The face value of the bond is $5000 and it has a coupon rate of 8.1% compounded semi-annually and a yield rate of 8.6% compounded semi-annually. A) $4809.32 B) $4998.32 C) $4908.32 D) $4888.32 E) $4898.32 Answer: C Diff: 2 Type: MC Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

64) A $1000 bond, with interest at 8% payable semi-annually on January 1 and July 1, was purchased on October 8 at 104 plus accrued interest. What is the purchase payment for the bond? Answer: Days of accrued interest = 99 Market Value = $1000 × 104% = $1040 Accrued interest = $1000 × 0.08 ×

= $22

Purchase payment for bond = $1040 + $22 = $1062 Diff: 2 Type: SA Page Ref: 627-629 Topic: 15.4 Finding the Yield Rate Objective: 15-4: Calculate the yield rate for bonds bought on the market by the method of averages. 65) Clarington issued 15 year bonds in the amount of $500 000. Interest on the bonds is 2.2% payable annually. What is the promised payment at the end of each year? A) $11 100 B) $693 000 C) $360 750 D) $374 675 E) $13 925 Answer: A Diff: 1 Type: MC Page Ref: 603-604 Topic: 15.1 Purchase Price of Bonds Objective: 15-1: Determine the purchase price of a bond when the market rate equals the bond rate, bought on or between interest dates. 66) Clarington issued 15 year bonds in the amount of $500 000. Interest on the bonds is 2.2% payable annually. What is the issue price of the bonds, if the bonds are sold to yield 2.8% compounded quarterly? A) $329 004 B) $132 950 C) $461 954 D) $330 425 E) $343 749 Answer: C Diff: 2 Type: MC Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

67) A $1000 bond, with interest at 8% payable semi-annually on January 1 and July 1, was purchased on February 8, 2012. The bond matures on July 1, 2015 and yields 2.2% compounded semi-annually. What is the purchase payment for the bond? Answer: FV = $1000; n = 3.5 × 2 = 7;

P/Y = 2;

PMT = $1000 × C/Y = 2;

= $40 I/Y = 2.2%;

Purchase price of the bond on January 1, 2012 = $1,000 × 1.011-7 + $40

= 1.1% = $1194.35

Number of days from January 1, 2012 and February 8, 2012 = 39 Number of days from January 1, 2012 and July 1, 2012 = 183 Interest = $1,000 × 0.04 ×

= $8.53

Purchase price of the bond = $1202.87 Diff: 2 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 68) Nick buys a $25 000, 5.4% bond with quarterly interest coupons, 3 years before maturity, to yield 3.5% compounded quarterly. Determine the premium or discount. A) $2481 discount B) $2481 premium C) $1347 premium D) $1347 discount E) $3829 premium Answer: C Diff: 2 Type: MC Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 69) Nick buys a $25 000, 5.4% bond with quarterly interest coupons, 3 years before maturity, to yield 7% compounded quarterly. Determine the premium or discount. A) $4698.55 premium B) $4698.55 discount C) $3624.60 discount D) $1073.95 premium E) $1073.95 discount Answer: E Diff: 2 Type: MC Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

70) Nick buys a $25 000, 5.4% bond with quarterly interest coupons, 3 years before maturity, to yield 7% compounded quarterly. Calculate the shortage of the actual interest received compared to the required interest based on the yield rate. A) $337.50 B) $437.50 C) $100 D) $1350 E) $1750 Answer: C Diff: 1 Type: MC Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 71) Nick buys a $25 000, 5.4% bond with quarterly interest coupons, 3 years before maturity, to yield 3.5% compounded quarterly. Construct a schedule of amortization of premium. Answer: FV = $25 000;

n = 3 × 4 = 12;

Premium = $25000 × (0.0135 - 0.00875)

= 1.35%;

= 0.875%

= $1347.16

Purchase price = $25000 + $1347.16 = $26 347.16 End of Interest Payment Interval 0 1 2 3 4 5 6 7 8 9 10 11 12

Bond Interest Interest on Received Book value at Amount of (Coupon) yield rate Premium b = 1.35% i = 0.875% Amortized $337.50 $337.50 $337.50 $337.50 $337.50 $337.50 $337.50 $337.50 $337.50 $337.50 $337.50 $337.50

$230.54 $229.60 $228.66 $227.71 $226.74 $225.78 $224.80 $223.81 $222.82 $221.81 $220.80 $219.78

$106.96 $107.90 $108.84 $109.79 $110.76 $111.72 $112.70 $113.69 $114.68 $115.69 $116.70 $117.72

Book Value of Remaining Bond Premium $26,347.16 $1,347.16 $26,240.20 $1,240.20 $26,132.30 $1,132.30 $26,023.46 $1,023.46 $25,913.66 $913.66 $25,802.91 $802.91 $25,691.18 $691.18 $25,578.48 $578.48 $25,464.79 $464.79 $25,350.11 $350.11 $25,234.42 $234.42 $25,117.72 $117.72 $25,000.00 $0.00

Diff: 2 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts.

72) Nick buys a $25 000, 5.4% bond with quarterly interest coupons, 3 years before maturity, to yield 7% compounded quarterly. Construct a schedule of accumulation of discount. Answer: FV = $25000;

n = 3 × 4 = 12;

Discount = $25000 × (0.0175 - 0.0135)

= 1.35%;

= 1.75%

= $1073.95

Purchase price = $25000 - $1073.95 = $23 926.05 End of Interest Payment Interval 0 1 2 3 4 5 6 7 8 9 10 11 12

Bond Interest Interest on Received Book value at Amount of (Coupon) yield rate i = Discount Book Value of Remaining b = 1.35% 1.75% Accumulated Bond Discount $23,926.05 $1,073.95 $337.50 $418.71 $81.21 $24,007.26 $992.74 $337.50 $420.13 $82.63 $24,089.88 $910.12 $337.50 $421.57 $84.07 $24,173.96 $826.04 $337.50 $423.04 $85.54 $24,259.50 $740.50 $337.50 $424.54 $87.04 $24,346.54 $653.46 $337.50 $426.06 $88.56 $24,435.11 $564.89 $337.50 $427.61 $90.11 $24,525.22 $474.78 $337.50 $429.19 $91.69 $24,616.91 $383.09 $337.50 $430.80 $93.30 $24,710.21 $289.79 $337.50 $432.43 $94.93 $24,805.14 $194.86 $337.50 $434.09 $96.59 $24,901.73 $98.27 $337.50 $435.77 $98.27 $25,000.00 $0.00

Diff: 2 Type: SA Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts. 73) Nick buys a $25 000, 5.4% bond with quarterly interest coupons, 3 years before maturity, to yield 7% compounded quarterly. The bond was sold a year later at 102. What is the gain or loss on the sale of the bond? A) $740.51 gain B) $740.51 loss C) $1240.50 gain D) $1240.50 loss E) $500 gain Answer: C Diff: 2 Type: MC Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts.

74) From his retirement fund, Jack buys a $50 000, 5.4% bond with quarterly interest coupons at 104.1, redeemable at par in 15 years. What is the approximate yield rate? A) 1.256% B) 5.024% C) 1.23% D) 4.92% E) 5.19% Answer: B Diff: 2 Type: MC Page Ref: 627-629 Topic: 15.4 Finding the Yield Rate Objective: 15-4: Calculate the yield rate for bonds bought on the market by the method of averages. 75) A $30 000, 3% bond redeemable at par is purchased 7 years before maturity to yield 5.5% compounded quarterly. If the bond interest is payable quarterly, what is the purchase price of the bond? A) $20 467.13 B) $225.00 C) $5199.75 D) $25 666.88 E) $34 719.51 Answer: D Diff: 2 Type: MC Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 76) A $63 000 bond bearing interest at 9.1% bond payable semi-annually matures in 20 years. If it is bought to yield 11.1% compounded semi-annually, what is the purchase price of the bond? A) $7261.24 B) $2866.50 C) $45 695.74 D) $52 956.98 E) $74 510.72 Answer: D Diff: 2 Type: MC Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

77) A $20 000, 3.6% bond with semi-annual coupons is purchased 5 years before maturity. Calculate the discount or premium if the bond is sold to yield 3% compounded semi-annually. A) $553.33 premium B) $553.33 discount C) $360 premium D) $544.64 premium E) $554.64 discount Answer: A Diff: 2 Type: MC Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 78) Nine $2000, 9% bonds with interest payable semi-annually and redeemable at par are purchased 7 years before maturity. Find the premium or discount and the purchase price if the bonds are bought to yield 4%. A) $5447.81 premium B) $5447.81 discount C) $4600.27 premium D) $11.14 discount E) $4600.27 discount Answer: A Diff: 2 Type: MC Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 79) Six $20 000, 2% bonds with interest payable semi-annually and redeemable at par are purchased 18.5 years before maturity. Find the premium or discount and the purchase price if the bonds are bought to yield 5.9%. A) $56 268.82 premium B) $56 268.82 discount C) $12 011.81 premium D) $8711.47 discount E) $12 011.81 discount Answer: B Diff: 2 Type: MC Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

80) Twenty $1 000 bonds redeemable at par bearing 6% coupons payable quarterly are sold eight years before maturity to yield 5.5% compounded annually. What is the purchase price of the bonds? A) $17 969.05 B) $20 229.82 C) $20 633.46 D) $13 031.98 E) $19 861.49 Answer: B Diff: 3 Type: MC Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 81) A $1000, 6% bond redeemable at par with semi-annual coupons was purchased 10 years before maturity to yield 5% compounded semi-annually. The bond was sold 3 years later at 102. Calculate the gain or loss on the sale of the bond. A) $38.45 Gain B) $38.45 Loss C) $76.48 Gain D) $76.48 Loss E) $58.45 Loss Answer: B Diff: 2 Type: MC Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts. 82) Find the gain or loss on the sale without constructing a bond schedule for six $3500, 3.5% bonds with interest payable semi-annually redeemable at par bought twenty-one years before maturity to yield 5.5% compounded semi-annually. The bonds were sold three years later at 102. A) $5180.68 Gain B) $5180.68 Loss C) $5612.65 Gain D) $5612.65 Loss E) $5153.98 Loss Answer: A Diff: 2 Type: MC Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts.

83) A 4.5% annuity bond of $50 000 with interest payable quarterly is to be redeemable at par in seven years with a yield 6% compounded quarterly. What is the gain or loss if the bond is sold six years after the date of purchase at 99.625? A) $535.20 Gain B) $535.20 Loss C) $4073.76 Gain D) $4073.76 Loss E) $916.87 Loss Answer: A Diff: 1 Type: MC Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts. 84) A 6.3% annuity bond of $12 000 with interest payable monthly is to be redeemable at par in nine years with a yield 7.9% compounded monthly. What is the gain or loss if the bond is sold eight years after the date of purchase at 101.325? A) $343.03 Gain B) $343.03 Loss C) $184.03 Gain D) $184.03 Loss E) $336.94 Gain Answer: A Diff: 1 Type: MC Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts. 85) Three 9.1% annuity bond of $15 000 with interest payable quarterly are to be redeemable at par in seventeen years with a yield 7.1% compounded quarterly. What is the gain or loss if the bond is sold six years after the date of purchase at 102? A) $5931.18 Gain B) $5931.18 Loss C) $6831.18 Gain D) $6831.18 Loss E) $5815.32 Loss Answer: B Diff: 1 Type: MC Page Ref: 621-626 Topic: 15.3 Bond Schedules Objective: 15-3: Construct bond schedules showing the amortization of premiums or accumulation of discounts.

86) A $15 000, 4.4% bond with semi-annual coupons redeemable at par on April 25, 2025, was purchased on June 25, 2016, at 93.725. What was the approximate yield rate? A) 2.63% B) 5.26% C) 2.55% D) 5.10% E) 4.54% Answer: B Diff: 2 Type: MC Page Ref: 627-629 Topic: 15.4 Finding the Yield Rate Objective: 15-4: Calculate the yield rate for bonds bought on the market by the method of averages. 87) A bond for which no security is offered by the issuer is called a: A) Premium B) Unsecure bond C) Debenture D) Mutual Fund E) Index Fund Answer: C Diff: 1 Type: MC Page Ref: 603-604 Topic: 15.1 Purchase Price of Bonds Objective: 15-1: Determine the purchase price of a bond when the market rate equals the bond rate, bought on or between interest dates. 88) When the market rate is greater than the bond rate, the bond is described as: A) Premium B) Discount C) Par D) Debenture E) Expensive Answer: B Diff: 1 Type: MC Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

89) Jack purchases a $4300 bond with an interest rate of 2.8% p.a. payable semi-annually and a maturity date of 9 years from now. If the yield rate is 4.2% p.a. compounded semi-annually, what is the purchase price of Jack's bond? Answer: i = 0.042/2 = 0.021 = 2.1%; n = 9 * 2 = 18 Periodic interest payment = 4300 * 0.028/2 = $60.20 Present Value of Principal PV = 4300(1.021)-18 = $2958.05 Present Value of Interest Payments PV = 60.20

= 60.20(14.861050) = $894.64

Purchase Price = 894.64 + 2958.05 = $3852.69 Diff: 1 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 90) The Government of Canada begins to sell $10 000 bonds that have an interest rate of 3.8% p.a. payable semi-annually. What is the price of the bonds if they mature in 5 years and are sold to yield 3.6% p.a. monthly? Answer: i = 0.036/12 = 0.003 = 0.3%; c = 12/2 = 6; p = 1.003^6 - 1 = 0.018136; n = 5 * 2 = 10 Periodic interest payment = 10 000(0.038/2) = $190 Present Value of Principal PV = 10000(1.018136)-10 = $8534.95 Present Value of Interest Payments PV = 190

= 190(9.070845) = $1723.46

Purchase Price = 8354.95 + 1723.46 = $10 078.41 Diff: 1 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate.

91) Wendy decided to redeem her $9100 bond 6 years and 3 months early. The bond has an interest rate of 4.4% payable semi-annually and the yield rate is 5.0% compounded semi-annually. What is the cash price? Answer: i = 0.05/2 = 0.025 The last interest date prior to her redemption is 6.5 years before maturity: n = 6.5 * 2 = 13 Periodic interest payment = 9100(0.044/2) = 200.20 Present Value of Principal PV = 9100(1.025^-13) = $6601.33 Present Value of Interest Payments PV = 200.20

= 200.20(10.983185) = $2198.83

Purchase Price on interest date preceding purchase date Purchase Price = 6601.33 + 2198.83 = $8800.16 Accrued interest and Accumulated Price n = 3/6 = 0.5 Cash Price = 8800.16(1.025)0.5 = 8909.48

Diff: 3 Type: SA Page Ref: 604-618 Topic: 15.2 Purchase Price of a Bond when Market Rate Does Not Equal Bond Rate Objective: 15-2: Determine the purchase price of a bond when the market rate does not equal the bond rate. 92) A $100 provincial government bond has a bond rate of 9.4% p.a. payable semi-annually and a yield rate of 6.8% p.a. compounded semi-annually. The bond is redeemable in 2 years. Determine the premium or discount. Answer: Since the bond rate b is greater than the yield rate i, the bond will sell at a premium b = 0.094/2 = 0.047; i = 0.068/2 = 0.034; n = 2 * 2 = 4 Periodic Interest Payment: 0.047(100) = $4.70 Required Interest: 0.034(100) = $3.40 Excess actual interest: 4.70-3.40 = $1.30 Premium = 1.30

= 1.30(3.681816) = $4.79

Contemporary Business Mathematics, 12e (Hummelbrunner/Coombs) Chapter 16 Investment Decision Applications 1) An obligation can be settled by making a payment of $7500.00 now and a final payment of $10 000.00 in five years. Alternatively the obligation can be settled by payments of $750.00 at the end of every three months for five years. If interest rate is 10% compounded quarterly, determine the preferred alternative. Answer: PV of 10 000: n = 5(4) = 20; i =

= 0.025; FV = 10 000; I/Y = 10;P/Y = C/Y = 4

PV = 10000(1.025)-20 = 6102.71 ALT. 1 = 7500 + 6102.71 = 13 602.71 = $13 603 Programmed solution:

For ALT. 2: PMT = 750; n = 5(4) = 20; i = 0.025; I/Y = 10; P/Y = C/Y = 4 PV = 750

= 11 691.87 = $11 692

Programmed solution:

ALT. 2 is better (less to pay) Diff: 2 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion. 2) You win a lottery and have a choice of taking $200 000.00 immediately or taking payments of $7311.15 at the end of every three months for ten years. Which offer is preferable if interest is 8% compounded quarterly? Answer: PMT = 7311.15; n = 10(4) = 40; i = 0.02; I/Y = 8; P/Y = C/Y = 4 PV = 7311.15

= $200 000.01 = $200 000

Either option is preferable. Diff: 1 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion.

3) An obligation can be settled by making a payment of $12 000 now and a final payment of $32 000 in 4 years. Alternatively, the obligation can be settled by payments of $2700 at the end of every three months for five years. Interest is 10% compounded quarterly. Compute the present value of each alternative and determine the preferred alternative according to the discounted cash flow criterion. Answer: ALT. 1 = 12000.00 + 32000.00(1.025)-16 = 12000.00 + 32000.00(.6736249) = 12000.00 + 21556.00 = $33 556

ALT. 2 = 2700.00

= 2700.00(15.5891623) = $42090.74 = $42 091

Since the present value of Alternative 1 is less than the present value of Alternative 2, Alternative 1 is preferable. Diff: 2 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion. 4) An obligation can be settled by making a payment of $16 000 now and a final payment of $30 000 in 3 years. Alternatively, the obligation can be settled by payments of $2500 at the end of every three months for four years. Interest is 12% compounded quarterly. Compute the present value of each alternative and determine the preferred alternative according to the discounted cash flow criterion. Answer: ALT. 1 = 16000.00 + 30000.00(1.03)-12 = 16000.00 + 30000.00(.701378802) = 16000.00 + 21041.40 = $37041.40 = $37 041.

ALT. 2 = 2500.00

= 2500.00(12.56110203) = $31402.76 = $31 403

At 12% compounded quarterly Alternative 2 is preferable. Diff: 2 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion.

5) An expenditure may be met by outlays of $1700 now and $2210 at the end of every six months for 6 years or by making monthly payments of $500 in advance for seven years. Interest is 11% compounded annually. Compute the present value of each alternative and determine the preferred alternative according to the discounted cash flow criterion. Answer: ALT. 1 p =

- 1 = .05356538

= 1700.00 + 2210.00 = 1700.00 + 2210.00(8.6876858) = 1700.00 + 19199.79 = $20899.79 = $20 900. BEST ALT. 2 p =

- 1 = .0087346

= 500.00

(1.0087346)

= 500.00(59.3435117)(1.0087346) = $29930.93 = $29 931 Diff: 2 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion. 6) An expenditure may be met by outlays of $3000 now and $1000 at the end of every six months for 5 years or by making monthly payments of $250 in advance for three years. Interest is 12% compounded annually. Compute the present value of each alternative and determine the preferred alternative according to the discounted cash flow criterion. Answer: ALT. 1 p

- 1 = .05833005

= 3000.00 + 1000.00 = 3000.00 + 1000.00(7.419712748) = 3000.00 + 7419.71 = $10419.71 = $10 420 ALT. 2 p = = 250.00

- 1 = .00948879 (1.00948879)

= 250.00(30.3747539)(1.009488793) = $7665.74 = $7666 BEST Diff: 2 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion.

7) A selection has to be made between two investment alternatives. The first alternative offers a net return of $42 430.00 after three years, $30 000.00 after five years and $26 000.00 after seven years. The second alternative provides a net return of $13 000.00 per year for seven years. Determine the preferred alternative according to the discounted cash flow criterion if money is worth 11%. Answer: ALT. 1 = 42430.00(1.11)-3 + 30000.00(1.11)-5 + 26000.00(1.11)-7 = 42430.00(.7311914) + 30000.00(.5934513) + 26000.00(.4816584) = 31024.45 + 17803.54 + 12523.12 = $64692.66 = $61 351.11.

ALT. 2 = 13000.00

= 13000.00(4.7121963) = $61258.55 = $61 259

Since at 11% the present value of Alternative 1 is greater than the present value of Alternative 2, Alternative 1 is preferable. Diff: 2 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion. 8) Donna and Keith want to sell their business. They have received two offers. If they accept Offer A they will receive $61 000 immediately and $20 000 in three years. If the accept Offer B they will receive $37000 now and $3000 at the end of every six months for 5 years. If interest is 6.67%, which offer is preferable? Answer: p =

-1 = .0328117

ALT. A = 61000.00 + 20000.00(1.0667)-3 = 61000.00 + 20000.00(.8238974) = 61000.00 + 16477.95 = $77477.95 = $77 478 BEST ALT. B = 37000.00 + 3000.00 = 37000.00 + 3000.00(8.4090961) = 37000.00 + 25227.29 = $62227.29 = $62 227 Diff: 2 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion. 9) Sean and Jessica want to sell their interest in a small business. They have received two offers. If they accept Offer A they will receive $40 000 immediately and $30 000 in two years. If the accept Offer B they will receive $50 000 now and $2000 at the end of every six months for 5 years. If interest is 8%, which offer is preferable? Answer: p =

-1 = .0392305

ALT. A = 40000.00 + 30000.00(1.08)-2 = 40000.00 + 30000.00(.857339) = 40000.00 + 25720.16 = $65720.16 = $65 720 ALT. B = 50000.00 + 2000.00 = 50000.00 + 2000.00(8.142056025) = 50000.00 + 16284.11 = $66284.11 = $66 284 BEST Diff: 2 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion.

10) A piece of property may be acquired by making an immediate payment of $125 000 and payments of $37 500 and $50 000 three and five years from now respectively. Alternatively, the property may be purchased by making quarterly payments of $11 150 in advance for five years. Which alternative is preferable if money is worth 12.2% compounded semi-annually? Answer: ALT. 1 = 125000.00 + 37500.00(1.061)-6 + 50000.00(1.061)-10 = 125000.00 + 37500.00(.7009833) + 50000.00(.5531541) = 125000.00 + 26286.87 + 27657.71 = $178944.58 = $178 945.

ALT. 2: I/Y = 12.2%; P/Y = 4; C/Y = 2; c =

= 0.5; n = 4(5) = 20; p =

PV of ALT. 2 = 11 150

(1.03004854)

-1 = .03004854

= 11150.00(14.87080004)(1.03004854) = $170791.75 = $170 792 Since the present value of the Alternative 2 is smaller than the present value of the Alternative 1, Alternative 2 is preferable. Diff: 2 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion. 11) A car costs $29 700. Alternatively, the car can be leased for three years by making payments of $540 at the beginning of each month and can be bought at the end of the lease for $14 750. If interest is 9.2% compounded bi-weekly, which alternative is preferable? Answer: FV = $14 750; PMT = $540; c = 26/12 = 13/6 ; n = 12(3) = 36; i = 0.092/26 = 0.00353846 p=

- 1 = 0.007682495

ALT. CASH = $29 700.00 ALT. LEASE = 540

(1.007682495) + 14750(1.007682495)-36

= 540(31.34623955)(1.007682495) + 14750(.7591827) = $28254.96 Since the present value of the lease is smaller than the present value of the cash payment, the lease is preferable Diff: 2 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion.

12) A new car costs $21 000. Alternatively, the car can be leased for three years by making payments of $360 at the beginning of each month and can be bought at the end of the lease for $10 000. If interest is 8% compounded semi-annually, which alternative is preferable? Answer: FV = $10 000; PMT = $360; I/Y = 8%; P/Y = 12; C/Y = 2; c = p=

; n = 3(12) = 36;

- 1 = .0065582.

ALT. CASH = $21 000.00 ALT. LEASE = 360.00

(1.0065581969) + 10000.00(1.0065581969)-36

= 360.00(31.97303733)(1.0065581969) + 10000.00(.7903145268) = 11585.78 + 7903.15 = $19488.93 = $19 489 BEST Diff: 2 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion. 13) A wireless telephone system with a disposable value of $5 000 after five years can be purchased for $15 000. Alternatively, a leasing agreement is available that requires an immediate payment of $2000 plus payments of $100.00 at the beginning of each month for five years. If money is worth 6% compounded monthly, should the telephone system be leased or purchased? Answer: P/Y = C/Y = 12; I/Y = 6%; i = 0.005; n = 5(12) = 60 PV of Purchase: = 15000.00 - 5000.00(1.005)-60 = 15000.00 - 5000.00(.7413721962) = 15000.00 - 3706.86 = $11 293.14 = $11 293

PV of Lease: = 2000 + 100

(1.005)

= 2000.00 + 100(51.72556075)(1.005) = 2000 + 5198.42 = $7198.42 At 6% monthly the present value of the decision to lease is smaller than the present value of the decision to buy. The wireless telephone system should be leased. Diff: 1 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion.

14) A telephone system with a disposable value of $4200 after five years can be purchased for $16 600. Alternatively, a leasing agreement is available that requires an immediate payment of $1900 plus payments of $150.00 at the beginning of each month for five years. If money is worth 9.6% compounded monthly, should the telephone system be leased or purchased? Answer: PMT = $150; P/Y = C/Y = 12; n = 5(12) = 60; I/Y = 9.6%; i = 0.008 PV of Purchase: = 16 600 - 4200(1.008)-60 = 16 600 - 4200(.6199663) = 16600.00 -2603.86 = $13 996

PV of Lease: = 1900 + 150

(1.008)

= 1900 + 150(47.5042142)(1.008) = 1900 + 7182.64 = $9083 At 9.6% monthly the present value of the decision to lease is smaller than the present value of the decision to buy. The telephone system should be leased. Diff: 2 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion. 15) A contract is estimated to yield net returns of $7000.00 quarterly for seven years. To secure the contract, an immediate outlay of $80 000.00 and a further outlay of $60 000.00 three years from now are required. If interest is 6% compounded quarterly, determine if the investment should be accepted or rejected. Answer: PV of 60 000: n = 3(4) = 12; i =

= 0.015; FV = 60 000; I/Y = 6; P/Y = C/Y = 4

PV = 60000(1.015)-12 = 50183.25 PVout = 80 000 + 50 183.25 = 130 183.25 = $130 183 Programmed solution:

For PVIN: PMT = 7000; n = 7(4) = 28; i = 0.015; I/Y = 6; P/Y = C/Y = 4 PVIN = 7000

= 159087.02 = $159 087

Programmed solution:

NPV = 159 087 - 130 183 = $28 904, accept the project. Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

16) An investment project requires an initial expenditure of $160 000.00 with a salvage value of $30 000.00 after ten years. It is estimated that it will have annual returns of $23 149.00 for ten years. Should the company undertake this project if it wants to achieve a 9% rate of return? Answer: FV = $30 000; n = 10(1) = 10; i =

= 0.09; PMT = 21 000;

PVout = 160 000 - 30000(1.09)-10 = $147 328 PVin = 23149

= 148562.35 = $148 562

NPV = 148 562 - 147 328 = $1234 > 0, accept the project. Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 17) Tamia Industries plans to replace the outdated equipment that will cost the company $100 000.00 now and $60 000.00 six years from now. This replacement will result in revenues of $6000.00 at the end of each quarter for twelve years. At an interest rate of 9% compounded annually and using the Net Present Value criterion, should the company replace this equipment or not? Answer: PV of $60 000 Outlay is: FV = 60 000; n = 6(4) = 24; i = 0.09; I/Y = 9; P/Y = 4; C/Y = 1; c =

;

PV = 60000(1.09)-6 = 35 776.04 PVOUT = 100 000 + 35 776.04 = $135 776.04 = $135 776 Inflow is: PMT = 6000; n = 12(4) = 48 p = (1.09)1/4 - 1 = 0.0217782 PV = 6000

= 177 553.47

PVIn = $177 553 Programmed solution:

NPV = $177 553 - 135 776 = $41 777 > 0; Replace the equipment. Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

18) Replacing old equipment at an immediate cost of $75 000 and an additional outlay of $10 000 six years from now will result in savings of $3120 per quarter for 11 years. The required rate of return is 11.4% compounded annually. Use the net present value method to determine whether the company should replace old equipment or not. Answer: Outlays: FV = $10 000; I/Y = 11.4%; P/Y = C/Y = 1; n = 6(1) = 6; PV = 10 000(1.114)-6 = $5232.25 PVOUT = 75 000 + 5232.25 = $80 232

Inflows: PMT = $3 120; P/Y = 4; C/Y = 1; c = p=

= 0.25; n = 4(11) = 44;

- 1 = .0273568

PV = 3120

= $79 266.63 = $79 267

PVIn = $79 267 NPV = 79 267 - 80 232 = -$965 < 0, The equipment shouldn't be replaced at the current rate of return. Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 19) Replacing old equipment at an immediate cost of $75 000 and an additional outlay of $10 000 six years from now will result in savings of $3120 per quarter for 11 years. The required rate of return is 8% compounded annually. Use the net present value method to determine whether the company should replace old equipment or not. Answer: Outlays: FV = 10 000; I/Y = 8%; P/Y = C/Y = 1; n = 6(1) = 6 PV = 10 000 (1.08)-6 = $6301.70 PVOUT = 75 000 + 6301.70 = $81 302

Inflows: PMT = $3120; P/Y = 4; C/Y = 1; c =

= 0.25; n = 11(4) = 44;

p = 1.081/4 - 1 = .0194265; PVIN = 3120

= 91 724.25 = 91 724

NPV = 91 724 - 81 302 = $10 422 Since NPV > 0, the old equipment should be replaced. Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

20) A company has two investment choices. Alternative A requires an immediate outlay of $4000.00 and offers a return of $14 000.00 after seven years. Alternative B requires an immediate outlay of $3600.00 in return for which $500.00 will be received at the end of every six months for the next seven years. If the rate of return is 6% compounded semi-annually, determine which alternative is preferable. Answer: Alterative A: PV of 14 000: n = 7(2) = 14; i =

= 0.03; FV = 14 000; I/Y = 6; P/Y = C/Y = 2

PV = 14000(1.03)-14 = $9255.65 Programmed solution:

NPVA = 9255.65 - 4000.00 = $5255.65 = $5256 Alternative B: PMT = 500; n = 7(2) = 14; i = 0.03; I/Y = 6; P/Y = C/Y = 2 PV = 500

= $5648.04

Programmed solution:

NPVb = 5648.04 - 3600.00 = $2048.04 = $2048 Since NPVA > NPVB, alternative A is preferred. Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

21) Project A requires an immediate investment of $18 000 and another $16 000 in three years. Net returns are $4500 after two years, $13 000 after four years, and $8900 after six years. Project B requires an immediate investment of $4000, another $6000 after two years, and $4000 after four years. Net returns are $3375 per year for 8 years. Determine the net present value at 11%. Which project is preferable according to the net present value criterion? Answer: ALT. A: PVIN = 4500 + 13 000 + 8900 = 3652.30 + 8563.50 + 4758.30 = $16 974. PVOUT = 18 000 + 16 000 = 18 000 + 11 699.06 = $29 699 NPVA = 16 974 - 29699 = -$12 725 ALT. B: PVIN = 3375

= $17 368

PVOUT = 4000 + 6000 + 4000 NPVB = 17 368 - 11 505 = $5863

= 4000 + 4869.73 + 2634.92 = $11 505

Since NPVB > NPVA , alternative B is preferable. Diff: 3 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 22) Suppose you are offered two investment alternatives. If you choose Alternative 1, you will have to make an immediate outlay of $26 000. In return, you will receive $1500 at the end of every three months for the next ten years. If you choose Alternative 2, you will have to make an outlay of $14 000 now and $8000 in two years. In return, you will receive $60 000 ten years from now. Interest is 8.22% compounded semi-annually. Compute the net present value for each alternative and determine which investment should be accepted or rejected according to the net present value criterion. Answer: ALT. 1: PMT = $1500; I/Y = 8.22%; P/Y = 4; C/Y = 2; n = 10(4) = 40; c = p= NPV1

= 0.5; i =

= 0.0411;

- 1 = .02034308 = -26000 + 1500 = -26000.00 + 1500.00(27.19160583) = -26000.00 + 40787.41 = $14787.41 = $14 787

ALT. 2: Outlays $14 000 now and $8000 in two years PVOUT = 14 000 + 8000(1.0411)-4 = $20 809.58 = $20 810 PVIN = 60000(1.0411)-20 = $26 810.34 = $26 810 NPV2 = 26 810 - 20 810 = $6000 Since NPV1 > NPV2, alternative 1 is preferred. Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

23) The SHREK Company has to make a decision about expanding its production facilities. Research indicates that the desired expansion would require an immediate outlay of $160 000 and an outlay of a further $60 000 in five years. Net returns are estimated to be $25 000 per year for the first five years and $20 000 per year for the following 9 years. Find the net present value of the project. Should the expansion project be undertaken if the required rate of return is 10% compounded annually? Answer: PVOUT = 160000 + 60000(1.10)-5 = $197 255 PVIN = 25000

+ 20000

(1.10)-5

= 25000(3.7907868) + 20000(5.7590238)(.6209213) = $166 287.68 = $166 288 NPV = 166 288 - 197 255 = -$30 967 Since NPV < 0, the project shouldn't be undertaken. Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 24) A company is considering a project that will require a cost outlay of $17 200 per year for 3 years. At the end of the project the salvage value will be $15 000. The project will yield returns of $60 000 in Year 4 and in Year 5. There are no returns after Year 5. Alternative investments are available that will yield a return of 14.2%. Should the company undertake the project? Answer: Outlays: PMT = $17 200; P/Y = C/Y = 1; I/Y = 14.2%; i = 0.0142 PVOUT = 17200

(1.142) -15000(1.142)-5 = 45449.8213 - 7722.5502 = 37 727.27

PVOUT = $37 727 PVIN = 60000(1.142)-4 + 20000(1.142)-5 = 35276.6093 + 10296.7336 = 45 573.34 PVIN = $45 573 NPV = 45 573 - 37 727 = $7 846 The company should undertake the project. Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

25) A company is considering a project that will require a cost outlay of $25 000 per year for 3 years. At the end of the project the salvage value will be $15 000. The project will yield annual net returns of $17 500 for 5 years. Alternative investments are available that will yield a return of 15%. Should the company undertake the project? Answer: Outlays: PMT = 25 000; P/Y = C/Y = 1; I/Y = 15%; i = 0.015 PVOUT = 25 000

(1.15) - 15 000

= 65642.7221 - 7457.6510 = 58 185.07

PVOUT = $58 185 PVIN = 17 500

= 58 662.71

PVIN = $58 663 NPV = 58 663 - 58 185 = $478. The net present value is slightly positive, which means that the net present value method does not provide a clear signal as to whether to undertake the project or not. Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 26) Big Sky Company expects a demand of 30 000 units per year for a special purpose component for six years. Net return per unit is $4.10. To produce the component, the company must buy a machine costing $245 000 with a life of six years and a salvage value of $47 000 after six years. The company estimates that repair costs will be $18 000 per year during Years 2 to 6. If Big Sky company requires a return on investment of 15.5%, should it market the component? Answer: Inflows: PMT = 30 000(4.10) = $123 000; I/Y = 15.5%; n = 6(1) = 6; P/Y = C/Y = 1 PVIN = 123000.00

= 459 290.62 = $459 291

Outlays: FV = $47 000; PMT = 19 600 PVOUT = 245 000 + 18 000

- 47000.00(1.155)-6

= 245 000 + 59 631.3170 - 19 797.2995 = $284 834 NPV = 459 291 - 284 834 = $174 457 Since the net present value is positive, the rate of return on investment is greater than 15.5%. Big Sky Company should market the component. Diff: 3 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

27) LD Winery expects a demand of 10 000 bottles per year for a special purpose wine for six years. Net return per unit is $6.10. To produce the wine, LD must buy equipment costing $200 000 with a life of six years and a salvage value of $10 000 after six years. The company estimates that repair costs will be $8000 per year during Years 2 to 6. Should LD invest in the equipment if it requires a return of 16% on its investment? Answer: Inflows: PMT = $10 000(6.10) = $61 000; P/Y = C/Y = 1; I/Y = 16%; n = 6 PVIN = 61000 PVOUT = 200 000 + 8 000

= 224 769 - 10 000

= 200 000 + 26 194.3492 - 4104.4225 = 222 089.9267 = $222 090 NPV = 224 769 - 222 090 = $2679 Since NPV > 0, the return on the investment is greater than 16%. LD Winery should invest in the equipment. Diff: 3 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 28) Replacing old equipment at an immediate cost of $5000 and $7000 six years from now will result in a savings of $3000 semi-annually for ten years. At 11% compounded annually, should the old equipment be replaced? Answer: PMT = $3 000; P/Y = 2; C/Y = 1; I/Y = 11; c = PVIN = 3000

= 0.5; n = 10(2) = 20; p =

- 1 = .053565

= 36 281.7688 = $36 282

PVOUT = 5000 + 7000 = 8742.48 = $8742 NPV = 36 282 - 8742 = $27 540 Since NPV > 0, the old equipment should be replaced. Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

29) Replacing old office equipment at an immediate cost of $3000 and $4000 four years from now will result in savings of $400 semi-annually for ten years. Should the old equipment be replaced a) at 8% compounded annually? b) at 12% compounded annually? Answer: a) PMT = 400; P/Y = 2; C/Y = 1; I/Y = 8%; n = 10(2) = 20; c = PVIn = 400

= 0.5; p =

- 1 = 0.039230

= 5473.36 = $5473

PVOUT = 3000 + 4000 = $5940 NPV = 5473 - 5940 = -$467 Since NPV is slightly negative, the net present value method doesn't provide a clear signal as to whether to replace the equipment or not. b) PMT = 400; P/Y = 2; C/Y = 1; I/Y = 8%; n = 10(2) = 20; c = PVIn = 400

= 0.5; p =

- 1 = 0.058301

= 4651.94 = $4652

PVOUT = 3000 + 4000 = 3000 + 2542.07 = $5542 NPV = 5542 - 4 652 = $890 Since NPV is positive, the return on the investment is greater than 12%. The old equipment should be replaced. Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 30) A company is considering a project that will require a cost outlay of $30 000 per year for three years. At the end of the project, the company expects to salvage the physical assets for $50 000. The project is estimated to yield net returns of $60 000 in Year 4, $40 000 in Year 5, and $15 000 for each of the following five years. Alternative investments are available yielding a rate of return of 14.5%. Compute the net present value of the project. Answer: Outlays: PMT = 30 000; P/Y = C/Y = 1; I/Y = 14.5% PVOUT = 30 000

(1.145) - 50 000(1.145)-10 = 79 084 -12 909.6699 = $66 174

Inflows: PVIN = 60000(1.145)-4 + 40000 (1.145)-5 + 15000

(1.145)-5

= 34 908.3486 + 20 325.0938 + 25 855.2358 = $81 089 NPV = 81 089- 66 174 = $14 915 Since NPV is positive, the rate of return on the project is greater than 14.5%. The project should be undertaken. Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

31) A new venture that requires outlays of $127 000 for each of the first two years will yield net returns of in each year for years 3 to 6 and $70 000 for each of the following four years. A residual value of can be recovered at the end of the last income period. Should the venture be undertaken if a yield of 11.56% is required? Answer: Outlays: PVOUT = 127 000 + 127 000 - 130 000 = 127 000 + 113 840.09 - 43 536.98 = $197 303 Inflows: PVIN = 85000

(1.1156)-2 + 70000

(1.1156)-6

= 209 379.10 + 111 321.34 = $320 700 NPV = 320 700 - 197 303 = $123 397 Since NPV > 0, the rate of return on the venture is greater than 11.56%. The venture should be undertaken. Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 32) A special chemical development project requires an immediate outlay of $110 000 and $50 000 at the end of each year for 3 years. Net returns are nil for the first 3 years and $60 000 per year thereafter for fourteen years. What is the net present value of the project at 17%? Answer: Inflows: PMT = 60 000; P/Y = C/Y = 1; I/Y = 17%; n = 14(1) = 14 PVIN = 60000

(1.17)-3 = $195 901

Outlays: PMT = 50 000; n = 3(1) PVOUT = 110 000 + 50000

= 110 000 + 110 479.2481 = $220 479

NPV = 195 901 - 220 479 = -$24 578 The net present value of the project at 17% is -$24 578. Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

33) A commitment on the project requires an initial outlay of $10 000.00 and a further outlay of $5000.00 after eleven years. Net returns are $5 000.00 per year for five years. What is the net present value of the project at 16%? Answer: PVOUT = 10 000 + 5000 = $10 977.08 PVIN = 5000

= $16 371

NPV = 16 371 - 10 977 = $5394 The net present value of the project at 16% is $5394. Diff: 1 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 34) A project requires an initial outlay of $350 000 and a further outlay of $100 000 after one year. Net returns are $105 000 per year for five years. What is the net present value of the project at 9.9%? Answer: PVOUT = 350 000 + 100 000 = $440 992 PVIN = 105 000

= $399 052

NPV = 399 052 - 440 992 = -$41 940 Since NPV < 0, the rate of return on the project is less than 9.9%. Diff: 1 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 35) A once in a lifetime project requires an immediate outlay of $100 000 and $25 000 at the end of each year for 3 years. Net returns are nil for the first 3 years and $30 000 per year thereafter for fourteen years. What is the net present value of the project at 16%? Answer: PVIN = 30 000

(1.16)-3 = $105 084

PVOUT = 100 000 + 25000

= 100 000 + 56 147.2385 = $156 147

NPV = 105 084 - 156 147 = -$51 063 Diff: 2 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

36) Assume that the net present value of a project is $ 3870 at 10%, and -$1853 at 12%. Use linear interpolation to compute the rate of return correct to the nearest tenth of a percent. Answer: The ROI is contained between 10% < ROI < 12% Rate 10% 12% x% d=

NPV $3870 -$1853 0 = 1.35%

The rate of discount for which the NPV = 0 is approximately 10% + 1.35% = 11.4% Diff: 1 Type: SA Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment. 37) Assume that the net present value of a project is $ 220 at 22%, and -$305 at 24%. Use linear interpolation to compute the rate of return correct to the nearest tenth of a percent. Answer: The ROI is contained between 22% < ROI < 24% Rate 22% 24% x% d=

NPV $220 -$305 0 = 0.84%

The rate of discount for which the NPV = 0 is approximately 22% + 0.84% = 22.8% Diff: 1 Type: SA Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment.

38) An investment of $180 000 yields annual net returns of $27 700 for seven years. What is the rate of return on the investment (correct to the nearest tenth of a percent) if part of the initial investment, in the amount of, $80 000 is recovered at the end of 7 years? a) Use linear interpolation to find the approximate value of the rate of return. b) Find the answer using Cash Flow and IRR. Answer: PVOUT = 180 000 - 80 000 (1 + i)-7; PVIN = 27 700 NPV = PVIN - PVOUT Index =

× 100%; Increase/Decrease in Rate = (Index - 100%)/4 rounded.

Calculations below show the attempts in approximating the ROI i = 12% i = 10% i = 8% PVOUT 143812 138947 133321 PVIN NPV Index Inc/Decr Rate

126416 -17396 88 -3 ROI < 12%

134855 -4092 97 -1 ROI < 10%

144216 10896 108 2 ROI > 8%

Hence, 8% < ROI < 10%. Use linear approximation to find the approximate value of ROI. d=

= 1.45%

The rate of return, correct to the nearest tenth of a percent, is 9.5% b) CF0 = 180 000± C01 = 27 700 F01 = 6 C02 = 107 700 F02 = 1 Compute IRR = 9.429578 = 9.4% Diff: 3 Type: SA Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment.

39) The introduction of a new product requires an immediate outlay of $145 000 and has a residual value of after 10 years. The anticipated net returns from the marketing of the product are expected to be per year for ten years. What is the rate of return on the investment (correct to the nearest tenth of a percent)? a) Use linear interpolation to find the approximate value of the rate of return. b) Find the answer using Cash Flow and IRR. Answer: PVOUT = 145 000 - 30 000(1 + i)-10; PVIN = 25 500 NPV = PVIN - PVOUT Index =

× 100%; Increase/Decrease in Rate = (Index - 100%)/4 rounded

Calculations below show the attempts in approximating the ROI Attempts i = 16% i = 10% i = 14% PVOUT 138 199 133 434 136 908 PVIN NPV Index Inc/Decr Rate

123 247 -14 952 89 -3 ROI < 16%

156 686 232 534 117 4 ROI > 10%

133 011 -38973897 97 -1 ROI < 14%

i = 12% 135 341 144 081 8740 106 2 ROI > 12%

Hence, 12% < ROI < 14%. Use linear approximation to find the approximate value of ROI. d=

= 1.38%

The rate of return, correct to the nearest tenth of a percent, is 13.4% b) CF0 = 145 000 ± C01 = 25 500 F01 = 9 C02 = 55 500 F02 = 1 Compute IRR = 13.353832 = 13.4% Diff: 3 Type: SA Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment.

40) A restaurant may be purchased for $250 000. Annual net income from the operation of the restaurant is expected to be $61 000 for each of the first 4 years and $30 000 for each of the next three years. After 7 years, the restaurant can be sold for $315 000. Determine the rate of return on the investment correct to the nearest tenth of a percent. a) Use linear interpolation to find the approximate value of the rate of return. b) Find the answer using Cash Flow and IRR. Answer: PVOUT = 250 000 - 315 000(1 + i)-7; PVIN = 61 000

+ 30 000

(1 + i)-4;

NPV = PVIN - PVOUT Index =

× 100%; Increase/Decrease in Rate = (Index - 100%)/4

Calculations below show the attempts in approximating the ROI Attempts i = 16% i = 28% i = 24% PVOUT 138 544 194 045 180 119 PVIN NPV Index Inc/Decr Rate

207 901 69 357 150 12 ROI > 16%

157 580 -364 644 81 -5 ROI < 28%

171 802 -8317 95 -1 ROI < 24%

i = 22% 171 695 179 768 8073 105 1 ROI > 22%

Hence, 22% < ROI < 24%. Use linear approximation to find the approximate value of ROI. d=

= 0.985%

The rate of return, correct to the nearest tenth of a percent, is 22 + 0.985 = 23.0% b) CF0 = 250 000± C01 = 61 000 F01 = 4 C02 = 30 000 F02 = 2 C03 = 345 000 F03 = 1 Compute IRR = 22.958989 = 23.0% Diff: 3 Type: SA Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment.

41) A project requiring an immediate investment of $150 000 and a further outlay of $60 000 after four years has a residual value of $50 000 after nine years. The project yields a negative net return of $10 000 in Year 1, a zero net return in Year 2, $60 000 per year for the following four years, and $70 000 per year for the last three years. Find the rate of return (correct to the nearest tenth of a percent). Answer: CF0 = 150 000 ± C01 = 10 000 ± F01 = 1 C02 = 0 F02 = 1 C03 = 60 000 F03 = 1 C04 = 0 F04 = 1 C05 = 60 000 F05 = 2 C06 = 70 000 F06 = 2 C07 = 120 000 F07 = 1 Compute IRR = 17.261192 = 17.3% Diff: 3 Type: SA Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment. 42) The Radium Hot Springs plans to install a swimming pool. Construction of the lift is estimated to require an immediate outlay of $420 000. The life of the pool is estimated to be 20 years with a salvage value of $20 000. Cost of preparing the area is expected to be $30 000 for each of the first 2 years of operation. Net cash inflows from the pool are expected to be $49 000 for each of the first five years and $90 000 for each of the following 15 years. Find the rate of return (correct to the nearest tenth of a percent). Answer: CF0 = 420 000 ± C01 = 19 000 F01 = 2 C02 = 49 000 F02 = 3 C03 = 90 000 F03 = 14 C04 = 110 000 F04 = 1 Compute IRR = 13.596638% = 13.6% Diff: 3 Type: SA Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment.

Use Excel's IRR function to answer the question. 43) The introduction of a new product requires an initial outlay of $610 000. The anticipated net returns from the marketing of the product are expected to be $92 300 per year for 12 years. Find the rate of return (correct to the nearest tenth of a percent). Answer:

Diff: 3 Type: SA Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment. 44) The owner of a music store is considering remodelling the store in order to carry a larger inventory. The cost of remodelling and additional inventory is $43 200. The expected increase in net profit is $7000 per year for the next 3 years and $10 000 each year for the following 7 years. After ten years, the owner plans to retire and sell the business. She expects to recover the additional $40 000 invested in inventory but not the $43 200 invested in remodeling. Compute the rate of return. Answer:

Diff: 3 Type: SA Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment.

45) A project requires an initial outlay of $100 000 and promises net returns of $18 500 per year over a twelve-year period. If the project has a residual value of $4000 after twelve years, what is the rate of return? Answer:

Diff: 3 Type: SA Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment. 46) Billy Bob wants to convert his farm into a ski resort. He asked you to determine his rate of return based on the following estimates. - Development cost for each of the first 2 years, $87 000. - Construction of a chalet in Year 3, $1 240 000. - Upon his retirement in fifteen years, improvements in the property will yield him $270 000. - Net returns from the operation of the golf course will be nil for the first three years and $200 000 per year afterwards until his retirement. Answer:

Diff: 3 Type: SA Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment.

47) A company buys equipment for $15 000 today and has annual net cash inflows of $6000 for 3 years. The discount rate is 12% compounded annually. What is the Net Present Value (NPV)? A) -$525.90 B) $552.90 C) $992.50 D) $652.90 E) -$652.90 Answer: A Diff: 1 Type: MC Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 48) A company needs to spend $15 000 for each of the next three years. Net returns beginning in Year 4 are estimated at $2500 per year for ten years. The required rate of return is 9.75% compounded annually. The NPV is? A) -$27 521.41 B) -$27 721.41 C) -$25 721.41 D) $25 217.41 E) $27 521.41 Answer: C Diff: 1 Type: MC Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 49) A company has the following net cash inflows. Today -$7000, Year 1 -$15 000, Year 2 -$9000, Year 3 + $12 000, Year 4 -$3000, Year 5 + $19 000. Compute the net present value if the required rate of return is 7.22% compounded annually. A) -$4709.89 B) -$3000.00 C) -$7009.89 D) -$7409.89 E) $4709.89 Answer: D Diff: 2 Type: MC Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

50) A company is looking to invest in a very risky project. They have a required rate of return of 27% compounded annually. The project has the following cash inflows: Year 1 $17500, Year 2 $15000, Year 3 $27500. It also has the following cash outflows: Immediately -$10 000, Year 1 -$15 000, Year 3 -$9500. What is the NPV? A) $9718.06 B) $7918.06 C) $7819.06 D) $9918.06 E) -$9718.06 Answer: B Diff: 2 Type: MC Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 51) A company has cash outflows of $21 275 starting today for each of the next 5 years. After that they will have cash inflows of $16 000 for each of the following 7 years. The discount rate is 9% compounded annually. What is the NPV? A) -$30 415.15 B) -$34 015.15 C) -$4015.15 D) -$341.51 E) -$41.51 Answer: A Diff: 2 Type: MC Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 52) A company spends $117 500 today and has positive cash inflows of $34 000 for each of the next 6 years. What is the Internal Rate of Return (IRR)? A) 14.87% B) 18.47% C) 17.48% D) 16.84% E) 15.87% Answer: B Diff: 2 Type: MC Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment.

53) A company has the following pattern of cash flows. Today -$47 000, Year 1 -$15 500, Year 2 + $16 000, Year 3 + $35 000, Year 4 + $57 000. What is the IRR? A) 9.301% B) 29.301% C) 19.301% D) 39.301% E) 49.301% Answer: C Diff: 3 Type: MC Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment. 54) A company is planning on investing the following monies. They spend Today -$26 000, Year 1 -$7000, Year 4 -$11 000, and Year 7 -$10 100. Their cash inflows are Years 1-3 inclusive + $12 000, Years 4-9 inclusive + $16 000. What is their IRR? A) 15.18% B) 25.18% C) 5.18% D) 35.18% E) 10.18% Answer: D Diff: 3 Type: MC Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment. 55) A company has an immediate cash outlay of -$325 000 and additional subsequent cash outlays of $6000 per year for the life of the project. They will receive cash inflows in year 3 for + $77 000 and increasing by $3000 per year until the project ends 7 years later (total of 10 years). What is the IRR? A) 9.79% B) 0.79% C) 7.79% D) 10.91% E) 3.79% Answer: D Diff: 3 Type: MC Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment.

56) What is the IRR for the following net annual cash flows today -$45 000, Year 1 + $12 000, Year 2 + $37 000, Year 3 + $12 000, Year 4 + $17 000? A) 26.722% B) 16.722% C) 6.722% D) 36.722% E) 46.722% Answer: A Diff: 3 Type: MC Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment. 57) Nick has a choice to pay $1499 for a TV now or pay $1599 3 years from now. What should be the minimum interest rate at which Nick should invest $1499 to make a decision to pay 3 years from now? A) 2.176% B) 6.528% C) 0.725% D) 6.67% E) 3% Answer: A Diff: 1 Type: MC Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion. 58) Molly needs to decide whether to buy a water heater for $2200 cash and enter a service contract requiring a payment of $30 at the end of every 3 months for 10 years or to enter a 10 years lease requiring a payment of $90 at the beginning of every 3 months. If the leased water heater can be bought after 10 years for $80, should Molly buy or lease the water heater, if money is compounded quarterly at 2.5%? Answer: Present value of cash payment for the water heater = $2200 For PMT = $30; n = 40; P/Y = 4; C/Y = 4; I/Y = 2.5%; i = 0.625% Present value of the service contract involves an ordinary annuity = $30

= $1058.85

Present value of decision to buy the water heater = $3258.85 For PMT = $100;

n = 40;

i = 0.625%

Present value of the lease payment annuity due = $90 × 1.00625 ×

= $3196.39

Present value of the purchase price after 10 years = $80 × 1.00625-40 = $62.35 Present value for the decision to lease = $3258.74 Both options result into the same economic value, within the rounding error. Diff: 2 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion. 59) Kevin needs to decide whether to buy a Honda Civic for $19 900 with a salvage price of $6500 after 5

years or lease the car for 5 years making monthly payments of $269 at the beginning of each month. If money is worth 5% compounded annually, should Kevin buy or lease? Answer: Present value of the salvage price = $6500 × 1.05-5 = $5092.92 Net present value if Kevin decides to buy = $19 900 - $5092.92 = $14 807.08 For monthly lease: PMT = $269;

P/Y = 12;

C/Y = 1;

I/Y = 5%;

;

n = 60;

i = 5%

-1 = 0.004074

PV = $269 × 1.004074 ×

= $14 351.23

PV for lease is smaller than PV to buy. Therefore Kevin should lease the Honda Civic. Diff: 2 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion. 60) A house is on sale in Markham. Marlene has an option to pay $575 000 lump sum or pay $6000 at the end of every month for the next 10 years. If money earns 5% compounded monthly, which option has a better economic advantage? A) Paying monthly has an advantage of $9312 in PV B) Paying lump sum has an advantage of $9312 in PV C) Paying monthly has an advantage of $85 129 in PV D) Paying lump sum has an advantage of $85 129 in PV E) Both options are the same. Choose any. Answer: A Diff: 1 Type: MC Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion.

61) If Ontario decides to build two new units of nuclear power plants, the overnight cost of construction is $15 billion. The annual net rate of return from the two units is $1 billion every year for the next 60 years, at which time the plants should be decommissioned and have a residual value of $0.96 billion. Should Ontario decide to build the two units of nuclear power plants if the required rate of return is 7%? Answer: PMT = $1 billion; n = 60; i = 7% PVIN = $1

= $14.04 billion

PVOUT = $15 billion - $0.96 billion = $14.9834329 billion NPV = $14.04 - $14.9834329 billion = -$0.94 billion Since NPV is negative, Ontario should not build the two units of nuclear power plants. Diff: 1 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 62) A local community church is contemplating installment of solar cells on its roof and sell the access electricity to the Ontario grid via a power purchase agreement. The project requires an initial investment of $75 000 with a residual value of $2000 after 10 years. It is estimated to yield annual net returns of $15 000 for 10 years. What is the NPV of the project given a rate of return of 7%? A) -$31 370 B) $32 387 C) $31 370 D) -$32 387 E) $1017 Answer: C Diff: 2 Type: MC Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 63) Pfizer is planning to embark on developing a new medicine for the cure of common cold. If undertaken, the project requires $10 million dollar on R&D, every year for 5 years. The medicine will be marketed in Year 6 and net returns are estimated to be $10 million for the next 5 years at the end of which Pfizer plans to sell the formula to its competitor for $1 million. If corporation requires a ROI of 15%, what is the NPV for the project? A) $16.6 million B) -$16.6 million C) $16.85 million D) -$16.85 million E) $0.123 million Answer: B Diff: 2 Type: MC Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

64) A firm invests $200 000 in machinery that yields net after-tax cash flows of $90 000 at the end of each of the next three years. The opportunity cost of capital is 12%. What is the net present value of the project (to the nearest thousand dollars)? A) -$16 000 B) -$8000 C) $0 D) $8000 E) $16 000 Answer: E Diff: 2 Type: MC Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 65) Saint Mary's is offered a contract, which offers a net present value of $15 000 if the required rate of return is 12%. Alternatively, net present value is -$12 000 if the required rate of return is 18%. At what rate of return, would the NPV be zero? A) 13% B) 13.7% C) 14% D) 14.5% E) 15% Answer: B Diff: 2 Type: MC Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment. 66) Saint Mary's is offered a contract, which requires an immediate investment of $25 million. The estimated returns are $5 million per year for 20 years. Compute the rate of return. A) 19.42% B) 99.9% C) 7.75% D) 12.45% E) 13.79% Answer: A Diff: 2 Type: MC Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment.

67) Korea Nuclear tritium removal facility (TRF) project requires an immediate investment of $33 million with a residual value of $7 million at the end of the project. It is expected to yield a net return of $1 million in Year 1 from the sale of immobilized tritium to ITER, $5 million dollars in Year 2, $8 million per year for the following six years, and $7 million per year for the remaining four years. Find the rate of return using Excel's IRR function. Answer: Initial Cash Flow Net Cash Flow (in $million) (in $million) -$33 New Cash Flow Period 1 $1 New Cash Flow Period 2 $5 New Cash Flow Period 3 $8 New Cash Flow Period 4 $8 New Cash Flow Period 5 $8 New Cash Flow Period 6 $8 New Cash Flow Period 7 $8 New Cash Flow Period 8 $8 New Cash Flow Period 9 $7 New Cash Flow Period 10 $7 New Cash Flow Period 11 $7 New Cash Flow Period 12 $14 Internal Rate of Return 16.38% Diff: 2 Type: SA Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment. 68) Suleiman must make a choice between two investment alternatives. Alternative 1 will provide him returns of $300 000 at the end of second year and $700 000 at the end of fifth year. Alternative 2 will provide him returns $180 000 at the end of each of the next six years. Alternative 3 will provide him returns $370 000 at the end of second, fourth and sixth years. Which alternative is preferable if money is worth 9.4%? A) Alternative 1 B) Alternative 2 C) Alternative 3 D) Alternative 2 or Alternative 3, as both are essentially same E) Alternative 1 or Alternative 3, as both are essentially same Answer: B Diff: 2 Type: MC Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion.

69) Suleiman must make a choice between two investment alternatives. Alternative 1 will provide him returns of $300 000 at the end of second year and $700 000 at the end of fifth year. Alternative 2 will provide him returns $180 000 at the end of each of the next six years. Alternative 3 will provide him returns $370 000 at the end of second, fourth and sixth years. Which alternative is preferable if money is worth 5.5556%? A) Alternative 1 B) Alternative 2 C) Alternative 3 D) Alternative 2 or Alternative 3, as both are essentially the same E) Alternative 1 or Alternative 3, as both are essentially same Answer: D Diff: 2 Type: MC Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion. 70) Jeremy has a number of outstanding credit cards debts. He seeks help from BDO to consolidate his debts and make it easy for him to make the debt payments. BDO offers him two alternatives. In the first alternative, the debt can be settled by making a payment of $30 000 now and a final payment of $42 000 in 3 years. Alternatively, the obligation can be settled by payments of $8400 at the end of every three months for three years. Which alternative is preferred if the interest is 8.5% compounded quarterly. A) Alternative 1 B) Alternative 2 C) Neither Alternative D) Don't take offer from the BDO E) Alternative 1 or Alternative 2, as both are essentially same Answer: B Diff: 2 Type: MC Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion. 71) Jeremy has a number of outstanding credit cards debts. He seeks help from BDO to consolidate his debts and make it easy for him to make the debt payments. BDO offers him two alternatives. In the first alternative, the debt can be settled by making a payment of $30 000 now and a final payment of $42 000 in 3 years. Alternatively, the obligation can be settled by payments of $8400 at the end of every three months for three years. Which alternative is preferred and by how much if the interest is 13.5% compounded quarterly. A) Alternative 1 B) Alternative 2 C) Neither Alternative D) Don't take offer from the BDO E) Alternative 1 or Alternative 2, as both are essentially same Answer: A Diff: 2 Type: MC Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion.

72) Jasmine has two investment choices. Alternative 1 requires an immediate outlay of $150 000 and offers a return of $417 000 after seven years. Alternative 2 requires an immediate outlay of $180 000 in return for which $25 000 will be received at the end of every six months for the next seven years. Alternative 3 requires an immediate outlay of $200 000 in return for which $60 000 will be received at the end of every year for the next seven years. The required rate of return on investment is 6.3% compounded semiannually. What is Jasmine's most preferred option? A) Alternative 1 B) Alternative 2 C) Alternative 3 D) Alternative 3 or Alternative 2, as both are essentially same E) Alternative 1 or Alternative 2, as both are essentially same Answer: C Diff: 3 Type: MC Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 73) Jasmine has two investment choices. Alternative 1 requires an immediate outlay of $150 000 and offers a return of $417 000 after seven years. Alternative 2 requires an immediate outlay of $180 000 in return for which $25 000 will be received at the end of every six months for the next seven years. Alternative 3 requires an immediate outlay of $200 000 in return for which $60 000 will be received at the end of every year for the next seven years. The required rate of return on investment is 7.4% compounded semiannually. What is Jasmine's most preferred option? A) Alternative 1 B) Alternative 2 C) Alternative 3 D) Alternative 3 or Alternative 2, as both are essentially same E) Alternative 1 or Alternative 2, as both are essentially same Answer: B Diff: 3 Type: MC Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

74) Jasmine has two investment choices. Alternative 1 requires an immediate outlay of $150 000 and offers a return of $417 000 after seven years. Alternative 2 requires an immediate outlay of $180 000 in return for which $25 000 will be received at the end of every six months for the next seven years. Alternative 3 requires an immediate outlay of $200 000 in return for which $60 000 will be received at the end of every year for the next seven years. The required rate of return on investment is 6.58% compounded semiannually. What is Jasmine's most preferred option? A) Alternative 1 B) Alternative 2 C) Alternative 3 D) Alternative 3 or Alternative 2, as both are essentially same E) Alternative 1 or Alternative 2, as both are essentially same Answer: D Diff: 3 Type: MC Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 75) CGI Inc. is developing a new software platform, which requires an immediate investment of $780 000 and another $260 000 in three years. Net returns expected are $450 000 after two years, $313 000 after four years, and $889 000 after six years. However, if they simply upgrade their current platform, it requires an immediate investment of $400 000, another $260 000 after two years, and $340 000 after four years. Net returns are $187 500 per year for 8 years. Determine the net present value at 5.9%. Which project is preferable according to the net present value criterion? A) New Software platform by $53 309 B) Upgrade by $53 309 C) Upgrade by $277 277 D) New Software platform by $70 203 E) New Software platform by $76 599 Answer: B Diff: 3 Type: MC Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

76) Simion Inc is considering to build a new office in Clagary, which requires an immediate investment of $78 million and another $35 million in one year. Being near to the customers, it is expected a revenue of $2 million in Year one, $10 million in year 2 and then a steady stream of $15 million revenue each year in the subsequent 12 years. However, if they simply expand their current offices in Toronto and build a satellite office in Calgary, it requires an immediate investment of $30 million, another $15 million after one year, and $10 million after two years. Net returns are $expected to be $10 million per year from year 2 to year 14. Determine the net present value at 7.4%. Which option is preferable according to the net present value criterion? A) Build a new office with NPV of $21 million B) Expand current office with NPV of $9 million C) Build a new office with NPV of -$21 million D) Expanding current office is $30 million better in NPV compared to building a new office E) Either B or D Answer: B Diff: 3 Type: MC Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 77) LG expects a demand of 10 000 headphones per year for a special purpose headphone for six years. Net return per unit is $6.10. To produce the headphones, LG must buy equipment costing $200 000 with a life of six years and a salvage value of $10 000 after six years. The company estimates that repair costs will be $8000 per year during Years 2 to 6. Should LG invest in the equipment if it requires a return of 16% on its investment? Calculate the NPV. A) Yes, NPV = $2679 B) Yes, NPV = $6783 C) No, NPV = -$5530 D) No, NPV = -$2679 E) No, NPV = -$6783 Answer: A Diff: 3 Type: MC Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 78) Assume that the net present value of a project is $74.3 million at 6%, and -$21.2 million at 9%. Use linear interpolation to compute the rate of return correct to the nearest hundredth of a percent. A) 2.33% B) 8.33% C) 6.67% D) 4.2% E) 10.2% Answer: B Diff: 1 Type: MC Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment.

79) Matthew purchased a motorcycle and is offered two alternatives for payment. The first option requires a $5550.00 now and a final payment of $12 325.00 in four years. The second alternative would require Matthew to make payments of $1085.00 at the end of every three months for four years. If interest rate is 8.2% compounded quarterly, determine the preferred alternative. Answer: n = 4(4) = 16; i = 8.2%/4 = 0.0205; PMT = 1085 Alternative 1 PV = 12325(1.0205)-16 = 8907.97 Total Payment = 5550 + 8907.97 = 14 459.97 Alternative 2 PV = 1085

= 14 673.63

Therefore, Alternative 1 is the better option as there is less to pay. Diff: 2 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion. 80) Aki meets with his grandparents, who offer him either $85 right now or $100 on his birthday 11 months from now. Which option should Aki choose to maximize the amount of money he earns? Suppose the interest rate is 12.5% compounded annually. Answer: FV = $100; i = 0.125; n = 11/12 PV = 100(1.125)-11/12 = 89.77

Since the PV of the $100 payment is larger than the $85 alternative, Aki should choose to receive $100 on his birthday. Diff: 1 Type: SA Page Ref: 651-657 Topic: 16.1 Discounted Cash Flow Objective: 16-1: Determine the discounted value of cash flows and choose between alternative investments on the basis of the discounted cash flow criterion. 81) Aafia is overseeing a project that will establish a new Faculty of Business building at her university. While analysts estimate that the building would result in $325 500 in annual net returns over the first four years, the project would require $1 100 500 in initial investment costs. If there is a 14.5% required rate of return, and the building will leave a residual value of $200 000 after four years, should Aafia pursue the development of the building? Answer: PVin = 325500

= 938773.85

PVout = 1100500 - 200000 = 984138.83 NPV = 938773.85 - 984138.83 = -45364.98 Since the NPV is negative, Aafia should not pursue development of the building. Diff: 1 Type: SA Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible.

82) Arjun is interested in opening a new branch for Habitat for Humanity, which will take $102 325 in initial startup costs but will yield $24 775 in annual net returns every year for 6 years. Assume that after 6 years, there are no further benefits nor any residual value left on Arjun's investment. If the required rate of return is 10%, should he start work on the project? What if the required rate of return is 17%? A) Yes if the rate of return is 17%, No if the rate of return is 10%. B) Yes if the rate of return is 10%, No if the rate of return is 17%. C) Yes for both rates of return. D) No for both rates of return. E) More information is required. Answer: B Diff: 2 Type: MC Page Ref: 659-668 Topic: 16.2 Net Present Value Objective: 16-2: Determine the net present value of a capital investment project and infer from the net present value whether a project is feasible. 83) If the present value of inflows is greater than the present value of outflows: A) NPV = 0, Profitability Index > 1, Rate of Return on Investment > i B) NPV > 0, Profitability Index > 1, Rate of Return on Investment = i C) NPV > 0, Profitability Index = 1, Rate of Return on Investment < i D) NPV > 0, Profitability Index > 1, Rate of Return on Investment > i E) NPV < 0, Profitability Index < 1, Rate of Return on Investment > i Answer: D Diff: 1 Type: MC Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment. 84) The annual net return on a project is $160.30 for 5 years, and the initial investment is $635. The required rate of return is 8.3%, and there is no residual value after 5 years. Which of the following are true? A) NPV = 0, Profitability Index = 1, Rate of Return on Investment = i B) NPV > 0, Profitability Index > 1, Rate of Return on Investment = i C) NPV > 0, Profitability Index = 1, Rate of Return on Investment < i D) NPV < 0, Profitability Index > 1, Rate of Return on Investment > i E) NPV = 0, Profitability Index < 1, Rate of Return on Investment > i Answer: A Diff: 2 Type: MC Page Ref: 671-681 Topic: 16.3 Rate of Return on Investment Objective: 16-3: Compute the rate of return on investment.