Solution Manual For Contemporary Business Mathematics with Canadian Applications, 13th Edition by Si by StudyGuide

Contents PART ONE Mathematics Fundamentals and Business Applications Chapter 1: Review of Arithmetic

Chapter 2: Review of Basic Algebra

Chapter 3: Ratio, Proportion, and Percent

Chapter 4: Linear Systems

135

Chapter 5: Cost-Volume-Profit Analysis and Break-Even

193

PART TWO Mathematics of Business and Management Chapter 6: Trade Discount, Cash Discount, Markup, and Markdown

225

Chapter 7: Simple Interest

263

Chapter 8: Simple Interest Applications

289

PART THREE Mathematics of Finance and Investment Chapter 9: Compound Interest—Future Value and Present Value

335

Chapter 10: Compound Interest—Further Topics

379

Chapter 11: Ordinary Simple Annuities

415

Chapter 12: Ordinary General Annuities

455

Chapter 13: Annuities Due, Deferred Annuities, and Perpetuities

493

Chapter 14: Amortization of Loans, Including Residential Mortgages

545

Chapter 15: Bond Valuation and Sinking Funds

611

Chapter 16: Investment Decision Applications

669

PART ONE Chapter 1

Mathematics Fundamentals and Business Applications Review of Arithmetic

Exercise 1.1 A. 1.

12  6  3  12  2  14

(3  8  6)  2  (24  6)  2  18  2  9

(7  4)  5  2  11 5  2  55  2  53

5  3  2  4  15  8  23

6(7  2)  3(5  3)  6(5)  3(2)  30  6  24

20  16 4 1    0.2 15  5 20 5

4(8  5)2  5(3  22 )  4(3)2  5(3  4)  4(9)  5(7)  36  35  1

(3  4  2)2  (2  2  72 )  (12  2) 2  (2  2  49)

 102  (2  98)  100  96  4 9.

250(1  0.08)10  250(2.158925)  539.73

10. (1  0.04)4  1  1.169859  1  0.17 11. 30  600  2500  12  600  18,000  2500  7200  8300 12.

1  [(1  0.40)(1  0.25)(1  0.05)]  1  [(0.6)(0.75)(0.95)]  1  [0.4275]  0.5725  0.57 13. 15  7  6(2  3)  3  15  7  6(5)  3

 15  7  30  3  15  7  10  18 14. 16  2  4  6(4  2)

 8  4  6(6)  32  36  68

15. (1  0.7)  4  20  5  (0.3)  4  4  (0.3)  16

 15.7 16. 50[(1  0.2)(1  0.175)(1  0.04)]  50[(0.8)(0.825)(0.96)]  50[(0.6336)]

 31.68 17. 7a  6[4  (3a  6)]  7a  6[4  3a  6]

 7a  6[2  3a]  7a  12  18a  25a  12 18. 6a  4b  2(16  2a  b)  6a  4b  32  4a  2b  2a  6b  32

Exercise 1.2

24 24 / 2 12 12 / 2 6 6 / 3 2       36 36 / 2 18 18 / 2 9 9 / 3 3

also

24 / 12 2  36 / 12 3

28 28 / 2 14 14 / 2 7 7/7 1       56 56 / 2 28 28 / 2 14 14 / 7 2

also

28 / 28 1  56 / 28 2

210 210 / 10 21 21 / 3 7     360 360 / 10 36 36 / 3 12

also

210 / 30 7  360 / 30 12

360 360 / 5 72 72 / 9 8     225 225 / 5 45 45 / 9 5

also

360 / 45 8  225 / 45 5

144 144 / 2 72 72 / 9 8 8/ 4 2 144 / 72 2       also  360 360 / 2 180 180 / 9 20 20 / 4 5 360 / 72 5

25 25 / 5 5   365 365 / 5 73

A. 1.

365 365 / 73 5   73 73 / 73 1

365 365 / 73 5   219 219 / 73 3

B. 1. 6

1 13  2 2

5 29 2. 4  6 6 3 15 3. 3  4 4

2 26 4. 8  3 3 5.

23 1  11 2 2

51 1  5 10 10

31 3  7 4 4

19 5  2 7 7

C. 1.

11  1.375 8

7  1.75 4

5  1.666667  1.6& 3

5  0.833333  0.83& 6

11  1.833333  1.83& 6

7  0.777778  0.7& 9

13  1.083333  1.083& 12

19  1.266667  1.26& 15

D. 1.

3 3  3.375 8

2 3  3.4 5

1 8  8.333333  8.3& 3

2 16  16.666667  16.6& 3

1 33  33.333333  33.3& 3

1 83  83.333333  83.3& 3

7 7  7.777778  7.7& 9

1  7.083333  7.083& 12

E. 1.

$5.63

$17.45

$18

$253.49

$57.70

$3.10

$13

$40

F. 1.

25, 000(15  8)  146, 000  25, 000(7)  146, 000  175, 000  146, 000  29, 000

(300  8000)  (180  8000)  63, 000  2, 400, 000  1, 440, 000  63, 000  897, 000

1  [(1  0.4)(1  0.25)(1  0.08)]  1  [(0.6)(0.75)(0.92)]  1  [0.414]  0.586

1  [(1  0.32)(1  0.15)(1  0.12)]  1  [(0.68)(0.85)(0.88)]  1  [0.50864]  0.49136

1500 

$54 $54 $54    $730 225 0.12  365 0.12  0.616438 0.073973

264 264 264    0.15 146 4400  365 4400  0.4 1760

45   $620 1  0.14    $620(1  0.017260)  $620(1.017260)  $630.70 365  

292   $375 1  0.16    $375(1  0.128)  $375(1.128)  $423 365  

10.

$250, 250 $250, 250 $250, 250    $220,364.90 330 1  0.15  365 1  0.135616 1.135616

11.

$2358 $2358 $2358    $2250 146 1  0.12  365 1  0.048 1.048

1500  1500  30,000  31,500 0.05

 (1  0.03) 24  1  1.032794   1000   1000[34.426470]  $34, 426.47 12. $1000   0.03  0.03   

 (1  0.02) 20  1  0.485947   70(1.02)  13. $70(1  0.02)   0.02  0.02     71.4[24.29737]  $1734.83 14. $50

[1  (1  0.075) 8 ] 50[1  (0.560702)] 50[0.439297]    50[5.857303] 0.075 0.075 0.075  $292.87

Exercise 1.3

64  0.64 100

A. 1.

64% 

300% 

2.5% 

2.5  0.025 100

0.1% 

0.1  0.001 100

0.5% 

0.5  0.005 100

85% 

250% 

4.8% 

4.8  0.048 100

7.5% 

7.5  0.075 100

10. 0.9% 

0.9  0.009 100

300 3 100

85  0.85 100 250  2.5 100

11. 6.25%  12. 99% 

13.

6.25  0.0625 100

99  0.99 100

225% 

225  2.25 100

14. 0.05% 

0.05  0.0005 100

1 8.25 15. 8 %   0.0825 4 100

16.

1 0.5 %  0.005 2 100

1 112.5 17. 112 %   1.125 2 100 3 9.375 18. 9 %   0.09375 8 100 19.

3 0.75 %  0.0075 4 100

1 162.5 20. 162 %   1.625 2 100 21.

2 0.4 %  0.004 5 100

22.

1 0.25 %  0.0025 4 100

23.

1 0.025 %  0.00025 40 100

1 137.5 24. 137 %   1.375 2 100 25.

5 0.625 %  0.00625 8 100

26. 0.875% 

27.

0.875  0.00875 100

1 2.25 2 %  0.0225 4 100

2 16.6& 28. 16 %   0.16& 3 100 2 116.6 29. 116 %   1.16 3 100 1 183.3& 30. 183 %   1.83& 3 100 1 83.3 31. 83 %   0.83 3 100

2 66.6& 32. 66 %   0.6& 3 100

B. 1.

25% 

25 1  100 4

1 62.5 625 5 62 %    2 100 1000 8

175% 

5% 

1 37.5 375 3 37 %    2 100 1000 8

75% 

4% 

4 1  100 25

8% 

8 2  100 25

40% 

175 7  100 4

5 1  100 20

75 3  100 4

40 2  100 5

1 87.5 875 7 10. 87 %    2 100 1000 8 11.

250% 

12.

2% 

250 5  100 2

2 1  100 50

1 12.5 125 1   13. 12 %  2 100 1000 8 14. 60% 

60 3  100 5

15.

2.25% 

16. 0.5% 

17.

2.25 225 9   100 10, 000 400

0.5 5 1   100 1000 200

1 1 1 %  8 8(100) 800

1 100 100 1 18. 33 %  %  3 3 3(100) 3 19.

3 3 3 %  4 4(100) 400

2 200 200 2 20. 66 %  %  3 3 3(100) 3 21. 6.25% 

6.25 625 1   100 10, 000 16

22. 0.25% 

0.25 25 1   100 10, 000 400

2 50 50 1 23. 16 %  %   3 3 3(100) 6 24. 7.5% 

7.5 75 3   100 1000 40

25. 0.75% 

26.

7 7 7 %  8 8(100) 800

27. 0.1% 

28.

0.75 75 3   100 10, 000 400

0.1 1  100 1000

3 3 3 %  5 5(100) 500

29.

2.5% 

2.5 25 1   100 1000 40

1 400 400 4 30. 133 %  %  3 3 3(100) 3 1 550 550 11 31. 183 %  %  3 3 3(100) 6 2 500 500 5 32. 166 %  %  3 3 3(100) 3 C. 1.

3.5  3.5(100)  350%

0.075  0.075(100)  7.5%

0.005  0.005(100)  0.5%

0.375  0.375(100)  37.5%

0.025  0.025(100)  2.5%

2  2(100)  200%

0.125  0.125(100)  12.5%

0.001  0.001(100)  0.1%

0.225  0.225(100)  22.5%

10. 0.008  0.008(100)  0.8% 11. 1.45  1.45(100)  145% 12. 0.0225  0.0225(100)  2.25% 13. 0.0025  0.0025(100)  0.25% 14. 0.995  0.995(100)  99.5% 15.

0.09  0.09(100)  9%

16.

3  3(100)  300%

17.

3  0.75(100)  75% 4

18.

3  0.12(100)  12% 25

19.

5 &  1.666667(100)  166.6% 3

20.

7  0.035(100)  3.5% 200

21.

9  0.045(100)  4.5% 200

22.

5  0.625(100)  62.5% 8

23.

3  0.0075(100)  0.75% 400

24.

5 &  0.833333(100)  83.3% 6

25.

9  0.01125(100)  1.125% 800

26.

7 &  1.166667(100)  116.6% 6

27.

3  0.375(100)  37.5% 8

28.

11  0.275(100)  27.5% 40

29.

4 &  1.333333(100)  133.3% 3

30.

9  0.0225(100)  2.25% 400

31.

13  0.65(100)  65% 20

32.

4  0.8(100)  80% 5

Exercise 1.4 A. 1.

Total weight  1 1 3  2 3 4  1 5 8  3 5 6  1.3& 2.75  1.625  3.83& 9.5416&ounces Total selling value of 4 pieces  $1569  9.5416& $14,970.88

1 3 1 1 3 Total hours  15  13  18  21  22 2 4 2 4 4  15.5  13.75  18.5  21.25  22.75  91.75 Total cost of labour  91.75  25.75  $2362.56

6  224, 400 = $122, 400 11 3.75 Property tax  122, 400   $4590 100

Assessed value 

Retail value  $0.90  2700  $0.90  2700  $2430 3 Discount   2430  $911.25 8 Credit received  2430  911.25  $1518.75

64  $0.75  $ 48.00 1 45.00 54  83 ¢  54  $0.83&  3 27.00 72  $0.375  & & 42  $1.33  42  $1.3  56.00 Total  $176.00

96  $0.875 2 330  16 ¢  330  $0.16& 3 144  $1.75 240  $1.66& 240  $1.6& Total

 $ 84.00 

55.00

 252.00  400.00  $791.00

7. Assessment Quiz 1

Mark 7

Weight 5%

Contribution to Final Grade 3.5

Quiz 2 Quiz 3 Quiz 4 Test 1 Test 2 Test 3 Exam

7.25 9

5% 5% 20% 20% 20% 25% 100%

6.5 38 41 43

50 50 50

79%

3.625 4.5 15.20 16.40 17.20 19.75 80.175

Michael’s final grade in physics is 80% . (His teacher did not count Quiz 4.) B. 1.

1100 1.088  $1196.80 1600 1.197  $1915.20 1400 1.277  $1787.80 Total cost  $4899.80 Average cost per litre 

$4899.80  $1.195073 4100  $1.195

(a) 56  60  70  54  240

Average number of litres  240  4  60 (b) Total cost  56  $2.080  $116.48

60  $1.985  $119.10 70  $2.122  $148.54 54  $2.075  $112.05 $496.17 Average cost per litre  $496.17  240  $2.067375  $2.067 (c) Average cost per km  $2.067375  8.75  $0.236271  $0.236

 3  4  5  2  2  6  4  2  4 1 2  6  12  10  12  8  4  12  58 Total hours  3  5  2  4  4  2  20 58 Grade-point average   2.9 20

Weighted hours

Weighted investment:

January1  February 28 : $7500  2  $15, 000 March1  July 31: 6600  5  33, 000 August1  August 31: 8100 1  8100 September1  December 31: 7800  4  31, 200 $87,300 Average investment balance  $87,300  12  $7275 5.

(a) Simple average of unit prices



10.00  10.60  11.25  9.50  9.20  12.15 62.70   $10.45 6 6

(b) Number of units purchased  Date

Amount invested Unit price

Amount Invested

Unit Price

Number of Units Purchased

February 1

200

10.00

200  20.000 10

March 1

200

10.60

200  18.868 10.60

April 1

200

11.25

200  17.778 11.25

May 1

200

9.50

200  21.053 9.50

June 1

200

9.20

200  21.739 9.20

July 1

200

12.15

200  16.461 12.15

Total number of units purchased (c) Average cost of units purchased 

1200  $10.35 115.899

(d) Value on July 31  115.899(11.90)  $1379.20

Exercise 1.5 A. 1.

(a) Annualsalary  $43, 056

Semi-monthly payment 

43, 056  $1794 24

43, 056  $828 52 828 Hourly rate   $23 36

(b) Weekly pay 

= $ 1794.00 Overtime pay  11 23 1.5  379.50 Gross pay

 $2173.50

115.899

(a) Annual salary  $43,875

Biweekly pay 

43,875  $1687.50 26

1687.50  $843.75 2 843.75 Hourly rate   $22.50 37.5

(b) Weekly pay 

 $1687.50 Overtime pay  8  22.50  1.5  270.00

 $1957.50

Gross pay 3.

(a) Monthly pay  $2657.20

Yearly pay  2657.20 12  $31,886.40 Weekly pay  31,886.40  52  $613.20 Hourly rate of pay  613.20  35  $17.52  $2657.20 Overtime pay  7.75 17.52 1.5  203.67

(b) Regular pay for May

 $2860.87

Gross pay 4.

(a) Semi-monthly pay  $1586

Yearly salary  1586  24  $38, 064 Weekly gross pay  38, 064  52  $732 Hourly rate  732  40  $18.30 (b) Gross pay Regular pay

 $1816.58  $1586.00

Overtime pay  $ 230.58 Number of overtime hours  ($230.58  1.5)  $18.30  8.4 5.

Total hours = 45 Regular hours = 40 Overtime hours = 5 At time-and-a-half, 5 overtime hours are equivalent to 5 1.5  7.5 regular hours

Rate of pay 

$917.70  $19.32 47.5

(a) Biweekly payment  $3942

Annual salary  3942  22  $86, 724 Daily pay  86, 724  200  $433.62 Hourly rate  433.62  7.5  57.816  $57.82  $3942.00 Less: two days  433.62  2  867.24

(b) Regular pay

 $3074.76

Gross pay 7.

Gross sales Less:returns

 $12, 660.00  131.20

Net sales  $12,528.80 Gross commission  12,528.80  0.0975  $1221.56 Less:drawings  720.00  $501.56

Amount due 8.

Net sales  $16, 244 1 Commission: 8 % on first $6000 4 3 9 % on next $6000 4 11.5% on $(16, 244  12, 000) Total commission

Gross sales Less:returns

 $24, 250  855

Net sales

 $23,395



$495.00



585.00



488.06

 $1568.06

Commission: 4.5% on first $10, 000  0.045 10, 000  $450.00 6% on next $5000  0.06  5000  300.00 8% on remaining $8395  0.08  8395  671.60 Total commission

 $1421.60

10. (a) Sales  $8125 Base salary on quota of $8500  $825

(b) Sales  $10,150

Base salary on quota of $8500  $825.00 1 Commission  6 % on $1650  0.065  $1650  107.25 2 Gross earnings  $932.25 11. (a) Commission at 6.5% on sales of $5830 = 0.065 × $5830 = $378.95. This is less than $540 guarantee, therefore weekly salary  $540 (b) Commission at 6.5% on sales of $8830  0.065  $8830  $573.95 This exceeds $540 guarantee, therefore weekly salary = $573.95 12. Gross sales Less: returns  3% of $31, 240

 $30,302.80

Net sales Rate of commission 

Commission: Sales for week Quota: Commission sales Rate of commission

Net sales 

1590.90  0.0525 30,302.80

 5.25%

 $566.25  450.00

13. Gross earnings Less: base salary

14.

 $31, 240.00  937.20

 $116.25  $6550  5000  $1550 116.25   0.075  7.5% 1550

$Commission $2036.88   $18,105.60 Rate 0.1125

Net sales  gross sales  returns 18,105.60  S  0.08S 0.92S  18,105.60 S  19, 680 Gross sales were $19,680

15. Gross earnings Less: Base salary

 $837.50  664.00

 $173.50 173.50 Commission sales   $1982.86 0.0875 Sales for week  $4800  $1982.86  $6782.86 Commission

16. Method A

Method B

Regular hours  40 17.60 Overtime pay  3.5 17.60 1.5 6 17.60  2

 $704.00  92.40  211.20

Gross earnings

 $1007.60

At regular rate: 49.5 17.60 Overtime premium: 3.5 17.60  0.5 6 17.60 1

 $ 871.20  30.80  105.60

Gross earnings

 $1007.60

Exercise 1.6 1. GST collected

GST paid 5% of purchases

GST payable

Month

5% of sales

January

$27,345

$7391.60

$19,953.40

February

12,200

3475.00

8725.00

March

29,400

43,300.00

(13,900.00)

April

32,515

22,500.00

10,015.00

May

7840

4904.90

2935.10

$109,300

$81,571.50

$27,728.50

5-month totals

(GST receivable)

Cook’s owes the government $27, 728.50.

Riza’s revenue of $28,350 includes 5% GST.

GST taxable revenue 

28,350  $27,000 1.05

GST collected  5% of $27,000  $1350 GST paid  5% of $8000  $400 Riza owes the Canada Revenue Agency $(1350  400)  $950

Savings on GST  5% of $780  0.05(780)  $39

Cost of shirt  $15.00 GST in Regina  5% of $15  0.05(15)  0.75 PST  6% of $15  0.06(15)  0.90  $16.65

Consumer pays 5.

At Blackcomb, B.C. Cost of ski pass

GST  5% of $214  0.05(214) PST  7% of $214  0.07(214)

 $214.00  10.70  

Amount paid at Blackcomb, B.C.

 $

At Mont Tremblant, Que. Cost of ski pass  $214.00 GST  5% of $214.00  0.05(214.00)  10.70 PST  9.975% of $214.00  0.09975(214.00)  21.35 Amount paid at Mont Tremblant

 $246.05

Difference  246.05  239.68

 $ 6.37

Total cost in Toronto Retail price HST  13% of $625  0.13(625) Total cost in Toronto

 $625.00  81.25  $706.25

Total cost in Calgary Retail price GST  5% of $625  0.05(625.00) PST 

 $625.00  31.25 nil

Total cost in Calgary

 $656.25

Difference  PST

$ 50.00

Purchase price of the first item = $70.56  0.25 = $17.64 Purchase price of the second item, including 5% GST = 70.56 – 17.64 = $52.92 Purchase price of the second item = $52.92 1.05  $50.40 GST paid = $52.92 – 50.40 = $2.52

 22.751  Property tax  125, 000    $2843.88  1000 

Property tax  307,500(0.019368)  $5955.66

2216  0.004626 479, 000 Semi-annual tax rate  0.004626(1000)  4.626305 The annual tax rate  2(4.626305)  9.252610

10. Semi-annual tax rate 

 9.25 mills 11. (a) Total expenditure  $(3, 050, 000  2, 000, 000  250, 000  700, 000  850, 000)  $6,850, 000

Total residential property tax  0.80(6,850, 000)  $5, 480, 000 (b) Tax rate per $1000 

5, 480,000 (1000)  10.96 500,000,000

 10.96  (c) Property tax  $375, 000    $4110 1000  

Business Math News Box

1. There are 52 weeks per year during which the employee works a 40-hour week. Total hours worked during the year is 52 × 40 = 2080. Hourly Rate Calculations Location Vancouver Calgary Toronto Montreal National Average

Financial Controller Human Resources Manager 99,500/2080 = $47.84 88,324/2080 = $42.46 106,082/2080 = $51.00 88,611/2080 = $42.60 98,500/2080 = $47.36 83,350/2080 = $40.07 99,758/2080 = $47.96 80,641/2080 = $38.77 99,234/2080 = $47.71 78,669/2080 = $37.82

Marketing Manager 78,663/2080 = $37.82 84,836/2080 = $40.79 77,823/2080 = $37.41 76,554/2080 = $36.80 75,450/2080 = $36.27

2. Dollar and percentage differences by job function:

Financial Controller National Average $99,234 National Average $99,234 National Average $99,234 National Average $99,234

–

Vancouver $99,500 Calgary $1,06,082 Toronto $98,500 Montreal $99,758

= = = = = = = =

$ difference $266 $ difference $6848 $ difference ($734) $ difference $524

– – – – – – – –

Vancouver $88,324 Calgary $88,611 Toronto $83,350 Montreal $80,641

= = = = = = = =

$ difference $9655 $ difference $9942 $ difference $4681 $ difference $1972

– – – – – – – –

Vancouver $78,663 Calgary $84,836 Toronto $77,823 Montreal $76,554

= = = = = = = =

$ difference $3213 $ difference $9386 $ difference $2373 $ difference $1104

– – – – – – –

= 266/99,234 = 6848/99,234 = –734/99,234 = 524/99,234

% difference 0.002681 0.27 % difference 0.069009 6.90 % difference –0.007397 –0.74 % difference 0.005280 0.53

Human Resources Manager National Average $78,669 National Average $78,669 National Average $78,669 National Average $78,669

= 9655/78,669 = 9942/78,669 = 4681/78,669 = 1972/78,669

% difference 0.122729 12.2 % difference 0.126378 12.6 % difference 0.059502 5.9 % difference 0.025067 2.5

Marketing Manager National Average $75,450 National Average $75,450 National Average $75,450 National Average $75,450

= 3213/75,450 = 9386/75,450 = 2373/75,450 = 1104/75,450

% difference 0.042584 4.26% % difference 0.124400 12.44% % difference 0.031451 3.15% % difference 0.014632 1.46%

3. Discrepancies between the national averages and specific metropolitan centres might be the result of many factors, including: - National average takes into account data supplied from all geographic locations. - Lack of supply and/or high demand for specific jobs in geographic locations might cause salaries to exceed the national average.

Review Exercise 1.

(a) 32  24  8  32  3  29 (b) (48 18) 15 10  30 15 10  2 10  8 (c) (8  6  4)  (16  4  3)  (48  4)  (16  12)  44  4  11 (d) 9(6  2)  4(3  4)  9(4)  4(7)  36  28  8 (e)

108 108 108    $1520.83 216 0.12  365 0.12  0.591781 0.071014

(f)

288 288 288    0.15 292 2400  365 2400  0.8 1920

225   (g) 320 1  0.10    320(1  0.061644)  320(1.061644)  339.73 365   150   (h) 1000 1  0.12    1000(1  0.049315)  1000(0.950685)  950.68 365  

(i)

660 660 660    625.45 144 1  0.14  365 1  0.055233 1.055233

(j)

1120 1120 1120    1250 292 1  0.13  365 1  0.104 0.896

(a) 185%  1.85 (b) 7.5%  0.075 (c) 0.4%  0.004 (d) 0.025%  0.00025

1 (e) 1 %  1.25%  0.0125 4 (f )

3 %  0.75%  0.0075 4

1 (g) 162 %  162.5%  1.625 2 3 (h) 11 %  11.75%  0.1175 4 1 8.3& (i) 8 %   0.083& 3 100 1 83.3& (j) 83 %   0.83& 3 100 2 266.6& (k) 266 %   2.6& 3 100

3 (l) 10 %  10.375%  0.10375 8 3.

(a) 50% 

50 1  100 2

1 37.5 375 3 (b) 37 %    2 100 1000 8 16 2 503 2 1 (c) 16 %  3  100  3 100 6 1 100  66 23 2 2 5  1  (d) 166 %  3 100 3 3 (e)

1 1 1 1 1 % 2    2 100 2 100 200

(f ) 7.5% 

7.5 75 3   100 1000 40

3 3 (g) 0.75%  %  4 400

(h)

5 5 1 %  8 800 160

(a) 2.25  2.25 100  225% (b) 0.02  0.02 100  2% (c) 0.009  0.009 100  0.9% (d) 0.1275  0.1275 100  12.75%

(e)

5 5  100  125% 4 4

(f )

11  1.375  1.375 100  137.5% 8

(g)

5  0.025  0.025 100  2.5% 200

(h)

7 28   28% 25 100

(a)

150% of 140  1.5 140  210

(b)

3% of 240  0.03  240  7.2

(c)

(d)

3 9 % of 2000 4  0.0975  2000  195 0.9% of 400  0.009  400  3.6

1 3 1 5 (a) 4  3  5  6 3 4 2 8 &  4.3& 3.75  5.5  6.625  20.2083kg (b) 20.2083&1.20  $24.25 (c) 20.2083& 4  5.052083& 5.05 kg

(d) 24.25  4  6.0625  $6.06 7.

56  $0.625



$ 35.00

1 180  83 ¢  180  $0.83& 3 126  $1.16&



$150.00



$147.00

144  $1.75



$252.00

Total



$584.00

(a)

30.45  20.20  16.40  15.50 82.55   20.6375  $20.64 4 4

(b) 30.45  2  20.20  6  16.40  9  15.50 13  30 

$ 60.90 $121.20 $147.60 $201.50 $531.20

Average rate 

$531.20  $17.71 30

January 1  May 31: 15, 000  5  $ 75, 000 June 1  July 31: 13, 000  2  26, 000 August 1  October 31: 11,500  3  34,500 November 1  December 31: 15,500  2  31, 000 12  $166,500 $166,500 Average monthly investment   $13,875 12 Total

10.

January 1  March 31: April 1  May 31: June 1  September 30: October 1  December 31:

12, 000  3  $ 36, 000 14, 400  2  28,800 12,960  4  51,840 15,840  3  47,520

12  $164,160 $164,160 Average monthly investment   $ 13, 680 12 Total

11. (a) Monthly remuneration 

34,944  $2192 12

(b) Weekly pay  34,944  52  $672 Hourly rate  672  35  $19.20

Regular gross pay Overtime pay

 3387.20  2912.00  475.20

Overtime hours  475.20  (19.20 1.5)  16.5

12. (a) Semi-monthly pay  31, 487.04  24  $1311.96 (b) Weekly pay  31, 487.04  52  $605.52 Hourly rate  605.52  36  $16.82

 

Overtime pay  12 16.82 1.5

$1311.96 $302.76

 $1614.72

Gross earnings 13. (a) Gross sales  11,160 Less: returns  120 Net sales  11, 040

Commission: 4% of $6000 8% of $3000 12.5% of $[11, 040  9000]

 $ 240  240  255

Gross earnings



$735

(b) Average hourly rate  735  43  $17.09 14. (a) Regular earnings  44 15.80  $ 695.20

Overtime pay  6.5 15.80 1.5  154.05 Gross earnings

 $849.25

(b) Overtime premium  6.5 15.80  0.5  $51.35 15. (a) Base salary on quota of $8000  $540.00

Commission  4.75% on $3340  158.65  $698.65

Gross earnings

(b) Hourly rate  698.65  35  $19.96 16. Gross earnings

Base salary

 $741.30  $675.00

Commission  $66.30 Commission sales  6560  5000  $1560 Rate of commission  66.30  1560  0.0425  4.25%

17.

Net sales  2101.05  0.105  $20, 010 Net sales  Gross sales  Returns 20, 010  Gross sales  8% of Gross sales 20, 010  92% of Gross sales 20, 010 Gross sales   $21, 750 0.92

18.

Hours worked  47 Regular hours  40 Overtime hours  7 7 overtime hours are equivalent to 7 1.5  10.5 regular hours. Total hours paid at regular rate  40  10.5  50.5 779.72 Hourly rate of pay   $15.44 50.5

19. (a) Annual salary  1413.75  24  $33,930

Weekly pay  33,930  52  $652.50 Hourly rate of pay  652.50  37.5  $17.40 (b) Gross earnings  $1552.55

Regular earnings  1413.75 Overtime pay  $138.80 Overtime hourly rate  17.40  1.5  $26.10 Overtime hours  138.80  26.10  5.318008  5.32 20. Gross earnings  $728.54 Less: base salary  $680.00

Commission  $ 48.54 Commission sales  48.54  0.06  $809 Net sales  5000  809  $5809 Gross sales  5809  136  $5945

 $731.92 21. Gross earnings Regular earnings  35 15.80  553.00 Overtime pay  $178.92 Overtime hours  178.92  (15.80  1.5)  7.549367 Number of hours worked  35  7.549367  42.55

22. GST collected  5% of $76, 000  0.05(76, 000)  $3800

 5% of $14,960  0.05(14,960) 

GST paid

GST remittance

748

$3052

23. GST collected:

Parts : 5% of $175, 000 Labour : 5% of $165, 650 Total : 5% of $340, 650 = 0.05(340, 650) GST paid : Parking fees : 5% of $ 2000 Supplies : 5% of $55, 000 Utilities : 5% of $ 4000 Other : 5% of $ 3300 Total :

 $17, 032.50

5% of $64,300 = 0.05(64,300) = $ 3215.00 GST remittance

$13,817.50

24. Amount paid in Kelowna, B.C.  Retail price  5% GST  7% PST = 1868  0.05(1868)  0.07(1868) = 1868  93.40  130.76  2092.16

Amount paid in Kenora, Ont.  Retail price  13% HST  1868  0.13(1868)  1868  242.84  2110.84 The difference = 2110.84  2092.16 = $18.68

 10.051  25. Property tax in Ripley  350, 000    $3517.85  1000   12.124  Property tax in Amberly  335, 000    $4061.54  1000 

The person in Amberly pays $543.69 more in property tax.

26. (a) Tax rate =

15,567,000 (1000)  15.957970 975,500,000

 15.957970  (b) Property tax = 435, 000    $6941.72  1000 

2, 000, 000 (1000)  2.050231 975,500, 000

 2.050231  (d) Additional property tax = 435, 000    $891.85  1000 

Self-Test 1.

45   (a) 4320 1  0.18    4320(1  0.022192)  4415.87 365   105   (b) 2160  0.15    2160(0.043151)  93.21 365   285   (c) 2880 1  0.12    2880(1  0.093699)  2610.15 365  

(d)

410.40 410.40   4623.33 0.24  135 0.088767 365

(e)

5124 5124   5489.46 270 1  0.09  365 0.933424658

(a) 175% 

(b)

175  1.75 100

3 3 1 3 %    0.00375 8 8 100 800

1 5 5 1 5 1   (a) 2 %  %   2 2 2 100 200 40 50 16 23 2 2 50 1 7 (b) 116 %  100%  16 %  1   1 3  1  1  3 3 100 100 300 6 6

(a) 1.125  1.125  100  112.5% (b)

9  0.0225  0.0225  100  2.25% 400

72  $1.25  $ 90.00 2 84 16 ¢ = 84  $0.16&  $ 14.00 3 40  $0.875  $ 35.00 48  $1.33& 48  $1.3&  $ 64.00 Total

$203.00

5  $9 6  $7 3  $8 6  $6

 $ 45  $ 42  $ 24  $ 36

Total 20

 $147

Average cost 

147  $7.35 20

1 3 5  1 Total size =  5  6  4  3  sq. metres 3 8 6  4 & & sq. metres  (5.25  6.3  4.375  3.83) = 19.7916&sq. metres Sales value = 25,120 19.7916& = $497,166.67

January 1  February 28 : March 1  July 31: August 1  September 30 : October 1  December 31:

7200  2  $14, 400 6720  5  33, 600 7320  2  14, 640 7440  3  22,320

Total

Average monthly balance = 9.

$84,960

84,960  $7080 12

Annual salary  2080  24 = $49,920 Weekly pay  49,920  52  $960 Hourly rate of pay  960  40  $24

10.

Net sales  0.885  5880  $5203.80 806.59 Commission rate   0.155  15.5% 5203.80

11.

Weekly pay  52,956.80  52 = $1018.40 Hourly pay  1018.40  38 = $26.80 Regular monthly pay  52,956.80  12  $4413.07 Overtime earnings  26.80  8.75 1.5  351.75  $4764.82

Gross pay

12. Total hours  8.25  8.25  9.5  11.5  7.25  44.75

Regular hours  8  8  8  8  7.25  39.25 Overtime hours  0.25  0.25  1.5  3.5  5.50 Regular pay  39.25  16.60  $651.55 Overtime pay  5.5  16.60  1.5  136.95  $788.50

Gross earnings 13.

Total hours  52.5 Regular hours  44.0 Overtime hours  8.5

At time-and-a-half, 8.5 overtime hours are equivalent to 8.5 1.5  12.75 regular hours

Hourly rate of pay  14.

983.15  $17.32 56.75

Base salary on first $4500  $600 Commission on next $2000 = 0.11 2000  220 Commission on additional sales = (8280  6500)  0.15  1780  0.15  267  $1087

Gross earnings

$6400  $20 GST 5% of $6420 $321.00 Manitoba PST 7% of $6420 449.40

15. Total value

Total purchase price

16.

Purchase price Less discount Net price Add shipping charge Total cost before taxes

 $6420.00 770.40 $7190.40

$ 17.95 2.50 $15.45 1.45 $16.90

HST

15% of $16.90  2.535

Final purchase price is 17.

18.

 $2.54 $19.44

= Assessed Value  Tax Rate 18 4502.50  Assessed Value  1000 4502.50(1000) Assessed Value =  $250,139 18 Property Tax

2 Assessed value   $390, 000  $260, 000 3 12.5 Property tax  $260, 000   $3250 1000

Challenge Problems 1.

Purchase price of the first item = $821.40  0.29 = $238.206 Purchase price of the second item, including 5% GST and 7% PST = $821.40 – 238.206 = $583.194 Purchase price of the second item = $583.194 / 1.12 = $520.708929 Total GST paid = $520.708929(0.05) = $26.035446 = $26.04 BC PST paid on second item = $520.708929(0.07) = $36.449625 = $36.45 BC PST paid on first item = ($238.206 / 1.07)(0.07) = $15.583570 = $15.58 Total BC PST paid = $36.45 + $15.58 = $52.03

2. Test score

Weight

Final grade contribution

Test 1

30%

60(0.30) = 18

Test 2

30%

50(0.30)  15

Final exam

40%

Final mark

Final exam contribution to final mark  70  (18  15)  70  33  37 37 Final examination mark required   92.5% 0.40 Case Study 1.

HST collected

13% of $280,000

$36,400

HST paid HST remittance

13% of $ 40,000

$ 5200 $31,200

(a) HST by Quick Method HST on sales = 185,000  0.13 = $24,050 Purchases: Goods for resale (185,000 × 47%) × 1.13 = $98,253.50 Other expenses (48,000 – 42,000) × 1.13 = Total taxable goods and expenses

6,780.00 105,033.50

Input tax credits = 13/113 × 105,033.50 =

$12,083.50

Remittance by Quick Method: $24,050.00 – 12,083.50 = $11,966.50 (b) HST by Standard Method HST collected 13% of $185,000

$24,050.50

HST paid on purchases and taxable services 13% of (47% of $185,000)

$11,303.50

13% of ($48,000 – $42,000)

780.00

Remittance by Standard Method

12,083.50 $11,966.50

Line 101 Line 103 Line 104 Line 105 Line 106 Line 107 Line 108 Line 109 Line 110 Line 111 Line 112 Line 113 Line 114 Line 115

13% of $486,530

13% of $239,690

63,248.90 – 31,159.70 3120 × (12)

32,089.20 – 37,440

$486,530.00 63,248.90 0.00 63,248.90 31,159.70 0.00 31,159.70 32,089.20 37,440.00 0.00 37,440.00 –5350.80 5350.80 0

Refund Claimed is $5350.80

Chapter 2

Review of Basic Algebra

Exercise 2.1 A. 1.

19a

a  10

3a 14

2 x  4 y

3p  q

14 f  4v

2c  3d

0.8x

10. 1.06x 11. 1.4x 12. 0.98x 13. 2.79x 14. 4.05 y 15.  x2  x  8 16. ax  x  2 17. 2 x  3 y  x  4 y  x  7 y 18. 4  5a  2  3a  2a  2 19. 12b  4c  9  8  8b  2c 15  4b  2c  2

20. a 2  ab  b2  3a 2  5ab  4b2  2a 2  6ab  5b2 21. 3m2  4m  5  4  2m  2m2  m2  6m  1 22. 6  4 x  3 y  1  5 x  2 y  9  14  9 x  y 23. 7a  5b  3a  4b  5b  10a 14b 24. 3 f  f 2  fg  f  3 f 2  2 fg  2 f  2 f 2  3 fg 25. 4b4 d  2ac 7  (5b 4 d )  3ac

 9b4 d  2ac7  3ac 26. (8t 2  6t  9)  (7t 2  6t  7)

 8t 2  6t  9  7t 2  6t  7  t 2  16 27.

18 y 12 1  3 y 2 5 4

 9 y  2.4  3.25 y  12.25 y  2.4 x 28. 1.3x  x 2   2 x  4 2

 x 2  (1.3  0.5  2) x  4  x 2  0.2 x  4 29.

k k  (1  0.05) (1  0.05)2  0.952381k  0.907029k  1.859410k

142  x  30. x 1  0.052   91  365    1  0.052   365  

B. 1.

12x

56a

10ax

27ab

2x2

24m 2

60xy

24abc

2 x  4 y

10. 10 x  20 11. 2ax 2  3ax  a 12. 24 x  12bx  6b2 x 13. 20 x  24  6  15x  35x  30 14. 24a  3b  14a  18b  10a  15b 15. 15ax  3a  5a  2ax  3ax  3a  20ax  5a 16. 24 y  32  4 y  2  1  y  21y  31 17. 3x 2  x  6 x  2  3x 2  5x  2 18. 5m2  2mn  15mn  6n2  5m2  17mn  6n2 19. x3  x2 y  xy 2  x2 y  xy 2  y3  x3  y3 20. a3  2a 2  a  a 2  2a  1  a3  3a 2  3a  1 21. 10 x 2  8x  5x  4  3x 2  21x  5x  35  7 x 2  3x  39

22. 2(2a 2  2a  3a  3)  3(3a 2  2a  3a  2)

 4a 2  10a  6  9a 2  3a  6  5a 2  13a  12 23. 3x 2 ( x 2  2 x  3)  4 x( x 2  2 x  3)  ( x 2  2 x  3)

 3x 4  6 x3  9 x 2  4 x3  8 x 2  12 x  x 2  2 x  3  3x 4  2 x3  16 x 2  14 x  3 24. (5b2  5b  5)(b3  4b  2)  5b2 (b3  4b  2)  5b(b3  4b  2)  5(b3  4b  2)  5b5  20b3  10b 2  5b 4  20b 2  10b  5b3  20b  10  5b5  5b 4  15b3  30b 2  10b  10

25. 4ab 26. 5 y 27. 4x 28. 6 29. 10m  4 30. 2 x  3 31. 2 x 2  3x  6 32. a 2  4a  3 C. 1. 3x  2 y  3  3(4)  2(5)  3  12  10  3  5 2.

1 1 (3x2  x  1)  (5  2 x  x 2 ) 2 4

1 1  [3(3) 2  (3)  1]  [5  2(3)  (3) 2 ] 2 4 1 1  (27  3  1)  (5  6  9) 2 4 1 1  (29)  (2) 2 4  14.5  0.5  14 3. ( pq  vq)  f  ( p  v)q  f  (12  7)2000  4500  10,000  4500  5500 4. F /C  13,000/0.65  20,000 5. (1  d1 )(1  d 2 )(1  d3 )  (1  0.35)(1  0.08)(1  0.02)  (0.65)(0.92)(0.98)  0.58604 6. C  0.38C  0.24C  (1  0.38  0.24)C  1.62C  1.62 ($25)  $40.50 7.

RP(n  1) 0.21 $1200  (77  1)   $378 2N 2  26

I 63 63    0.125 219 Pt 840  365 840  0.60

I $198 $198    $3000 146 rt 0.165  365 0.165  0.40

10.

2NC 2  52  60 2  52  60    0.13 P(n  1) 1800(25  1) 1800  26

76   11. P(1  rt )  $880 1  0.12   365  

 $880(1  0.024986)  $880(1.024986)  $901.99 256   12. FV(1  rt )  $1200 1  0.175   365  

 $1200(1  0.122740)  $1200(0.877260)  $1052.71 13.

P $1253 $1253 $1253     $1400.06 284 1  dt 1  0.135  365 1  0.105041 0.894959

14.

S $1752 $1752 $1752     $1600.08 228 1  rt 1  0.152  365 1  0.094948 1.094948

t   15. S 1  r  365  

for

S  3240, r  0.125, t  290

  290    3240 1  (0.125)    365     3240 (1.099315)  3561.780822 16. (SP  X )  FC  (VC  X ) for SP  13, X  125, FC  875, VC  L

 (13 125)  875  (4 125)  1625  875  500  250 17. (1  i)m  1 for i  0.0275, m  2  (1  0.0275) 2  1  1.055756  1

 0.055756  (1  i ) n  1  PmT 18.   for PmT  500, i  0.025, n  2 i  

 (1  0.025) 2  1   500   0.025    0.050625   500   0.025   500 (2.025)  1012.50 19. 1  [(1  d1 )(1  d 2 )] for d1  0.15, d 2  0.04

 1  [(1  0.15)(1  0.04)]  1  [(0.85)(0.96)]  1  (0.816)  0.184

 ( FV )(i )  20.   for n  (1  i)  1 

FV  10, 000, i  0.0075, n  20

 (10, 000)(0.0075)    20  (1  0.0075)  1  75      0.161184   465.306319 Exercise 2.2 A. 1.

16 81

625 1296



1 64



8 27

0.25

10. 113.379904 11. 0.001 12. 335.544320 13. 1

14. 1 15.

1 9

16. 512 17. 

18.

1 125

1 167.9616

19. 125 20.

81 16

21.

1 1.01

22. 1 23. 11.526683 24.

1 1  1 0 (1.07) 1

25.

1 1   0.781198 10 (1  0.025) 1.280085

26. 100(1  0.0225)7  100(1.168539)  116.853901 27. 425(1  0.16)4  425(0.552291)  234.723717 0.5

 1500  28.    1  2.738613  1  1.738613  200  (1  0.03) 29. 0.03



2.093778  69.792598 0.03

1  (1.01) 20  1  0.819544 0.180456    18.045553 30.  0.01  0.01 0.01 

B. 1. 25  23  253  28 2. (4)3  (4)  (4)31  (4)4 3. 47  44  474  43 4. (3)9  (3)7  (3)97  (3) 2 5. (23 )5  235  215 6

6. (4)3   (4)36  (4)18 7. a 4  a10  a 410  a14 8. m12  m7  m127  m5 9. 34  36  3  3461  311 10. (1)3 (1)7 (1)5  (1)375  (1)15

67  63 11.  6 7  3 9  6 9 6 12.

( x 4 )( x5 )  x 4 57  x 2 7 x 4

3 3 3 13.        5 5 5 5

47

1 1 1 14.         6 6 6 6

311  11 5

5 3

1 62



1 6  4

 3  3   3   3  15.              2  2   2   2  8

87

 3  3  3 16.             4  4  4

 

(3)11  211

3 4

17. (1.025)80 (1.025)70  (1.025)8070  1.025150 18. 1.005240 1.005150  1.005240150  1.00590

19. 1.0420   1.04204  1.0480 3

 3 5   3 53 315 20.          15 7  7    7  21. (1  i)100 (1  i)100  (1  i)100100  (1  i)200 22. (1  r )2 (1  r )2 (1  r )2  (1  r )222  (1  r )6 2

23. (1  i)80   (1  i)802  (1  i)160 3

24. (1  r )40   (1  r )403  (1  r )120 25. (ab)5  a5b5 26. (2 xy)4  16 x4 y 4 27. (m3n)8  m24 n8 4

 a3b2  a12b8 28.    x4  x  29. 23  25  24  2354  24 30. 52  53  52( 3)  55 8

b8 a 31.    8 a b

 1 i  32.    i 

n

in  (1  i)n

Exercise 2.3

5184  72.0000

A. 1.

205.9225  14.3500

2. 3.

2187  3.0000

1.1046221  1.0100

4.3184  1.075886  1.0759

0.00001526  0.500002  0.5000

1.0825  1.0133

1.15  1.011715  1.0117 1

B. 1. 3025 2  55 1

2. 24014  7 2

3. 525.21875 5  12.25 4

4. 21.6 3  60.154991 5.

1.1257  1.071122

1.095  1.015241  1  

7. 4 3  

1 4

8. 1.06

1 3

 1    12 



1  0.629961 1.587401



1 1.06

1 12



1  0.995156 1.004868

1.0360  1 5.891603  1   163.053437 0.03 0.03

10.

1  1.0536 1  0.172657   16.546852 0.05 0.05

11. 2.158925 12. 0.589664  3.536138  1  13. 26.50(1.043)    26.50(1.043)(58.979962)  1630.176673 0.043  

 2.653298  1  14. 350 (1.05)    350 (1.05)(33.065954)  12,151.73813 0.05  

 1  0.520035  15. 133    133(8.570795)  1139.915716 0.056    1  0.759412  16. 270    270 (6.873956)  1855.967995 0.035    1  0.581251  17. 5000 (0.581251)  137.50    0.0275 

 2906.252832  137.50(15.227252)  2906.252832  2093.747168  5000  1  0.623167  18. 1000 (0.623167)  300   0.03  

 623.166939  300(12.561102)  623.166939  3768.330608  4391.497547 19. 112.55  100(1  i) 4

(1  i ) 4  1.1255 (1  i )  1.12550.25 (1  i )  1.029998 i  0.029998 20. 380.47  300(1  i)12

(1  i )12  1.268233 (1  i )  1.2682330.083 (1  i )  1.019999 i  0.019999

21. 3036.77  2400(1  i)6

(1  i )6  1.265321 (1  i )  1.2653210.16 (1  i )  1.04 i  0.04 22. 1453.36  800(1  i)60

(1  i )60  1.8167 (1  i )  1.8167 0.016 (1  i )  1.01 i  0.01 Exercise 2.4 A. 1. 29  512 9  log 2 512

2. 37  2187 7  log 3 2187

3. 53 

1 125

3  log5

1 125

4. 105  0.00001 5  log10 0.00001

5. e2 j  18

2 j  log e 18 or 2 j  ln 18 6. e3 x  12

 3x  log e 12 or 3x  ln 12 B. 1. log 2 32  5

25  32

2. log3

1  4 81 34 

1 81

3. log10 10  1

101  10 4. ln e 2  2

e2  e2 C. 1. ln 2  0.693147 2. ln 200  5.298317 3. ln 0.105  2.253795 4. ln 300(1.1015 )   ln 300  ln 1.1015

 ln 300  15(ln 1.10)  5.703782  15(0.095310)  5.703782  1.429653  7.133435  2000   ln 2000  ln 1.099 5. ln  9 1.09 

 ln 2000  9(ln 1.09)  7.600902  9(0.086178)  7.600902  0.775599  6.825303 1.01120   ln 850  ln 1.01120  ln 0.01 6. ln 850    0.01 

 ln 850  120(ln 1.01)  ln 0.01  6.745236  120(0.009950)  (4.605170)  6.745236  1.194040  4.605170  10.156367

Business Math News Box 1.

Total amount invested  $1200 Total number of shares purchased  10 + 10.225 + 9.615 + 10.395 + 9.524 + 9.302 + 10.132 + 9.302  9.009  8.696  8.849  8.888  113.937 Average cost per share 

1200  $10.53 113.937

$10.53 is less than the current $11.25 cost per share. 2. Number of shares purchased  Share price  Amount invested 10 shares  $17  $170 a  16 15 shares   240  b 16.50  20 shares  330 c $740 Average cost per share 

740  $16.44 45

3. Amount invested  Share price  Number of shares purchased $5000  $25  200 156.25 5000  32  5000





250 606.25

Average cost per share 

15, 000  $24.74 606.25

The first $5000 allocation purchased 200 shares at $25 per share. The second $5000 allocation only bought 156.25 shares because the price rose to $32 per share in the second month. The third $5000 allocation bought 250 shares at $20 per share. After three months, the couple owned 606.25 shares at an average cost of $24.74. Their investment is worth $15,156.25. (i.e., 606.25 shares  $25 current value). If they had invested $15,000 all at once, they would only have 600 shares. At the current share price, their investment would only be worth $15,000, the same as the original lump sum. 4. Answers will vary. However, markets tend to go up in the long term.

Exercise 2.5 A. 1. 15 x  45 x  3

2. 7 x  35

x  5 3. 0.9 x  72

x  80 4. 0.02 x  13

x  650 5.

1 x3 6 x  18

1 6.  x  7 8

x  56 7.

3 x  21 5 1 x  7 5 x  35

4 8.  x  32 3 1 x8 3 x  24

9. x  3  7

x  4 10. 2 x  7  3x

x7 11. x  6  2

x  8

12. 3x  9  2 x

x9 13. 4  x  9  2 x

x5 14. 2 x  7  x  5

x  12 15. x  0.6 x  32

1.6 x  32 x  20 16. x  0.3x  210

0.7 x  210 x  300 17. x  0.04 x  192

0.96 x  192 x  200 18. x  0.07 x  64.20

1.07 x  64.20 x  60 B. 1. 3x  5  7 x 11

4 x  16 x4 LS: 3 x  5  3(4)  5  12  5  17 RS: 7 x  11  7(4)  11  28  11  17

2. 5  4 x  4  x

 3x  9 x3 LS: 5  4 x  5  (4)(3)  5  12  7 RS:  4  x  4  3  7

3. 2  3x  9  2 x  7  3x

 3x  7  5 x  7  8x  0 x0 LS:  2  3x  9  2  3(0)  9  7 RS:  2 x  7  3x  2(0)  7  3(0)  7 4. 4 x  8  9 x  10  2 x  4

 5x  8  6  2 x  7 x  14 x  2 LS:  4 x  8  9 x  4(2)  8  9(2)  8  8  18  2 RS:  10  2 x  4  10  2(2)  4  10  4  4  2 5. 3x  14  4 x  9

 x  5 x5 LS:  3x  14  3(5)  14  15  14  29 RS:  4 x  9  4(5)  9  20  9  29

6. 16 x 12  6 x  32

10 x  20 x  2 LS:  16 x  12  16(2)  12  32  12  44 RS:  6 x  32  6(2)  32  12  32  44 7. 5  3  4 x  5x  12  25

4 x  5 x  12  25  5  3  x  21 x  21 LS:  5  3  4 x  8  4(21)  8  84  92 RS:  5 x  12  25  5(21)  13  105  13  92 8. 3  2 x  5  5x  36  14

 2 x  5 x  36  14  3  5  3x  24 x 8 LS:  3  2 x  5  3  2(8)  5  3  16  5  18 RS:  5 x  36  14  5(8)  36  14  40  36  14  18

x  50  100  0.34 x  0.21x x  50  100  0.55 x x  0.55 x  150 (1  0.55) x  150 0.45 x  150

x  333.3 CHECK: L.S.

R.S.

333.3  50

100  0.34(333.3)  0.21(333.3)

 283.3

 100  113.3  70  283.3

10.



23 x 32

6 12 x 8  x 1.125 9 0.8 x  0.8 x  0

1  0.25 

x  all real numbers CHECK: L.S. 1

1  0.25 

R.S. 23 (1) 32 8  (1) 9  0.8

6 12

1 1.125  0.8 

Exercise 2.6 A. 1. 12 x  4(9 x  20)  320 12 x  36 x  80  320

 24 x  240 x  10 LS  12(10)  4[9(10)  20]  120  4[90  20]  120  440  320 RS  320

2. 5( x  4)  3(2  3x)  54 5 x  20  6  9 x  54

14 x  26  54 14 x  28 x  2

LS  5[2  4]  3[2  3(2)]  5(6)  3(2  6)  30  24  54 RS  54 3. 3(2 x  5)  2(2 x  3)  15

6 x  15  4 x  6  15 2 x  9  15 2 x  6 x  3 LS  3[2(3)  5]  2[2(3)  3]  3[65]  2[6  3]  3(11)  2(9)  33  18  15 RS  15 4. 17  3(2 x  7)  7 x  3(2 x  1)

17  6 x  21  7 x  6 x  3  6 x  38  x  3  7 x  35 x5 LS  17  3[2(5)  7]  17  3[10  7]  17  9  8 RS  7(5)  3[2(5)  1]  35  3[10  1]  35  27  8 5. 4 x  2(2 x  3)  18

4 x  4 x  6  18 8 x  24 x3

LS  4(3)  2[2(3)  3]  12  2[6  3]  12  6  18 RS  18 6. 3(1  11x)  (8 x  15)  187

 3  33x  8 x  15  187 33x  8 x  187  3  15 41x  205 x5 LS  3[(1  11(5)]  [8(5)  15]  3[54]  25  162  25  187 RS  187 7. 10 x  4(2 x  1)  32

10 x  8 x  4  32 2 x  28 x  14 LS  10(14)  4[2(14)  1]  140  4[27]  140  108  32 RS  32

8. 2( x  4)  12(3  2 x)  8

 2 x  8  36  24 x  8  2 x  24 x  8  8  36  26 x  52 x2 LS  2(2  4)  12[3  2(2)]  4  8  36  48  8 RS  8

65   9. x 1  0.12    1225.64 365   1.021370 x  1225.64 x  1199.996245

x  1200 CHECK:

L.S.

65   1200 1  0.12   365    1200 1.021370 

R.S. 1225.64

 1225.643836 10. x 

x x 1000   3148  2 1.25 (1.25) (1.25)3

x  0.80 x  0.64 x  3148  512 2.44 x  3660

x  1500 CHECK: L.S. 1500 1500 1500   1.25 (1.25) 2  1500  1200  960

3660

 3660 B. 1.

1 x  x  15 4 4 x  x  60 3 x  60 x  20

5 x  x  26 8 8 x  5 x  208 13x  208 x  16

2 1 7 5 x   x 3 4 4 6

8 x  3  21  10 x 18 x  18 x  1 4.

5 2 1 1  x  x 3 5 6 30

50  12 x  5 x  1  17 x  51 x3 5.

3 113 2 x4  x 4 24 3

18 x  96  113  16 x 34 x  17 x

1 2

3 2 31 2 x  x 2 3 9

36  27 x  12 x  62  39 x  26 x

2 3

1 2 1  x  15  x 3 3 2   1   3 1  x   3 15  x  3   3   3  x  45  2 x 42  3x x  14

3x  2 2 x  1  5 3 3(3 x  2)  5(2 x  1) 9 x  6  10 x  5

x  1 9.

21 2 11 1  x  x 8 5 4 10 1  21 2  11 40   x   40  x   10  8 5  4 105  16 x  110 x  4 109  126 x x

10.

109 126

2 1 3 1  x x   x 3 12 4 24 1   2 3 1  24   x  x   24   x  12   3  4 24  16 x  2 x  18  x 18 x  18  x 18  19 x x

C. 1.

18 19

3 1 55 (2 x  1)  (5  2 x)   4 3 12 9(2 x  1)  4(5  2 x)  55 18 x  9  20  8 x  55 26 x  29  55 26 x  26 x  1

4 53 3 7 (4  3x)   x  (2 x  3) 5 40 10 8

32(4  3x)  53  12 x  35(2 x  3) 128  96 x  53  12 x  70 x  105  96 x  181  58 x  105  38 x  76 x2

2 3 20 (2 x  1)  (3  2 x)  2 x  3 4 9

24(2 x  1)  27(3  2 x)  72 x  80 48 x  24  81  54 x  72 x  80 102 x  105  72 x  80 30 x  25 x

5 6

4 3 11 (3x  2)  (4 x  3)   3x 3 5 60

80(3x  2)  36(4 x  3)  11  180 x 240 x  160  144 x  108  11  180 x 96 x  52  11  180 x  84 x  63 x

3 4

2 3 (5 x  1)   ( x  2) 3 5 2   3  15  (5 x  1)   15   ( x  2)  3   5  10(5 x  1)  9( x  2) 50 x  10  9 x  18 59 x  8 x

8 59

y  mx  b

D. 1.

y  b  mx x

r

y b m

M S

Sr  M S

M r

PV 

PMT i

PMT  PVi 4.

I  P rt t

I Pr

S  P(1  rt ) S  1  rt P S  1  rt P S 1 r P t SP r P t SP r Pt

PV  FV(1  i) n

PV  (1  i )  n FV 

 PV  n  FV   1  i 1

 FV  n  PV   1  i 1

 FV  n i 1  PV 

S for t (1  rt ) P(1  rt )  S P  Prt  S Prt  S  P SP t Pr

7. P 

N  L(1  d ) for d N  L  dL dL  L  N

d

LN L

f  (1  i ) m  1 for i (1  f )  (1  i) m 1

(1  f ) m  ((1  i) m ) m 1

(1  f ) m  1  i 1

i  (1  f ) m  1 10. FV  PV (1  i)n

for n

FV  (1  i ) n PV  FV  ln    n ln (1  i )  PV 

 FV  ln   PV  n  ln 1  i  Exercise 2.7 1. Let the cost be $x. 3   Selling price  $  x  x  4  

3 x  49.49 4 4 x  3x  197.96 7 x  197.96 x  28.28

x 

The cost was $28.28.

2. Let the regular selling price be $x. 1   Sale price  $  x  x  3  

1  x  x  576 3 3x  x  1728 2 x  1728 x  864 The regular selling price was $864. 3. Let the price be $x. Total  $ x  0.05x  x  0.05 x  $57.75 1.05 x  $57.75 x  55

The price was $55. 4. Let the regular price be $x. Sale price  $( x  0.40 x)  x  0.40 x  11.34 0.60 x  11.34 x  18.90

The regular selling price was $18.90. 5. Let the last month’s index be x. This month’s index  x 

1 x 12

1 x  176 12 12 x  x  2112 11x  2112 x  192

x 

Last month the index was 192.

6. Let the original hourly wage be $x. 1   New hourly wage  $  x  x  8  

1  x  x  15.75 8 8 x  x  126 9 x  126 x  14 The hourly wage before the increase was $14. 7. Let Vera’s sales be $x. Tai’s sales  $(3x  140) Total sales  $( x  3x  140)  x  3 x  140  940 4 x  1080 x  270

Tai’s sales  3  270   140  $670. 8. Let the shorter piece be x cm. Length of longer piece  (2 x  15) cm. Total length  ( x  2 x  15) cm  x  2 x  15  90 3 x  75 x  25

The longer piece is 2(25) cm + 15 cm  65 cm. 9. Let the cost of a ticket be $x. Total  $( x  18.40) 1.05  2

 ( x  18.40) 1.05  2  345.24 ( x  18.40)  2.10  345.24 ( x  18.40)  164.40 x  146

The cost per ticket is $146.

10. Let Ken’s investment be $x. 2  Martina’s investment  $  x  2500  3  2   Total investment  $  x  x  2500  3   x 

2 x  2500  55, 000 3 5x  52,500 3 x  31,500

Martina’s investment is

2  31,500  2500  $23,500. 3

11. Let the number of chairs produced by the first shift be x. Number of chairs produced by the second shift 

4 x  60. 3

4 Total production  x  x  60  2320. 3 x 

4 x  60  2320 3 7 x  2380 3 x  1020

Production by the second shift is

4 1020  60  1300. 3

12. Let the number of type A lights be x. Number of type B lights  60  x. Value of type A lights  $40 x. Value of type B lights  $(60  x)50.

 40 x  50(60  x)  2580 40 x  3000  50 x  2580  10 x  420 x  42

The number of type B lights is 18.

13. Let the number of units of Product A be x; then the number of units of Product B is 60  x. The number of hours for Product A is 4x; The number of hours for Product B is 3(60  x).  4 x  3(60  x)  200 4 x  180  3 x  200 x  20

Production of Product A is 20 units. 14. Let the number of dimes be x. Number of nickels  3x  4 Number of quarters 

3 x 1 4

Value of the dimes  10x cents Value of nickels  5(3x  4) cents 3  Value of quarters  25  x  1 cents 4 

3  10 x  5(3 x  4)  25  x  1  880 4  75 10 x  15 x  20  x  25  880 4 75 25 x  x  875 4 175 x  3500 x  20 Alick has 20 dimes, 56 nickels, and 16 quarters.

15. Let the number of $12 tickets be x. Number of $8 tickets  3x  10 Number of $15 tickets 

4 x 3 5

Value of the $12 tickets  $12x Value of the $8 tickets  $8(3x  10) 4  Value of the $15 tickets  $15  x  3  5 

4  12 x  8(3x  10)  15  x  3   1475 5  12 x  24 x  80  12 x  45  1475 48 x  1440 x  30 30 $12 tickets, 100 $ 8 tickets, Sales were and 21 $15 tickets. 16. Let the number of medium pizzas be x. Number of large pizzas  3x  1 Number of small pizzas  2 x  1 Value of medium pizzas  $15x Value of large pizzas  $18(3x  1) Value of small pizzas  $11(2 x  1)

15 x  18(3x  1)  11(2 x  1)  539 15 x  54 x  18  22 x  11  539 91x  546 x6 6 medium pizzas, 17 large pizzas, Sales were and 13 small pizzas.

17. Let the taxable income (in dollars) be x. Then x – 50,197 is the amount by which his income exceeds $50,197. 7529.55 + 0.205 (x – 50,197 ) = 10,121.16 7529.55 + 0.205x – 10,290.385 = 10,121.16 0.205x = 12,881.995 x = $62,839 His taxable income is $62,839. 18. Let the amount invested at 3% be $x. Then the amount invested at 4.5% is (3000 – x). 0.03x + 0.045 (3000 – x) = 128.25 0.03x + 135 – 0.045x = 128.25 –0.015x  – 6.75 x  $450 at 3% 3000 – 450 = $2550 at 4.5% 19. Let x be the number on the second shift. Then 2x is the number on the second shift. And x  12 is the number on the third shift. x + 2x + (x – 12) = 196 4x – 12 = 196 4x = 208 x = 52 on the second shift 2x = 2(52) = 104 on the first shift x – 12 = 52 – 12 = 40 on the third shift 20. Let x be the number of options received by each employee. Then 1.5x is the number received by each team leader. And 3x is the number received by each senior manager. 421x + 22(1.5x) + 7(3x) = 171,000 Copyright © 2025 Pearson Canada Inc.

421x + 33x + 21x = 171,000 475x = 171,000 x = 360 options for each employee 1.5x = 1.5(360) = 540 options for each team leader (2)(540) = 1080 options for each senior manager Check: 421(360) + 22(540) + 7(1080) = 171,000

21. Let the amount of money spent on recreational players be $x. If twice as much money was spent on rep players, then the amount spent on recreational players can be determined by $x + $2x = $4320 $3x = $4320 $x = $1440 And therefore, the amount spent on rep players was $4320  $1440 = $2880. Let the number of Youth Large shirts purchased for recreational players be y. $10y + $8(50) + $8(50) = $1440 $10y + $400 + $400 = $1440 $10y = $640 y = 64 64 Youth Large shirts were purchased for recreational players. Let the number of Adult Small and Adult Medium shirts be z. For rep players, the cost of shirts is given by $8(50 – 10) + $10(3 × 64) + $16z + $16z = $2880 Therefore, the number of shirts of each Adult size ordered can be calculated as $320 + $1920 + $16(2z) = $2880 $2240 + $16(2z) = $2880 $16(2z) = $640 2z = 40 z = 20 20 Adult Small and 20 Adult Medium shirts were purchased for rep players. (50 + 50 + 64) + (40 + 192 + 20 + 20) = 436 There are a total of 436 players in the organization.

Review Exercise 1. (a) 3 x  4 y  3 y  5 x  2 x  7 y (b) 2 x  0.03x  1.97 x (c) (5a  4)  (3  a)  5a  4  3  a  6a  7 (d) (2 x  3 y )  (4 x  y )  ( y  x)  2 x  3 y  4 x  y  y  x  x  3 y (e) (5a 2  2b  c)  (3c  2b  4a 2 )

 5a 2  2b  c  3c  2b  4a 2  9a 2  4b  4c (f ) (2 x  3)  ( x 2  5 x  2)  2 x  3  x 2  5 x  2   x 2  3x  1 2. (a) 3(5a)  15a (b) 7m(4 x)  28mx (c) 14m  (2m)  7 (d) (15a 2b)  (5a)  3ab (e) 6(3 x)(2 y )  36 xy (f ) 4(3a)(b)(2c)  24abc (g) 4(3 x  5 y  1)  12 x  20 y  4 (h) x(1  2 x  x 2 )  x  2 x 2  x3 (i) (24 x  16)  (4)  6 x  4 (j) (21a2 12a)  3a  7a  4 (k)

4(2a  5)  3(3  6a)  8a  20  9  18a  26a  29

(l)

2a( x  a)  a(3x  2)  3a(5 x  4)

 2ax  2a 2  3ax  2a  15ax  12a  14ax  2a 2  10a (m)

(m  1)(2m  5)  2m2  2m  5m  5  2m 2  7 m  5

(n)

(3a  2)(a 2  2a  3)  3a3  2a 2  6a 2  4a  9a  6  3a3  8a 2  5a  6

(o)

3(2 x  4)( x  1)  4( x  3)(5 x  2)  3(2 x 2  4 x  2 x  4)  4(5 x 2  15 x  2 x  6)  6 x 2  18 x  12  20 x 2  52 x  24  14 x 2  34 x  36

(p)

 2a(3m  1)(m  4)  5a(2m  3)(2m  3)  2a(3m2  m  12m  4)  5a(4m2  6m  6m  9)  6am2  26am  8a  20am 2  45a  26am2  26am  37a

(a) for x  2, y  5,

3xy  4 x  5 y  3(2)(5)  4(2)  5(5)  30  8  25  47

1 2 (b) for a   , b  , 4 3  5(2a  3b)  2(a  5b)  10a  15b  2a  10b  12a  5b 1 1  1 2  12     5    3  3  6 3 3  4 3

2NC 2 12  432 12  48 16     0.16 P(n  1) 1800  (35  1) 100  36 100 (d) for I  600, r  0.15, P  7300,

365 I 365  600 2    200 rP 0.15  7300 0.01

(e) for A  $720, d  0.135, t 

280 , 365

280   A(1  dt )  $720 1  0.135    $720(1  0.103562)  645.435616  $645.44  365 

(f ) for S  2755, r  0.17, t 

219 , 365

S 2755 2755 2755     2500 219 1  rt 1  0.17  365 1  0.034  3 1  0.102

4. (a) (3)5  243 4

16 2 (b)    81 3 (c) (5)0  1 (d) (3)1  

1 3

4

625 2 5 (e)       16 5 2 (f ) (1.01)0  1 (g) (3)5 (3)4  (3)9  19,683 (h) 47  42  45  1024 5

(i) (3)2   (3)10  59,049 (j) (m3 )4  m12 3

6

16 2 2 2 2 (k)           81 3 3 3 3 5

25  5  5  5 (l)             16  4  4  4

(m) (1.0350 )(1.03100 )  1.03150 (n) (1  i)180  (1  i)100  (1  i)80 5

(o) (1.05)30   1.05150 (p) (2 xy)4  16 x4 y 4 4

 a 2b  81  3  (q)    2   8 4 ab a b  3  (r) (1  i)  n 

5. (a)

1 (1  i) n

0.9216  0.96

(b) 6 1.075  1.012126 (c) 14.9744581/ 40  1.07 (d) 1.085/12 

1  0.968442 1.085/12

(e) ln 3  1.098612 (f ) ln 0.05  2.995732 (g) ln(5.1) / ln(1.015)  1.629241 / 0.014889  109.428635

 5500   ln 5500  ln 1.1016 (h) ln  16  1.10  

 ln 5500  16ln 1.10  8.612503  16(0.095310)  8.612503  1.524963  7.087540

  1  1.0172   72 (i) ln 375(1.01)     ln 375  ln 1.01  ln (1  1.01 )  ln 0.01  0.01   

 ln 375  ln 1.01  ln (1  0.488496)  ln 0.01  ln 375  ln 1.01  ln 0.511504  ln 0.01  5.926926  0.009950  0.670400  (4.605170)  9.871647

6. (a) 9 x  63

x  7 (b) 0.05x  44

5 x  4400 x  880 1 (c)  x  3 7

 x  21 x  21 (d)

5 x  15 6 1 x  3 6 x  18

(e)

x  8  5 x  8  8  5  8 x3

(f )

x  9  2 x  9  9  2  9 x  11

(g) x  0.02 x  255

1.02 x  255 x  250 (h) x  0.1x  36

0.9 x  36 9 x  360 x  40 (i) 4 x  3  9 x  2

 5x  5 x  1 (j) 9 x  6  3x  15  4 x  7

6x  6  8  4x 2 x  14 x7 1 (k) x  x  26 3

2 x  26 3 1 x  13 3 x  39 3 (l) x  x  77 8

11 x  77 8 1 x7 8 x  56 7. (a) 9(3x  8)  8(9  7 x)  5  4(9 x  11)

 27 x  72  72  56 x  5  36 x  44 29 x  49  36 x  7 x  49 x  7 Check LS  9[3(7)  8]  8[9  7(7)]  9(29)  8(58)  203 RS  5  4[9(7)  11]  5  4(52)  5  208  203 (b) 21x  4  7(5 x  6)  8 x  4(5 x  7)

21x  4  35 x  42  8 x  20 x  28  14 x  38  12 x  28  2 x  10 x5

Check LS  21(5)  4  7[5(5)  6]  105  4  7(19)  101 133  32 RS  8(5)  4[5(5)  7]  40  4(18)  40  72  32

(c)

5 1 5 2 x   x 7 2 14 3 5  1  5 2  42  x   42    42    42  x  7  2  14  3  6(5 x)  21(1)  3(5)  14(2 x) 30 x  21  15  28 x 2 x  6 x  3 5 1 30  7 23 Check LS  (3)    7 2 14 14 5 2 5 23 RS   (3)   2   14 3 14 14

(d)

4x 9 x 2  3 8 6

8(4 x)  24(2)  3(9)  4( x) 32 x  48  27  4 x 36 x  21 x

7 12

4 7  28 7 18 11 Check LS      2    2     3  12  36 9 9 9 9 1  7  9 7 81  7 88 11 RS           8 6  12  8 72 72 72 9 (e)

7 3 (6 x  7)  (7 x  15)  25 5 8 56(6 x  7)  15(7 x  15)  40(25) 336 x  392  105 x  225  1000 231x  617  1617 x7 7 3 Check LS  [6(7)  7]  [7(7)  15] 5 8 7 3  (35)  (64)  7(7)  24  49  24  25 5 8 RS  25

(f )

5 3 1 1 (7  6 x)  (3 15 x)  (3 x  5)  9 4 12 2

20(7  6 x)  27(3  15 x)  3(3 x  5)  18 140  120 x  81  405 x  9 x  15  18 285 x  59  9 x  33 276 x  92 x

Check LS 

1 3

5  1  3   1  7  6      3  15      9  3  4   3 

5 3  (7  2)  (3  5) 9 4  5  6  1 RS 

1   1  1 3    5  12   3   2

1 1 (6)  12 2 1 1     1 2 2 

(g)

5 2 16 (4 x  3)  (3x  4)  5 x  (1  3x) 6 5 15

25(4 x  3)  12(3 x  4)  150 x  32(1  3 x) 100 x  75  36 x  48  150 x  32  96 x 64 x  123  246 x  32  182 x  91 x

1 2

Check LS 

5  1  2  1  4     3  3    4 6   2   5   2  

5 2 3   (2  3)     4  6 5 2  5 25 25 31  (5)       1   6 52 6 6  1  16   1  RS  5     1  3      2  15   2  5 16  3  5 16  5     1        2 15  2  2 15  2  5 8 15  16 31     2 3 6 6 8. (a) I  P rt I Pt

r

(b)

S  P(1  rt ) S  1  rt P S  1  rt P S 1 P t r SP t P r SP t Pr

r

D L

é(1  p) n  1ù ú (d) FV  PMT ê ê ú p ë û é FVp ù ú PMT  ê êë(1  p) n  1ú û

9. Let the size of the workforce be x. Number laid off 

1 x 6

1 Number after the layoff  x  x 6

1  x  x  690 6 5 x  690 6 5 x  4140 x  828  the number laid off is

1  828  138. 6

10. Let last year’s average property value be $x. 2   Current average value  $  x  x  7  

2  x  x  346,162.50 7 9 x  346,162.50 7 1 x  38, 462.50 7 x  269, 237.50  Last year’s average value was $269, 237.50.

11. Let the quoted price be $x.

x 

1 x  $12,957 20 21 x  12,957 20 1 x  617 20

 The gratuities were $617.

12. Let the value of the building be $x.

1 Value of the land  $ x  2000 3 1 Total value of the property  $ x  x  2000 3

1  x  x  2000  790, 000 3 4 x  792, 000 3 1 x  198, 000 3 x  594, 000 The value assigned to land is $(790,000  594,000)  $196,000. 13. Let the cost of power be $x. 3  Cost of heat  $  x  22  4  1  Cost of water  $  x  11 3 

3 1 Total cost  x  x  22  x  11  2010  10% of 2010. 4 3 12 x  9 x  4 x  12(2010  201  11) 25 x  26 400 x  1056

3 Cost of heat  1056  22  $814 4 Cost of power  $1056

1 Cost of water  1056  11  $341 3

14. Let the amount allocated to newspaper advertising be $x. Amount allocated to TV advertising  $(3x  1000)

3 Amount allocated to direct selling  [ x  3x  1000] 4

3  x  3x  1000  [4 x  1000]  87,500 4 3 4 x  [4 x  1000]  86,500 4 16 x  12 x  3000  346, 000 28 x  343, 000 x  12, 250 The amount allocated to newspaper advertising is $12,250; the amount allocated to TV advertising is $37,750; the amount allocated to direct selling is $37,500. 15. Let the number of minutes on Machine B be x. Time on Machine A 

4 x  3 minutes 5

5 4  Time on Machine C   x  x  3  minutes 6 5 

Total time  x 

x

4 5 4  x  3   x  x  3  minutes 5 6 5 

4 5 4  x  3   x  x  3   77 5 6 5 

4   30 x  24 x  90  25  x  x  3   30(77) 5   54 x  90  25 x  20 x  75  2310 99 x  165  2310 99 x  2475 x  25 Time on Machine B is 25 minutes; time on Machine A is time on Machine C is

5 (25  17)  35 minutes. 6

4 (25)  3  17 minutes; 5

16. Let the number of pairs of superlight poles be x. Number of pairs of ordinary poles  72  x Value of superlight poles  $130x Value of ordinary poles  $56(72  x) Total value of all poles  $130 x  56(72  x)

130 x  56(72  x)  $6030 130 x  4032  56 x  6030 74 x  1998 x  27 The number of pairs of superlight poles is 27; the number of pairs of ordinary poles is 45. 17. Let the number of $2 coins be x. Number of $1 coins 

3 x 1 5

 

 

3 5

Number of quarters  4  x  x  1 Value of the $2 coins  $2x

3 5

 

Value of the $1 coins  $  x  1

1 4

 

3 5

 

3 5

Value of the quarters  $ (4)  x  x  1  x  x  1 3 3 Total value  2 x  x  1  x  x  1  107 5 5 10 x  3x  5  5 x  3x  5  535 21x  10  535 21x  525 x  25

3 5

 

The number of $2 coins is 25; the number of $1 coins is   25  1  16; the number of quarters is 4(25  16)  164.

18. Let $x represent Jaime’s monthly savings. $2975 ¸ 2 = $1487.50 Jaime has $1487.50 after paying for school and transportation. 0.30(1487.50) + 900 + x  1487.50 1346.25 + x  1487.50 x  141.25 Jaime has $141.25 left over for savings each month. 19. Let x represent the total valuation of Baldwin Industries. Then Inspire Inc.’s stake is 0.49x and Crown Company’s stake is 0.24x. 0.80(0.49x) = $19,600,000 0.392x = $19,600,000 x = $50,000,000 0.24(50,000,000) = $12,000,000 Crown Company’s stake in Baldwin Industries is worth $12 Million. Self-Test 1. (a) 4  3x  6  5x  2  8x (b) (5x  4)  (7 x  5)  5x  4  7 x  5  2 x  9 (c)

2(3a  4)  5(2a  3)

 6a  8  10a  15  16a  7 (d)

 6( x  2)( x  1)

 6( x 2  2 x  x  2)  6( x 2  x  2)  6 x 2  6 x  12

2. (a) For x  3, y  5

2 x 2  5 xy  4 y 2  2(3)2  5(3)(5)  4(5)2  18  75  100  7 2 3

(b) For a  , b  

3 4

3(7 a  4b)  4(5a  3b)  21a  12b  20a  12b  a  24b 2  3  24    3  4 2   18 3 2  18 3 

2NC (2)(12)(400) 2(12)(400)    0.192 P(n  1) 2000(24  1) 2000(25) (d) For I = 324, P = 5400, r = 0.15 I 324   0.4 P r 5400  0.15

(e) For S = 1606, d = 0.125, t =

240 365

240   S(1  dt )  1606 1  0.125   365    1606(1  0.082192)  1606(0.917808)  1474

(f ) For S = 1566, r = 0.10, t =

292 365

S 1566  292 1  rt 1  0.10  365 1566 1  0.08  1450 

3. (a) (2)  8 3

2 4 (b)     

3

(d) (3) (3)  (3)  2187 2

2

1 1 9 4   (e)    2 16 16 3 4   9 3 (f) ( x3 )5   x15

4. (a) 10 1.35  1.3510  1.35

0.10

 1.030465

1  1.0340 1  0.306557 0.693443    23.114772 (b) 0.03 0.03 0.03 (c) ln 1.025  0.024693 (d) ln (3e0.2 )

 ln 3  ln e0.2  ln 3  0.2ln e  1.098612  0.2  0.898612

 600  11   1.06 

(e) ln 

 ln 600  ln 1.0611  ln 600  11ln 1.06  6.396930  11(0.058269)  6.396930  0.640958  5.755972

  1.075 1  ln (f ) 250     0.07   ln 250  ln (1.075  1)  ln 0.07  ln 250  ln 0.402552  ln 0.07  5.521461  0.909932  ( 2.659260)  5.521461  0.909932  2.659260  7.270789 5. (a)

1 1   81  3 

n 2

1 1   34  3  4

n2

1 1     3 3

n2

Since the bases are common

4 n2 n6

5 1  40   2 2

(b)

1 1  8  2 2 1 1   16  2 

n1

n 1

1 1     2 2 4  n 1 n5 2 3

6. (a)  x  24

 3 x  24     2 x  36 (b) x  0.06 x  8.46

0.94 x  8.46 x9 (c) 0.2 x  4  6  0.3x

0.5 x  10 x  20 (d) (3  5x)  (8 x  1)  43

3  5 x  8 x  1  43  13x  39 x  3 (e) 4(8 x  2)  5(3x  5)  18

32 x  8  15 x  25  18 17 x  33  18 17 x  51 x3

(f ) x 

3 1 3 x   x  x  1  103 10 2 5

3 3 3 x  x   103 10 5 2 20 x  3x  6 x  15  1030 29 x  1015

2x 

x  35

4 5 4  x  x  3   x  x  3   77 5 6 5 

(g)

4   30 x  24 x  90  25  x  x  3   30(77) 5   54 x  90  25 x  20 x  75  2310 99 x  165  2310 99 x  2475 x  25 (h)

2 3 9 5 (3x  1)  (5 x  3)  x  (7 x  9) 3 4 8 6

16(3x  1)  18(5 x  3)  27 x  20(7 x  9) 48 x  16  90 x  54  27 x  140 x  180  42 x  38  113x  180 71x  142 x2 7. (a) I  P rt P

I rt

(b) S 

P 1  dt

S 1  P 1  dt P  1  dt S dt  1 

P S

1

P S

d

S P d S t

d

S P St

8. Let the regular selling price be $x. 1 5

Reduction in price  $ x x 

1 x  1920 5 4 x  1920 5 x  2400

The regular selling price is $2400. 9. Let the floor space occupied by shipping be x. Floor space occupied by weaving  2 x  400 Total floor space  x  2 x  400  x  2 x  400  6700 3 x  6300 x  2100

The floor space occupied by weaving is 2(2100)  400  4600 square metres.

10. Let the number of units of Product A be x. Number of units of Product B  95  x Number of hours for Product A  3x Number of hours for Product B  5(95  x)

 3x  5(95  x)  395 3x  475  5 x  395  2 x  80 x  40 11. The number of units of Product B is 95  40  55. Let the sum of money invested in the bank be $x. 2 3

Sum of money invested in the credit union  $ x  500 Yield on the bank investment  $

1 x 12

12 93

 

Yield on the credit union investment  $  x  500 



1 12  x   x  500   1000 12 93  2  3 x  4  x  500   36, 000 3  8 3 x  x  2000  36, 000 3 17 x  34, 000 3 17 x  102, 000 x  6000

The sum of money invested in the credit union certificate is

2  $   6000  500   $4500. 3 

Challenge Problems 1. Counting a nickel as a quarter overstates the total by $0.20; for x nickels, the total must be reduced by $0.20x. Counting a toonie as a loonie understates the total by $1; for x toonies, the total must be increased by $1x. The total adjustment  0.20 x  1x  $0.80 x The clerk must increase the total by $0.80x. 2. There are 5 tires, so each tire is idle at some point. Therefore, the number of rotations is 5. The distance per rotation 

4000  800 km; each tire will be used for four rotations 5

for a total distance of 3200 km. (See table below.) Distance Rotation

Tire A

Tire B

Tire C

Tire D

Tire E

travelled

800

—

800

—

800

—

800

—

800

—

800

Total

3200

4000

3. The lowest possible two-digit number is 10; the highest possible two-digit number is 99. For a difference in value of $17.82, the two-digit numbers must differ by 18, such as 10 and 28, 11 and 29, etc. The lowest possible correct value of the cheque is $10.28; the largest possible correct value of the cheque is $81.99. In either case the difference between is $17.82. (a) FALSE than 70.

In the possible correct cheque value $81.99, the x-value 81 is greater

(b) TRUE

In the possible correct cheque value $18.36, the y-value 36 equals 2x.

A cheque cannot have zero cents.

(d) FALSE

Let the correct amount be $A;

then the incorrect amount is $2A; the difference is $A;

A  17.82 For the correct value $17.82, the incorrect cheque value $82.17 is unequal to 2($17.82). (e) FALSE

In the possible correct amount $10.28, the sum of the digits is 1  0  2  8  11, which is not divisible by 9.

Case Study 1. $72,924 – $48,946 = $23,978. 2. The contributions continue until the 65th year. Therefore, total contributions (65 – 45) × 12 months per year × $250 = $60,000. 3. The contributions continue until the 65th year. a. Total contributions (65 – 45) × 12 months per year × $100 = $24,000. Total value of TFSA = $29,275. Therefore, interest earned is $29,275 – 24,000 = $5275. b. Total contributions (65 – 45) × 12 months per year × $250 = $60,000. Total value of TFSA = $72,924. Therefore, interest earned is $72,924 – 60,000 = $12,924. 4. Annual salary of $48,000 ÷ 12 months = $4000 per month. a. $150 ÷ $4000 = 0.0375 or 3.75% of salary b. $250 ÷ $4000 = 0.0625 = 6.25% of salary

Chapter 3

Ratio, Proportion, and Percent

Exercise 3.1 A. 1.

(a) 12 : 32  3:8 (b) 84 : 56  3: 2 (c) 15: 24 : 39  5:8:13 (d) 21: 42 : 91  3: 6 :13

(a)

12 dimes 120 cents 24   5 quarters 125 cents 25

(b)

15 hours 15 hours 5   3 days 72 hours 24

(c)

6 seconds 6 3   50 metres 50 25

(d) $72 per dozen =

72 6  12 1

(e) 40 :12 :14  20 : 6 : 7 (f) 2 : 24 : 5000  1:12 : 2500 3.

(a)

1.25 125 5   4 400 16

(b)

2.4 24 2   8.4 84 7

(c)

0.6 : 2.1: 3.3 = 6 : 21: 33 = 2 : 7 :11

(d)

5.75 : 3.50 :1.25 = 575 : 350 :125 = 23 :14 : 5

(e)

1 2 :  5: 4 2 5

(f )

5 7 :  25 : 21 3 5

(g)

(h)

3 2 3 : : 8 3 4  9 :16 :18

2 4 5 : : 5 7 14  28 : 40 : 25

(i)

(j)

(k)

(l)

(m)

2 3 5 : : 5 4 16  32 : 60 : 25 3 1 17 : : 7 3 21  9 : 7 :17

5 1 8 :11 8 2 69 92  : 8 8  69 : 92 3 7 1 :3 4 16 28 55  : 16 16  28 : 55 1 1 2 :4 5 8 11 33  : 5 8  88 :165  8 :15

(n)

1 5 5 :5 4 6 21 35  : 4 6  63 : 70  9 :10

B. 1.

2. 3.

Food Cost 40% 8   Beverage Cost 35% 7 Commissions $25, 000 1   Sales $875, 000 35

Supervisors : office employees : production workers  6 : 9 : 36  2 : 3:12

Direct material : direct labour : overhead  $4.25 : $2.75 : $3.25  425 : 275 : 325  17 :11:13

5. 6.

Instructors 8 1    1: 29 Students 232 29 Lecture : study : travel  20 : 45 : 5  4 : 9 :1

C. 1.

A : B : C := 9 : 2 :1 Total number of shares is 12. Value of each share = $30, 600  12 = $2550 A receives 9  2550  $22,950 B receives 2  2550  $ 5100 C receives 1 2550  $ 2550

Department A : Department B : Department C : Department D  1000 : 600 : 800 : 400  10 : 6 : 8 : 4  5:3: 4: 2

Total number of parts = 14 $21, 000 Value of each part =  $1500 14 Allocation: Department A : Department B : Department C : Department D : 3.

5 1500  $7500 3 1500  $4500 4 1500  $6000 2 1500  $3000

5 1 1 Ratio = : : = 15 : 8 : 4 8 3 6 Total number of parts = 27 Value of each part = $9450  27 = $350 Distribution: Manufacturing: 15  350  $5250 Selling: 8  350  $2800 Administration: 4  350  $1400

Northern : Eastern : Western  $10.8 : $8.4 : $14.4  108 : 84 :144  9 : 7 :12

Total number of parts = 28 Value of each part: $588, 000  $21, 000 28 Allocation: Northern Division: 9  21, 000  $189, 000 Eastern Division: 7  21, 000  $147, 000 Western Division: 12  21, 000  $252, 000 5.

Raw Materials : Work-in-Process : Finished Goods  1/ 3 :1/ 6 : 3 / 8  8 / 24 : 4 / 24 : 9 / 24  8: 4:9

Total number of parts = 21 Value of each part: $11,550, 000  $550, 000 21 Allocation: Raw Materials: 8  550, 000  $4, 400, 000 Work -in -Process: 4  550, 000  $2, 200, 000 Finished Goods: 9  550, 000  $4,950, 000 6.

New : Used : Servicing : Administration  1/ 8 : 1/ 4 : 1/2 : 1/16  2 /16 : 4 /16 : 8/16 : 1/16  2 : 4 : 8 :1

Total number of parts = 15 Value of each part: $480, 000  $32, 000 15 Allocation: New: Used: Servicing: Administration:

2  32, 000  $64, 000 4  32, 000  $128, 000 8  32, 000  $256, 000 1 32, 000  $32, 000

Total allocated

$480, 000

Exercise 3.2 A. 1.

3: n  15 : 20

15n  20  3 20  3 n 15 n 4 2.

n : 7  24 : 42 42n  24  7 24  7 n 42 n 4

3 : 8  21: x 3 x  21 8 21 8 x 3 x  56

7 : 5  x : 45 5 x  45  7 45  7 x 5 x  63

1.32 :1.11  8.8 : k 1.32k  1.11 8.8 1.11 8.8 k 1.32 k  7.4

2.17 :1.61  k : 4.6 1.61k  4.6  2.17 4.6  2.17 k 1.61 k  6.2

m : 3.4  2.04 : 2.89 2.89m  2.04  3.4 2.04  3.4 m 2.89 m  2.4

3.15 : m  1.4 :1.8 1.4m  1.8  3.15 1.8  3.15 m 1.4 m  4.05

10.

3 7 15  : 4 8 16 15 3 7 t  16 4 8 3 7 16 t   4 8 15 7 t 10

3 5 4 :t  : 4 8 9 5 3 4 t  8 4 9 3 4 8 t   4 9 5 8 t 15

11.

12.

B. 1.

9 3 8 : t: 8 5 15 3 8 9 t  5 15 8 8 9 5 t   15 8 3 t1 16 4 15 :  :t 7 9 14 16 15 4 t  7 14 9 15 4 7 t   14 9 16 5 t 24 Let the number of months to earn $8.75 per share be x.

 $1.25 : 3  8.75 : x 1.25 x  3(8.75) x  21 2.

Let the tax assessment for a tax of $3854 be $x. $32 tax $3854 tax  $1000 assessment $ x assessment 32 x  3,854, 000 x  120, 437.50

The assessment is $120,437.50.

Let the distance travelled on 75 litres be x km.  9l : 72 km  75l : x km 9 x  72(75) x  600

The car can travel 600 km. 4.

Let the supervision cost for 16,000 hours be $x.

$85 $x  64 hours 16, 000 hours 64 x  85 16, 000 x  21, 250 The cost is $21,250. 5.

(a) Let the total value before selling be $x. 5 : 6  300,000: x 5 x  1,800,000 x  360,000

Total value was $360,000. (b) Let the value of the partnership be $y. 2 : 5  360,000: y 2 y  1,800, 000 y  900,000

The value is $900,000. 6.

Let the original value of the slightly damaged part be $x. 1: 3  13,000 : x x  39,000

Let the total value be $y.

3 39,000  8 y 3 y  8  39,000 y  104,000 (a) 104,000 The value was $104,000. (b)

5 104,000  65,000 8

The value was $65,000.

Let last year’s profit be $x.

4 $128,000 dividend  9 $ x profit 4 x  9(128,000) x  288,000 Let revenue be $y.

2 288,000  7 y 2 y  7(288,000) y  1,008,000 Last year’s revenue was $1, 008, 000. 8.

Let material cost be $x.

1 15  3 x x  45 Let total cost be $y.

5 45  8 y 5 y  8(45) y  72 Total cost is $72.

Exercise 3.3 A. 1.

0.40  90  36

0.001 950  0.95

2.50 120  300

0.07  800  56

0.03  600  18

0.15  240  36

0.005 1200  6

3.00  80  240

0.0002  2500  0.5

10. 0.005  500  2.5 11. 0.0025  800  2 12. 0.00875  3600  31.5 B. 1.

1  48  $16 3

11  400  11 50  $550 8

2  72  2  24  $48 3

3  24  3  3  $9 8

5 160  5  40  $200 4

5  720  5 120  $600 6

5  90  5  30  $150 3

7  42  7  7  $49 6

1  54  $9 6

10.

3 180  3  45  $135 4

11.

4  45  4 15  $60 3

12.

1  440  $110 4

C. 1.

R

36  0.60  60% 60

R

54 3   75% 72 4

R

920  1.15  115% 800

R

490  3.5  350% 140

R

6 1   5% 120 20

R

11 1   2.5% 440 40

R

132  6  600% 22

R

30 2 2   66 % 45 3 3

R

150 5 2   166 % 90 3 3

10.

R

39 13 2   216 % 18 6 3

D. 1.

$60  30% of x 60  0.3x x  $200 $36 12   2.4  240% $15 5

R

x  0.1% of $3600  0.001  3600  $3.60

150% of x  $270 1.5 x  270 x  $180

1 % of $612 2 1%  $6.12 1 %  $3.06 2 x

250% of x  $300 2.5 x  300 x  $120

80  40% of x 80  0.4 x x  200

10. E. 1.

$120  2  200% $60

1 x  % of $880 8 1%  $8.80 1 %  $1.10 8

R

180  0.40  40% 450

Let the reduction be $x.

x  40% of 70 x  28 The reduction is $28. 2.

Let labour cost be $x.

1 x  37 % of 72 2 3 x   72 8 x  27

The labour cost is $27. 3.

Let waste be $x. x  6% of 25, 000  0.06  25, 000  1500

The waste is $1500.

Let the deduction be $x.

2 x  16 % of 55, 800 3 1   55, 800 6  9300 The deduction is $9300 5.

Let the budgeted sales be $x. 90% of x  40,500 0.9 x  40,500 x  45, 000

The sales budget is $45,000. 6.

Let gross wages be $x. 5.7% of x  123.12 0.057 x  123.12 x  2160

The gross wages are $2160. 7.

Let the original cost be $x. 300% of x  540, 000 3x  540, 000 x  180, 000

The original cost is $180, 000. 8.

Let the prize be $x.

25% x  280 1 x  280 4 x  1120 The prize is $1120. 9.

Let Shari’s portion be $x.

x  0.5% of 1, 200, 000 x  6000

Shari’s portion is $6000.

10. Let the percentage be x. x% of 45  18 x  40%

Exercise 3.4 A. 1.

x  120  40% of 120  120  0.40 120  168

x  900  20% of 900  900  0.2  900  720

x  $1200  5% of $1200  1200  0.05  1200  $1140

x  $24  200% of $24  24  2  24  $72

B. 1.

1 x  48  83 % of 48 3 5  48   48 6  88 2 x  $66  16 % of $66 3 1  66   66 6  $55

Increase = 15 15 1 R=   50% 30 2

Decrease  $18 $18 R  0.2  20% $90

Increase  $160 $160 R  2  200% $80

Decrease  $55 R

$55 1 1   33 % $165 3 3

Decrease  $6 $6 R  0.02  2% $300

Increase  $25 $25 R  0.0125  1.25% $2000

x  5% of x  $4.18 0.95 x  4.18 x  $4.40

The original amount is $4.40. 8.

x  7% of x  $749 1.07 x  749 x  700

The original amount is $700. Business Math News Average Rents in Canada Soar above $2K for First Time Ever, New Data Suggests 1. Let $x be the base rental rate from one year ago.

Vancouver:

x(1.205)  $2633 x  2633 /1.205  2185.06  $2185

Toronto:

x(1.23)  $2532 x  2532 /1.23  $2058.54  $2059

Montreal:

x(1.06)  $1572 x  1572 /1.06  1483.02  $1483

Let $x be the October 2022 two-bedroom price.

Burlington:

x(1  0.009)  $2541 x(0.991)  $2541 x  $2564.08  $2564

Victoria:

x(1  0.01)  $2698 x(0.990)  $2698 x  $2725.25  $2725

Kitchener:

x(1  0.015)  $2332 x(0.985)  $2332 x  $2367.51  $2368

Ottawa:

x(1  0.055)  $2308 x(0.945)  $2308 x  $2442.33  $2442

London:

x(1  0.004)  $2153 x(0.996)  $2153 x  $2161.65  $2162

Hamilton:

x(1  0.009)  $2127 x(0.991)  $2127 x  $2146.32  $2146

Kingston:

x(1  0.003)  $2123 x(0.997)  $2123 x  $2129.39  $2129

Answers will vary. Explanations may include supply and demand factors such as

rising interest rates pushing up monthly mortgage payments; acceleration in Canada’s population growth (new immigrants tend to rent); improvement in the employment rate leading to young adults seeking rentals; and a lack of affordable rental housing inventory.

Exercise 3.5 A. 1.

Let the number of absentees be x.

1 x  2 % of 1200 4 1%  12 2%  24 1 %3 4 1 2 %  27 4 The number absent is 27. 2.

Let the profit be $x.

1 x  33 % of 1575 3 1  1575 3  525 The profit is $525. 3. 4. 5.

92.50 1   12.5% 740 8 8.72 Penalty   0.05  5% 174.40 Let the weekly sales be $x. 2 16 % of x  720 3 1 x  720 6 x  4320 Increase 

Weekly sales must be $4320. 6.

Let the amount collected be $x. 75% of x  2490

3 x  2490 4 x  3320 The amount collected was $3320. 7.

(a) Let the policy value be $x.

3 % of x  675 8 1 % of x  225 8 1% of x  1800 Face value is $180, 000. (b) Let appraised value be $y. 80% of y  180, 000 0.8 y  180, 000 y  225, 000 Appraised value is $225, 000. 8.

(a) Let the assessed value be $x.

1 3 % of x  7200 3 1 x  7200 30 x  216, 000 Assessed value is $216, 000. (b) Let the market value be $y. 40% of y  216, 000 0.4 y  216, 000 y  540, 000

Market value is $540, 000. B. 1.

Let the selling price be $x.

1 7.92  83 % of 7.92  x 3 5 7.92   7.92  x 6 7.92  6.60  x x  14.52 The article sold for $14.52.

Let the cash payment be $x.

1 x  840  2 % of 840 2  840  21  819 Cash payment is $819. 3.

Let the reduced price be $x.

1 x  195  33 % of 195 3 1  195  195 3  195  65  130 The reduced price was $130. 4.

Let the December price be x dollars. x  1.691  7.1% of 1.691  1.691 (1  0.071)  1.691 (0.929)  1.570939

The December price was $1.571 per litre. 5.

Let the cost before harmonized sales tax be $x. 13% of x  9.62 0.13 x  9.62 x  74

Total cost  74  9.62  $83.62. The total cost of the shoes was $83.62. 6.

Let the monthly salary be $x.

1 37 % of x  1880 2 3 x  1880 8 3 x  15, 040 x  5013.33 The monthly salary is $5013.33.

Let the face value be $x.

1 4 % of x  225 2 0.045 x  225 x  5000 The face value is $5000. 8.

Let the amount of the purchase be $x.

1 2 % of x  432 4 0.0225 x  432 x  19, 200 The amount of the purchase was $19, 200. 9.

Decrease  6540  1090  5450 5450 1 Rate   0.83  83 % 6540 3 1 Decrease in profit was 83 %. 3

10.

Increase  16.12  15.50  0.62 0.62 Rate   0.04  4% 15.50 The raise was 4%.

11. Gain in value  443, 625  136,500  307,125 307,125 Rate   2.25  225% 136,500 12.

Reduction  6%  5.75%  0.25% 0.25 Rate   0.0416  4.16% 6 The reduction in rate is 4.16%.

13. Let the marked price be $x. 1 x  33 % of x  64.46 3 1 x  x  64.46 3 2 x  64.46 3 1 x  32.23 3 x  96.69 The marked price was $96.69. 14. Let the index ten years ago be x. x  125% of x  279 5 x  x  279 4 9 x  279 4 1 x  31 4 x  124 The index ten years ago was 124. 15. Let the invoice amount be $x. x  5% of x  646 0.95 x  646 x  680

The invoice amount was $680. 16. Let April sales be $x.

2 x  16 % of x  24,535 3 1 x  x  24,535 6 7 x  24,535 6 1 x  3505 6 x  21, 030 April sales were $21,030.

17. Let the second quarter working capital be $x.

x  75% of x  78, 400 7 x  78, 400 4 1 x  11, 200 4 x  44,800 Working capital at the end of the second quarter was $44,800. 18. Let the proceeds be $x. x  8% of x  88, 090 0.92 x  88, 090 x  95, 750 Fees  8% of 95, 750 = $7660

19. Let the compensation before vacation pay be $x.

x  4% of x  70, 200 1.04 x  70, 200 x  67,500 Vacation Pay  4% of 67,500  $2700. 20. Let the price of the car be $x.

x  15% of x  24, 725 x  0.15 x  24, 725 1.15 x  24, 725 x  21,500 Sales tax = 15% of 21,500 = $3225.

Exercise 3.6 A. 1.

Let the number of US$ be x. C$1 C$750  US$0.8168 US$ x 1 750  0.8168 x x  750(0.8168)

x  612.60 C$750 will buy US$612.60.

Let the number of C$ be x. C$1 C$ x  R12.3407 R1000

1 x  12.3407 1000 1000 x 12.3407 x  81.032680 R1000 will buy C$81.03. 3.

Let the number of C$ be x. C$1 C$ x  R60.589335 R137, 790

1 x  60.589335 137, 790 137, 790 x 60.589335 x  2274.162606 The flight costs C$2274.16

Let the cost per U.S. gallon be C$x. C$1.2864 C$ x  US$1 US$4.319 1.2864 x  1 4.319 x  1.2864(4.319)

x  5.555962 U.S. gallon costs $5.556 5.556 Cost per litre   $1.462 3.8 B. 1.

US$1  C$1.2830 US$350 converts to 350(1.2830)  C$449.05

C$1 = 0.7379€ C$200 converts to 200(0.7379)  147.58

US$1  0.9747 CHF US$175 converts to 175(0.9747)  170.57 CHF

1 GBP  159.7668 JPY 250 GBP converts to 250(159.7668) = 39,942 yen

1 €  C$1.3552 550 € converts to 550(1.3552)  C$745.36

Exercise 3.7 1.

2.58 (100)  103.6145 2.49 220 Simple price index for a bus pass  (100)  110.5528 199 1600 Simple price index for clothing  (100)  96.9697 1650 Simple price index for bread 

Interpretation: The price of bread increased 3.61% from 2019 to 2021. The price of a bus pass increased 10.55% from 2019 to 2021. The price of clothing decreased 3.03% from 2019 to 2021.

1492.85 (100)  100.00 1492.85 1725.60 Gold price index for 2016  (100)  115.5910 1492.85 1642.10 Gold price index for 2017  (100)  109.9977 1492.85 1616.80 Gold price index for 2018  (100)  108.3029 1492.85 1984.20 Gold price index for 2019  (100)  132.9136 1492.85 2507.15 Gold price index for 2020  (100)  167.9439 1492.85 Gold price index for 2015 

Interpretation: Relative to 2015, the price of gold increased 15.59% in 2016, nearly 10% in 2017, 8.30% in 2018, 32.91% in 2019, and 67.94% in 2020. 3.

(a) (i) Purchasing power of dollar in 2015 relative to 2002 

1 (100)  0.7899 126.6

(ii) Purchasing power of dollar in 2021 relative to 2002  (b) Purchasing power of dollar in 2021 relative to 2015  4.

Real income 

Nominal income (100) CPI

1 (100)  0.7062 141.6

126.6  0.8941 141.6

$60, 000 (100)  $47,393.36 126.6 $70, 000 2019 : Real income  (100)  $51, 470.59 136.0 $75, 000 2021 : Real income  (100)  $52,966.10 141.6 2015 : Real income 

Inflation rate from 2016 to 2021 CPI 2021  CPI 2016 (141.6 128.4) 2021  (100)  (100)  10.280374% CPI 2016 128.4 She would have to earn 10.280374% more, or $74,000(0.102804) = $7607.48 more. Therefore, the total she must earn in 2021  74,000  7607.48  $81,607.48

Factor increase in index 

Index in 2022 20,197.60   1.033495 Index in 2021 19,543

Therefore, value of portfolio on May 17, 2021 = $279,510 (1.033495)  $288,872.29 Exercise 3.8 1.

Federal tax = 0.15(49,450) = $7417.50

Federal tax = 0.15(50,197) + 0.205(100,392 − 50,197) + 0.26(106,300 − 100,392) = $7529.55 + $10,289.98 + $1536.08 = $19,355.61

Total income = $32,920 + $17,700 = $50,620 Federal tax = 0.15(50,197) + 0.205(50,620 − 50,197) = $7529.55 + $86.72 = $7616.27

(a)

Percent increase in pay before federal taxes 

($105,000  $87,000) $18,000   20.690% $87,000 $87,000

(b) Federal taxes on $87,000 = 0.15(50,197) + 0.205(87,000 − 50,197) = $7529.55 + $7544.62 = $15,074.17 Federal taxes on $105,000 = 0.15(50,197) + 0.205(100,392 − 50,197) + 0.26(105,000 − 100,392) = $7529.55 + $10,289.98 + $1198.08 = $19,017.61

Percent increase in pay after federal taxes ($105,000  19,017.61)  ($87,000  $15,074.17)  ($87,000  $15,074.17) ($85,982.39  71,925.83) $14,056.56    19.543% 71,925.83 71,925.83 Review Exercise 1.

(a) 25 dimes:3 dollars  250 : 300  5 : 6 (b) 5 hours:50 minutes  300 : 50  6 :1 (c) $36.75 : 30 litres = 3675 : 3000 = 49 : 40 (d) $21: 3.5 hours  210 : 35  30 : 5  6 :1 (e) 1440 words:120 lines:6 pages  240 : 20 :1 (f ) 90 kg:24 ha: 18 weeks  15: 4 : 3

(a) 5 : n  35 : 21 5 35  n 21 21 5  35n 21 5 n 35

n3 (b) 10 : 6  30 : x 10 30  6 x 10 x  (30)(6) (30)(6) x 10

x  18 (c) 1.15 : 0.85  k :1.19 0.85k  1.15 1.19 1.15 1.19 k 0.85 k  1.61

(d) 3.60 : m  10.8 : 8.10 10.8m  3.60  8.10 3.60  8.10 m 10.8 m  2.7 (e)

5 15 6 :  :t 7 14 5 5 15 6 t  7 14 5 3 3 7 t   7 1 5 9 t 5 9 5 45  : 8 4 64 45 9 5 y  64 8 4 9 5 64 y   8 4 45 y2

(f ) y :

(a)

(b)

(c)

(d)

2 66 % of $168 3 2  168  2  56  $112 3 1 37 % of $2480 2 3   2480  3  310  $930 8 125% of $924 5   924  5  231  $1155 4

1 183 % of $720 3 11   720  11120  $1320 6

(a) 1% of $2664  $26.64 1 %  $6.66 4 (b) 1% of $1328  $13.28

1 %  $1.66 8 5 %  $8.30 8 (c) 1% of $5400  $54.00 1 %  $18.00 3 2 %  $36.00 3 2 1 %  $90.00 3 (d) 1% of $1260  $12.60 2%  $25.20 1 %  $2.52 5 1 2 %  $27.72 5 5.

(a) Rate 

55 5   0.625  62.5% 88 8

(b) Rate 

63 7   1.75  175% 36 4

3 (c) x  % of $64 4 1%  $0.64 1 %  $0.16 4 3 %  $0.48 4  x  $0.48 (d) 450% of $5  x x  4.50  5  $22.50

1 (e) $245  87 % of x 2 7 x  245 8 1 x  35 8 x  $280

1 (f) 2 % of x  $9.90 4 0.0225 x  9.90 x  $440 (g) Rate 

$1.25 125 1    0.02  2% $62.50 6250 50

(h) Rate 

$30  5  500% $6

2 (i) 166 % of x  $220 3  2 1   x  220 3 5 x  220 3 1 x  44 3 x  $132

(j)

1 $1.35  % of x 3 1% of x  3($1.35)  $4.05 x  $405

(a) $8  125% of $8  x

8  10  x x  $18 1 (b) x  $2000  2 % of $2000 4  2000  45  $1955

Rate of decrease 

$20 1 2   16 %. $120 6 3

(d) Increase  $975  $150  $825

$825  5.50  550%. $150 (e) $98  x  75% of x 1.75 x  98 x  56 Rate of increase 

The amount is $56. (f) x  15% of x  $289 0.85 x  289 x  340 The price was $340. (g) x  250% of x  $490 x  2.5 x  490 3.5 x  490 x  140 The sum of money is $140. 7.

Total number of parts  $4000  $6000  $5000  $15, 000 Value of each part  $4500 / $15, 000  0.30 Distribution of profit: to D : 4000  0.30  $1200 to E : 6000  0.30  $1800 to F : 5000  0.30  $1500

Total number of parts  80  140  160  380 Rental allocation per part  11, 400 / 380  30 Allocation:

Department A : 80  30  $2400 Department B : 140  30  $4200 Department A : 160  30  $4800

1 1 3 1 8 6 9 1 Ratio  : : :  : : :  8: 6 : 9 :1 3 4 8 24 24 24 24 24 Total number of parts  24 Value of each part: 189, 000  7875 24 Distribution of estate: First beneficiary :

8  7875  $63, 000

Second beneficiary : 6  7875  $47, 250 Third beneficiary :

9  7875  $70,875

Fourth beneficiary : 1 7875  $7875 10.

1 1 2 15 10 12 : :  : :  15 :10 :12 2 3 5 30 30 30 15 x  10 x  12 x  185, 000 x  5, 000 Ratio 

Total number of parts  37 Value of each part: 185, 000  5000 37 Allocation of fire loss: Company 1: 15  5000  $75, 000 Company 2 : 10  5000  $50, 000 Company 3 : 12  5000  $60, 000 11. Let the number of minutes for the 176 L tank be x.

220 litres 20 minutes  176 litres x minutes 220 20  176 x 220 x  20(176) 20(176) x 220 x  16 It will take 16 minutes to heat the 176 L tank.

12. Let the variable cost for sales of $350,000 be $x. $130, 000 variable cost $ x variable cost  $250, 000 sales $350, 000 sales 130, 000 x  250, 000 350, 000 25 x  13(350, 000) 13(350, 000) x 25 x  182, 000 The variable cost for sales of $350,000 is $182, 000. 13. (a) Net income 

2 of gross profit 7

2 G  $4200 7 1 G  2100 7 G  14, 700 Gross profit is $14,700. (b) Gross profit 

2 of net sales 5

2 N  $14, 700 5 1 N  7350 5 N  36, 750 Net sales are $36,750. 14.

Faculty 5 x   Support 4 192 192(5)  4 x x  240 The number of faculty members is 240. 4 of total  240 9 4 T  240 9 4T  240  9 240  9 T 4 T  540 Employment is 540.

1 15. (a) Number eligible  62 % of 94,800 2 5   94,800 8  59, 250

1 (b) Number voting  33 % of 59, 250 3 1   59, 250 3  19, 750 16.

1  37 % of 150, 000 2 3  150, 000  $56, 250 8

Bonds

Common

Preferred

1  56 % of 150, 000 4  0.5625 150, 000  $84,375  (100%  37.5%  56.25%) of 150, 000  6.25% of 150, 000  0.0625 150, 000  $

Total

9375

= $150, 000

17. (a) May order

2  $51,120  16 % of $51,120 3 1  51,120   51,120 6  51,120  8520  $42, 600 (b) Decrease  $8520

18. (a) Appraised value

1  $120, 000  233 % of $120, 000 3 7  120, 000  120, 000 3  120, 000  280, 000  $400, 000 (b) Gain = $280,000

 $103.95

50.00%

Labour cost



22.2%

Overhead Total

 57.75 27.7%  $207.90 100.00%

19. (a) Material cost

(b) Overhead rate  20. (a) Not passed 

46.20

$57.75  125% $46.20

180 3   7.5% 2400 40

(b) Scrapped as a percent of production not passed



30 1 2   16 % 180 6 3

21. (a) $Change  $56.25  $51.75  $4.50

Change 

$4.50  0.08  8% $56.25

(b) New price as a percent of old price



$51.75  0.92  92% $56.25

22. (a) Increase in pay = $16.80  $6.30  $10.50

Percent change =

$10.50 2  1.6  166 % $6.30 3

(b) Current pay as a percent of old pay

$16.80 2  2.6  266 % $6.30 3

23.

1 Bad debts = 2 % of sales 4 $7875 = 0.0225S S  350, 000 Sales are $350, 000.

24.

List price = 240% of cost $396 = 2.40C C  165

Cost is $165.

1 25. 87 % of asking price = sale price 2 7 A  $191,100 8 A  $218, 400 List = C  160% of C $218, 400  2.60C C  84, 000 Cost was $84,000. 26. 77.5% of list price = sale price

0.775L  $15,500 L  20, 000 1 List  C  33 % of C 3 4 $20, 000  C 3 C  15, 000 Cost was $15,000. 27. (a) Let revenue be $x.

3  9 % of x  $29, 250 4 0.0975 x  29, 250 x  $300, 000 (b) After -tax income  $29, 250  15% of $29, 250  0.85(29, 250)

 $24,862.50

Alice's dividend 5  Total dividend 8 x 5   $18, 646.88 8 x  $11, 654.30 $11, 654.30 Alice's dividend = $29, 250  0.398438  39.84%

28. (a) Let the asking price be $x.

2  91 % of x  770, 000 3 11 x  $770, 000 12 1 x  70, 000 12 x  $840, 000 Let the cost be $y. y  320% of y  $840, 000 4.2 y  840, 000 y  200, 000

The original cost was $200, 000. (b) Gain  $770,000  $200,000  $570,000 (c) Percent gain 

$570,000  2.85  285% $200,000

29. (a) Exchange rate US$ per C$ =

$310.61  0.817395 $380

C$1 = US$0.817395 (b) C$725 = 725(0.817395) = US$592.61 30. Let the value of the coupon be US$x.

US$ x US$1  C$316 C$1.25 x 1  316 1.25 316 x  252.80 1.25 The coupon has a value of US$252.80 $1 (100) 151.2  0.661376

31. (a) Purchasing power of dollar =

(b) Real income = $62,900(0.661376) = $41,600.53 32.

Federal tax  0.15(50,197) + 0.205(100,392  50,197) + 0.26(102, 450  100,392)  $7529.55 + $10, 289.98 + $535.08 = $18,354.61

33. (a) Percent increase in pay = (b)

$6000  8.8235% $68,000

Federal tax on $68, 000 = 0.15(50,197) + 0.205(68,000  50,197) = $7529.55 + $3649.62  $11,179.17 Net pay after federal tax  $68, 000  11,179.17 = $56,820.83

Federal tax on $74, 000 = 0.15(50,197) + 0.205(74,000  50,197) = $7529.55 + $4879.62  $12, 409.17

Net pay after federal tax  $74,000  12, 409.17 = $61,590.83 Increase in net pay after federal tax  $61,590.83  $56,820.83 = $4770 Percent increase 

$4770 (100)  8.395% $56,820.83

Self-Test

5 (a) 125% of $280   280  5  70  $350 4 (b)

3 % of $20, 280 :1%  $202.80 8 1 %  $25.35 8 3 %  $76.05 8

1 5 (c) 83 % of $174  174  5  29  $145 3 6 1 (d) 1 % of $1056 :1%  $ 10.56 4 1 %  $ 2.64 4 1 1 %  $13.20 4 2.

(a) 65 : 39  x :12 65 x  39 12 65 12 x  20 39 (b)

7 35 6 :  :x 6 12 5 7 35 6 x  6 12 5 7 6 x   3 2 7

Total in sample = 24 + 36 + 20 = 80 36 9 Brand Y preference =   0.45  45% 80 20

Total number of square metres = 40 + 80 + 300 = 420 25, 200 Price per square metre =  $60 420 Amount paid by Department B = 80  60 = $4800

Beverage sales $9.60 96 4    Food sales $12.00 120 5 Let the budget for beverage sales be $x.

x 4  42,500 5 4 x   42,500 5 x  34, 000

The monthly budget for beverage sales is $34,000. 6.

2 Let the marked price be $x; then the price reduction is 16 % of x. 3 2 x  16 % of x  $60 3 1 x  x  60 6 5 x  60 6 x  60 

6 5

x  72 The marked price is $72. 7.

1 1 1 1 15 10 6 5 : : :  : : :  15 :10 : 6 : 5 2 3 5 6 30 30 30 30 Total number of shares = 15 +10 + 6 + 5 = 36 Value of each share = $40,500  36 = $1125 First bonus :

15 1125  $16,875

Second bonus : 10 1125  $11, 250

Third bonus :

6  1125  $6750

Fourth bonus :

5 1125  $5625

Raise  $18.24  $16.00  $2.24 $2.24 Percent raise =  0.14  14% $16.00

Let the price of the bicycle before taxes be $x; then the amount of HST is 13% of $x; the sales value is x  13% of x.

x  13% of x  $282.50 x  0.13x  282.50 1.13x  282.50 282.50 x 1.13 x  $250 HST  13% of $250  $32.50 10.

Price reduction  $220  $209  $11 $11 % reduction =  0.05  5% $220

11. Let the index 10 years ago be x; then the increase is 100% of x. x  100% of x  360 x  x  360 2 x  360 x  180

The index 10 years ago was 180. 2 12. Let Braid’s interest in the racehorse be $x; then the amount sold by him is $ x. 3

2 x  $18, 000 3 x  18, 000 

3 2

x  $27, 000 Let the value of the racehorse be $y; then the interest held by Braid before selling is 3 $ y. 8 

3 y  $27, 000 8 y  27, 000 

8 3

y  72, 000

The value of the racehorse is $72,000.

13. (a) Let the number of Brazilian reals be x. x BRL 1 BRL  C$1 C$0.2631 x 1  1 0.2631 x  3.8008 BRL

C$1 costs 3.8008 BRL (b) To buy C$500 costs 500(3.8008)  1900.42 BRL 14. Let the number of Canadian dollars be x.

C$ x C$1  $800 US$0.8150 x 1  800 0.8150 800 x  $981.60 0.8150 US$800 costs C$981.60.

15. Purchasing power of dollar relative to 2002



1 (100)  0.789266 126.7

16. Federal tax  $7529.55 + 0.205 (52, 707  $50,197) = $7529.55 + $514.55  $8044.10

Challenge Problems 1.

Let each reduction in price be x percent; then the net price after the first reduction is $25(1  x); and the net price after the second reduction is $25(1  x)(1  x).

$25(1  x)(1  x)  $16 16 (1  x) 2  25 4 1 x  5 1  0.80  x x  0.20  20% The two consecutive price reductions are 20% each.

Let the price per blue pair of socks be $x.; then the price per pair of black socks is $2x. Let the number of pairs of blue socks be y. The value of the original order is $[4(2 x)  xy]; the value of the interchanged order is $[ y(2 x)  4 x]; the increase in value is 50% of $[4(2 x)  xy];

4(2 x)  xy  0.50[4(2 x)  xy]  2 xy  4 x 8 x  xy  4 x  0.50 xy  2 xy  4 x 8 x  0.50 xy y  16 The number of pairs of blue socks in the original order is 16. The ratio of black socks : blue socks  4 :16  1: 4 3.

Let the amount of the annual salary be $x; then the amount of salary after the 10% decrease is $0.90x; the increase needed is $0.10x. The percent increased based on the reduced salary



0.10 x (100)  11.1% 0.90 x

Case Study Gross business income = $42,350 Total business expenses = $4849 Total eligible home expenses = $20, 770

(a) Eligible home expense claim based on area



45 ($20,770)  $2709.13 345

(b) Eligible home expense claim based on number of rooms

1  ($20,770)  $2596.25 8

Gross business income Less: Business expenses Eligible home expenses

$42,350.00 $4849 2709.13

Net business income 3.

$34, 791.87

Federal tax: 15% of $34, 791.87 $5218.78 Less non-refundable tax credit claim:15% of $12,138 1820.70 Basic federal tax

Basic federal tax as a percent of gross business income $3398.08 =  0.080238  8.02% $42,350

Basic federal tax as a percent of taxable income $3398.08   0.097669  9.77% $34, 791.87

Chapter 4

Linear Systems

Exercise 4.1 A.

7,558.13

A(4, 3) B(0, 4) C(3, 4) D(2, 0) E(4,3) F(0,3) G( 4, 4) H( 5, 0)

(a)

$3398.08

(b)

(a) x

−2

−1

−5

−3

−1

−3

−2

−1

−6

−4

−2

(b)

(d)

(a)

4 x  5 y  11 5 y  4 x  11 4 x 11 y  5 5 4 11 Slope, m   ; y-intercept, b  5 5

(b)

2 y  5 x  10 2 y  5 x  10 5 x 10 y  2 2 Slope, m 

(c)

5 ; y -intercept, b  5 2

1 y  2x 2 1  y  2x 1 2 y  4 x  2

1

Slope, m  4 ; y-intercept, b  2

(c)

(d)

3y  6  0 3 y  6 y  2

Slope, m  0 ;(line is parallel to x-axis); y-intercept, b  2 (e)

2x  y  3 2x  y  9  y  2 x  9 y  2x  9

Slope, m  2 ; y-intercept, b  9

(f )

0.15 x  0.3 y  0.12  0 15 x  30 y  12  0 30 y  15 x  12 15 x 12 y  30 30 1 2 y   x 2 5 1 2 Slope, m   ; y -intercept, b  2 5

(g)

1 2 x  0 2 1 2 x 2 x4

Slope is undefined (h)

no y-intercept Line is parallel to y-axis

( x  2)( y  1)  xy  2 xy  2 y  x  2  xy  2 2 y   x  4 x y  2 2 1 Slope, m  ; y -intercept, b  2 2

2. x 0 4 2 y 2 0 1

4. x 0 4 4 y 0 2 2

6. x 0 3 6 y 2 0 2

For y  2 x  3 slope, m  2 y-intercept, b  3

10.

For y  3x  9 slope, m  3 y -intercept, b  9

For y  3x  20 x 0 20 40 y 20 80 140 or m  3 b  20

2 For y   x  40 5 x 0 50 100 y 40 20 0 2 or m   5 b  40

For 3x  4 y  1200 x 0 200 400 y 300 150 0 3 or m   4 b  300

For 3 y  12 x  2400  0 x 0 100 200 300 y 800 1200 1600 2000 or m  4 b  800

x = 5 for values of y from 0 to 4

y = 4 for values of x from 0 to 5

Exercise 4.2 A.

x y4

x 0 y 4 x  y 4 x –4 y 0

4 0 0 4

x y  3 x 0 3 y –3 0 x y  5 x 0 5 y 5 0

x  2 y 1

x 3 1 5 y 2 0 2 y  4  3x x 0 2 1 y 4 2 1

2 x  3 y  10 x 5 2 y 0 2 3x  4 y  2 x 6 2 y 5 2

1 4 2 1

3x  4 y  18 x 6 2 2 y 0 3 6 2 y  3x x 0 2 2 y 0 3 3

4 x  5 y x 0 5 5 y 0 4 4 2x  y  6 x 0 3 1 y 6 0 4

5x  2 y  20 x 4 2 y 0 5

6 5

3 y  5 x x 0 3 3 y 0 5 5

For

y  4x  0

x 0 5000 10,000 y 0 20,000 40,000 For y  2 x  10000  0 x 0 5000 10,000 y 10,000 20,000 30,000

For 4 x  2 y  200 x 0 20 40 y 100 60 20 For x  2 y  80 x 0 40 80 y 40 20 0

50 0

For 3x  3 y  2400 x 0 400 800 y 800 400 0

For 2 y  5 x x 0 4000 8000 y 0 10,000 20,000

2 x  y  4 and 2 y  4 x  1

2x  y  4

equation ①

4 x  2 y  1

equation ②

4x  2 y  8

multiply ① by 2 and add to

07

② result is impossible

This result is a contradiction and impossible. This is an inconsistent system and therefore, there is no solution for this system of linear equation.

x  2  0 and 2 x  3  0 equation ① x2  0 equation ② 2x  3  0 rearrange ① x2 rearrange ② x  3/2 These two lines are parallel to the y axis and do not intersect. Therefore, there is no solution for this system.

y  3x  5  0 and 3 y  9 x  2  0

y  3x  5  0

equation ①

3 y  9x  2  0

equation ②

3 y  9 x  15  0

multiply ① by 3 and add the result to ②

result is impossible 17  0 This result is a contradiction and impossible. This is an inconsistent system and therefore, there is no solution for this system of linear equation. 4.

y  x  3 and 2 y  2 x  6  0 yx3

equation ①

2 y  2x  6  0

equation ②

2 y  2 x  6

multiply ① by 2

2 y  2x  6

rearrange ② and add to above

00 The result indicates that the original equations are equivalent and the system is consistent and dependent. Any real number value for x will result a value for y that satisfies both equations. Therefore, there are many solutions for this system.

Exercise 4.3 A.

x  y  9 x  y  7 (1)  (2) 

In (1)

(1) (2)

2 x  16 x  8  8  y  9 y  1 ( x, y )  (8, 1)

Check: In (1) LS  8  1  9  RS In (2) LS  8  (1)  7  RS 2.

x  3y  0 x  2 y  10 (1)  (2) 

In (2)

(1) (2)

5 y  10 y  2 x  4  10 x6 ( x, y )  (6, 2)

Check: In (1) LS  6  3(2)  0  RS In (2) LS  6  2(2)  10  RS 3.

5x  2 y  74 7 x  2 y  46 (1)  (2) 

In (1)

(1) (2)

12 x  120 x  10 5(10)  2 y  74 2 y  24 y  12 ( x, y )  (10,12)

Check: In (1) LS  5(10)  2(12)  50  24  74  RS In (2) LS  7(10)  2(12)  70  24  46  RS

3x  9 y  17 3x  6 y  13 (1)  (2) 

In (1)

(1) (2)

 3 y  30 y  10 3 x  9(3)  17 3 x  90  17 3x  73 x  73/3 ( x, y )  (73/3,10)

Check : In (1) LS  3(73 / 3)  9(10)  73  90  17  RS In (2) LS  3(73 / 3)  6(10)  73  60  13  RS 5.

y  3x  12 x  y Rearrange

(3)  (4)  In (2)

(1) (2) 3x  y  12 x y 0 4 x  12 x  3 3  y y3

(3) (4)

( x, y )  (3,3) Check: In (1)

In (2)

3x  10  2 y 5 y  3x  38 Rearrange

(3)  (4)  In (1)

LS  3 RS  3(3)  12  9  12  3 LS  3 RS  3 (1) (2) 3x  2 y  10 3x  5 y  38

7 y  28 y  4 3 x  10  2(4) 3x  18 x6 ( x, y )  (6, 4)

Check:

(3) (4)

LS  3(6)  18 RS  10  2( 4)  10  8  18 LS  5( 4)  20 RS  3(6)  38  18  38  20

In (1) In (2)

4 x  y  13 x  5 y  19 To eliminate y (1)  5 

(3)  (2)  In (1)

(1) (2) 20 x  5 y  65 x  5 y  19

(3) (2)

21x  84 x  4 4( 4)  y  13  16  y  13 y3 ( x, y )  ( 4,3)

Check: In (1) In (2) 2.

LS  4(4)  3  16  3  13  RS LS  4  5(3)  4  15  19  RS

4 x  5 y  20 2 x  8 y  10 To eliminate x (2)  2  4 x  16 y  20 (1)  (1)  4 x  5 y  20 (3)  (4)  11 y  0

In (1)

(1) (2) (3) (4)

y0 4 x  5(0)  20 4 x  20 x5 ( x, y )  (5, 0)

Check: In (1) In (2) 3.

LS  4(5)  5(0)  20  0  20  RS LS  2(5)  8(0)  10  0  10  RS

7 x  5 y  22 4x  3 y  5 To eliminate y (1)  3  21x  15 y  66 (2)  5  20 x  15 y  25

(1) (2) (3) (4)

(3)  (4) 

41x  41 x  1  4  3y  5 3y  9 y3

In (2)

( x, y )  ( 1,3)

Check: In (1) In (2) 4.

LS  7(1)  5(3)  7  15  22  RS LS  4(1)  3(3)  4  9  5  RS

8x  9 y  129 6 x  7 y  99 To eliminate x (1)  3 

(2)  ( 4)  (3)  (4)  In (2)

(1) (2)

24 x  27 y  387 24 x  28 y  396  y  9 y9 6 x  7(9)  99 6 x  63  99 6 x  36 x6

(3)

(4)

( x, y )  (6,9) Check: In (1) In (2)

LS  8(6)  9(9)  48  81  129 RS  129 LS  6(6)  7(9)  36  63  99 RS  99

12 y  5 x  16 6 x  10 y  54  0 Rearrange 5x  12 y  16 6 x  10 y  54 To eliminate x (3)  6  30 x  72 y  96 (4)  5  30 x  50 y  270

(1) (2) (3) (4) (5) (6)

(5)  (6) 

122 y  366 y3

In 1

12(3)  5 x  16 36  16  5 x 5 x  20 x4 ( x, y )  (4,3)

Check: In (1) In (2)

LS  12(3)  36 RS  5(4)  16  20  16  36 LS  6(4)  10(3)  54  24  30  54  0 RS  0

3x  8 y  44  0 7 x  12 y  56 Rearrange 3x  8 y  44 7 x  12 y  56 To eliminate y (3)  3  9 x  24 y  132 (4)  ( 2)  14 x  24 y  112

(5)  (6)  In (1)

(1) (2) (3) (4) (5) (6)

 5 x  20 x4 3(4)  8 y  44  0 12  8 y  44  0  8 y  56 y7 ( x, y )  (4, 7)

Check: In (1)

In (2)

LS  3(4)  8(7)  44  12  56  44  0 RS  0 LS  7(4)  28 RS  12(7)  56  84  56  28

0.4 x  1.5 y  16.8

(1)

1.1x  0.9 y  6.0 To eliminate decimals (1) 10  4 x  15 y  168 (2) 10  11x  9 y  60 To eliminate y (3)  3  12 x  45 y  504 (4)  5  55 x  45 y  300

(2) (3) (4)

67 x  804 x  12 In (3) 4(12)  15 y  168 48  15 y  168 15y  120 y8

Add:

( x, y )  (12,8)

1.4 x  2.5 y  1.7 1.5x  0.8 y  7.5 (1)  0.8  1.12 x  2 y  1.36 (2)  2.5  3.75x  2 y  18.75 To eliminate y (3)  (4)  4.87 x  20.11 x  4.13

In (1)

(1) (2) (3) (4)

1.4(4.13)  2.5 y  1.7 5.78  2.5 y  1.7  2.5 y  4.08 y  1.63 ( x, y )  (4.13,1.63)

2.4 x  1.6 y  7.60 3.8x  0.6 y  7.20 To eliminate decimals (1)  5  12 x  8 y  38 (2)  5  19 x  3 y  36 To eliminate y (3)  3  36 x  24 y  114 (4)  ( 8)  152 x  24 y  288

(1) (2) (3) (4)

Add: In (3)

116 x  174 x  1.5 12(1.5)  8 y  38 18  8 y  38 8 y  20 y  2.5 ( x, y )  (1.5, 2.5)

2.25x  0.75 y  2.25 1.25 x  1.75 y  2.05 To eliminate decimals (1) 100  225x  75 y  225 (2) 100  125 x  175 y  205 To simplify (3)  75  3x  y  3 (4)  (5)  25 x  35 y  41 To eliminate y (5)  35  105 x  35 y  105 (6)  (1)  25 x  35 y  41

Add: In (5)

(1) (2) (3) (4) (5) (6)

80 x  64 x  0.8 3(0.8)  y  3 2.4  y  3 y  0.6 ( x, y )  (0.8, 0.6)

3x 2 y 13   4 3 6 4 x 3 y 123   5 4 10 To eliminate fractions (1) 12  9 x  8 y  26 (2)  20  16 x  15 y  246 To eliminate y (3) 15 135 x  120 y  390 (4)  8  128 x  120 y  1968

(1) (2) (3) (4)

Add: In (3)

263x  1578 x6 9(6)  8 y  26 54  8 y  26  8 y  80 y  10 ( x, y )  (6,10)

9 x 5 y 47   5 4 10 2x 3 y 5   9 8 36 To eliminate fractions (1)  20  36 x  25 y  94 (2)  72  16 x  27 y  10 To eliminate x (3)  4  144 x  100 y  376 (4)  (9)  144 x  243 y  90 Add: In (3)

(1) (2) (3) (4)

 143 y  286 y  2 36 x  50  94 36 x  144 x4 ( x, y )  (4, 2)

x 2y 7   3 5 15

(1)

3x 7 y   1 2 3 To eliminate fractions (1) 15  5 x  6 y  7 (2)  6  9 x  14 y  6 To eliminate y (3)  7  35 x  42 y  49 (4)  3  27 x  42 y  18 Add: 62 x  31 1 x 2

(2) (3) (4)

 1 In (3) 5    6 y  7  2 5  6y  7 2 5  12 y  14 12 y  9 3 y 4

1 3 ( x, y)   ,  2 4 8.

x 3 y 2   4 7 21 2 x 3 y 7   3 2 36 To eliminate fractions (1)  84  21x  36 y  8 (2)  36  24 x  54 y  7 To eliminate y (3)  3  63x  108 y  24 (4)  ( 2)  48 x  108 y  14 Add: 15 x  10 2 x 3  2 In (3) 21    36 y  8  3 14  36 y  8 36 y  6 1 y 6

 2 1 ( x, y)    ,   3 6

(1) (2) (3) (4)

Business Math News Box SOLUTIONS: 1. Year

2009

5000

2010

5000

2011

5000

2012

5000

2013

5500

2014

5500

2015

10,000

2016

5500

2017

5500

2018

5500

2019

5500

2020

6000

2021

6000

2022 Total Contribution 2.

Maximum Contribution To the TFSA account

6000 $81,500

Assuming that you were 18 when the TFSA was introduced: y  6000(71  x)  81, 500

Exercise 4.4 A. 1. B be y.

Let the number of units of Product A be x and the number of units of Product

Let the number of units of Product 1 be x and the number of units of Product 2

be y.

Let the number of tax returns be x and the total cost be $y.

Let monthly sales be $x and the total salary be $y.

Let x represent the number of employees at the larger location, y, the number of employees at the smaller location. (1)  x  y  24 (2) 2x  3 y  3 (1)  3  3x  3 y  72 From (2)  2x  3y  3 5x  75 x  15 y9

Add:

The number of employees is 15 at the larger location, and 9 at the smaller one.

Check: Sum  15  9  24 2 15  30  (3  9)  3 2. special.

Let x represent the number of orders for the 1st special, y those for the 2nd 7 x  4 y  18 3 1 x  y  14 4 2 (2)  4 

(3)  2 

(1) (2)

3x  2 y  56 6 x  4 y  112

(3)

Add: (1) and (4) In (1)

13 x  130 x  10 7(10)  4 y  18 4 y  52 y  13

(4)

There are 10 orders for the 1st special, 13 for the 2nd special.

Check : 7 10  4 13  70  52  18 3 1 10  13  7.5  6.5  14 4 2 3. brand be y

Let the number of jars of Brand X be x and the number of jars of the No-Name x  y  140 2.25 x  1.75 y  290 (2)  4  9 x  7 y  1160 (1)  7  7 x  7 y  980 Subtract:

(1) (2)

2 x  180 x  90 y  50

Sales were 90 jars of Brand X and 50 jars of No-Name brand.

Check : Totalsold  90  50  140 Value: 90  2.25  $202.50 50 1.75  87.50  $290.00 4.

Let Nancy’s sales be $x and Andrea’s sales be $y. x  y  1940 x  3 y  240 (1)  3  3x  3 y  5820 From (2)  x  3 y  240

(1) (2)

4 x  5580 x  1395 y  545 Nancy’s sales were $1395; Andrea’s sales were $545.

Total  1395  545  $1940 3  545  240  1635  240  $1395 5.

Let Kaya’s investment be $x, and Fred’s investment be $y.

 x  y  55,000 2 y  x  2500 3 (2)  3   2 x  3 y  7500

(1)  2 

(1) (2)

2 x  2 y  110, 000

5 y  117,500 y  23,500 x  31,500

Add:

Kaya’s investment is $31,500 and Fred’s investment is $23,500.

Check : Total  31,500  23,500  $55, 000 2  31,500  2500  21, 000  2500  $23,500 3 6.

Let x represent the number of rooms in the larger hotel and y represent the number of rooms at the smaller hotel. x + y = 180 (1) x = 3y – 80 (2) (1)  x  y  180  (2)  x  3 y  80  x  y  180   x  3 y  80

–1 × (2) → Add (1) and (2)

→

(1) (2)

4y = 260 y = 65

x + 65 = 180 x = 115 The number of rooms in the larger hotel is 115 and there are 65 rooms in the smaller one.

Let the number of chairs produced by the first shift be x and the number of chairs produced by the second shift be y.  x  y  2320 (1) 4 (2) y  x  60 3 (2)  3   4 x  3 y  180 (1)  4  4 x  4 y  9280

Add:

7 y  9100 y  1300 x  1020

The first shift produced 1020 chairs and the second shift 1300 chairs.

Check : Total  1020  1300  2320 4  1020  60  1360  60  1300 3

Let the number of Type A lights be x and the number of Type B lights be y. (1)  x  y  60 (2) 40 x  50 y  2580 (2)  10  4 x  5 y  258 (1)  (4)   4 x  4 y  240

y  18 x  42

Add:

The number of Type A is 42, and the number of Type B is 18.

Check : Total  42  18  60 Value: 42  40  $1680 18  50  900 $2580 9.

Let the number of units of Product A be x and the number of units of Product B be y. (1)  x  y  60 (2) 4 x  3 y  200 (1)  4  4 x  4 y  240  4 x  3 y  200

Subtract:

y  40 x  20

The number of units of Product A is 20, and of Product B is 40. Check : Total number  20  40  60 Number of hours: Product A:20  4  80 Product B:40  3  120 Total hours  200

10.

Let the number of quarters be x and the number of loonies be y.  25 x  100 y  8575

(1)

3 y 1 4 x  4 y  343 4x  3 y  4 4 x  16 y  1372 4x  3 y  4 x

(1)  25  (2)  4  (3)  4  Rearrange (4) (5)  (6)



(2) (3) (4) (5) (6)

19 y  1368 y  72

Substitute (2) 3 x  (72)  1  54  1  55 4 Marysia has 55 quarters and 72 loonies. Value: 55  0.25  $13.75 72 1.00  72.00 $85.75 11.

Let the number of $12-tickets be x and the number of $15-tickets be y. (1) 12 x  15 y  675 4 (2) y  x 3 5 Substitute (2) in (1). 4  12 x  15  x  3   675 5  12 x  12 x  45  675

24 x  720 x  30 Substitute in (2). 4 y  (30)  3 5 y  21 The club bought 30 of $12-tickets and 21 of $15-tickets.

Check: Value  30(12)  21(15)  360  315  $675

12.

4a  2b  180 a  2b Rearrange (2) a  2b  0 Add (1) to (3) 5a  180 a  36 a  2b

36  2b b  18 Check: In (1): In (2)

4(36)  2(18)  180 36  2(18)

Exercise 4.5 A.

x y 4

Solution For x  y  4 Set x  0; solve for y Set y  0; solve for x

x  y  2

Solution For x  y  2 Set x  0; solve for y Set y  0; solve for x x

(1) (2) (3)

–2

x  2y  4

Solution For x  2 y  4 Set x  0; solve for y Set y  0; solve for x

–2

3x  2 y  10

Solution For 3x  2 y  10 Set x  0; solve for y Set y  0; solve for x

–3.33

2 x  3 y 2x  3 y  0 Solution For 2 x  3 y  0 Set x  0; solve for y

Set x  3; solve for y x

–2

4 y  3x 4 y  3x  0 Solution Set x  0; solve for y Set y  3; solve for x

x  2 Solution

x  2 is a line parallel to the y axis

y5 Solution

y  5 is a line parallel to the x axis

3x  2 y  6  0

Solution

x0 y0

y3 x2

(0,3) (2, 0)

y  3 and x  y  2 Solution

y  3 is a line parallel to the x axis For x  y  2 Set x  0; solve for y Set y  0; solve for x

x  3 and x  2 y  4 Solution

x  3 is a line parallel to the y axis For x  2 y  4 Set x  0; solve for y Set y  0; solve for x x

–2

3x  y  6 and x  2 y  8

Solution For 3x  y  6 Set x  0; solve for y Set y  0; solve for x x

–6

For x  2 y  8

5x  3 y and 2 x  5 y  10

Solution For 5x  3 y or 5x  3 y  0 Set x  0; solve for y Set y  5; solve for x x

–3

For 2 x  5 y  10

–2

2 y  3x  9, x  3 and y  0

Solution

x  3 is a line parallel to the y axis y  0 is the x axis For 2 y  3x  9 Set x  0; solve for y Set y  0; solve for x

4.5

–3

2 x  y  6, x  0 and y  0

Solution

x  0 is the y axis y  0 is the x axis For 2 x  y  6 Set x  0; solve for y Set y  0; solve for x

y  3x, y  3 and 2 x  y  6

Solution

y  3 is a line parallel to the x axis For y  3x or y  3x  0 x

–3

For 2 x  y  6 Set x  0; solve for y Set y  0; solve for x x 0 3 8.

y –6 0

2 x  y, x  3 y and x  2 y  6

Solution For 2x  y or 2 x  y  0 x

For x  3 y or x  3x  0 x

–1

For x  2 y  6

–6

Solution Change the inequalities to equation and then draw a line for each equation. Next, determine which part of the line is feasible

x y 6 x  0 y  6

(0, 6)

y0 x6

(6, 0)

x y 6 x0 y6 y0 x6

(0, 6) (6, 0)

Review Exercise 1.

(a)

(b)

2x  y  6 x 3 0 y 0 6

5 4

3x  4 y  0

4

3

(c)

5 x  2 y  10 x 0 2 y 5 0

4 5

(d)

y  3

(e)

5 y  3x  15 x 0 5 2.5 y 3 0 1.5

(f )

5x  4 y  0 x 0 4 y 0 5

4 5

(g)

x  2

(h)

3 y  4 x  12 x 3 0 6 y 0 4 4

(i)

3x  2 y  6 x 0 2 y 3 0

(a)

1 3/2

3x  y  6 and x  y  2 x 0 2 4 y 6 0 6 x y

0 2

2 0

4 2

(b)

x  4 y  8 and 3x  4 y  0 x 0 4 4 y 2 3 1

x y

(c)

0 0

4 4 3 3

5 x  3 y and y  5 x 0 3 3 y 0 5 5

(d)

2 x  6 y  8 and x  2 x 4 2 1 y 0 2 1

(e)

y  3x  2 and y  3 x 0 5/3 1 y 2 3 5

(f )

y  2 x and x  4 x 0 4 2 y 0 8 4

(g)

x  2 and 3x  4 y  12 x 0 2 4 y 3 4.5 0

(h)

y  2 and 5x  3 y  15 x 3 0 4.2 y 0 5 2

(i)

(a)

y  2 is a line horizontal to the x axis and 2 points above it x  3 is a line parallel to the y axis and three point to the right of it

For 7 x  3 y  6 3 y  7 x  6 7 x y 2 3 7 Slope, m   ; y-intercept, b  2 3

(b)

(c)

(d)

(e)

(f )

(g)

For 10 y  5 x 5x y 10 1 y x 2 1 Slope, m  ; y -intercept, b  0 2 2 y  3x For 4 2 2 y  3x  8 2 y  3x  8 3 y  x4 2 3 Slope, m  ; y -intercept, b  4 2

For 1.8 x  0.3 y  3  0 18 x  3 y  30  0 3 y  18 x  30 y  6 x  10

Slope, m  6 ; y-intercept, b  10 1 For x  2 3 x  6 Line is parallel to the Y-axis. Slope, m is undefined. Line is parallel to y axis. There is no y-intercept. For 11x  33 y  99 33 y  11x  99 11x 99 y  33 33 1 y  x 3 3 1 Slope, m  ; y -intercept, b  3 3 For xy  ( x  4)( y  1)  8 xy  xy  4 y  x  4  8 4 y   x  4 1 y  x 1 4

1 ; y -intercept, b  1 4 For 2.5 y  12.5  0 25 y  125  0 y5

Slope, m 

(h)

Line is parallel to the x-axis; slope, m  0; y-intercept, b  5 4.

(a)

3x  2 y  1 5x  3 y  2 To eliminate y (1)  3  9 x  6 y  3 (2)  ( 2)  10 x  6 y  4

(1) (2)

 x 1 x  1 3( 1)  2 y  1 2y  2 y 1

Add: In (1)

( x, y )  (1,1)

(b)

4 x  5 y  25 3x  2 y  13 To eliminate y (1)  2  (2)  5 

Add: In (2)

(1) (2) 8 x  10 y  50 15 x  10 y  65

23 x  115 x5 15  2 y  13 2 y  2 y  1 ( x, y )  (5, 1)

(c)

y  10 x 3 y  29  x Rearrange

(1) (2) 10 x  y  0 x  3 y  29

To eliminate y (3)  3 

(3) (4)

30 x  3 y  0 x  3 y  29

29 x  29 x  1 y  10( 1) y  10

Subtract: In (1)

( x, y )  ( 1,10)

(d)

2 y  3x  17 3x  11  5 y Rearrange 3x  2 y  17 3x  5 y  11

Add: In (2)

(e)

(1) (2)

7 y  28 y4 3x  11  20 3x  9 x  3 ( x, y )  (3, 4)

2 x  3 y  13 3x  2 y  12 To eliminate y (3)  2  4 x  6 y  26 (2)  (3)   9 x  6 y  36 Add: In (1)

(1) (2)

5 x  10 x2 2(2)  3 y  13 3 y  9 y  3 ( x, y )  (2, 3)

(f )

2 x  3 y  11 y  13  3x From (1)  (2)  3 

(1) (2) 2 x  3 y  11  9 x  3 y  39

Add: In (2)

7 x  28 x  4 y  13  3( 4) y  13  12 y 1 ( x, y )  ( 4,1)

(g)

2a  3b  14  0 a b2  0 Rearrange and multiply (2)  2 2a  3b  14 2a  2b  4 Subtract: 5b  10 b  2 In (2) a22  0 a4 (a, b)  (4, 2)

(1) (2)

(h)

a  c  10 8a  4c  0 (1)  4 

(1) (2)

Subtract: In (1)

(i)

4a  4c  40 8a  4c  0 4a  40 a  10 10  c  10 c  20 (a, c)  (10, 20)

3b  3c  15 2b  4c  14 (1)  2  6b  6c  30 (2)  3  6b  12c  42 Add: 6c  12 c2 3b  6  15 In (1) 3b  9 b  3 (b, c )  ( 3, 2)

(1) (2)

(j)

48a  32b  128 16a  48b  32 (1) 16  3a  2b  8 (2) 16  a  3b  2 (4)  3  (3)  Subtract:

In (4)

(1) (2) (3) (4)

3a  9b  6 3a  2b  8 11b  2 2 b 11  2  a  3   2  11 

a2

6 11

2  6 ( a , b)   2 ,    11 11  (k)

0.5m  0.3n  54 0.3m  0.7n  74 (1)  10  (2)  10  (3)  3  (4)  5  Subtract:

In (3)

(l)

5m  3n  540 3m  7n  740

(1) (2) (3) (4)

15m  9n  1620 15m  35n  3700 26n  2080 n  80 5m  240  540 5m  300 m  60 (m, n)  (60,80)

3 5 3 m n  4 8 4 5 2 7 n m  6 3 9 To eliminate fractions (1)  8  6m  5n  6 (2)  18  15n  12m  14 (3)  2  12m  10n  12 (4)  12m  15n  14 Subtract: 5n  2 2 n 5

(1) (2) (3) (4)

 2 6m  5    6  5

In (3)

6m  4 2 m 3 2 2 (m, n)   ,  3 5 5.

(a)

Let the number of announcements be x and the total cost be $y.

(b)

Let the number of units of Product A be x and the number of units of Product

(a)

Let x represent the number of sports tires sold, y the number of all-season tires

B be y.

6. sold.

6 x  5 y  93 3 2 x y 0 4 3 (1)  12  To eliminate y (3)  5  (1)  8  Add:

(1) (2) 9x – 8y = 0

45x – 40y = 0 48x + 40y = 744 93x  744 x 8 9(8)  8 y  0 In (3) 8 y  72 y9 8 sports tires were sold, 9 all-season tires were sold.

(3)

(b)

Let the number of $2.50-tickets be x and that of $3.50-tickets be y. (1) x y  450 (2) 2.5x  3.5 y  1300 To eliminate x (1)  2  5 x  7 y  2600 (2)  5  5 x  5 y  2250 2 y  350 y  175 x  275 The number of $2.50-tickets is 275 and

Subtract:

the number of $3.50-tickets is 175.

(c)

Let the price of a jacket be $x and that of a pair of pants be $y. (1) x  2 y  175 (2) x  3y Substitute (2) in (1) 3 y  2 y  175 5 y  175 y  35 x  105 The price of the jacket is $105.

(d)

3x  2 y  236 x  12  2 y Rearrange (2) 3x  2 y  236 x  2 y  12 Add (1) and (2) 4 x  248 x  62 62  12  2 y 50  2 y y  25

(1) (2) (1) (2)

Let X be the number of doors to produce Let Y be the number of windows to produce 3x  5 y  100 20 x  10 y  1000

Self-Test 1.

(a)

y  x  2 x 0 2 2 y 2 0 4 x y 4 x 0 4 2 y 4 0 2

(b)

3x  2 y and x  2 x 0 2 2 y 0 3 3

(a)

For  x  55  y x 0 25 55 y 55 30 0

(b)

For 3x  2 y  600 x 0 100 200

y 300 150

(a)

For 4 y  11  y y

11 3

Slope, m  0 ; y -intercept, b   (b)

(c)

11 3

2 1 For x  y  1 3 9 6x  y  9 y  6x  9

Slope, m  6 ; y-intercept, b  9 For x  3 y  0 3y  x 1 y x 3 1 Slope, m   ; y -intercept, b  0 3

(d)

(e)

For  6 y  18  0 6 y  18 y  3

Line is parallel to the x-axis; slope, m  0 ; y-intercept, b  3 1 For 13  x  0 2 26  x  0 x  26 Line is parallel to the y-axis; slope, m is undefined ; there is no y -intercept

(f )

For ax  by  c by  ax  c a c y   x b b a c Slope, m   ; y-intercept, b  b b

(a)

6x  5 y  9 4 x  3 y  25 To eliminate y (1)  3  18 x  5 y  27 (2)  5  20 x  15 y  125

(1) (2)

38 x  152 x4 24  5 y  9 5 y  15 y  3

Add: In (1)

( x, y )  (4, 3)

(b)

12 – 7x = 4y 6 – 2y = 3x Rearrange 12  7 x  4 y 6  3x  2 y To eliminate y (1)  2  12  6 x  4 y 12  7 x  4 y Subtract: In (1)

(1) (2) (3) (4)

0 x 12  4 y y 3 ( x, y )  (0,3)

(c)

0.2a  0.3b  0 0.7a  0.2b  250

(1) (2)

To eliminate b from (1) and (2) (1)  10  2a  3b  0 (2)  10  7a  2b  2500 (3)  2  4a  6b  0 (4)  3  21a  6b  7500 Add: 25a  7500 a  300 In (3) 600  3b  0 b  200 (a, b)  (300, 200) (d)

4 3 17 b c   3 5 3 5 4 5 b c  6 9 9 (1)  15  20b  9c  85 (2) 18  15b  8c  10 (3)  3  60b  27c  255 (4)  4  60b  32c  40 Subtract: 59c  295 c5 In (4) 15b  40  10 15b  30 b  2 (b, c)  (2,5)

(3) (4)

(1) (2) (3) (4)

Let the amount invested at 4% be $x, and the amount invested at 6% be $y.

x  y  12000

(1)

The amount of interest earned at 4% is $0.04x; the amount of interest earned at 6% is $0.06y.

0.04 x  0.06 y  560

(2)

To eliminate x from (1) and (2) (2) 100  4 x  6 y  56, 000 (1)  4  4 x  4 y  48, 000

2 y  8000 y  4000 x  4000  12, 000 In (1) x  8000 The amount invested at 4% is $8000; Subtract:

the amount invested at 6% is $4000. 6.

Let Eyad’s share of the profit be $x, and Rahia’s share of the profit be $y. x  y  12, 700

(1)

2 y  2200 5 To eliminate fractions (2)  5  5 x  2 y  11, 000 To eliminate y from (1) and (3) Rearrange (3)  5 x  2 y  11, 000 (1)  2  2 x  2 y  25, 400 x

(2) (3)

7 x  36, 400 x  5200 5200  y  12, 700 y  7500

Add: In (1)

Rahia s share is $7500.

Challenge Problems 1.

Let the amount invested at 5% be $x; let the amount invested at 7% be $y; then the annual returns are $0.05x and $0.07y. x  y  60, 000 0.05 x  0.07 y (2) 100  5x  7 y

(1) (2)

5x  7 y  0 (1)  5  5 x  5 y  300,000 Subtract 12 y  300, 000 y  25, 000 x  35, 000 The total annual return  0.05(35, 000)  0.07(25, 000)  1750  1750  $3500 3500 The average annual return   0.0583  5.83%. 60,000 2.

Let the price per new bottle be $x; let the refund per returned bottle be $y. For September 3216x – 1824y = 574.56 (1) For October 5208x – 2232y = 944.88 (2) To eliminate y (1)  2232  7,178,112x – 4,071,168y = 1,282,417.92 (3) (2)  1824  9,499,392x – 4,071,168y = 1,723,461.12 (4) (4) – (3)  Substitute x  0.19in(1)

2,321,280x = 441,043.20 x = 0.19

3216(0.19)  1824 y  574.56 611.04  1824 y  574.56 1824 y  36.48 y  0.02 Polar Bay charges 19 cents for each new bottle and refunds 2 cents for each returned bottle. 3.

Let the current weekly pay for each painter be $x; let the current weekly pay for each helper be $y. Then the current weekly payroll is (1) 16 x  24 y  29760 The decrease in weekly wages for each painter is $0.10x; The decrease in weekly wages for each helper is $0.08y. Then the total decrease in the weekly payroll is 16(0.10 x)  24(0.08 y )  2688 1.60 x  1.92 y  2688 160 x  192 y  268,800 (2) (3) (1) 10  160 x  240 y  297,600 (3)  (2)  48 y  28,800 y  600 Substitute in (1) 16 x  24(600)  29, 760 16 x  14, 400  29, 760 16 x  15,360 x  960 960 The hourly rate for painters before the paycut   $24 40 600 The hourly rate for helpers before the paycut   $15 40 The hourly rate for painters after the paycut  0.90(24)  $21.60

The hourly rate for helpers after the paycut  0.92(15)  $13.80 3. a.

Occupancy rate  84 /120  0.7

70% of the rooms are sold. b.

REVPAR = ADR  Occupancy rate  = 180  0.7 = $126

c. Revenue = ADR (Occupancy rate) (total number of rooms available) Revenue = 180  x 120 = $21,600x d. Revenue  180 120x 15, 000  21, 600x x  0.694

To get a minimum of $15,000 in revenue, you should sell at least 69.4% of the available rooms. Case Study 1. Name of Amount Company Invested Barrick Gold $7125 TELUS 7125 Air Canada 7125 Bank of Montreal 7125 2.

Price per Share $25.60 $28.92 $17.95 $127.04

Number of Shares 7125/$25.60 = 278 7125/$28.92 = 246 7125/$17.95 = 396 7125/$127.04 = 56

Let the number of Barrick Gold shares be x; let the number of TELUS shares be y. (1) y  3x The amount invested in Barrick Gold is $25.60x; the amount invested in TELUS is $28.92y. Substitute (1): 25.60x  28.92(3x)  28,500

25.60x  86.76x 112.36x x Substitute in (1): y  3(253) y  759

 28,500  28,500  253.65

Amarjit will get 253 shares of Barrick Gold and 759 shares of TELUS.

Let the number of Bank of Montreal shares be x; let the number of Air Canada shares be y. x:2  y :3 x y  2 3 3x  2 y (1) x  2/3 y The amount invested in Bank of Montreal shares is $127.04x; the amount invested in Air Canada shares is $17.95y. (2) 127.04x  17.95y  28,500 Substitute (1) in (2): 127.04(2/3 y )  17.95 y  28,500 102.64  28,500 y  277.66 Substitute in (1): x  2/3(277.66)  185 Amarjit will buy 185 shares of Bank of Montreal and 277 shares of Air Canada.

PART 1 COMPREHENSIVE CASE QUESTIONS 1. CHAPTER 1

(a) i. Let monthly sales be $x and total salary be $y. y = 2000 + 0.02x y = 2000 + 0.02 (120,000) y = 4400 Karen would receive $4400 per month.

ii. To consider either job 61,200 / 12 = 2000 + 0.02x 5100 = 2000 + 0.02x 3100 = 0.02x x = 155,000 A monthly sales level of $155, 000 is necessary for Karen to consider either position.

(b) Cost of food items Cost of wine Total cost of meal

$120 27 $147

HST on food = 13% of $120 = 0.13(120) = $15.60 HST on wine = 13% of $27 = 0.13(27) = $3.51 Total cost including taxes = $147 + 15.60 + 3.51 = $166.11 Tip = 15% of $147 = 0.15(147) = $22.05 Total amount spent = $166.11 + $22.05 = $188.16

(c) i. First gross pay = 0.08 × 63,000 = $5040 Last gross pay = 0.12 × 63,000 = $7560 Gross pay for all other pay periods = 0.04 × 63, 000 = $2520

ii. Gross pay for a regular pay period = $2520 Less two days of pay = 2/200 of $63,000 = $630 Gross pay = $2520 – $630 = $1890

2. CHAPTER 2

(a) i. Annual interest = $5000 (0.03) (1) = $150 ii. Interest paid = $5000 (0.03)(9/12) = $112.50 Amount of interest saved = $150 – $112.50 = $37.50 (b) i. Let the number of roses be represented by x; then the number of daisies is (50 – x). The value of x dozen roses is $18x. The value of (50 – x) dozen daisies is $10(50 – x). The total value is $[18x +10(50 – x)]. Since the total budgeted value is given as $700, 18 x  10(50  x)  700

18 x  500  10 x  700 8 x  200 x  25 25 dozen roses and 25 dozen daisies can be purchased.

ii. Let the regular selling price of the dress be represented by $x. The reduction in price is $1/4x, and the reduced price is $(x – 1/4x). Since the reduced price is given as $900, x – 1/4x = 900 3/4x = 900 x = 4(900)/3 x = 1200 The regular selling price of the dress was $1200. 3. CHAPTER 3

(a) Divide the total wedding budget into 4 + 2 + 2 = 8 parts. Each part has a value of $20,000 / 8 = $2500.

Karen’s parents will pay 4 of the 8 parts: Sean’s parents will pay 2 of the 8 parts: Sean and Karen will pay 2 of the 8 parts:

4 × 2500 = $10, 000. 2 × 2500 = $5000. 2 × 2500 = $5000.

(b) i. Cost to host the reception at the Olympic Club = $2000 + $65(120) = $9800. Cost to host the reception at the Fairmont Hotel = $105 × 120 = $12,600. The Olympic Club is more affordable. ii. New budget for the reception = $9800 + $2000 = $11,800 Let x represent the maximum number of guests that will allow Sean and Karen to host the reception at the Fairmont Hotel. To stay within the budget, $105x = $11,800 Copyright © 2025 Pearson Canada Inc.

x = 112.380952

Karen and Sean can host 112 guests at the Fairmont Hotel without exceeding the new reception budget.

(c) Let the balance in the Bond Fund be $x; let the balance in the Equity Fund be $y; the total amount available in the mutual funds is x + y = (1.17) × $4000 x + y = $4680 (1) $750 less than three times the balance in the Bond Fund is y = 3x – 750 (2) Substitute (2) into (1) x + (3x – 750) = 4680 4x = 5430 x = 1357.50 y = 3x – 750 y = 3(1357.50) – 750 y = 3322.50 The balance in the Bond Fund is $1357.50, and the balance in the Equity Fund is $3322.50. Karen will get one-third of the balance in each of these funds, so she will get $452.50 from the Bond Fund and $1107.50 from the Equity Fund.

4. CHAPTER 4 (a) Assets Chequing/savings account(s) Investments Automobile(s) Cash value of life insurance RRSP(s) Other Total Assets

Sean $3200 17,000 n/a 3100 4000 600 $27,900

Karen $800 n/a 3000 n/a n/a n/a $3800

Total Joint Assets $4000 17,000 3000 3100 4000 600 $31,700

$11,000 n/a 5000 n/a 475 $16,475 ($12,675)

Total Joint Liabilities $18,500 n/a 6350 n/a 765 $25,615 $6085

Liabilities Student loans Car loans Other debts Line of credit Credit cards Total Liabilities Net Worth (Assets – Liabilities)

$7500 n/a 1350 n/a 290 $9140 $18,760

(b) Monthly Income Gross monthly income Paycheck deductions Net Income Monthly Expenses Housing (rent, utilities, cable, phone) Transportation (lease, gas, repairs, insurance, parking) Medical and dental Living expenses (groceries, clothes, entertainment, miscellaneous) Credit payments Total Monthly Expenses The Difference

Sean

Karen

Total

$5250 1313 $3937

$5100 1020 $4080

$10,350

$1700

$ 780

$ 2480

1140

435

1575

100 425

200 460

300 885

200 $3565

780 $2655

980 $ 6220

$ 372

$1425

$ 1797

$ 8017

(c) é1  (1  (fixed rate  5% /12)) 24 ù ú Loan Amount  Monthly Payment ê ê ú (fixed rate  5% /12) ë û é1  (1  (0.035  0.05 /12)) 24 ù ú $7500  Monthly Payment ê ê ú (0.035  0.05 /12) ë û é 0.6023 ù ú $7500  Monthly Payment ê êë0.03917 ú û

$7500  Monthly Payment 15.3779 é $7500 ù ú Monthly Payment  ê êë15.3779 ú û Monthly Payment  487.713  $487.713

(d) i. Sean’s annual salary = $63,000 Karen’s annual salary = $61,200 Total gross household income = $63,000 + $61,200 = $124,200 2.5 × $124,200 = $310,500 3 × $124,200 = $372,600 Sean and Karen can afford a home in the range of $310,500 to $372, 600.

(Monthly mortgage payment + Monthly property taxes  Monthly heating)  100% (Gross montly income) ($2300  $250  $180) GDS   0.263768  100%  26.4% $10,350

ii. GDS 

(Monthly mortgage payment + Monthly property taxes  All other monthly debts)  100% (Gross montly income) ($2300  $250  $180  $425  $750) TDS   0.377295  100%  37.73% $10,350

iii. TDS 

iv. Yes, since the ratios are within the guidelines, I would approve the mortgage.

Chapter 5

Cost-Volume-Profit Analysis and Break-Even

Exercise 5.1

A. 1.

(a)

(b)

  where x represents the  number of units per period.  

TR = 120 x

TC  2800  (35  15) x

To break even, TR = TC 120 x  2800  50 x 70 x  2800 x  40

Break-even volume is 40 units. 2.

Break-even volume in sales dollars  40  120  $4800

Break-even volume as a percent of capacity



40  0.40  40% 100

(c)

(a)

Let the volume be x units 1.

TR = 12x

TC  1200  8x

(b)

To break even, 12 x  1200  8 x 4 x  1200 x  300

Break-even volume is 300 units. 2.

Break-even volume in sales dollars  300  12  $3600

Break-even volume as a percent of capacity



300  0.30  30% 1000

(c)

(a)

Expressing the functions in terms of sales. Let X represent the number of units sold. Assuming that the price per unit is $1, then TR = $1X .

Expressing the functions in terms of sales. Let X represent the number of units sold.

Total variable cost 324, 000   0.45 Total revenue 720, 000 Total variable cost = $0.45 X Total cost  TC  FC  TVC = 220, 000 + 0.45 X (b)

Break -even point : TR = TC 1X = 220, 000 + 0.45 X 0.55 X = 220, 000 X = 400, 000

The break-even point in units is 400, 000 units 2. $400, 000

Since each unit sells for $1, then the break-even sales in dollars is

400,000  0.50  50% 800,000

Break-even as a percent of capacity 

Expressing the functions in terms of sales.

(c)

(a)

Let X represent the number of units sold. Assuming that the price per unit is $1, then TR = $1X . 2.

Expressing the functions in terms of sales. Let X represent the number of units sold.

Total variable cost 48, 000   0.40 Total revenue 120, 000 Total variable cost = $0.40 X Total cost  TC  FC  TVC = 43, 200 + 0.40 X (b)

Break -even point : TR = TC 1X = 43, 200 + 0.40 X 0.60 X = 43, 200 X = 72, 000 The break-even point in units is 72, 000 units

2. $72,000

Since each unit sells for $1, then the break-even sales in dollars is

Break-even as a percent of capacity 

72, 000  0.48  48% 150, 000

(c)

B. 1.

Let the volume be x units

Revenue = 6.95 x Total cost = 1800 + 3.95 x To break even, 6.95 x  1800  3.95 x 3x  1800 x  600

Break-even volume is 600 books. 2.

Let the volume be x units

Revenue = 18 x Total cost = 280 + 4 x To break even, 18 x  280  4 x 14 x  280 x  20

Break-even volume is 20 scarves. 3.

Let x represent the volume in units

Revenue = 30 x Total cost  630  16 x

At break even, 30 x  630  16 x 30 x  16 x  630 14 x  630 x  45

Break-even volume is 45 games.

Let x represent the volume in units.

Revenue = 69 x Total cost  243  42 x At break even, 69 x  243  42 x 69 x  42 x  243 27 x  243 x9

Break-even volume is 9 hats. 5.

Let X represent the quantity in units. Assume the unit price is $1. The variable cost per unit is then Total variable cost 259, 000   $0.35 Total revenue 740, 000 Total cost  0.35 X  11, 700 At break even, 1X  0.35 X  11, 700 0.65 X  11, 700 X  18, 000

Total revenue  118, 000 units = $18, 000. Break-even revenue is $18,000. 6.

Let X represent the quantity in units. Assume the unit price is $1. The variable cost per unit is then Total variable cost 36,550   $0.43 Total revenue 85, 000 Total cost  0.43 X + 27,360

At break even, 1X  0.43 X  27,360 0.57 X  27,360 X  48, 000

Total revenue  1 48,000 units = $48,000. Break-even revenue is $48,000.

Let P represent the price per unit.

Revenue = 60 P

60 P  (3  60)  417  0 60 P  597 P  $9.95 To break even, the price to charge is $9.95. 8.

Let P represent the charge per participant.

Revenue = 90 P Total fixed cost = 200 +150 + 280 = 630

90 P  (4  90)  630  0 90 P  990 P  $11 To break even, the charge per participant is $11. 9.

(a)

Let FC represent the fixed cost. Total cost = (2  80) + FC

(22  80)  (2  80)  FC  0 1760  160  FC FC  1600 To break even, the fixed cost must be $1600. (b)

Substitute the desired profit into the formula:

(22  80)  (2  80)  FC  900 1760  160  900  FC FC  700 To realize a $900 profit, the fixed cost must be $700. 10. (a)

Let FC represent the fixed cost. Total cost = (0.60  200) + FC

(2  200)  (0.60  200)  FC  0 400  120  FC FC  280 To break even, the fixed cost must be $280.

(b)

Substitute the new quantity into the formula:

(2  300)  (0.60  300)  FC  0 600  180  FC FC  420 To break even, the fixed cost can be $420. 11. (a)

Let X represent the quantity of units.

(35 X )  (8 X )  756  0 27 X  756 X  28 To break even, the quantity is 28 unfinished boxes. (b)

Let VC represent the unit variable cost.

(30 100)  (VC 100)  756 3000  (VC 100) (VC 100) VC $10.44  $8  $2.44

   

12 100 1200 + 756 1200  756  3000 10.44

To make the desired profit, an additional $2.44 can be spent on each unit. 12. (a)

Let VC represent the unit variable cost.

FC  120  300  420 (40  20)  (VC  20)  420 800  (VC  20) (VC  20) VC

   

0 420 420  800 19

To break even, the variable cost per child is $19. (b)

(40  20)  (VC  20)  420 800  (VC  20) (VC  20) VC

   

200 620 620  800 9

To make the desired profit, the variable cost per child is $9.

13. (a)

At the break-even point, the total revenue equals to total cost, TR = TC TR = FC + TVC TR = SP(X) = 65X and TC = FC + VC(X) SP(X) = FC + VC(X)

FC 2690 = = 827.6923 SP - VC 10 - 6.75

Therefore, John has to sell 828 ―BBQ Meals‖ to break even. (b)

SP( X ) - FC - VC( X ) = PFT 10( X ) - 2690 - 6.75( X ) = 2500

Solve for X.

3.25( X ) = 5190

5190 = 1596.92 3.25

To make a $2500 profit, John needs to sell 1597 ―BBQ Meals‖. Exercise 5.2 A. 1.

(a)

Contribution margin is $150  ($61  $17)  $72

The contribution margin is $72. (b)

Contribution rate is

$72 / $150  0.48  48% (c)

Let X represent the volume in units. To break even,

150 X  78 X  5904 (150  78) X 72 X X

   

0 5904 5904 82

Or: Break-even volume  Fixed cost / Contribution margin = $5904 / $72 = 82 The break-even volume is 82 units.

(d)

At the break-even volume, the sales dollars are 82  $150  $12,300

The sales dollars at break-even are $12,300. 2.

(a)

Contribution margin is $499  $60  $439 The contribution margin is $439.

(b)

Contribution rate is

$439 / $499  0.879760  87.976% (c)

Let X represent the volume in units. To break even,

FC  1800  834  2634 499 X  60 X  ($1800  $834) (499  60) X 439 X X

   

0 2634 2634 6

Or: Break-even volume  Fixed cost/Contribution margin = 2634/439 = 6 The break-even volume is 6 cakes. (d)

At the break-even volume, the sales dollars are

6  $499  $2994 The sales dollars at break-even are $2994. 3.

(a)

Contribution margin is

$99  $53  $46 The contribution margin is $46. (b)

Contribution rate is

$46 / $99 = 0.46 = 46.47% (rounded) (c)

Let X represent the volume in units. To break even,

99 X  53 X  500 (99  53) X 46 X X

   

0 500 500 10.87

Or: Break-even volume  Fixed cost/Contribution margin = 500/46 = 10.87 The break-even volume is 11 cell phones.

(d)

At the break-even volume, the sales dollars are

11 $99  $1089 The sales dollars at break-even are $1089. 4.

(a)

Contribution margin is $(100*0.80)  $38  $42

The contribution margin is $42. (b)

Contribution rate is

$42 / $80  0.525  52.5% (c)

Let X represent the number of rooms. To break even,

80 X  38 X  720 (80  38) X 42 X X

   

0 720 720 17.14

Or: Break-even volume  Fixed cost/Contribution margin = 720/42 = 17.14 Remember that we always round up to find break-even point. The break-even volume is 18 rooms. (d)

At the break-even volume, the sales dollars are

18  $80  $1440 The sales dollars at break-even are $1440. B. 1.

(a)

Contribution margin is

$1,020,000  $581, 400  $438,600 (b)

Contribution rate is

$438,600 / $1,020,000  0.43  43%

(c)

Let X represent the volume in units. Assume price per unit is $1. Break-even volume  Fixed cost/Contribution rate = $160, 000 / 0.43 = 372, 093.0233

The break-even volume is 372,094 units. At the break-even volume, the sales dollars are 372, 094  $1  $372, 094

The sales dollars at break-even are $372, 094. 2.

(a)

Contribution margin is

$765,000  $497, 250  $267,750 (b)

Contribution rate is

$267,750 / $765,000  0.35  35% (c)

Let X represent the volume in units. Assume price per unit is $1. Break-even volume  Fixed cost/Contribution rate = $152,100 / 0.35 = 434,571.4286

The break-even volume is 434,572 units. At the break-even volume, the sales dollars are 434,572  $1  $434,572

The sales dollars at break-even are $434,572. 3.

(a)

Contribution margin is $130,000  $32,500  $97,500

(b)

Contribution rate is $97,500 / $130,000  0.75  75%

(c)

Let X represent the volume in units. Assume price per unit is $1.

Break-even volume  Fixed cost/Contribution rate = $85,000 / 0.75 = 113,333.33

The break-even volume is 113,334 units At the break-even volume, the break-even sales at $1 per unit are $113,334. 4.

(a)

Contribution margin is $1, 436,000  $545,680  $890,320

(b)

Contribution rate is $890,320 / $1, 436,000  0.62  62%

(c)

Let X represent the volume in units. Assume price per unit is $1. Break-even volume  Fixed cost/Contribution rate = $650, 000 / 0.62 = 1, 048,387.097

The break-even volume is 1,048,388 units At the break-even volume the break-even sales at $1 per unit are $1, 048,388. C. 1.

Contribution margin is

$19.99  $7  $12.99 Let X represent the volume in units. Break-even volume  Fixed cost/Contribution margin

= 346/12.99  26.64 The break-even volume is 27 watches. 2.

Let PFT represent the profit.

(369.60 16)  (168 16)  2465  PFT (201.60) 16  2465  PFT PFT = 760.60 The profit is $760.60. 3.

Contribution margin is

$21.99  $6.59  $15.40

Let X represent the volume in units. Break-even volume  Fixed cost/Contribution margin

= 2602.60/15.40  169 The break-even volume is 169 oil changes.

Let P represent the price per unit.

(30 P  30  8)  300  0 30 P  540 P = 18 The price per unit is $18. 5.

(a)

Let X represent the volume in units. Break-even volume  Fixed cost/Contribution margin

= 140/(3  1.25) X  80 The break-even volume is 80 chocolate bars. (b)

Let PFT represent the profit.

(3 1000)  (1.25 1000)  140  PFT (1.75) 1000  140  PFT PFT = 1610 The profit is $1610. 6.

Contribution margin is $50  2(12)  5  $21

Let X represent the volume in units. To break even, 50 X  29 X  395  0 21X  395 X  18.8

Or: Break-even volume  Fixed cost/Contribution margin = 395/21 = 18.809524 The break-even volume is 19 jobs.

Contribution margin is $902,000  $613,360  $288,640. Contribution rate is $288, 640 / $902, 000  0.32  32% Let X represent the volume in units. Assume price per unit is $1. Break-even volume  Fixed cost/Contribution rate = $232, 400 / 0.32 = 726, 250

The break-even volume is 726,250 units At the break-even volume the break-even sales at $1 per unit are $726, 250. 8.

Contribution margin is $298, 000  $184, 760  $113, 240. Contribution rate is $113, 240 / $298, 000  0.38 Let X represent the volume in units. Assume price per unit is $1. Break-even volume  Fixed cost/Contribution rate = $76, 660 / 0.38 = 201, 736.8421

The break-even volume is 201,737 units At the break-even volume the break-even sales at $1 per unit are $201,737. 9.

Net Income (NI) = $178, 000, Contribution Rate (CR) = 0.27, FC = $151, 200, S = Sales NI = S  TC = S  (FC  TVC) 178, 000  S  (151, 200  TVC) 329, 200 = S  TVC Contribution Rate (CR) = CM / S = (S  TVC) / S 0.27 = 329, 200 / S S  329, 200 / 0.27 S  1, 219, 259.26 Last year’s sales were $1, 219, 259.26.

10.

Net Income (NI) = $224, 000, Contribution Rate (CR) = 0.403, FC = $315, 000, S = Sales NI = S  TC = S  (FC  TVC) 224, 000  S  (315, 000  TVC) 539, 000 = S  TVC Contribution Rate (CR) = CM / S = (S  TVC) / S 0.403 = 539, 000 / S S  539, 000 / 0.403 S = 1,337, 468.98 Last year’s sales were $1,337, 468.98. 11. Contribution Margin (CM) = S  TVC, TVC = TC  FC 145, 600 = 416, 000  (TC  FC)  416, 000  (TC  79,800) 145, 600  416, 000  79,800  TC TC = 350, 200 TVC = 350, 200  79,800 = 270, 400 CM = S  TVC = 416, 000  270, 400  145, 600 Contribution Rate (CR) = CM / S = 145, 600 / 416, 000 = 0.35 Break-even volume = FC / CR = 79,800 / 0.35 = 79,800 / 0.35 = 228, 000 The break-even volume is 228,000 units. 12.

To break even, net income = NI = 0. Net Income (NI) = S  TC = 192, 000  TC = 192, 000  (FC  TVC)  192, 000  (FC + 99,840) At break -even NI  0 192, 000  FC  99,840  0 FC  92,160

The most the fixed cost would have to be to break even is $92,160. 13. (a)

At break-even, total revenue equals total cost, TR = TC TR = FC + TVC TR = SP(X) = 65X and TC = FC + VC(X)

SP(X) = FC + VC(X)

X

FC 2285   45.34 SP  VC 65  14.60

Therefore, Peter has to make 46 consultations to break even.

(b)

SP( X ) - FC - VC ( X ) = PFT 65( X ) - 2285 - 14.60( X ) = 1000

Solve for X.

50.40( X ) = 3285

3285 = 65.1786 50.40

To make a $1000 profit, Peter needs to make 66 consultations. (c)

A Contribution Margin is the contribution of each unit of product needed to cover costs before the break-even point. The profit increases by the amount of contribution margin after the break-even point. Mathematically, it is calculated by subtracting the unit variable cost from the unit selling price of a product. Contribution Margin (CM) = 65 - 14.60 = 50.40

A Contribution Margin of $50.40 means that this amount is used to cover the fixed cost of Peter’s consultation business before the break-even point and then for each additional consultation, he makes $50.40 in profit. To calculate the break-even point, divide the fixed cost by contribution margin.

Break-even Point =

2285 = 45.34 50.40

which is the same answer as we got in part (a). Exercise 5.3 1.

(a)

Let X represent the volume in units.

To break even, 14 X  8 X  984  0 6X = 984 X = 164 The break-even volume is 164 units. (b)

Let X represent the volume in units.

To break even, 12 X  8 X  984  0 4X = 984 X = 246 The break-even volume is 246 units.

(c)

Let X represent the volume in units.

To break even, 14 X  8 X  1500  0 6X = 1500 X = 250 The break-even volume is 250 units. (d)

Let P represent the price per units

To break even, 300P  8(300)  1500  0 300P = 3900 P = 13 The break-even price is $13. 2.

(a)

Let X represent the volume in units.

To break even, FC  190  321  511 9.99 X  2.69 X  511  0 7.30 X = 511 X = 70 The break-even volume is 70 calendars. (b)

Let X represent the volume in units.

To break even, 9.99 X  2.69 X  430.70  0 7.30 X = 430.70 X = 59 The break-even volume is 59 calendars. (c)

Let PFT represent the profit in dollars.

New unit price is 75% of $9.99 = $7.49. (7.49 120)  (2.69 120)  511  PFT [(7.49  2.69) 120]  511  PFT PFT = $65 The profit is $65.

(d)

Let P represent the price per units

To break even, 200 P  2.69  200  412  0 200 P = 950 P = $4.75 The break-even price is $4.75. 3.

(a)

Let X represent the volume in units.

To break even, FC  2500  900  3400 5.69 X  2.20 X  3400  0 3.49 X = 3400 X = 974.212034 The break-even volume is 975 sandwiches. (b)

Let X represent the volume in units.

To break even, 5.69 X  2.40 X  3400  0 3.29 X = 3400 X = 1033.43465 The break-even volume is 1034 sandwiches. (c)

Let PFT represent the profit in dollars.

(5.69  1600)  (2.20  1600)  3490  PFT ((5.69  2.20)  1600)  3490 = PFT PFT = $2094 The profit is $2094. (d)

Let X represent the volume in units.

5.49 X  2.20 X  3400  1000 3.29 X = 4400 X = 1337.386018 The break-even volume is 1338 sandwiches. 4.

(a)

Let X represent the volume in units.

5 X  3.75 X  190  100 1.25 X = 290 X = 232 The break-even volume is 232 magazines.

(b)

Let X represent the volume in units.

5 X  4 X  190  150 1X = 340 X = 340 The break-even volume is 340 magazines. (c)

Let PFT be the profit.

(4.50  300)  (3.75  300)  190  PFT 0.75  300  190 = PFT PFT = 35 The profit would be $35. (d)

Let P represent the price per unit.

200 P  ((0.70  5)  200)  190  0 200 P = 890 P = 4.45 The break-even price is $4.45. 5.

(a)

Contribution Margin (CM) = S  TVC = 946, 000  227, 040  $718,960 The contribution margin is $718,960. Contribution Rate (CR) = CM / S = 718,960 / 946, 000 = 0.76 = 76% The contribution rate is 76%.

(b)

Break-even volume  FC/CR = 588 000/0.76 = 773 684.2105 The break-even volume is 773 685 meals.

(c)

New fixed cost  $588, 000  $23, 000  $611, 000 Break-even volume  FC/CR = 611 000/0.76 = 803 947.3684 The break-even volume is 803,948 meals.

(d)

New sales = $946,000 + 0.05($946,000) = $993,300 Assume variable cost also increases by 5%

Net income = Sales  Total cost = $993,300  $588, 000  (1.05)($227,040)  $166,908

Net income is $166,908.

(a)

Contribution Margin (CM) = S  TVC = $1, 238, 000  $841,840  $396,160 The contribution margin is $396,160. Contribution Rate (CR) = CM / S = 396,160 /1, 238, 000 = 0.32 The contribution margin is 32%.

(b)

Break-even volume  FC / CR = 218,000 / 0.32 = 681, 250 The break-even volume is 681, 250 backpacks.

(c)

CM = S  TVC = $1, 238, 000  ($841,840  $56, 000) = $452,160 CR = CM / S = 452,160 /1, 238, 000 = 0.365234249 Break-even volume = FC / CR = 218, 000 / 0.365234249 = 596,877.2116

The break-even volume is 596,878 backpacks. (d)

New TVC = $841,840 + 0.05($841,840) = $883,932 New FC = $218, 000 + $15, 000 = $233, 000 New sales = S = $1, 238, 000 + 0.10($1, 238, 000) = $1,361,800 New income = S  TC = S  (FC + TVC)  $1,361,800  ($233, 000  $883,932)  $244,868 Net income will be $244,868. 7.

(a)

Fixed cost = $1,200,000 Total revenue (sales) = $2,300,000 Total variable cost = $750,000 Contribution Margin (CM) = 2,300,000 – 750,000 = $1,550,000

Contribution rate 

1,550, 000  0.673913 2,300, 000

Contribution rate is 67.39%.

1, 200,000  $1,780,645.28 0.673913

(b)

Break-even =

(c)

Fixed cost = 1,200,000 + 250,000 = $1,450,000 Break-even =

(d)

1, 450,000  $2,151,613.04 0.673913

Fixed cost = $1,200,000 Total revenue (sales) = 2,300,000(1.15) = $2,645,000 Total variable cost = 750,000(1.15) = $862,500

Contribution Margin (CM) = 2,645,000 – 862,500 = $1,782,500 Net income = 1,782,500 – 1,200,000 = $582,500

Business Math News Box 1. Contribution Margin (CM)  $520,325 – 713,379  $ 193, 054  Contribution Rate (CR) = CM/Sales  193,054 / 520,325  0.371 2. At break-even, total revenue equals total cost. TR = TC TR – TC = 0 520,325 – 822,506   $302,181 (The company must sell $302,181 more cannabis to break even.) 3. Fixed Cost = 109,127 + $80,000 = $189,127 Total Cost = 189,127 + 713,379 = $902,506 Sales Revenue = $520,325 Additional sales needed to break even  902,506 – 520,325  $382,181 4. Sales  520,325 1.2   $624,390 No, the company will not reach the break-even point if they sell 20% more cannabis. Review Exercise 1.

(a)

(b)

Contribution margin per unit = 185  157 = $28

Contribution rate =

To break even, 185 x  3136  157 x 28 x  3136 x  112

28 = 0.1513514 = 15.14% 185

Or, break-even volume  Fixed cost/Contribution margin = 3136/28 = 112 Break-even volume is 112 units. 2.

Break-even volume as a percent of capacity

 3. (c)

112  0.35  35% 320

Break-even volume in dollars  112(185)  $20, 720

Let the number of units be x. Then:

Revenue = 185 x Total cost = 3136 +157 x

(d)

If fixed cost is $2688

185 x  2688  157 x 28 x  2688 x  96 96 Output level =  30% 320 2.

If fixed cost is $4588 and VC is 80% of SP

185 x  4588  148 x 37 x  4588 x  124 124 Output level =  38.75% 320 3.

If SP is $171

171x  3136  157 x 14 x  3136 x  224 224 Output level =  70% 320 2.

(a)

(b)

Contribution margin = 19,360  13,552 = $5808

Contribution rate =

x  4800  0.70 x 0.30 x  4800 x  16, 000

5808 = 0.30 = 30% 19,360

Break-even volume is $16,000.

Break-even point as a percent of capacity



16, 000 32, 000

 50% (c)

Let x represent the sales volume in dollars.

Revenue = x 13,552 Variable cost =  0.70 x 19,360 Total cost = 4800 + 0.70 x

(d)

FC  4800  600  4200 x  4200  0.70 x 0.30 x  4200 x  14, 000 Break-even point is a sales volume of $14,000, or at

14, 000  43.75% of capacity. 32, 000

Let x represent the sales volume in dollars.

VC  0.55 x x  5670  0.55 x 0.45 x  5670 x  12, 600 Break-even point is a sales volume of $12,600, or at 3.

(a)

(b)

12,600  39.375% of capacity. 32,000

Contribution margin = 400,000  260,000 = $140, 000

Contribution rate =

x  105, 000  0.65 x 0.35 x  105, 000 x  300, 000

140,000 = 0.35 = 35% 400,000

Or, break-even volume = 105,000 / 0.35 = 300,000 Break-even volume as a percent of capacity

 2. (c)

300, 000  60% 500, 000

Break-even volume is a sales volume of $300, 000.

Let x represent the sales volume in dollars. Revenue = x 260, 000 Variable cost =  65% of revenue = 0.65 x 400, 000 Total cost = 105, 000 + 0.65 x

(d)

FC  105, 000  11, 200  $93,800 VC = 72% of revenue = 0.72 x x  93,800  0.72 x 0.28 x  93,800 x  335, 000 Break-even volume is $335, 000.

(a)

(b)

Contribution margin per unit = 640  360 = $280

Contribution rate =

280 = 0.4375 = 43.75% 640

Let the number of units be x. Revenue = 640 x Total cost = 26,880 + 360 x

To break even 640 x  26,880  360 x 280 x  26,880 x  96

Or, break-even volume = 26,880 / 280 = 96 The break-even point is 96 units. 2.

Break-even point as a percent of capacity

 3.

96  0.64  64% 150

Break-even point in dollars

 96(640)  $61, 440

(c)

For FC  $32, 200 640 x  32, 200  360 x 280 x  32, 200 x  115

Break-even point in dollars is 115(640)  $73,600.

(d)

VC = 60% of 640 x = 384 x 640 x  23,808  384 x 256 x  23,808 x  93

Break-even point  5.

93  0.62  62% of capacity. 150

FC = 1500 + 2000 +1700 = 5200 Let x = number of units 9.99 x  3.50 x  5200  0 6.49 x  5200 x  801.232666

They need to sell 802 meals to break even. 6.

(a)

Let P = price for each job. 80P  80  52  1840  0 80P  1840  4160 80P  6000 P  75

They must charge 75 for each job to break even. (b)

Let P = price for each job. 80P  80  52  1840  1200 80P  7200 P  90

They must charge 90 for each job to break even. (c)

Let PFT = profit. 90  90  90  52  1840  PFT 8100  4680  1840  1580 The profit would be 1580.

(d)

Let P = price for each job. 100P  100  52  1840  0 100P  7040 P = 70.40

They minimum price they could charge would be 70.40

(a)

Contribution Margin (CM) = S  TVC = $2,995, 200  $778, 752  $2, 216, 448 The contribution margin is $2, 216, 448. Contribution Rate (CR) = CM / S = 2, 216, 448 / 2,995, 200 = 0.74 The contribution margin is 74%.

(b)

Break-even volume = FC / CR = 1,962, 000 / 0.74 = 2, 651,351.351 The break-even volume is 2,651,352 units.

(c)

New TVC = $778, 752 + 0.45($778, 752) = $1,129,190.40 New Sales = S = $2,995, 200 + 0.20($2,995, 200) = $3,594, 240 CM = S  TVC = $3,594, 240  $1,129,190.40  $2, 465, 049.60 CR = CM / S = 2, 465, 049.6 / 3,594, 240  0.68583 Break-even volume = FC / CR = 1,962, 000 / 0.68583 = 2,860, 753.34 The break-even volume is 2,860, 754 units.

(d)

New TVC = $778, 752  0.07($778, 752)  $724, 239.36 New FC = $1,962, 000 + $9, 000 = $1,971, 000 Net income = S  TC = S  (FC + TVC)  $2,995, 200  ($1,971, 000  $724, 239.36)  $299,960.64 Net income will be $299,960.64. Self-Test 1.

(a)

1. 2.

(b)

Contribution margin per unit = 10  (2.60  2.40  (0.2 10)) = $3 3 Contribution rate = = 0.3 = 30% 10 To break even, NI = 0 10 x  18, 000  7 x

3x  18, 000 x  6000 Or, break-even volume  18, 000 / 3  6000

Break-even volume is 6000 jump drives. 2.

Break-even volume, in sales dollars,  10  6000  $60,000

Break-even volume, as a percent of capacity, 

6000  40% 15, 000

(c)

Let the number of jump drives be x. Revenue = 10 x Cost = 18, 000  2.60 x  2.40 x  (0.2)(10) x  18, 000  7 x

(d)

Fixed cost = 18, 000  1600  19, 600 Variable cost = 2.10 x  2.40 x  2 x  6.50 x 10 x = 19, 600 + 6.50 x 3.50 x = 19, 600 x = 5600 (units)

(e)

Revenue = (10 +10% of 10) x  11x Variable cost = 2.60 x  2.40 x  ((0.2)(11)) x  7.20 x Fixed cost = 18, 000 + 2900 = 20,900 11x = 20,900 + 7.20 x 3.80 x = 20,900 x = 5500 (units)

(a)

Let Revenue be x. 270, 000  Revenue = 0.45 x 600, 000 Total cost = 275, 000 + 0.45 x

Variable cost =

Contribution margin = 600,000  270,000 = $330,000

Contribution rate =

330, 000 = 0.55 = 55% 600, 000

(b)

x  275, 000  0.45 x  NI To break even, NI = 0 x  275, 000  0.45 x 0.55 x  275, 000 x  $500, 000 Or, break-even volume = 275, 000 / .55 = 500, 000

Break-even volume is $500, 000 2.

(c)

Break-even volume as a percent of capacity 500,000   0.625  62.5% 800,000

x  (275, 000  40, 000)  0.4 x 0.6 x  315, 000 x  $525, 000

Challenge Problems 1.

Profit = 0.7 x  24.5 Cost = 0.9 x  24.5

Revenue = Cost + profit = 0.9 x  24.5  0.7 x  24.5  1.6 x To break even, Revenue = Cost 1.6 x  0.9 x  24.5 0.7 x  24.5 x  35 The break-even point is 35 million units. 2.

Target volume = 1500 units Selling price = $479 Fixed cost = $3000 × 12 months = $36,000 Variable cost per unit = $190

(a)

For break-even, TR = TC Let X represent the break-even volume 479X = 36,000 + 190X 289X = 36,000 X = 124.567474 The break-even volume is 125 guitars.

(b)

Break-even as a percent of capacity = 125/2000 × 100 = 6.25%

(c)

PFT = (SP × X ) – FC – (VC × X ) PFT = (479 × 1500) – 36,000 – (190 × 1500) PFT = 718,500 – 36,000 – 285,000 PFT = $397,500

(d)

PFT = (479 × 1200) – 36,000 – (190 × 1200) PFT = 574,800 – 36,000 – 228,000 PFT = $310,800

(e)

(397,500 – 310,800) / 397,500 × 100 = 0.218113 = 21.81% 1 – 0.218113 = 0.781887 = 78.19%

If only 100 units are sold each month, then Clapton Guitar Company will miss its expected profit by 21.81%. It will earn only 78.19% of its expected profit this year. 3.

FC1 = fixed cost for first piece of equipment VC1 = variable cost for first piece of equipment FC2 = fixed cost for second piece of equipment VC2 = variable cost for second piece of equipment Let X represent the break-even volume.

For break-even volume, FC1 + (VC1  X ) = FC2 (VC2  X ) 125,000 + (0.15X ) = 75,000 + (0.25X ) 50,000 = 0.25X – 0.15X 50,000 = 0.10X X = 500,000 The break-even volume is 500,000. If the manufacturer anticipates that it will continue to produce less than 500,000 units, then it should chose the second piece of equipment, since it has lower fixed cost. Case Study 1.

(Selling Price  Volume) – (Variable Cost per unit  Volume) – Fixed Cost = Profit ($35  (60/10)  3  90) – ($15  3  90) – (3  $6500) – (2  $995) – ($100  90) = Profit $56,700 – $4050 – $19,500 – $1990 – $9000 = $22,160 He expects to make $22,160 profit this summer.

For break-even, profit = 0 (SP  (60/10)  3  90) – $4050 – $19,500 – $1990 – $9000 = 0 1620 SP = $34,540 SP = $21.32 He would break even if the half-hour rental was $21.32 per Segway.

For a profit of $300 per day (SP  (60/10)  3  90) – $4050 – $19,500 – $1990 – $9000 = ($300 × 90) 1620 SP – $34,540 = $27,000 1620 SP = $61,540 SP = $37.99 He needs to charge $2.99 more than the rate that he currently charges per half-hour tour if he wants to realize a profit of $300 per day.

PART TWO

Mathematics of Business and Management

Chapter 6

Trade Discount, Cash Discount, Markup, and Markdown

Exercise 6.1 1.

37.5% of 125.64  $47.115 125.64  47.115  78.525  $78.53

2 16 % of 49.98  $8.33 3 49.98  8.33  $41.65

17.5% of List  560 0.175L  560 560 List   $3200 0.175

List 

762.50  $1220 0.625

2 16 % of List  14.82 3 1 L  14.82 6 L  88.92 Net  88.92  14.82  $74.10

25% of List  44.75

0.25L  44.75 L  $179 Net price  L – A  179 – 44.75  $134.25 7.

0.83L  355 355 0.83 List  $426 L

63.31  0.65L 63.31 0.65 List  $97.40 L

Discount  975  820  $155 155 Rate   15.90% 975

10.

Discount  1136  760  $376 376 Rate   33.10% 1136

11. Single rate  12.

769.99  449.79 320.20   0.415849  41.58% 769.99 769.99

Net price  399.99  120  42  $237.99

Single rate  13. (a)

120  42 162   0.405010  40.50% 399.99 399.99

Single rate  1  (0.7)(0.875)  1  0.6125  0.3875  38.75%

(b)

Single rate  1  (0.6)(0.8)(0.97)  1  0.5173  0.4826  48.27%

14. (a)

Single rate  1  (0.83)(0.925)  1  0.77083  0.22916  22.92%

(b) 15. (a)

Single rate  1  (0.75)(0.916)(0.98)  1  0.67375  0.32625  32.63% Net  (0.7)(0.8)(0.95)(599)  $318.67

(b)

Discount  599  318.67  $280.33

(c)

Single rate 

280.33  0.467997  46.80% 599

Or: Net price factor  1   0.7 0.8 0.95  1  0.532  0.468  46.80%

16. (a)

Net  (0.83)(0.9)(0.92)(1074) = $741.06

(b)

Discount  1074  741.06  $332.94

(c)

Single rate  1  (0.83)(0.9)(0.92)  1  0.69  31.00%

17. (a)

Net  (0.64)(0.9)(0.98)(786.20)  $443.79

(b)

Discount  786.20  443.79  $342.41

(c)

Single rate  1  (0.64)(0.9)(0.98)  1  0.56448  0.43552  43.55%

18. (a)

Net  (0.816)(0.908)(0.97)(1293.44)  $931.27

(b)

Discount  1293.44  931.27  $362.17

(c)

Single rate  1  (0.816)(0.908)(0.97)  1  0.719991  0.280009  28.00%

19.

Net  750(0.8)(0.95)(0.98)  $558.60 Additional discount  558.60  474.81  $83.79 83.79 Additional rate   15.00% 558.60

20.

Net  440(0.75)(0.85)  $280.50 Additional discount  280.50  274.89  $5.61 5.61 Additional rate   2.00% 280.50

21.

Net  180(0.7)(0.875)(0.95)  $104.74 Additional discount  104.74  99.50  $5.24 5.24 Additional rate   5.00% 104.74

22.

Net  1260(0.6)(0.83)  $700 Additional discount  700  682.50  $17.50 17.50 Additional rate   2.50% 700

23. 113.40  (0.75)(0.875)(0.96)L 113.40 L (0.75)(0.875)(0.96)

List  $180 24. 564.48  (0.6)(0.9)(0.98)L 564.48 L (0.6)(0.9)(0.98)

List  $960 25.

Net  0.8(85)  $68 Reduction needed  68  57.80  $10.20 10.20 Additional discount rate   15.00% 68

26.

Net  0.6(66)  $39.60 Reduction needed  39.60  35.64  $3.96 3.96 Additional discount rate   10.00% 39.60

27. Galaxy net  0.75(299)  $224.25 Brilliants net  (0.65)(0.9)(350)  $204.75 Additional Galaxy discount  224.25  204.75  $19.50 19.50 Additional Galaxy discount rate   0.086957  8.70% 224.25 28.

Net  0.76(125)  $95 Reduction needed  95  87.40  $7.60 7.60 Additional discount rate   8.00% 95

Exercise 6.2 1.

(a)

May 23

(b)

(2499)(0.98)  $2449.02

(a)

July 1

(b)

(6200)(0.98)  $6076

(a)

(842)(0.95)  $799.90

(b)

(842)(0.98)  $825.16

(c)

No discount: $842

(a)

(2412)(0.97)  $2339.64

(b)

(2412)(0.99)  $2387.88

(c)

No discount: $2412

5. Invoice date

Amount

Rate of discount

Net amount paid

July 25

$929

—

August 10

763

(0.97)(763)

740.11

August 29

864

(0.97)(864)

838.08

$ 929.00

Amount remitted

$2507.19

6. Invoice date

Amount

Rate of discount

Net amount paid

March 30

$394.45

—

$ 394.45

April 15

595.50

(0.98)(595.50)

583.59

May 10

865.20

(0.95)(865.20)

821.94

Amount remitted

$1799.98

$5275 – $3000 = $2775 To reduce the debt to $3000, a partial payment equivalent to $2775 must be paid. Allow 4% discount on partial payment of $2275 Amount paid  (0.96)(2275)  $2184

Allow 5% discount on partial payment of $740 Amount paid  (0.95)(740)  $703

(a)

September 10

(b)

5(980)(0.75)(0.95)

$3491.25

4(696)(0.83)(0.875)(0.96)

1948.80

Amount of invoice

$5440.05

Less: Discount of 3%

163.20 $5276.85

Amount due (c)

Partial payment credit  5440.05  2000  $3440.05 Cash discount  (0.03)(3440.05)  $103.20

10. (a) (b)

July 18 100 @ 34.30

$3430

25 @ 63.60

1590

40 @ 54.50

2180

List

$7200

Net  (0.6)(0.925)(0.95)(7200)  $4218 Amount due July 15  (0.95)(4218)  $4007.10 (c)

Partial payment credit  4218  2500  $1718

Amount paid  (0.95)(1718)  $1632.10

11. (a)

Allow discount of 3% on partial payment of $1200 Amount paid  (0.97)(1200)  $1164

(b)

Allow discount of 1% on partial payment of $740.95 Amount paid  (0.99)(740.95)  $733.54

Amount paid on October 25 is $600. Allow 5% on partial payment of $2600 Amount paid  (0.95)(2600)  $2470

(b)

Allow 2% on partial payment of $3000 Amount paid  (0.98)(3000)  $2940

Unpaid balance  $4000 Allow 3% on unknown partial payment $P

0.97P  1867.25 P  1925 Credit to the account is $1925. (b) 14. (a)

Amount owing  5325  1925  $3400 Allow 5% on unknown partial payment $P

0.95P  5966 P  6280 Reduction in the amount due  $6280 (b)

Amount owing  13, 780  6280  $7500

15. List price  $1.12  50  $56

$56  54.04  $1.96 Discount rate 

1.96  3.50% 56

(a)

The discount was $1.96

(b)

The discount rate was 3.50%

16. $26, 465  24,877.10  $1587.90 Discount rate 

1587.90  6.00% 26, 465

(a)

The discount was $1587.90.

(b)

The discount rate was 6.00%.

Exercise 6.3 1.

Cost = $4

S CEP  C  1.1C  1.3C  3.4C  (3.4)(4)  $13.60 Selling price is $13.60 per pizza. 2.

Cost = $25; Selling price = $50

S  C  30% of S  P S  C  0.3S  P 50  25  0.3(50)  P P  50  25  15  $10 Profit should be $10 per sale. 3.

S  $14.10 S  C  260% of C  110% of C S  C  2.6C  1.1C 14.10  4.7C C  $3

Cost of buying is $3 per piece. 4.

Cost  25(0.6)(0.90)(0.96)  $12.96 S  C  E  P  C  0.35C  0.15C  1.5C  1.5(12.96)  $19.44

Selling price  $19.44

   Cost   66 %  (96)  3   $64

S  C  32% of C  27.5% of C  C  0.32C  0.275C  1.595C  (1.595)(64)  $102.08 Selling price should be $102.08 6.

Cost  (0.6)(0.75)(55)  $24.75 Markup  54.45  24.75  $29.70

29.70  1.2  120.00% 24.75

(a)

Markup based on cost 

(b)

Markup based on selling price 

29.70  0.54  54.55% 54.45

Cost  (0.5)(0.9)(1240)  $558 Markup  1395  558  $837

837  1.5  150.00% 558

(a)

Markup based on cost 

(b)

Markup based on selling price 

(a)

15% of cost = 3.42

837  0.6  60.00% 1395

0.15 C  3.42 C  22.80 Cost was $22.80

(b)

Selling price  22.80  3.42  $26.22

(c)

Markup based on selling price 

(a)

18% of selling price = 6.57

3.42  0.130435  13.04% 26.22

0.18 S  6.57 S  36.50 Selling price is $36.50 (b)

Cost  36.50  6.57  $29.93

(c)

Markup based on cost 

10. (a)

6.57  0.219512  21.95% 29.93

S  CM

S  C  0.4C  1.4C 382.20  1.4C 382.20 C  273 1.4 Cost is $273 (b)

M  S  C  382.20  273  109.20 %M on selling price 

M 109.20   0.285714  28.57% S 382.20

Rate of markup on selling price is 28.57% 11. (a)

C  40% of S  S 0.99  0.4S  S 0.99  0.6S S  1.65

Selling price is $1.65 per litre (b)

Markup  (0.4)(1.65)  $0.66 Rate of markup based on cost 

12. (a)

0.66  0.6  66.67% 0.99

S  CM

(b)

%M on cost 

M (39  20.28)   0.923077 C 20.28

Rate of markup on cost is 92.31% 13. (a)

C  90% of C  S C  0.9C  444.98 1.9C  444.98 C  234.20

Cost is $234.20 (b)

Markup  (0.9)(234.20)  $210.78 Markup based on selling price 

14. (a)

210.78  0.473684  47.37% 444.98

C  331/ 3% of C  S 4 / 3C  S 4 / 3(45)  S  $60

Selling price is $60 (b)

Markup  60  45  $15 Markup based on selling price 

15. (a)

15  0.25  25.00% 60

C  40% of S  S 84  0.4 S  S 84  0.6S S  $140

Selling price is $140 (b)

Markup  140  84  $56 Markup based on cost 

16. (a)

56  0.6  66.67% 84

2 C  16 % of S  S 3 3.24  0.16 S  S 3.24  0.83S S  $3.89 Selling price is $3.89

(b)

Markup  3.89  3.24  $0.65 Markup based on cost 

17. (a)

0.65  0.200617  20.06% 3.24

C  0.375 of C  $23.10 1.375 C  23.10 C  16.80

Cost is $16.80 (b)

Markup  23.10  16.80  $6.30 Markup based on selling price 

18. (a)

6.30  0.27  27.27% 23.10

C  0.5 of C  $42.90 1.5 C  42.90 C  28.60

Cost is $28.60 (b)

Markup  42.90  28.60  $14.30 Markup based on selling price 

14.30  0.3  33.33% 42.90

19. M = 289.80 and represents 31.5% of the selling price 289.80 = 0.315S S = 289.80/0.315 S = $920 C = 920 – 289.80 = $630.20 20. (a)

Total cost = $4500 Cost per tree = 4500/100 = $45 Revenue: On 50 trees sold at a markup of 35% of cost S = 45 + 0.35(45) = 45 + 15.75 = $60.75 Total revenue = 50(60.75) = $3037.50 On 40 trees sold at $52 each Total revenue = 40(52) = $2080 On 10 remaining trees sold at 33% below cost S = (1 – 0.33)(45) = $30.15 Total revenue = 10(30.15) = $301.50 Total sales revenue = $5419 Total sales revenue – Total cost = Total markup 5419 – 4500 = $919

(b)

Percent markup realized based on cost = 919/4500 = 0.204222 = 20.42%

(c)

Percent of gross profit realized based on selling price = 919/5419 = 0.169588 = 16.96%

21. (a)

Cost per case = 19,200/240 = $80 Unit prices: Good quality = 2.05(80) = $164 Seconds = 1.28(80) = $102.40

Substandard = 0.65(80) = $52

(b)

Total revenue = 164(134) + 102.40(79) + 27(52) = 21,976 + 8089.60 + 1404 = $31,469.60 Total cost = 19,200 + 0.34(19,200) = $25,728 Profit = 31,469.60 – 25,728 = $5741.60

(c)

Total markup realized = $31,469.60 – 19,200 = $12,269.60 Average rate of markup based on selling price = 12,269.60/31,469.60 = 0.389887= 38.99%

Exercise 6.4 1.

(a)

S  C  (0.21)S  (0.11)S

S  149.50  0.32S 0.68S  149.50 149.50 S  $219.85 0.68 The regular selling price is $219.85. (b)

Sale price = S – 0.20S Sale price  (0.80)219.85 Sale price  $175.88 The sale price is $175.88.

(c)

Total cost  C  0.21S Total cost  149.50 + (0.21)(219.85) Total cost  $149.50 + 46.17 Total cost  $195.67 Profit  175.88  195.67 Loss  $19.79 The piano benches were sold at an operating loss of $19.79.

(a)

S  C  (0.27)S  (0.18)S

S  44  0.45S 0.55S  44 44 S  $80 0.55 The regular selling price is $80 (b)

Sale price  S  0.4S Sale price  (0.6)80 Sale price  $48 The sale price is $48.

(c)

Total cost  C  0.27S Total cost  44  (0.27)80 Total cost  44  21.60 Total cost  $65.60 Profit  48  65.60 Profit  $17.60 The microwave ovens were sold at an operating loss of $17.60.

Markdown

 2.49  1.99

 $0.50 0.50 Rate of markdown   0.200803  20.08% 2.49 The rate of markdown is 20.08% 4.

 279  239  $40 40 Rate of markdown   0.143369  14.34% 279 Markdown

The rate of markdown is 14.34%

 125  105  $20 20 Rate of markdown   16.00% 125 Markdown

The rate of markdown is 16.00% 6.

 1299  935  $364 364 Rate of markdown   0.280216  28.02% 1299 Markdown

The rate of markdown is 0.280216 = 28.02% 7.

(a)

Markdown  (0.50)14  $7

(b)

Rate of markdown 

7  0.21  21.21% (19  14)

The rate of markdown is 21.21% 8.

(a)

Regular price  225  2 12  225  24  $249 Markdown  249  199  $50

(b)

Rate of markdown 

50  0.200803  20.08% 249

The rate of markdown is 20.08% 9.

(a)

S CEP

29  C  0.43C  0.2C 29  1.63C 29 Cost   $17.79 1.63 The cost is $17.79

(b)

C  E  17.79  0.43(17.79)  1.43(17.79)  $25.44

The new price would be $25.44 (c)

Markdown  29  25.44  $3.56 3.56 Rate of markdown   0.122759  12.28% 29 The rate of markdown is 12.28%

(d)

Markdown  29  17.79  $11.21 11.21 Rate of markdown   0.386552  38.66% 29 The rate of markdown is 38.66%

10. (a)

S CEP 3849  C  0.31C  0.17C 3849  1.48C 3849 Cost   $2600.68 1.48 The cost is $2600.68

(b)

C  E  2600.68  0.31(2600.68)  1.31(2600.68)  $3406.89

The new price would be $3406.89 (c)

Markdown  3849  3406.89  $442.11 442.11 Rate of markdown   0.114864  11.49% 3849 The rate of markdown is 11.4864%

11. (a)

M = 37% of S 56.24 = 0.37S S = $152

Regular selling price is $152 (b)

C = 152 – 56.24 = $95.76

(c)

Rate of markup based on cost = 56.24/95.76 = 0.587302 = 58.73%

(d)

Break-even price = C + E = 1.25C = 1.25(95.76) = $119.70

(e)

Sale price = 0.87(152) = $132.24 Profit = 132.24 – 119.70 = $12.54 profit

12. Regular selling price = C + E + P S = 286.99 + 0.23S + 0.17S S = 286.99 + 0.40S 0.60S = 286.99 S = $478.32 Sale price = 487.32(0.70) = $334.82 Total cost = 286.99 + 0.23(478.32) =286.99 + 110.01 =$397 Profit = 334.82 – 397 = –$62.18(a loss) The sale resulted in a $62.18 loss. 13. (a)

C = $334.89(0.73)(0.82)(0.94) = $188.44 S = C + 0.55C + 0.21C S = 1.76C S = 1.76(188.44) S = $331.65

Regular selling price is $331.65. (b)

Overhead E = 0.55(188.44) = $103.64 Acceptable loss = 0.20(103.64) = $20.73 Total cost = 188.44 + 103.64 = $292.08 Profit = Sale price – Total cost –20.73 = Sale price – 292.08 Sale price = 292.08 – 20.73 Sale price = $271.35

(c)

Maximum rate of markdown = 331.65 – 271.35 / 331.65 = 60.30 / 331.65 = 0.181818 = 18.18%

Exercise 6.5 1.

(a)

C  240(0.45)(0.75)  $81 S  C  M  C  2.3C  3.3C  3.3(81)  267.30

The regular selling price is $267.30

(b)

Sale price  0.60(267.30)  $160.38

(c)

Profit  S  C  E  160.38  81  0.25(267.30)  $12.56

(a)

Cost  (0.6)(0.8)(0.95)(420)  $212.80 S  Cost  Markup S  212.80  0.60S 0.40S  212.80 S  $532

Sale price  532  0.45(532)  532  239.40  $292.60 (b)

Markup realized  292.60  212.80  $79.80 Rate of markup realized based on cost 

(a)

79.80  0.375  37.50% 212.80

Markdown  125  105  $20

Rate of markdown 

20  0.16  16.00% 125

Cost  (0.6)(0.85)(120)  $68

(b)

Total cost  C  E  68  12% of 125  68  15  $83 Profit  Revenue  Total cost  105  83  $22 Operating profit of $22 (c)

Rate of markup based on cost

 (d)

Rate of markup based on sale price

 4.

(a)

105  68  0.544118  54.41% 68

105  68  0.352381  35.24% 105

Markdown  558  432.45  $125.55 Rate of markdown 

(b)

125.55  0.225  22.50% 558

Cost  (0.625)(0.96)(620)  $372 Regular markup  558  372  $186 Rate of markup based on selling price 

(c)

186  0.3  33.33% 558

Total cost  Cost  Expense

 372  0.15(558)  372  83.70  $455.70 Profit  432.45  455.70  $23.25

Operating loss of $23.25 (d)

Rate of markup realized based on cost



432.45  372  0.1625  16.25% 372

Regular selling price  C  E  P S  33.45  15% of S  10% of S S  33.45  0.25S 0.75S  33.45 S  $44.60 Sale price  S  15% of S  0.85(44.60)  $37.91 Total cost  33.45  15% of 44.60  33.45  6.69  $40.14 Profit  37.91  40.14  $2.23 Loss of $2.23

(a)

Clearance sale price  (0.6)(125)  $75

(b)

Inventory sale price

 Cost  Overhead  C  0.5C  1.5C  1.5(42)  $63 (c)

Total revenue  120(125)  60(75)  20(63)  15, 000  4500  1260  $20, 760

Total cost  200(42)  50% of (200  42)  8400  4200  $12, 600 Total profit  20, 760  12, 600  $8160

(d)

Average rate of markup, including overhead and profit

Total markup realized Total purchase cost 20, 760  8400 12,360    1.471429  147.14% 8400 8400 

(a)

Cost per pan



4950  $8.25 600

Unit selling prices: Normal quality  (1.8)(8.25)  $14.85

Seconds  (1.2)(8.25)  $9.90 Substandard  (0.8)(8.25)  $6.60 (b)

Total revenue  (360)(14.85)  (190)(9.90)  (50)(6.60)  5346  1881  330  $7557

Total cost

1  4950  33 % of 4950 3  4950  1650  $6600 Profit  7557  6600  $957 Operating profit of $957 (c)

Total markup realized  7557  4950  $2607 Average rate of markup, including overhead and profit, based on selling price

 8.

(a)

2607  0.344978  34.50% 7557

Cost  (0.6)(0.83)(24)  $12

1 Regular selling price  C  25% of C  33 % of C 3 1  158 %(12)  $19 3 Regular selling price is $19

(b)

Break-even price  Cost  Expenses  12  25% of 12  $15

Maximum amount of markdown  19  15  $4

(c)

Rate of markdown to break-even price

 9.

(a)

4  0.210526  21.05% 19

Cost  (0.6)(0.8)(0.9)(75)  $36

Regular selling price  C  75% of C  25% of C  2C  2(36)  $72 Regular selling price is $72

Overhead  (0.75)(36)  $27

(b)

Acceptable loss  (0.3)(27)  9 Sale price  Total cost  Loss  36  27  9  $54 (c)

Maximum rate of markdown



72  54  0.25  25.00% 72

10. Sale price  Cost  Markup Sale price  C  30% of Sale price Sale price  36.75  0.3 Sale price 0.7 Sale price  36.75 Sale price  $52.50

Regular selling price  Discount  S R  25% of R  S 0.75R  52.50 R  $70

The regular selling price should be $70 11.

Regular selling price = C  E  P S  36.40  24% of S  20% of S S  36.40  0.44S 0.56S  36.40 S  $65 Sale price  65(1  0.15)  65(0.85)  $55.25 Total cost  36.40  24% of 65  36.40  15.60  $52 Profit  55.25  52  $3.25 Operating profit of $3.25

12. Cost = 0.75(264) = $198 1 C  33 % of Sale price  Sale price 3 2 198  Sale price 3 Sale price  $297 Regular selling price  Discount  Sale price S  20% of S  Sale price 0.8S  297 S  $371.25

The regular selling price is $371.25. 13.

Cost  198(0.40)(0.83)  $66 Regular selling price  C  0.45S  0.25S 0.3S  $66 S  $220 New regular selling price  Discount  S N  0.375 N  220 0.625 N  220 N  352

Sale price  N  0.55N  0.45(352)  $158.40 Total cost  66  .045S  66  0.45(220)  165 Profit  158.40  165  6.60 Operating loss of $6.60

14. Cost  (0.6)(0.83)(500)  $250 Regular selling price  C  20% of S  17.5% of S S  C  0.375S 0.625S  250 S  $400 New regular selling price  Discount  S N  0.36 N  400 0.64 N  400 N  $625

Sale price  N  0.54N  0.46N  0.46(625)  $287.50 Total cost  250  0.2(400)  250  80  330 Profit  287.50  330  $42.50 Operating loss of $42.50

15. Cost  (0.6)(0.95)(900)  $570 Regular selling price  C  0.15S  0.09S S  570  0.24S 0.76S  570 S  $750

New regular selling price  Discount  S

N  0.25 N  750 0.75 N  750 N  $1000

Sale price  N  0.4N  0.6(1000)  $600 Total cost  570  0.15(750)  570  112.50  682.50 Profit  600  682.50  $82.50 Operating loss of $82.50

16.

Regular selling price  C  0.15C  0.1C 16,800  1.25C C  $13, 440

New regular selling price  Discount  S

S  0.2S  16,800 0.8S  16,800 S  $21, 000

Normal profit  0.1C  0.1(13, 440)  $1344 Required profit  0.25(1344)  $336 Normal overhead  0.15(13, 440)  $2016 Normal commission  0.5(2016)  $1008 Reduction in overhead  0.3(1008)  $336 Overhead to be recovered  2016  336  $1680

Sale price  C  E  P  13, 440  1680  336  $15, 456 Markdown  21, 000  15, 456  $5544 5544 Rate of markdown   0.264  26.40% 21, 000

CHAPTER 6 BMN BOX Indigenous entrepreneurship: Making a business case for reconciliation Answers: 1. Company valuation = $100,000 ÷ 0.20 = $500,000 2. Company valuation = $125,000 ÷ 0.50 = $250,000 3. Company valuation = $125,000 ÷ 0.30 = $416,667 (*rounded to the nearest dollar) 4. (i) At regular selling price

(ii) At the sale price

Markup

C+M=S

C + M = SR

23.40 + M = 156.00

23.40 + M = 124.00

M = $132.60

M = $100.60

Overhead

E = 30% of regular selling price

E = 0.30(156.00)

E = $46.80

Profit

E+P=M

46.80 + P = 132.60

46.80 + P = 100.60

P = $85.80

P = $53.80

Review Exercise 1.

(a)

Net price  (0.75)(0.8)(0.95)(56)  $31.92

(b)

Discount  56  31.92  $24.08

(c)

Single rate of discount 

Discount  168 105  $63

Rate of discount  3.

24.08  0.43  43.00% 56

63  0.375  37.50% 168

Single rate

 1  (0.65)(0.88)(0.95)  1  0.5434  0.4566  45.66% 4.

40% of List  1.44 0.4L  1.44 L  $3.60 Net price  3.60  1.44  $2.16

Regular price  (0.85)(112)  $95.20

Additional discount  95.20  80.92  $14.28

Additional discount percent  6.

14.28  0.15  15.00% 95.20

Net price  (0.83)(List) 387.50  0.83L L  $465 List price was $465.

Net price  (0.8)(0.85)L 20.40  (0.8)(0.85)L L  $30 List price was $30.

(a)

Last day of discount period is June 10.

(b)

Net invoice  (0.8)(0.85)(4000)  $2720 Net amount paid  (0.95)(2720)  $2584

Last day of discount period is September 12

Amount paid after discount  (0.98)(25,630)  $25,117.40 10.

11.

Invoice no.

Rate of discount

Net amount paid

312

—

429

(0.98)(784)

768.32

563

(0.95)(873)

829.35

Remittance

$2520.67

Net invoice  (0.66)(0.875)(8400)  $4900

Last day for discount is July 11: allow 3%

12.

(a)

Pay  (0.97)(2000)  $1940

(b)

Pay  (0.97)(2900)  $2813

Net invoice  (0.75)(0.8)(16, 000)  $9600

(a)

Original balance Payment Sept. 30

$9600 4600

Payment Oct. 20

$5000 3000

Balance due

$2000

923

(b)

Payment Sept. 30: Allow 5% Paid  (0.95)(4600)

$4370

Payment Oct. 20: Allow 2%

13.

Paid  (0.98)(3000) Final payment

2940 2000

Total paid

$9310

Net invoice  (0.85)(0.925)(4000)  $3145

(a)

Allow 3% discount Payment  97% of credit 1595.65  0.97C C  $1645 Amount due is reduced by $1645.

(b) 14. (a)

Amount owing  3145  1645  $1500

Cost  (0.8)(6)  $4.80 S  C  0.45C  0.2C S  1.65C S  1.65(4.80) S  $7.92 Selling price is $7.92 per bag.

(b)

Amount of markup  7.92  4.80  $3.12

(c)

Rate of markup based on selling price 

(d)

Rate of markup based on cost 

(e)

Break-even price  Cost  Overhead  4.80  0.45(4.80)  4.80  2.16

3.12  0.39  39.39% 7.92

3.12  0.65  65.00% 4.80

 $6.96

(f )

For a selling price of $7.50

Profit  7.50  6.96  $0.54 Operating profit of $0.54

15. (a)

Gross Profit  35% of Regular selling price 31.50  0.35S S  $90

Regular selling price is $90. (b)

Cost  90  31.50  $58.50

(c)

Rate of markup based on cost 

(d)

Total Cost  C  E  58.50  0.28C  58.50  0.28(58.50)

31.50  0.538462  53.85% 58.50

 $74.88 (e)

Sale price  (0.76)(90)  $68.40 Profit  68.40  74.88  $6.48 Operating loss of $6.48

16. (a)

C  0.35C  Selling price 1.35C  8.91 C  6.60 Cost was $6.60

(b)

Markup as a percent of selling price

 17. (a)

8.91  6.60 (100)  25.93% 8.91

S  C  0.3S 0.7S  54.25 S  77.50 Selling price was $77.50

(b)

Markup as a percent of cost

 18. (a)

77.50  54.25 (100)  42.86% 54.25

Cost  (0.625)(0.82)(1800)  $922.50

C  1.2C  Regular selling price 2.2C  S S  2.2(922.50) S  $2029.50 Sale price  0.6S  0.6(2029.50)  $1217.70 (b)

Realized rate of markup based on cost

 19.

1217.70  922.50 295.20   0.32  32.00% 922.50 922.50

Cost  (0.6)(195)  $130 Regular selling price  C  0.35S 0.65S  130 S  $200 New regular selling price  Discount  S

N  0.16 N  200 0.83 N  200 N  $240 New regular selling price is $240.

20. (a)

Cost  (0.6)(0.85)(36)  $18.36

C  0.4S  S 18.36  0.6S S  $30.60 Markdown  30.60  22.95  $7.65 7.65 Rate of markdown   0.25  25.00% 30.60

(b)

Total Cost  C  0.25(30.60)  18.36  7.65  $26.01 Profit  22.95  26.01  $3.06 Operating loss of $3.06

(c)

Rate of markup realized based on cost

 21. (a)

22.95  18.36 4.59   0.25  25.00% 18.36 18.36

Cost  (0.75)(0.8)(21)  $12.60

C  0.2S  0.17S  S 12.60  0.63S S  $20

Sale price  0.8(20)  $16 Total cost  12.60  0.2(20)  12.60  4  16.60 Profit  16  16.60  $0.60 Operating loss of $0.60

(b)

Markup realized  16  12.60  $3.40

Rate of markup realized on cost

 22. (a)

3.40  0.269841  26.98% 12.60

Cost  (0.6)(0.83)(0.92)(250)  $115

S  C  0.65C  0.55C  2.2C  2.2(115)  $253 Regular selling price is $253

(b)

Break-even price

 CE  115  0.65 115   115  74.75  $189.75

(c)

23. (a)

Maximum reduction  253  189.75  $63.25 63.25 Maximum rate of markdown   0.25  25.00% 253 Cost  (0.6)(0.916)(1080)  $660

C  0.18S  0.153S  S C  0.3S  S 0.6S  660 S  $990 New regular selling price  Discount  S

N  0.25 N  990 0.75 N  990 N  $1320 Sale price  0.625(1320)  $825 Total cost  660  0.18(990)  660  178.20  $838.20 Profit  825  838.20  $13.20 Operating loss of $13.20

(b)

Markup realized  825  660  $165 Rate of markup realized based on cost 

24.

165  0.25  25.00% 660

Net price  (0.6)(0.85)(1860)  $1054 Reduction required  1054  922.25  $131.75 131.75 Additional discount   0.125  12.50% 1054

25. (a)

Cost  (0.6)(0.8316)(0.9)(180)  $80.84

S  C  0.45S  0.2125S S  80.84  0.6625S 0.3375S  80.84 S  $239.53 (b)

Sale price  0.70(239.53)  $167.67

(c)

Profit/loss realized  Sale price  C  E  167.67  80.84  0.45(239.53)  20.96 Loss of $20.96 37.5% of Cost  42 0.375C  42 C  $112

26. (a)

Regular selling price  112  42  $154 (b)

Rate of markup based on regular selling price



42  0.27  27.27% 154

(c)

Markdown  154  138.95  $15.05 15.05 Rate of markdown   0.097727  9.77% 154

(d)

Profit/loss  Sale price  C  E  138.95  112  0.175(154)  138.95  112  26.95  $0.00 This is the break-even point.

27. (a)

Cost per sweater 

3100  $12.40 250

Revenue: on 50 sweaters sold at a markup of 150% of cost S  12.40  1.5(12.40)  2.50(12.40)  $31each Total revenue  50(31)

 $1550

on 120 sweaters sold at a markup of 75% of cost S  12.40  0.75(12.40)  1.75(12.40)  $21.70 each Total revenue  120(21.70) 

2604

on 60 sweaters sold for $15 each Total revenue  60(15)

on remaining 20 sweaters sold 20% below cost



900

(b)

S  0.8(12.40)  $9.92 each Total revenue  20(9.92)

 198.40

Total sales revenue Total cost of buying

$5252.40 3100

Markup realized

$2152.40

Percent markup realized based on cost

 (c)

2152.40 (100)  69.43% 3100

Percent of gross profit realized based on selling price



2152.40 (100)  40.98% 5252.40

Self-Test 1.

Net price = 0.625  0.875  0.916  590  $295.77

Amount of discount = 270  168.75  101.25 101.25 Rate of discount   0.375  37.50% 270

Single rate  1  (0.60)(0.90)(0.916)  1  0.495  0.505  50.50%

Store’s net price  (0.75)(0.85)(1020)  650.25 Competitor’s price  (0.75)(927)  695.25 Additional discount needed  45 45 Additional percent discount   0.064725  6.47% 695.25

5. Invoice Date

Net price

Discount

Pay

March 21

(0.8)(0.9)(850)

 612

Nil

April 10

 0.7   0.83  960 

 560

548.80

April 30

 0.6   0.75 0.951040

 494

474.24

Remittance

$1635.04

612

Reduction needed  3200  1200  $2000 Discount allowed: 3% Payment = (0.97)(2000) = $1940

Discount allowed on partial payment: 4% Payment  96% of reduction in debt 1392 = 0.96R R = 1450

Debt is reduced by $1450 8.

C + M = Regular selling price C + 0.2S = S 1270 + 0.2S = S 0.8S = 1270 Regular selling price = $1587.50

C + M = Regular selling price C + 0.4S = S 180 = 0.60S S = 300 Sale price = 0.8  300 = $240

10.

Cost = (0.65)(0.875)(350) = 199.06 C + M = Regular selling price

199.06 +1.5(199.06) = S S = 497.65 Sale price = (0.7)(497.65) = $348.36 11.

Markup = 2520  900  1620 Rate of markup based on cost 1620 =  1.8  180.00% 900

12.

C+M =S C + 0.4 C = S 1.4 C = 1904 C = $1360

13.

Markup = 0.45 Regular selling price 90 = 0.45S S = 200 Cost = 200  90  $110

14.

Markdown = 1560  1195  365 Rate of markdown 365 =  0.233974  23.40% 1560

15. Cost = (0.75)(0.85)(1480) = 943.50 C + 0.4S + 0.1S = S 943.50 + 0.5S = S 943.50 = 0.5S S = 1887 Sale price = (0.55)(1887) = 1037.85

Total cost = 943.50 + (0.4)(1887) = 943.50 + 754.80 = 1698.30 Operating profit = 1037.85  1698.30  $660.45 Operating loss of $660.45 16. Cost = (0.625)(0.875)(830) = 453.91

C + 0.2C + 0.15S = S 1.2C + 0.15S = S 1.2(453.91) = 0.85S 544.692 = 0.85S S = 640.81 New regular selling price  Discount = S N  0.3N = S 0.7N = 640.81 N = 915.44

Sale price = 0.5  915.44 = 457.72 Total cost = 1.2C = 1.2(453.91) = 544.69 Operating profit = 457.72  544.69  86.97 Operating loss of $86.97

Challenge Problems 1.

Number of dozens of saleable roses = 90% of 12 = 10.8 Since roses are sold in full dozens only, $117 buys 10 full dozens. 117 Cost per dozen =  $11.70 10 Selling price = Cost + Markup based on selling price S = C + 0.55 S 0.45 S = 11.70 11.70 S=  26 0.45 Rose Bowl Florists must charge $26 per dozen.

(a)

Cost = List price  Trade discount C = L  0.25L C = 0.75L

Sale price = Cost + Markup Sale price = C + 0.25 Sale price 0.75 Sale price = 0.75L Sale price = L Regular selling price = Sale price + Discount S = Sale price + 0.2S 0.8S = L L S=  1.25L 0.8 The goods should be marked at 125.00% of the list price. (b)

Cost = 0.75L Sale price = C + 0.25C Sale price = 1.25C Sale price = 1.25(0.75L) Sale price = 0.9375L Regular selling price = Sale price + Discount S = 0.9375L + 0.20S 0.80S = 0.9375L 0.9375L S=  1.171875L 0.80

The goods should be marked at 117.19% of the list price. 3.

Invoice amount = 0.75(2500) = $1875 Amount paid = 0.98(1875) = $1837.50 The extra profit = 1875  1837.50 = $37.50

Selling price = Invoice amount + Markup S = C + 0.6C S = 1.6(1875) S = 3000

Extra profit based on selling price =

37.50  0.0125  1.25% 3000

Case Study 1.

Edward’s cost per camera = 0.98(170) = $166.60 Staples’ cost per camera = 0.965(0.97)(170) = $159.13

Edward’s: 0.15(400)  $60 0.18(400) = $72

Overhead Profit

Minimum selling price = Cost + Overhead + Profit = 166.60 + 60 + 72 = $298.60 Staples’: 0.25(400)  $100 0.35(400) = $140

Overhead Profit

Minimum selling price = Cost + Overhead + Profit = 159.13 +100 +140 = $399.13

(a)

Edward’s extra profit = 400  298.60 = $101.40 Staples’ extra profit = 400  399.13 = $0.87

(b)

Extra profit as a percent of manufacturer’s suggested retail price:

101.40  0.2535  25.35% 400 0.87 For Staples’ =  0.002175  0.22% 400 For Edward’s =

Edward’s markdown = $101.40 101.40 Rate of markdown =  0.2535  25.35% 400

Chapter 7

Simple Interest

Exercise 7.1 1.

P  24, 000; r  0.015;

Time in days from April 1 to June 30  90  90  I  24, 000(0.015)    $88.77  365 

P  1500; r  0.0105

Time in days from July 17 to December 1  137  137  I  1500(0.0105)    $5.91  365 

P  8100; r  0.077

 90  I  8100(0.077)    $153.79  365 

P  1800; r  0.052

 220  I  1800(0.052)    $56.42  365 

P  1700; r  0.034

 160  I  1700(0.034)    $25.34  365 

P  200; r  0.084

 58  I  200(0.084)    $2.67  365 

P  3000; r  0.08

 28  I  3000(0.08)    $18.41  365 

P  21, 000; r  0.02

 35  I  21, 000 (0.02)    $40.27  365 

2028 is a leap year so February 2028 has 29 days Number of days = 30 + 30 + 31 + 31 + 29 + 31 + 30 + 31 + 30 + 31 + 3 = 307 days Alternatively, (365 – 275) + 216 + 1 = 307 days I = 635  0.065  307/365 = $34.72

10. Number of days in 2025 = (365 – 89) = 276 days Number of days in 2026 = 365 days Number of days in 2027 = 365 days Number of days in 2028 = 89 + 1 = 90 days (because 2028 is a leap year) 276 + 365 + 365 + 90 = 1096 days I = 5000  0.025  1096/365 = $375.34 11.

 163  I  15,160(0.0825    365  I  558.53178 I  $558.53

12.

 13  I  7500 (0.06375)   12  I  517.96875 I  $517.97 13.

* Note: 2024 is a leap year.

 366  I  245 (0.014)    365  I  3.439397 I  $3.44

14.

* Note: 2024 is a leap year. Number of days calculation  4  366  1  371 days  371  I  68, 450(0.019)    365  I  1321.928904 I  $1321.93

Exercise 7.2

I  148.32; r  0.0675; t  P

15 12

I 104.50 104.50    0.095  9.50% 15 Pt 880  12 1100

P  650; I  23.70; t  r

225 365

I 39.27 39.27    $2316.53 225 rt 0.0275  365 0.016952

P  880; I  104.50; t  r

I 148.32 148.32    $3296 8 rt 0.0675  12 0.045

I  39.27; r  0.0275; t  P

8 12

7 12

I 23.70 23.70    0.062505  6.25% Pt 650  127 379.16.

P  1387; I  63.84; t 

200 365

r

I 63.84 63.84    0.084  8.40% 200 Pt 1387  365 760

P  2400; I  22.74; t  r

I 22.74 22.74    0.038004  3.80% 91 Pt 2400  365 598.356164

P  1290; I  100.51; r  0.085

t (months) 

I 20.95  12   12  0.250030 yr  12  3.000358  3 months Pr 2660  0.0315

P  564; I  15.09; r  0.0775

t (days) 

10.

I 100.51  12   12  0.916644 yr  12  10.999726  11 months Pr 1290  0.085

P  2660; I  20.95; r  0.0315

t (months) 

91 365

I 15.09  365   365  0.345230 yr  365  126.008922  126 days Pr 564  0.0775

P  1200; I  12.22; r  0.169

t (days) 

I 12.22  365   365  0.060256 yr  365  21.993590  22 days Pr 1200  0.169

11. Number of days = 180;

Calculator: 6.1825 DT1

12.1525

180

DT2

DBD

I  39.96; r  0.0925; t  P

180 365

I 39.96 39.96    $876 180 rt 0.0925  365 0.045616

12. Number of days = 176;

Calculator: 9.1026 DT1

3.0427

176

DT2

DBD

I  42.49; P  740.48; t  r

176 365

I 42.49 42.49    0.119002  11.90% 176 Pt 740.48  365 357.053370

13.

I  2000; r  0.06; t 

1 12

2000  $400, 000 12 (0.06)

P 1

14.

P  7800; I  88.47; t 

r

15.

16.

88.47 88.47   0.034500  3.45% 120 7800  365  2564.383562

P  3200; I  168; t 

r

7 ; 12

168 168   0.09  9.00% 7 3200  12  1866.6&

P  360; I  3.20; r  0.135

t (days) 

17.

120 ; 365

I 3.20  365   365  0.065844 yr  365  24.032922  24 days Pr 360  0.135

P  3448; I  3827.66  (3448  344.80)  34.86; r  0.09;

t

34.86  0.112336 years 3448(0.09)

Number of days overdue  0.112336(365)  41.002514  41 days

Exercise 7.3 1.

P  2500; r  0.0345; t 

180 365

180   S  P(1  rt )  2500 1  0.0345  365    2500(1  0.017014)  $2542.53 2.

P  800; r  0.0275; t 

210 365

210   S  P(1  rt )  800 1  0.0275  365    800(1  0.015822)  $812.66 3.

P  26,750; r  0.013; t 

215 365

215   S  P(1  rt )  26, 750 1  0.013  365    26, 750(1  0.007658)  $26 954.84 4.

P  13,500; r  0.0365; t 

270 365

270   S  P(1  rt )  13,500 1  0.0365  365    13,500(1  0.027)  $13,864.50

(a)

P  50, 000; r  0.0395; t  1; S  P(1  rt )  50, 000[1  (0.0395 1)] Amount received  $51,975

(b)

First six months: P  50,000; r  0.0385; t 

6 12

6  S  50, 000 1  0.0385    $50,962.50 12   Amount received after six months  $50,962.50 Second six months: P  50,962.50; r  0.0385; t  6  S  50,962.50 1  0.0385    $51,943.53 12   Amount received after one year  $51,943.53 (c)

I  51,943.53  50, 000  $1943.53

r

1943.53  0.038871  3.89% 50, 000(1)

6 12

(a)

P  45, 000; r  0.013; t  1; S  45, 000[1  0.013(1)]  $45,585

Amount received  $45,585

(b)

First six months: P  45,000; r  0.011; t 

6 12

  6  S  45, 000 1  0.011    $45, 247.50  12    Amount received after six months  $45, 247.50 Second six months: P  45, 247.50; r  0.011; t 

6 12

  6  S  45, 247.50 1  0.011    $45, 496.36  12    Amount received after one year  $45, 496.36 (c)

I  45, 496.36  45, 000  $496.36

r 7.

496.36  0.011030  1.10% 45, 000(1)

P  5000; I  5113.87  5000  113.87; r  0.0875;

t

113.87  0.260274 years 5000(0.0875)

Number of days interest  0.260274(365)  95.000114  95 days

August 15, 2025: Day # 227 Number of days between: 95 Day #: Calculator:

(a)

322

 November 18, 2025

8.1525

November 18, 2025

DT1

DBD

DT2

r = 0.0275; t = 1 year S = 7000(1 + 0.0275  1) = $7192.50

(b)

For the first six months: P = 7000; r = 0.026; t = 6 months S1 = 7000(1+ 0.026  6/12) = $7091 For the second six month term: P = 7091; r = 0.029; t = 6 months S2 = 7091(1 + 0.029  6/12) = $7193.82

(c)

Interest described in part b) is I = 7193.82 – 7000 = 193.82; t = 1 year r = I/Pt = 193.82/(7000)(1) = 0.027689 = 2.77%

S  P(1  rt )   1   4, 000, 000 1  (0.006)     12     $4, 002, 000

10.

I  4368.55  4300 I  $68.55 68.55 4300 (0.04375) t  0.364385 years t

Number of days of interest  0.364385  365  133 days 133  (365  238)  6 days into 2026 They repaid the loan on January 6, 2026.

11. Future value from bond = S = P(1  rt )

  5  = 449  1  0.07      12    = $462.10 Future value of conference = $499 Pierre’s true savings = 499  462.10  $36.90

$275, 000 = $11, 000 25 $151, 773.99 Interest to be paid every year = = $6070.96 25 Total annual installment = $17, 070.96 Using S = P (1 + rt ): 17,070.96  11, 000 1  r (1)  17, 070.96 1  r   11, 000 1  r  1.551905 r  0.551905 r  55.19%

12. Principal to be paid every year =

Exercise 7.4

S  1241.86; r  0.039; t  P

4 12

S 1760 1760    $1704.60 4 1  rt 1  0.0975  12 1  0.0325

S  708.13; r  0.053; t  P

93 365

S 480.57 480.57    $475 93 1  rt 1  0.046  365 1  0.011721

S  1760; r  0.0975; t  P

S 1241.86 1241.86    $1222 1  rt 1  0.039  125 1  0.01625

S  480.57; r  0.046; t  P

80 365

S 708.13 708.13    $700 80 1  rt 1  0.053  365 1  0.011616

S  657.58; r  0.0475; t  P

5 12

162 365

S 657.58 657.58    $644 162 1  rt 1  0.0475  365 1  0.021082

S  1750; r  0.189; t  P

181 365

S 1750 1750    $1600.04 181 1  rt 1  0.189  365 1  0.093723

S  7345.64; r  0.0625; t = P

11 12

7345.64 7345.64   $6947.60 11 1  0.0625  12 1  0.057292

Interest  7345.64  6947.60  $398.04

S  10,000; r  0.0206; t  P

181 365

10, 000 10, 000   $9898.88 1  0.0206  181 1  0.010215 365

S  23,520.18; r  0.065; t  P

23,520.18 23,520.18   $23, 000 127 1  0.065  365 1  0.022616

10. S  10,000; r  0.0135; t  P

127 365

68 365

10, 000 10, 000   $9974.91 68 1  0.0135  365 1  0.002515

11. S = 499.99; r = 0.035; t = 6/12 = 0.5 years P = S / (1 + rt) = 499.99/(1 + 0.035(6/12)) = $491.39 12. P = 4965.50; S = 5000; r = 0.0245 S = P + I , so I = 5000 – 4965.50 = $34.50 Using t = I/Pr 34.50 / (4965.50)(0.0245) = 0.283618 years = 0.283618(365) days = 103.52 days = 104 days 2028 is a leap year so May 5, 2028 is the 126th day of 2028. 104 days prior to May 5, 2028 = 126 – 104 = day 22 Monica purchased the investment on January 22, 2028.

Business Math News Box 1.

I  P rt 6  2300(0.10)    12   $115 2. (a)

Calculate the amount owed after one year:

S  P(1  rt )  1500 (1  (0.08)(1))  1500 (1.08)  $1620 Let $x be the size of each payment.

x x x   3 6 1  (0.08)  12   1  (0.08)  12   1  (0.08)  129   x x x 1620    1.02 1.04 1.06 1620  0.980392 x  0.961538 x  0.943396 x 1620  2.885327 x x  $561.46 1620 

(b)

Interest  (3  561.46)  1500  $184.38

Exercise 7.5 1.

Let the size of the single payment be $x. The focal date is today. The equation of equivalence is x

600 600  3 1  0.10  12 1  0.10  126

x  585.37  571.43 x  $1156.80

Let the size of the single payment be $x. The focal date is today. The equation of equivalence is

x

700 800   2  1  0.083  125 1  0.083    12 

x  690.45  773.26 x  1463.71 The single payment today is $1463.71. 3.

Let the size of the single payment be $x. The focal date is 90 days from now. The equation of equivalence is

150  60    1000 1  0.08    1200  1  0.08  x 365  365    1032.88  1215.78  x x  2248.66 The single payment 90 days from now is $2248.66.

Let the size of the single payment be $x. The focal date is 75 days from now.

The equation of equivalence is

166  30    2200 1  0.09    1800 1  0.09   x 365  365    2290.05  1813.32  x x  4103.37

The single payment 75 days from now is $4103.37.

Let the size of the single payment be $x.

The focal date is one month from now.

The equation of equivalence is

5 1 400   500 1  0.105    800  1  0.105    x 12  12  1  0.105  122   x  521.88  807  393.12 x  1722

The size of the final payment is $1722.

Let the size of the single payment be $x.

The focal date is today.

The equation of equivalence is

6 1 700   1200 1  0.098    1500 1  0.098    x 12  12  1  0.098  122   x  1258.80  1512.25  688.75 x  3459.80

The size of the final payment is $3459.80.

Let the size of the single payment be $x.

The focal date is 125 days from now.

The equation of equivalence is

220  125    400 1  0.06    700  1  0.06   365  365   

95    600 1  0.06   x 365   414.47  714.38  609.37  x x  519.48 The size of the final payment is $519.48. 8.

Let the size of the single payment be $x. The focal date is 92 days from now. The equation of equivalence is 292  155    4000 1  0.083    6000 1  0.083   365  365   

92    5000 1  0.083   x 365   4265.60  6211.48  5104.60  x x  5372.48 The size of the payment is $5372.48. 9.

Let the size of the equal payments be $x. The focal date is today. The equation of equivalence is

1100 

x x  4 1  0.085  12 1  0.085  126

x x  1.0283& 1.0425 1100  0.972447 x  0.959233 x 1100  1.931680 x x  569.45 1100 

The size of the equal payments is $569.45. 10. Let the size of the equal payments be $x. The focal date is today. The equation of equivalence is

2300 

x x  90 1  0.0925  365 1  0.0925  135 365

x x  1.022808 1.034212 2300  0.977700 x  0.966919 x 2300  1.944620 x x  1182.75 2300 

The size of the equal payments is $1182.75. 11. Let the size of the equal loans be $x. The focal date is today. The equation of equivalence is

   7   5  800  x 1  0.11    x 1  0.11    12    12     &  x(1.04583) & 800  x(1.06416) 800  2.11x x  379.15 The size of the equal loans is $379.15. 12. Let the size of the equal loans be $x. The focal date is today. The equation of equivalence is

   45    190   2950  x 1  0.125     x 1  0.125    365    365     2950  1.015411x  1.065068 x 2950  2.080479 x x  1417.94 The size of the equal loans is $1417.94.

13. Let the size of the equal payments be $x. The focal date is today. The equation of equivalence is 60  30    800 1  0.0725    800  1  0.0725   365  365   

 x

x 60 1  0.0725  365

x 1.011918 1614.30  x  0.988223x 1614.30  1.988223x x  811.93

809.53  804.77  x 

The size of the equal payments is $811.93. 14. Let the size of the equal payments be $x. The focal date is today. The equation of equivalence is

14, 000 

x x x   240 260 1  0.07  120 1  0.07  365 1  0.07  365 365

x x x   1.023014 1.046027 1.049863 14, 000  0.977504 x  0.955998 x  0.952505 x 14, 000  2.886007 x x  4850.99 14, 000 

The size of the equal payments is $4850.99. 15. Let the size of the equal payments be $x. The focal date is today. The equation of equivalence is

4000 

x x x   8 4 1  0.085  12 1  0.085  12 1  0.085  12 12

x x x   1.028333 1.056667 1.085 4000  0.972447 x  0.946372 x  0.921659 x 4000  2.840479 x x  1408.21 4000 

The size of the equal payments is $1408.21. 16. Let the size of the equal payments be $x. Focal date is September 30. Equation of equivalence is

213  153   102    1500 1  0.0675    x 1  0.0675    x 1  0.0675   365  365   365    51    x 1  0.0675    400 365   1559.09  1.028295 x  1.018863x  1.009432 x  400 1159.09  3.056589 x x  379.21 The size of the equal payments is $379.21. 17. Let the size of the final payment be $x. The focal date is on September 30, 2026. 500(1 + 0.0625 (156/365)) + 1625 = 1200(1 + 0.0625(86/365)) + x / (1 + 0.0625 (107/365)) 513.36 + 1625 = 1217.67 + x / (1.018322) 920.69 = 0.982008x x = 937.56 The size of the final payment is $937.56. 18. Let the size of the first payment be $x. Using a focal date of June 7, 1475 = x / (1 + (0.046)(30/365)) + 2x / (1 + (0.046)(73/365)) + 3x / (1 + (0.046)(148/365)) 1475 = x / (1.003781) + 2x / (1.009200) + 3x / (1.018652) 1475 = 0.996233x + 2x(0.990884) + 3x(0.981689) 1475 = 0.996223x + 1.981768x + 2.945068x 1475 = 5.923070x x = $249.03

The first payment is $249.03. The second payment is $498.06. The third payment is $747.09.

Review Exercise

P  1975; r  0.055; t  I  1975  0.055 

215 365

215  $63.98 365

2. No. of days  2  30  31  31  30  31  30  31  3  219

I  $34.44; r  0.0825; t  P

124.29  0.726046 years  265 days 2075  0.0825 156 365

24.87 24.87   $1225.04 0.0475  156 0.020301 365

No. of days  30  31  30  31  31  28  31  29  241

I  148.57; r  0.075; t  P

3 12

18.70  0.11  11.00% 680  123

I  24.87; r  0.0475; t  P

39.39  0.074999  7.50% 284 675  365

I  124.29; r  0.0825; P  2075

t 6.

284 ; P  675 365

P  680; I  698.70  680  18.70; t  r

34.44 34.44   $695.76 219 0.0825  365 0.0495

I  39.39; t  r

219 365

241 365

148.57 148.57   $3000.17 241 0.075  365 0.049521

P  1545;I  58.93; t 

150 365

r

58.93 58.93   0.092813  9.28% 150 1545  365 634.931507

No. of days  30  31  31  30  31  30  0  183

I  1562.04  1500  62.04; P  1500; t  r

10.

62.04 62.04   0.082494  8.25% 183 1500  365 752.054795

I  51.04; P  2500; r  0.035

t (months)  11.

51.04 12  7 months 2500  0.035

I  3195.72  3100  95.72; P  3100; r  0.0575

t (days)  12.

183 365

95.72  365  196 days 3100  0.0575

P  4200; r  0.045; t 

11 12

11   S  P(1  rt )  4200 1  0.045   12    4200(1  0.04125)  $4373.25 13. No. of days  21  31  31  30  31  30  14  188

P  1550; r  0.065; t 

188 365

188   S  1550 1  0.065    1550(1  0.033479)  $1601.89 365  

14. S  1516.80; r  0.08; t 

P

8 12

S 1516.80 1516.80 1516.80     $1440 . 8 1  rt 1  0.08  12 1  0.053. 1.053

15. No. of days  30  31  31  29  31  30  30  212 (2028 is a leap year).

S  3367.28; r  0.09; t  P

212 365

S 3367.28 3367.28    $3200 212 1  rt 1  0.09  365 1  0.052274

16. S  3780; r  0.05; t  P

9 12

S 3780 3780    $3643.37 9 1  rt 1  0.05  12 1  0.0375

17. No. of days  30  31  31  30  14  136

S  1785; r  0.075; t  P

136 365

S 1785 1785    $1736.47 136 1  rt 1  0.075  365 1  0.027945

18. Let the size of the single payment be $x. The focal date is today.

1750 1600  x 4 1  0.09  12 1  0.09  129 1699.03  1498.83  x x  3197.86 The size of the single payment is $3197.86. 19. Let the size of the single payment be $x. The focal date is 30 days from now.

75  1200  1450 1  0.07  x  30 365  1  0.07  365  1470.86  1193.14  x x  2664 The size of the single payment is $2664. 20. Let the size of the final payment be $x. The focal date is one month from now.

3 1   800 1  0.0775    1200  1000 1  0.0775   12  12    x  1  0.0775  122 x 1.012916& 2015.50  1006.46  0.987248 x 1009.04  0.987248 x x  1022.07

815.50  1200  1006.46 

The size of the final payment is $1022.07.

21. Let the size of the equal payments to $x. The focal date is today.

10, 000 

x x  90 1  0.065  365 1  0.065  180 365

x x  1.016027 1.032055 10, 000  0.984225  0.968941x 10, 000  1.953166 x  5119.89 10, 000 

The size of the equal payments is $5119.89. 22. Let the size of the equal payments be $x. The focal date is today. 4000 4000 4000 x    x 9 4 11 12 1  0.0735  12 1  0.0735  12 1  0.0735  12 1  0.0735  12

3904.34  3791.02  3747.51  x 

x 1.0735

11, 442.87  x  0.931532 x 11, 442.87  1.931532x

x  5924.25 The size of the equal payments is $5924.25. 23. Let the size of the single payment be $x. The focal date is today.

2 1200 1400  x  1000 1  0.0825     12  1  0.0825  122 1  0.0825  124  x  1013.75  1183.72  1362.53 x  3560 The size of the single payment is $3560. 24. Let the size of the single payment be $x. The focal date is 5 months from now.

8 5 3    700 1  0.09    1000 1  0.09    800  1  0.09    x 12  12  12     742  1037.50  818  x x  961.50

The size of the final payment is $961.50. 25. Let the size of the equal payments be $x. The focal date is today.

5000 x x  x  6 12 1  0.065  12 1  0.065  12 1  0.065  12 12 5000 x x  x  1.065 1.0325 1.065 4694.84  x  0.968523 x  0.938967 x 4694.84  2.907490 x x  1614.74 The size of the equal payments is $1614.74. 26. Let the size of the instalments be $x. The focal date is the date of the loan. 5000 

x x x   60 120 1  0.069  365 1  0.069  365 1  0.069  180 365

x x x   1.011342 1.022685 1.034027 5000  0.988785 x  0.977818 x  0.967092 x 5000  2.933695 x x  1704.34 5000 

The size of the instalments is $1704.34.

Self-Test

 173  I = (1290)(0.035)    $21.40  365 

I = 8818.75  8500  318.75 318.75 t  0.75 years = 0.75(12) months = 9 months 8500  0.05

r

81.25  0.065  6.50% 2500  126

10, 000 10, 000   $9797.92 1  0.0825  123 1  0.020625

P

10   S  6000 1  0.0375    6000(1  0.03125)  $6187.50 12  

P

PV 

No. of days = 17  31  31  29  31  30  19  188 96.06 96.06 r   0.074600  7.46% 188 2500  365 1287.671233

t

10.

No. of days = 25  31  30  31  30  5  152 55.99 55.99 P   $1378.97 152 0.0975  365 0.040603

11.

P

67.14 67.14   $4781.68 82 0.0625  365 0.014041 4400 4400   $4306.81 243 1  0.325  365 1  0.021637

689.72  0.983558 years  365 = 358.998645 = 359 days 8500  0.0825

7500 7500   $7432.80 88 1  0.0375  365 1  0.009041

12.

No. of days = 24  30  31  31  29  31  30  31  30  31  3  301 (2028 is a leap year) 301 I = 835  0.075   $51.64 365 13. Let the single payment be $x.

115  40  655   x  1725 1  0.085    510 1  0.085   208 365  365  1  0.085  365   655  1725(1.026781)  510(1.009315)  1.048438  1771.20  514.75  624.74  $2910.69 14. Let the size of the final payment be $x.

12  7   1010 1  0.0775    1280  1  0.0775   12  12   

3   615 1  0.0775    x 12   1010(1.0775)  1280(1.045208)  615(1.019375)  x 1088.28  1337.87  626.92  x x  $1799.23

15. Let the size of the equal payments be $x.

3320 

x x x   92 235 1  0.0875  365 1  0.0875  365 1  0.0875  326 365

x x x   1.022055 1.0563356 1.078151 3320  0.978421x  0.946669 x  0.927514 x 3320  2.852604 x 3320 

x  $1163.85 Challenge Problems 1.

Let the size of the final payment be $x. Time period September 28, 2027 to October 31, 2028:

399 days

Time period February 17, 2028 to October 31, 2028:

257 days

Time period July 2, 2028 to October 31, 2028:

121 days

Time period October 1, 2028 to October 31, 2028:

30 days

   399    257   37,500 1  0.07     6350 1  0.07    365    365        121    30    8250 1  0.07     7500 1  0.07    x  365    365     40,369.52  6662.98  8441.45  7543.15  x x  17, 721.94 The payment required to pay off the balance is $17 721.94. 2.

Assume the invoice amount to be $100 The amount of discount is 2% of $100  $2. The amount paid is $98.

P  98; r

I = 2; t 

60 98  365 

60 365

 0.124150  12.42%

The highest annual rate of interest is 12.42%.

Case Study 1.

Plan 1:

  12   4000 1+ 0.04     $4160  12   

Plan 1 costs $4160. Plan 2: Installments =

4000  $2000 2

  12   Payment 1 = (2000 + 60) 1+ 0.04    = $2142.40  12      6  Payment 2 = 2000 1+ 0.04    = 2040.00  12    Total

= $4182.40

Plan 2 costs $4182.40. Plan 3: Payments 1 to 12 = $355 each

Value of payments at the focal date one year from now:

  12    355 1  0.04     369.20  12      11   Payment 2  355 1  0.04     368.02  12    Payment 1

  10   Payment 3  355 1  0.04     366.83  12      9  Payment 4  355 1  0.04     365.65  12      8  Payment 5  355 1  0.04     364.47  12      7  Payment 6  355 1  0.04     363.28  12      6  Payment 7  355 1  0.04     362.10  12      5  Payment 8  355 1  0.04     360.92  12      4  Payment 9  355 1  0.04     359.73  12   

  3  Payment 10  355 1  0.04     358.55  12      2  Payment 11  355 1  0.04     357.37  12      1  Payment 12  355 1  0.04     356.18  12    Total  = 4352.30 Plan 3 costs $4352.30. Plan 1 has the lowest cost. 2.

Focal date today:

Plan 1: Cost = $4000 Plan 2 : Payment 1: = $2060 2000 2000 Payment 2 :   1982.16 6 1.009 1+ 0.018  12 Plan 2 costs $2060  $1982.16  $4042.16 Plan 3 : Cost = Present value of the 12 payments  1 1 1 355 1+ + + 3 1 2  1+ 0.025  12  1+ 0.025  12  1+ 0.018  12  1 1 1 1 + + + + 5 6 4 1+ 0.018  12  1+ 0.018  12  1+ 0.018  12  1+ 0.018  127  +

 1 1 1 1 + + +  10 1+ 0.018  128  1+ 0.018  129  1+ 0.018  12  1+ 0.018  1211  

= 355(1+ 0.997921+ 0.995851+ 0.99552 + 0.994036 + 0.992556 + 0.99108 + 0.989609 + 0.988142 + 0.98668 + 0.985222 + 0.983768) = 355(11.90038) = $4224.64 Plan 3 costs $4224.64.

Plan 1 has the lowest cost 3.

Various answers.

Chapter 8 Simple Interest Applications

Exercise 8.1 1.

Due date is September 25, 2027 No. of days = 123

123 365 123   S  620 1  0.0525    620(1  0.017692)  $630.97 365   P  620; r  0.0525; t 

No. of days = 90 90 365 90   S  350 1  0.045    350(1  0.011096)  $353.88 365   P  350; r  0.045; t 

150 365 836.85 836.85 P   $820 1  0.05  150 1.020548 365

S  836.85; r  0.05; t 

The face value is $820 4.

Due date is June 1, 2028. Note: 2028 is a leap year.

213 365 6000 6000 P   $5748.41 213 1  0.075  365 1.043767

S  6000; r  0.075; t 

The face value is $5748.41 5.

Due date is January 31, 2026 No. of days = 95 95 P = 3600; r = 0.0581; t = 365 æ 95 ö ÷ S = 3600 çç1 + 0.0581´ ÷= 3600(1 + 0.015122) = $3654.44 çè ø 365 ÷ The maturity value is $3654.44. Due date is November 30, 2025

S  3654.44; r  0.0627; t  P

62 365

3654.44 3654.44   $3615.93 62 1  0.0627  365 1.010650

The present value is $3615.93 6.

Due date is October 1, 2026 No. of days = 183 183 P = 19,300; r = 0.08; t = 365 æ ö 183 ÷ S = 19,300 çç1 + 0.08´ ÷= 19,300(1 + 0.040110) = $20, 074.12 çè ø 365 ÷ The maturity value is $20, 074.12. Due date is June 20, 2026

S  20, 074.12; r  0.072; t  P

103 365

20, 074.12 20, 074.12   $19, 674.38 103 1  0.072  365 1.020318

The present value on June 20, 2026 is $19,674.38 7.

Maturity date is October 10, 2025 Number of days = 122

122 365 2500 2500 P   $2449.64 122 1  0.0615  365 1.020556

S  2500; r  0.0615; t 

The present value on July 20, 2025 $2449.64 8.

Maturity date is August 18, 2028 because 2028 is a leap year. Number of days = 78

78 365 7200 7200 P   $7114.25 78 1  0.0564  365 1.012053

S  7200; r  0.0564; t 

The present value is $7114.25

Maturity date is February 10, 2027; purchase date is September 10, 2026. Number of days = 153

153 365 549 549 P   $527.34 153 1  0.098  365 1.041079

S  549; r  0.098; t 

The cash value is $527.34 10. Maturity date is July 1, 2028; purchase date is November 1, 2027. Note: 2028 is a leap year. Number of days = 243

243 365 1980 1980 P   $1749.40 243 1  0.198  365 1.131819

S  1980; r  0.198; t 

The cash value is $1749.40

Exercise 8.2

364 365 50, 000 50, 000 P   49,330.94 364 1  0.0136  365 1.013563

S  50, 000; r  0.0136; t 

The price is $49,330.94 2.

91 ; r  0.0053 365 100, 000 100, 000 P   99,868.04 91 1  0.0053  365  1.001321

S  100, 000; t 

The price is $99,868.04

91 ; P  4966.20 365 Amount of yield = 5000  4966.20 = 33.80 33.80 33.80 Yield rate    0.027299 91 4966.20  365  1238.148493

S  5000; t 

The rate of return was 2.73% 4.

182 365 Amount of yield = 10, 000  9822 = 178 178 178 Yield rate    0.036345 182 9822  365  4897.545205

S  10, 000; P  9822; t 

The rate of return was 3.63% 5.

(a)

91 365 Amount of yield = 100, 000  99,326.85 = $673.15 673.15 673.15 Yield rate    0.027183  2.72% 91 99,326.85  365  24, 763.68

S  100, 000; P  99,326.85; t 

The original yield on the T-bill is 2.72% (b)

Time to maturity at the date of sale is 91  42 = 49 days. 49 S  100, 000; r  0.0252; t  365 100, 000 100, 000 P   $99, 662.84 49 1  0.0252  365  1.003383 The price of the T-bill is $99, 662.84

(c)

The investment grew from $99,326.85 to $99, 662.84 in 42 days. 42 S  99, 662.84; P  99,326.85; t  365 The amount of yield = 99, 662.84  99,326.85 = $335.99 335.99 335.99 Yield rate    0.029397  2.94% 42 99,326.85  365  11, 429.39 The rate of return realized on the T-bill is 2.94%

(a)

364 365 25, 000 25, 000 P   $24, 292.60 1  0.0292  364 1.029120  365

S  25, 000; r  0.0292; t 

The price of the T-bill on April 1 is $24, 292.60 (b)

The time period April 1 to August 15 is 136 days. The number of days to maturity is 364  136  228 days.

S  25,000; P  24,377.64; t 

228 365

The amount of yield = 25, 000  24,377.64 = $622.36 622.36 622.36 Yield rate    0.040870  4.087% 228 24,377.64  365  15, 227.68 The yield rate on August 15 on the T-bill is 4.09% (c)

The time period April 1 to October 1 is 183 days. The number of days to maturity is 364  183  181 days.

181 365 25, 000 25, 000 P   $24, 448.96 181 1  0.04545  365  1.022538

S  25, 000; r  0.04545; t 

The market value of the T-bill on October 1 is $24, 448.96 (d)

The time period April 1 to November 20 is 233 days. The number of days to maturity is 364  233  131 days.

131 365 25, 000 25, 000 P   $24,591.79 1  0.04625  131 1.016599  365

S  25, 000; r  0.04625; t 

The amount of yield = Price on November 20  Price on April 1 = 24,591.79  24, 292.60 = 299.19 299.19 299.19 Yield rate    0.019293  1.9293% 233 24, 292.60  365  15,507.33 The rate of return realized on the T-bill on November 20 is 1.93%

If purchased for $995,000, each T-bill would effectively pay $5000 interest when it matures at $1,000,000. The time remaining until maturity (when the T-bills will each have a market value of $995,000) is t = I/Pr = 5000/(1,000,000(0.025)) = 0.20 year = 0.20  365 = 73 days The T-bills’ price will first exceed $995,000 at 73 days before their maturity date.

Present value of $50,000 discounted at 0.96% for 182 days: P = S / (1 + rt) = 50,000 / (1 + (0.0096)(182/365)) = $49,761.80 Days remaining to maturity = 182 – 53 = 129 days Selling price = 50,000 / (1 + (0.0105)(129/365)) = $49,815.14 Effective interest on initial investment = 49,815.14 – 49,761.80 = $53.34 Rate of return: r = I / Pt = 53.34 / 49,761.80 (53/365) = 0.007382 = 0.74%

Price = present value of $750,000 discounted at 2.15% for 30 days Using the present value formula P = S / (1 + rt)

P  750, 000/ 1   0.0215  30 / 365    $748,677 10. Interest earned by the mutual fund = $2,000,000 – $1,994,000 = $6000 P = $1,994,000 Rearranging the formula I = Prt,

r  6000 / 1,994, 000  60 / 365   0.018305  1.83% 11. Interest earned by the investor = $3,000,000 − $2,976,878.22 = $23,121.78 P = $2,976,878.22 Rearranging the formula I = Prt r = 23,121.78 / (2,976.878.22)(90/365) r = 0.0315 = 3.15% 12. Present value of the $5,000,000 discounted at 4.68% for 30 days Using the present value formula P = S / (1 + rt) P = 5,000,000 / (1 + (0.0468)(60/365)) P = $4,961,827.91 Interest paid = $5,000,000 − $4,961,827.91 = $38,172.09

Exercise 8.3 1. March 10

Original balance

June 30

Partial payment

$10,000.00 $2500.00

Less interest due

10,000  0.055 

112 365

168.77

Balance Sept. 4

2331.23

$7668.77

Partial payment

$4000.00

Less interest due

7668.77  0.055 

66 365

Balance Nov. 15

76.27

3923.73

$3745.04

Interest due

3745.04  0.055 

72 365

Final Payment

40.63

$3785.67

2. Aug 12

Original balance

Nov 1

Partial payment

$20,000.00 $7500.00

Less interest due

20,000  0.0675 

81 365

299.59

Balance

Dec 15

7200.41

$12,799.59

Partial payment

$9000.00

Less interest due

12,799.59  0.0675 

44 365

Balance Feb 20

104.15

8895.85

$3903.74

Interest due

3903.74  0.0675 

67 365

Final Payment

48.37

$3952.11

3. March 10

Original balance

June 30

Partial payment

$6000.00 $2000.00

Less interest due

6000  0.11

112 365

202.52

Balance Sept. 5

1797.48

$4202.52

Partial payment

$2500.00

Less interest due

4202.52  0.11

67 365

Balance Nov. 15

84.86

2415.14

$1787.38

Interest due

1787.38  0.11 

71 365

Final Payment

38.25

$1825.63

4. Aug. 12

Original balance

Nov. 1

Partial payment

$15,000.00 $6000.00

Less interest

15,000  0.105 

81 365

349.52

Balance Dec. 15

5650.48

$9349.52

Partial payment

$5000.00

Less interest

9349.52  0.105 

44 365

Balance Feb. 20

118.34

4881.66

$4467.86

Interest due

4467.86  0.105 

67 365

Final Payment

86.11

$4553.97

5. Date

Interest period

Principal

Rate

Interest

June 10

May 10–June 10

8000

0.08

8000  0.08 

31  $ 54.36 365

July 10

June 10–July 10

8000

0.08

8000  0.08 

30  $ 52.60 365

Aug. 10

July 10–July 20

8000

0.08

8000  0.08 

10  $ 17.53 365

July 20–July 31 incl.

6000

0.08

6000  0.08 

12  $ 15.78 365

Aug. 1–Aug. 10

6000

0.095

6000  0.095 

9  $ 14.05 365 $ 47.36

Sept. 10

Aug. 10–Sept. 10

6000

0.095

6000  0.095 

31  $ 48.41 365

Oct. 10

Sept. 10–Sept. 30 incl.

6000

0.095

6000  0.095 

21  $ 32.79 365

Oct. 1–Oct. 10

3000

0.085

3000  0.085 

9 $ 365

6.29

$ 39.08

Nov. 10

Oct. 10–Nov. 10

3000

0.085

3000  0.085 

31  $ 21.66 365

Dec. 1

Nov. 10–Dec. 1

3000

0.085

3000  0.085 

21  $ 14.67 365

Total interest $278.14

6. Date

Interest period

Principal

Rate

Interest

July 31

July 20–July 31 incl.

12,500

9.5%

12,500  0.095 

Aug. 31

Aug. 1–Aug. 31 incl.

12,500

9.5%

12,500  0.095 

Sept. 30

Sept. 1–Sept. 14 incl.

12,500

8.5%

12,500  0.085 

14  $ 40.75 365

Sept. 15–Sept. 30 incl.

6500

8.5%

6500  0.085 

16  $ 24.22 365

12  $ 39.04 365 31  $100.86 365

$ 64.97

Oct. 31

Oct. 1–Oct. 31 incl.

6500

8.5%

6500  0.085 

31  $ 46.92 365

Nov. 30

Nov. 1–Nov. 10

6500

8.5%

6500  0.085 

9  $ 13.62 365

Nov. 10–Nov. 30 incl.

3500

8.5%

3500  0.085 

21  $ 17.12 365 $ 30.74

Dec. 30

Dec. 1–Dec. 30

3500

3500  0.09 

29  $ 25.03 365

Total interest $307.56

7. March 25

Original balance

May 15

Partial payment

$20,000.00 $600.00

Less interest

20,000  0.07 

51 365

195.62

Balance June 30

404.38 $19,595.62

Partial payment

$800.00

Less interest

19,595.62  0.07 

46 365

172.87

Balance Oct. 10

$18,968.49

Partial payment

$400.00

Less interest

18,968.49  0.07 

1 = $ 3.64 365

18,968.49  0.085 

62 = 273.87 365

18,968.49  0.095 

39 = 192.54 365

Unpaid interest Oct. 31 incl. 18,968.49  0.095 

627.13

470.05 $

22 365

Payment

70.05 108.61

$178.66

8. May 10

Original balance

June 25

Partial payment

$12,000.00 $300.00

Less interest

12,000  0.075 

46 365

113.42

Balance Sept. 20

186.58 $11,813.42

Partial payment

$150.00

Less interest

11,813.42  0.075  11,813.42  0.06 

Nov. 5

37 = $89.81 365

50 365

= 97.10

186.91

Unpaid interest

$ 36.91

Partial payment

$200.00

Less interest

11,813.42  0.06 

42 365

= $81.56

11,813.42  0.05 

4 365

Unpaid interest

6.47

= 36.91

124.94

Balance

$11,738.36

Accrued Interest Dec. 31

57 365

$ 91.66

Accrued Interest on December 31

$ 91.66

11,738.36  0.05 

75.06

inclusive

9. Mar 12

Original balance

June 17

Partial payment

$15,000.00 $1500.00

Less interest  97  15, 000(0.125)    365 

498.29

1001.71

Balance Sept. 10

$13,998.29

Partial payment

$1850.00

Less interest  45  13,998.29(0.125)    365   40  13,998.29  (0.0975)    365 

215.73

149.57

$1484.70

Balance Nov. 8

$12,513.59

Partial payment

$3000.00

Less interest  21  12,513.59  (0.0975)    365 

70.20

 38  12,513.59  (0.0745)    365 

97.06

Balance

$9680.85

Incl. Interest Dec. 31

 53  9680.85(0.0745)    365 

Payment due on Dec. 31

104.73

$9785.58

Exercise 8.4 1.

5 equal monthly payments to principal  $2500  5  $500 Annual rate of interest is 6% Month Loan amount owing

Interest for the month (I = P rt )

$2500

1 I = 2500(0.06)    $12.50  12 

2500  500  $2000

1 I = 2000(0.06)    10.00  12 

2000  500  1500

1 I = 1500(0.06)     12 

7.50

1500  500  1000

1 I = 1000(0.06)     12 

5.00

1000  500  500

1 I = 500(0.06)     12 

2.50

Total interest: $37.50 2.

4 equal monthly payments to principal = $900  4  $225 Annual rate of interest is 12% Month Loan amount owing

Interest for the month (I = P rt )

$900

1 I = 900(0.12)    $9.00  12 

675

1 I = 675(0.12)    6.75  12 

450

1 I = 450(0.12)    4.50  12 

225

1 I = 225(0.12)    2.25  12 

Total interest: $22.50 The loan balance after the second payment was $450. He paid total interest of $22.50 on this loan.

Two equal installments to principal  $1998  2  $999 Annual rate of interest is 3.75% Interest period

Line of credit balance Interest (I = P rt )

15 days

$1998

30 days

$ 999

 15  I  1998  0.0375    $3.08  365   15  I  999  0.0375    1.54  365 

Total interest: $4.62 Total interest paid was $4.62. 4.

(a)

Interest earned (on positive balances) March 10 to March 15 inclusive: 6 days at 1.25% on $572.29  6  I  572.29(0.0125)    $0.12  365 

March 16 to March 19 inclusive: 4 days at 1.25% on $307.29  4  I  307.29(0.0125)    $0.04  365 

Total interest earned  0.12  0.04  $0.16

(b)

Line of credit interest charged (on negative balances up to $1000) March 1: 1 day at 8% on $527.71  1  I  527.71(0.08)    $0.12  365 

March 2 to March 9 inclusive: 8 days at 8% on $1000  8  I  1000(0.08)    $1.75  365 

March 20 to March 21 inclusive: 2 days at 8% on $692.71  2  I  692.71(0.08)    $0.30  365 

March 22 to March 26 inclusive: 5 days at 8% on $776.21  5  I  776.21(0.08)    $0.85  365 

March 27 to March 31 inclusive: 5 days at 8% on $941.21  5  I  941.21(0.08)    $1.03  365 

Total line of credit interest charged  0.12  1.75  0.30  0.85  1.03  $4.05

(c)

Overdraft interest charged on negative balance in excess of $1000 March 2 to March 4 inclusive: 3 days at 18% on $127.71  3  Overdraft interest = 127.71(0.18)    $0.19  365 

March 5 to March 9 inclusive: 5 days at 18% on $427.71  5  Overdraft interest = 427.71(0.18)    $1.05  365 

Total overdraft interest  0.19  1.05  $1.24 (d)

Two transactions caused an overdraft or an overdraft to continue Service charge  2(5)  $10

(e)

The account balance on March 31

 941.21  0.16  4.05  1.24  10  $956.34

(a)

Balance April 1

Interest payment April 30 Balance May 1 Balance May 23 Interest payment May 31

$25,960.06

 30   25,960.06(0.06)    $128.02  365  25,960.06  200  $25, 760.06 25, 760.06  5000  $30, 760.06  22   9   25, 760.06(0.06)    30, 760.06(0.06)    365   365   93.16  45.51  $138.67 30, 760.06  200  $30,560.06

Balance June 1 Interest payment June 30

 19   11  = 30,560.06(0.06)    30,560.06(0.055)    365   365  = 95.45  50.65  $146.10

30,560.06  200  $30,360.06 30,360.06  5000  $35,360.06

Balance July 1 Balance June 19 Interest payment July 31

 18   13   30,360.06(0.055)    35,360.06(0.055)    365   365   82.35  69.27  $151.62

35,360.06  200  $35,160.06 35,160.06  10,500  $45, 660.06

Balance August 1 Balance August 5

 4   27  Interest payment August 31 = 35,160.06(0.055)    45, 000(0.055)    365   365   27  660.06(0.17)    365  Balance September 1 Interest due September 30

 21.19  183.08  8.30  $212.57 45, 660.06  200  $45, 460.06  9   21   45, 000.06(0.055)    45, 000(0.05)    365   365   30   460.06(0.17)    365   61.03  129.45  6.43  $196.91

(b)

Balance September 30

45, 460.06  200  $45, 260.06

Exercise 8.5 1.

Payment number

Balance before payment

Amount paid

Interest paid 0.708333%

Principal repaid

Balance after payment 1200.00

1200.00

180.00

8.50

171.50

1028.50

180.00

7.29

172.71

855.79

180.00

6.06

173.94

681.85

180.00

4.83

175.17

506.68

180.00

3.59

176.41

330.27

180.00

2.34

177.66

152.61

153.69

1.08

152.61

—

$1233.69

$33.69

$1200.00

Totals

Payment number

Balance before payment

Amount paid

Interest paid 0.645833%

Principal repaid

Balance after payment 3400.00

3400.00

800.00

21.96

778.04

2621.96

800.00

16.93

783.07

1838.89

800.00

11.88

788.12

1050.77

800.00

6.79

793.21

257.56

259.22

1.66

257.56

—

$3459.22

$59.22

$3400.00

Totals

Payment number (date)

Balance before payment

Amount paid

Interest paid( daily ) 7.5%pa

Principal repaid

0 Mar. 15

Balance after payment 900.00

1 Apr. 15

900.00

135.00

5.73

129.27

770.73

2 May 15

770.73

135.00

4.75

130.25

640.48

3 June 15

640.48

135.00

4.08

130.92

509.56

4 July 15

509.56

135.00

3.14

131.86

377.70

5 Aug. 15

377.70

135.00

2.41

132.59

245.11

6 Sept. 15

245.11

135.00

1.56

133.44

111.67

7 Oct. 15

111.67

112.36

0.69

111.67

—

$922.36

$22.36

$900.00

Totals

Payment number (date)

Balance before payment

Amount paid

Interest paid( daily ) 6%pa

Principal repaid

0 Feb. 08

Balance after payment 700.00

1 Mar. 08

700.00

100.00

3.22

96.78

603.22

2 Apr. 08

603.22

100.00

3.07

96.73

506.29

3 May 08

506.29

100.00

2.50

97.50

408.79

4 June 08

408.79

100.00

2.08

97.92

310.87

5 July 08

310.87

100.00

1.53

98.47

212.40

6 Aug. 08

212.40

100.00

1.08

98.92

113.48

7 Sept. 08

113.48

114.06

0.58

113.48

—

$714.06

$14.06

$700.00

Totals

Business Math News Box

15  0.15  15% interest for a two-week period 100 Cost of the loan  $100  0.15  $15

Number of two-week periods per year:

365  26.071429 14 Simple annual interest rate = 0.15  26.071429  3.910714  391.1% 2.

21  0.21  21% interest for a two-week period 100 Cost of borrowing $300  $300  0.21  $63 Amount paid back on due date  $300  63  $363

 365  Simple annual interest rate  0.21    5.475  14   547.50% 3.

Borrow from family or friends, a bank or credit union, or your credit card.

Review Exercise 1.

(a)

Due date: October 30

(b)

Interest period :

(c)

Maturity value = 1600 + 34.76 = $1634.76

June 30October 30 = 122 days 122 I = 1600  0.065   $34.76 365

P = 1250; r  0.08; t 

45 365

45   S  P(1  rt )  1250 1  0.08    1250(1  0.009863)  $1262.33 365   120 365 S 1533.29 1533.29 P    $1500 1+ rt 1  0.0675  120 1  0.022192 365

S = 1533.29; r  0.0675; t 

Due date: February 28, 2026

Interest period : July 31, 2025February 28, 2026 = 212 days S 3275 3275 P    $3133.93 212 1+ rt 1  0.0775  365 1  0.045014

P  $5000; r  0.06; t 

150 365

150   S  P(1  rt )  5000 1  0.06   365    5000(1  0.024658)  $5123.29 6.

Due date: September 18, 2026

June 18September 18 = 92 days 92 S = 950.89; r  0.045; t  365 S 950.89 950.89 P    $940.23 92 1+ rt 1  0.045  365 1  0.011342 Interest period :

Maturity value Due date: October 29

July 31October 29 : 90 days 90 P = 800; r  0.08; t  365 90   S  P(1  rt )  800 1  0.08    800(1  0.019726)  $815.78 365   Interest period :

Present value:

Discount period : October 20October 29 : 9 days 9 S = 815.78; r  0.08; t  365 S 815.78 815.78 P    $814.17 9 1+ rt 1  0.08  365 1  0.001973

Maturity value: Due date: October 1, 2025 Interest period :

June 1, 2025 October 1, 2025 = 122 days

122   S  P(1  rt )  1850 1  0.05    1850(1  0.016712)  1880.92 365  

Proceeds:

Discount period : August 28, 2025 October 1, 2025 = 34 days S 1880.92 1880.92 P    $1869.60 34 1+ rt 1  0.065  365 1  0.006055 Proceeds = $1869.60 9.

Due date: July 13, 2025

March 13, 2025July 13, 2025 = 122 days 122 S = 1300; r  0.07; t  365 S 1300 1300 P    $1270.28 122 1+ rt 1  0.07  365 1  0.023397 Interest period :

10. Maturity value: Due date: February 6, 2027

September 6, 2026 February 6, 2027 = 153 days 153 P = 7000; r  0.055; t  365 153   S  P(1+ rt ) = 7000 1+ 0.055    7000(1  0.023055)  $7161.38 365   Interest period :

Proceeds:

November 28, 2026 February 6, 2027 = 70 days 70 S = 7161.38; r  0.065; t  365 S 7161.38 7161.38 P    $7073.21 70 1+ rt 1  0.065  365 1  0.012466 Discount period :

Proceeds = $7073.21

183 365 500, 000 500, 000 P   497,381.59 183 1.005264 1  0.0105  365 The price is $497,381.59

11. S  500, 000; r  0.0105; t 

182 365 Amount of yield = 25, 000  24, 756.25  $243.75 243.75 243.75 Rate of return =  (365)  0.019746  1.97% 24, 756.25  182 24, 756.25(182) 365 

12. S  25, 000; P = 24, 756.25; t 

13. (a)

(b)

364 365 Amount of yield = 1, 000, 000  971,578  $28, 422 28, 422 28, 422 Yield rate =   0.029334  2.93% 364 971,578  365  968,916.14 S  1, 000, 000; P = 971,578; t 

Time period, April 7 to May 16, is 39 days. The time to maturity  364  39  325 days

325 365 Amount of yield = 1, 000, 000  983,500  $16,500 16,500 16,500 Yield rate =   0.018842  1.88% 983,500  325 875, 719.18 365  S  1, 000, 000; P = 983, 500; t 

(c)

Gain from April 7 to May 16  983,500  971,578  $11,922

 39  P  971,578; I = 11, 922; t     365  11,922 11,922 Rate of return =   0.114842  11.48% 39 971,578  365  103,812.44 14. (a)

91 365 Amount of yield = 100, 000  99, 600 = $400 400 400 Yield rate    0.016108  1.6108% 91 99, 600  365  24,831.78

S  100, 000; P  99, 600; t 

The original yield on the T-bill is 1.61%

(b)

Time to maturity at the date of sale is 91  42 = 49 days. 51 S  100, 000; r  0.014; t  365 100, 000 100, 000 P   $99,804.77 51 1  0.014  365  1.001956 The price of the T-bill is $99,804.77

(c)

The investment grew from $99, 600 to $99,804.77 in 40 days. 40 S  99, 600; P  99,804.77; t  365 The amount of yield = 99, 804.77  99, 600 = $204.77 204.77 204.77 Yield rate    0.018760  1.8760% 40 99, 600  365  10,915.07 The annual rate of return realized on the T-bill is 1.88%

15. April 2

Original balance

May 14

Partial payment

$15,000.00 $2800.00

Less interest due

15,000  0.09 

155.34

42 365

Balance June 19

2644.66

$12,355.34

Partial payment

$2400.00

Less interest due

12,355.34  0.09 

109.67

36 365

Balance August 3

2290.33

$10,065.01

Interest due

10,065.01 0.09 

111.68

45 365

Final Payment

$10,176.69

16. Date

Interest period

Principal

Rate

Interest

Mar. 31

Mar. 10–Mar. 31 incl.

15,000

5.5%

15,000  0.055 

22  $ 49.73 365

Apr. 30

Apr. 1–Apr. 30 incl.

15,000

5.5%

15,000  0.055 

30  $ 67.81 365

May 31

May 1–May 31 incl.

15,000

5.5%

15,000  0.055 

31  $ 70.07 365

June 30

June 1–June 19 incl.

15,000

6.25%

15,000  0.0625 

19  $ 48.80 365

June 20–June 30 incl.

11,000

6.25%

11,000  0.0625 

11  $ 20.72 365  $ 69.52

July 31

July 1–July 31 incl.

11,000

6.25%

11,000  0.0625 

31  $ 58.39 365

Aug. 31

Aug. 1–Aug. 31 incl.

11,000

6.25%

11,000  0.0625 

31  $ 58.39 365

Sept. 30

Sept. 1–Sept. 30 incl.

8000

6.25%

8000  0.0625 

30  $ 44.10 365

Oct. 31

Oct. 1–Oct. 31 incl.

8000

8000  0.06 

31  $ 40.77 365

Nov. 15

Nov. 1–Nov. 15

8000

8000  0.06 

14  $ 18.41 365

Total interest $477.19

17. May 25

Original loan balance

July 10

Partial payment

$20,000 $5000

Less interest

46 365

20,000  0.075 

189.04

Balance Sept. 15

4810.96 $15,189.04

Partial payment

$8000

Less interest July 10–July 31: incl.

22 365

 $ 68.66

45 365

= 149.81

15,189.04  0.075  Aug. 1–Sept. 15:

15,189.04  0.08 

218.47

Balance

$7407.51

Interest Payment Oct. 31

Sept. 15–Sept. 30: incl.

7407.51 0.08 

7781.53

16 365

= $ 25.98

31 365

Oct. 1–Oct. 31:

7407.51  0.085 

53.48

Payment due

$79.46

18. (a)

 $8195

Balance July 1

Balance July 15

- $8195 + 300 = - $7895

Interest charged July 31: æ14 ö ÷ 8195(0.08) çç ÷ ÷= çè365 ø æ17 ö ÷ 7895(0.08) çç ÷ ÷= çè365 ø

$25.15 29.42 $54.57

Balance July 31 Balance August 15 Balance August 20

- 7895 - 54.57 = - $7949.57 - 7949.57 + 300 = - $7649.57 - 7649.57 - 3000 = - $10, 649.57

Interest charged August 31: æ14 ÷ ö 7949.57(0.08) çç ÷ çè365 ÷ ø æ5 ÷ ö 7649.57(0.08) çç ÷ çè365 ÷ ø æ12 ÷ ö 10, 000(0.08) çç ÷ çè365 ÷ ø æ12 ö÷ 649.57(0.16) çç ÷ çè365 ø÷

= $24.39 =

8.38

= 26.30 =

3.42 $62.49

Balance August 31 Balance September 15

- 10, 649.57 - 62.49 = - $10, 712.06 - 10, 712.06 + 300 = - $10, 412.06

Interest charged September 30 : æ14 ö÷ 10, 000(0.08) çç ÷ = $30.68 çè365 ø÷ æ16 ÷ ö 10, 000(0.075) çç ÷ ÷= 32.88 çè365 ø æ14 ö÷ 712.06(0.16) çç ÷= 4.37 çè365 ø÷ æ16 ÷ ö 412.06(0.16) çç ÷ ÷ = 2.89 çè365 ø $70.82

Balance September 30 Balance October 15 Balance October 25

- 10, 412.06 - 70.82 = - $10, 482.88 - 10, 482.88 + 300 = - $10,182.88 - 10,182.88 - 600 = - $10,782.88

Interest charged October 31: æ 31 ÷ ö 10, 000(0.08) çç = $67.95 ÷ ÷ çè365 ø æ14 ö ÷ 482.88(0.16) çç ÷ ÷ = 2.96 çè365 ø æ10 ÷ ö 182.88(0.16) çç = 0.80 ÷ çè365 ÷ ø æ7 ö ÷ 728.88(0.16) çç ÷ ÷ = 2.40 çè365 ø $74.11 Balance October 31 Balance November 15

- 10, 782.88 - 74.11 = - $10,856.99 - 10,856.99 + 300 = - $10,556.99

Interest charged November 30 : æ 30 ÷ ö 10, 000(0.075) çç = $61.64 ÷ çè365 ø÷ æ14 ÷ ö 856.99(0.16) çç = 5.26 ÷ ÷ çè365 ø æ16 ÷ ö 556.99(0.16) çç = 3.91 ÷ çè365 ø÷ $70.81 (b) 19.

Balance November 30

 10,556.99  70.81  $10,627.80

Interest paid to the unsecured line of credit:

3996  0.05 

6  $99.90 12

Interest paid to the secured line of credit:

3996  0.035 

9  $104.90 12

The unsecured line of credit is his best option because he will pay less interest.

20. Monthly payment = 3000/3 = $1000 Monthly rate of interest = 0.0625/12 = 0.005208 Month

Loan Amount Owing

Interest for Month

3000

15.63

3000 – 1000 = 2000

10.42

2000 – 1000 = 1000

5.21 Total Interest = $31.26

21.

Payment number

Balance before payment

Interest paid Amount paid

0.75%

Principal repaid

Balance after payment 3000.00

3000.00

500.00

22.50

477.50

2522.50

500.00

18.92

481.08

2041.42

500.00

15.31

484.69

1556.73

500.00

11.68

488.32

1068.41

500.00

8.01

491.99

576.42

500.00

4.32

495.68

80.74

81.35

0.61

80.74

—

$3081.35

$81.35

$3000.00

Totals

22. Payment number (date)

Balance before payment

Interest paid

Amount paid

(daily)

Principal repaid

Balance after payment

7.8%pa

0 Dec 2/27

2400.00

1 Jan 2/28

2400.00

292.00

15.90

276.10

2123.90

2 Feb 2/28

2123.90

292.00

14.07

277.93

1845.97

3 Mar 2/28

1845.97

292.00

11.44

280.56

1565.41

4 Apr 2/28

1565.41

292.00

10.37

281.63

1283.78

5 May 2/28

1283.78

292.00

8.23

283.77

1000.01

6 Jun 2/28

1000.01

292.00

6.62

285.38

714.63

7 Jul 2/28

714.63

292.00

4.58

287.42

427.21

8 Aug 2/28

427.21

292.00

2.83

289.17

138.04

9 Sep 2/28

138.04

138.95

0.91

138.04

—

$2474.95

$74.95

$2400.00

Totals

Self-Test

Due date: June 10, 2026 Interest period:

January10–June 10 = 151 days

I  565  0.0825 

151  $19.28 365

Interest period: 120 days

120   S  1140 1  0.0775    1140(1.025479)  $1169.05 365  

Due date: September 20, 2026 Interest period: May 20–September 20  123 days

P 4.

1190.03 1190.03   $1160.69 1  0.075  123 1.025274 365

Due date: March 7, 2026 Maturity value: $1800 Discount period: December 20, 2025–March 7, 2026  77 days

P 5.

1800 1800   $1766.46 77 1  0.09  365 1.018986

Due date: September 14, 2025  180 days = March 13, 2026 180   Maturity value  1665 1  0.05   365    1665(1  0.024658)  $1706.05

Discount period: October 18, 2025–March 13, 2026  146 days

PV 

1706.05 1706.05   $1666.06 146 1  0.06  365 1  0.024000

91 365 25, 000 25, 000 P   $24,920.47 91 1  0.0128   365  1.003191

S = 25, 000; r  0.0128; t 

(a)

S = 100, 000; r  0.0385; t 

(b)

Time to maturity = 182  67  115 days 115 S = 100, 000; P = 98, 853.84; t  365 Amount of yield = 100, 000  98,853.84  $1146.16 1146.16 1146.16 Yield rate   (365)  0.036800  3.68% 115 98,853.84  365  98,853.84(115)

182 365 100, 000 100, 000 P   $98,116.43 1  0.0385   182 1  0.019197 365 

8. Date June 5

Interest period

Principal 6000

Rate

Interest

July 5

June 5–June 30 incl.

6000

8.5%

6000  0.085 

July 1–July 4 incl.

6000

9.5%

July 5–July 14 incl.

6000

9.5%

July 15–Aug. 4 incl.

4500

9.5%

Sept. 5

Aug. 5–Sept. 4 incl.

4500

9.5%

Oct. 5

Sept. 5–Sept. 30 incl.

4500

9.5%

Oct. 1–Oct. 4 incl.

4500

10%

Oct. 5–Oct. 9 incl.

4500

10%

Oct. 10–Nov. 4 incl.

2500

10%

Dec. 5

Nov. 5–Dec. 4 incl.

2500

10%

Dec. 30

Dec. 5–Dec. 30

2500

10%

Aug 5

Nov. 5

26  $ 36.33 365 4 $ 6.25 6000  0.095   365 $ 42.58 10  $15.62 365 21 $ 24.60 4500  0.095   365 $ 40.22 6000  0.095 

31  $36.31 365 26 4500  0.095   $30.45 365 4 $ 4.93 4500  0.10   365 $35.38 4500  0.095 

5  $6.16 365 26 $17.81 2500  0.10   365 $ 23.97 4500  0.10 

30  $ 20.55 365 25 2500  0.10   $17.12 365

2500  0.10 

Total interest $216.13

9. Mar. 1

Original balance

Apr. 15

Partial payment

$24,000.00 $600.00

Less interest

24,000  0.07 

45 365

207.12

Balance July 20

$23,607.12

Partial payment

$400.00

Less interest

23,607.12  0.07 

96 365

434.63

Unpaid interest Oct. 10

$ 34.63

Partial payment

$400.00

Less interest July 20–July 31:

12 365

= $54.33

70 365

= 384.83

Unpaid interest, July 20

34.63

23,607.12  0.07  Aug. 1–Oct. 10:

23,607.12  0.085 

Unpaid interest Nov. 30

Interest

22 365

= $120.95

30 365

= 145.52

Unpaid interest, Oct. 10

73.79

Accrued interest

$340.26

23,607.12  0.085  Nov. 1–Nov. 30:

23,607.12  0.075 

Balance March 1

473.79 $73.79

Oct. 10–Oct. 31:

10. (a)

392.88

$7265

Balance March 15 - 7265 + 200 = - $7065 Interest charged March 31: æ14 ö ÷ 7265(0.09) çç = ÷ ÷ çè365 ø æ17 ö ÷ 7065(0.09) çç = ÷ ÷ çè365 ø

$25.08 29.61 $54.69

Balance March 31 Balance April 10 Balance April 15 Interest charged April 30 :

- $7065 - 54.69 = - $7119.69 - $7119.69 - 3000 = - $10,119.69 - 10,119.69 + 200 = - $9919.69 æ9 ÷ ö 7119.69(0.09) çç = $15.80 ÷ çè365 ø÷ æ 5 ö÷ 10, 000(0.09) çç = 12.33 çè365 ÷ ø÷ æ16 ÷ ö 9919.69(0.09) çç = 39.14 ÷ ÷ çè365 ø æ5 ÷ ö 119.69(0.18) çç ÷= 0.30 çè365 ø÷ $67.57

Balance April 30 Balance May 15 Interest charged May 31:

- 9919.69 - 67.57 = - $9987.26 - 9987.26 + 200.00 = - $9787.26 æ14 ö÷ 9987.26(0.09) çç ÷= $34.48 çè365 ø÷ æ17 ÷ ö 9787.26(0.085) çç = 38.75 ÷ çè365 ÷ ø $73.23

Balance May 31 Balance June 15 Balance June 20 Interest charged June 30 :

- 9787.26 - 73.23 = - $9860.49 - 9860.49 + 200 = - $9660.49 - 9660.49 - 500 = - $10,160.49 æ14 ÷ ö 9860.49(0.085) çç = $32.15 ÷ çè365 ÷ ø æ5 ÷ ö 9660.49(0.085) çç = 11.25 ÷ çè365 ÷ ø æ 11 ÷ ö 10, 000(0.085) çç = 25.62 ÷ çè365 ø÷ æ 11 ÷ ö 160.49(0.18) çç = 0.87 ÷ çè365 ø÷ $69.89

(b)

Balance June 30

10,160.49  69.89  $10, 230.38

11.

Payment number

Balance before payment

Amount paid

Interest paid

0.54166675%

Principal repaid

Balance after payment 4000.00

4000.00

750.00

21.67

728.33

3271.67

750.00

17.72

732.28

2539.39

750.00

13.76

736.24

1803.15

750.00

9.77

740.23

1062.92

750.00

5.76

744.24

318.68

320.41

1.73

318.68

—

$4070.41

$70.41

$4000.00

Totals

Challenge Problems 1.

Maturity value of promissory note: P  12, 000; r  0.11; t 

12 12

  12   S  12, 000 1  0.11    12, 000(1  0.11)  $13,320  12   

MacDonald’s Furniture’s proceeds:

365  100 265  365 365 13,320 13,320 P   $12,171.24 265 1  0.13  365  1  0.094384

S  13,320; r  0.13; t 

MacDonald’s gain  12,171.24  12, 000  $171.24 Royal Bank’s proceeds:

265  25 240  365 365 13,320 13,320 P   $12,575.79 240 1  0.09  365  1  0.059178

S  13,320; r  0.09; t 

Royal Bank’s gain  12,575.79  12,171.24  $404.55 Friendly Finance Company’s gain, if the note is held to maturity,  13,320  12,575.79  $744.21

Interest  $6; original principal owed by son  $114. The son owes $114 for the first month, $104 for the second month, $94 for the third month, and so on until he owes $4 for the last (the 12th) month.

114  104  94  84  74  64  54  44  34  24  14  4 The average loan balance = 12 708 114 + 4 118     $59  alternatively,   $59  12 2 2   12 P  59; t  12 6(12) r  0.101695  10.17% 59(12)

Case Study 1. Date

Interest Paid

Principal Repaid

April 20

Balance 8500

May 20

 30  8500(0.0625)    43.66  365 

850

7650

June 20

 31  7650(0.0625)    40.61  365 

850

6800

July 20

 30  6800(0.0625)    34.93  365 

850

5950

Aug. 20

 31  5950(0.0625)    31.58  365 

850

5100

Sept. 20

 31  5100(0.0625)    27.07  365 

850

4250

Oct. 20

 30  4250(0.0625)    21.83  365 

850

3400

Nov. 20

 31  3400(0.0625)    18.05  365 

850

2550

Dec. 20

 30  2550(0.0625)    13.10  365 

850

1700

Jan. 20

 31  1700(0.0625)    9.02  365 

850

Feb 20

 31  850(0.0625)    4.51  365 

850

—

Total interest paid  $244.36

Date

Interest Paid

Principal Repaid

April 20

Balance 8500

May 20

 30  8500(0.10)    69.86  365 

850

7650

June 20

 31  7650(0.10)    64.97  365 

850

6800

July 20

 30  6800(0.10)    55.89  365 

850

5950

Aug. 20

 31  5950(0.10)    50.53  365 

850

5100

Sept. 20

 31  5100(0.10)    43.32  365 

850

4250

Oct. 20

 30  4250(0.10)    34.93  365 

850

3400

Nov. 20

 31  3400(0.10)    28.88  365 

850

2550

Dec. 20

 30  2550(0.10)    20.96  365 

850

1700

Jan. 20

 31  1700(0.10)    14.44  365 

850

Feb 20

 31  850(0.10)    7.22  365 

850

—

Total interest paid $391 3.

The difference  391  244.38  $146.62. Borrowing from Shannon’s parents saves $146.62

Various answers.

QUESTIONS 1. CHAPTER 5

(a)

Q

1000  95.23 (12  1.50)

Break-even would be 96 passes.

(b)

ii.

Total revenue = $12  96 passes = $1152

iii.

Percent of capacity =

(39.99  VC )1600  49,920  0 49,920 (39.99  VC )  1600 (39.99  VC )  31.20 VC  39.99  31.20

96  3  57.60% 500

VC  $8.79

ii.

Contribution rate  85% 

x 39.99

85%(39.99)  $33.99 VC  39.99  33.99  $6 (39.99  6) x  49,920  0 49,920 (33.99)  x 49,920 x 33.99 x  1468.67 rounded to 1469 units

(c)

Break-even volume 

2340 2340   518.85 rounded to 519 units 7.99  3.48 4.51

ii.

( P  2.78)3000  2340  0 2340 ( P  2.78)  3000 ( P  2.78)  0.78 P  $3.56

iii.

æ7.99 - 3.56 ö 4.43 ÷ Discount = çç = ÷ ÷ 7.99 = 55.44% çè 7.99 ø

iv.

Sale price  7.99(1  0.25)  $5.99 (5.99  2.78)950  2340  $709.50 profit

2. CHAPTER 6 (a)

Net Price  9800(1  0.30)(1  0.20)(1  0.05) Net Price  9800(0.70)(0.80)(0.95) Net Price  9800(0.532) Net Price  $5213.60

(b)

(c)

Amount paid  30,120(0.98)  $29,517.60

ii.

Amount paid  20, 000(0.98)  $19, 600

Selling price  12  (0.45)12  (0.60)12  12  5.40  7.20  $24.60

ii.

Markup  5.40  7.20  12.60

Markup based on cost 

12.60  105.00% 12

iii. Sale price  12  (0.45)12  (0.50)12  12  5.40  6  $23.40 (d)

Sale price  22.99(1  0.30)  $16.09

ii.

School price  16.09  5  11.09

Markdown rate 

22.99  11.09  51.76% 22.99

3. CHAPTER 7 æ ö 60 ÷ = 4242(1.009863) = $4283.84 (a) Invoice A: $4242: S = 4242 çç1 + 0.06´ ÷ çè ø 365 ÷

Invoice B: $12,567: P 

12,567 12,567   $12,505.33 30 (1  0.06  365 ) 1.004932

Invoice C: $18,451: P 

18, 451 18, 451   $18, 035.93 140 (1  0.06  365 ) 1.023014

Total cash needed: $34,825.10 (b) Original payment A: $11,000: æ 35 ö ÷ S = 11, 000 çç1 + 0.072´ ÷= 11, 000(1.006904) = $11, 075.95 çè ø 365 ÷ Original payment B: $16,000: P 

16, 000 16, 000   $15, 788.22 68 (1  0.072  365 ) 1.013414

Today’s equivalent value of original payments = $26,864.17 Today’s equivalent value of replacement payment: $26,864.17 – 6000 = $20,864.17 Final payment: æ ö 90 ÷ S = 20,864.17 çç1 + 0.072´ = 20,864.17(1.017753) = $21, 234.58 ÷ ÷ çè 365 ø (c) Original payment A: $18,000: P 

18, 000 18, 000   $17, 737.58 60 (1  0.09  365 ) 1.014795

Original payment B: $16,000: P 

16, 000 16, 000   $15,540.18 120 (1  0.09  365 ) 1.029589

Today’s equivalent value of original payments = $33,277.76 Today’s equivalent value of replacement payment: Replacement payment A: P 

xA xA   0.981843 xA 75 (1  0.09  365 ) 1.018493

Replacement payment B: P 

xB xB   0.975936 xB 100 (1  0.09  365 ) 1.024658

Replacement payment C: P 

xC xC   0.953003xC 200 (1  0.09  365 ) 1.049315

0.981843 xA  0.975936 xB  0.953003 xC  33, 277.76 2.910781x  33, 277.76

Replacement payments are each: x = $11,432.59

(d)

r

15,150  15, 000 150   7.30% 50 15, 000  365 2054.79

4. CHAPTER 8

(a)

183   S  10, 000 1  0.05    $10, 000(1.025068)  $10, 250.68 365  

(b) Interest charged on negative balances: Dates

Explanation

Equation

Interest

Mar 1–4

4 days at 8% on $4000

æ5 ÷ ö I = 4000(0.08) çç ÷ çè365 ÷ ø

$4.38

Mar 5–6

2 days at 8% on $6185

æ2 ö ÷ I = 6185(0.08) çç ÷ ÷ çè365 ø

2.71

Mar 7–11

5 days at 8% on $6606

æ5 ö ÷ I = 6606(0.08) çç ÷ ÷ çè365 ø

7.24

Mar 12–14

3 days at 8% on $606

 3  I  606(0.08)    365 

0.40

Mar 15–20

6 days at 8% on $5734

æ6 ö ÷ I = 5734(0.08) çç ÷ ÷ çè365 ø

7.54

Mar 21–22

2 days at 8% on $11 867

æ2 ö ÷ I = 11,867(0.08) çç ÷ çè365 ÷ ø

5.20

Mar 23–26

4 days at 8% on $9867

æ4 ö ÷ I = 9867(0.08) çç ÷ ÷ çè365 ø

8.65

Mar 27–31 inclusive

5 days at 8% on $4867

æ5 ö ÷ I = 4867(0.08) çç ÷ ÷ çè365 ø

5.33

31 days

Total Interest

$41.45

æ2 ö ÷ I = 1867(0.19) çç ÷ ÷ çè365 ø

$1.94

Interest charged in overdrafts: Mar 21–22

2 days at 19% on $1867

Overdraft service charge: 1 $10  $10 Balance = – 4867 – 41.45 – 1.94 – 10 = –$ 4920.39

Balance Before Payment

Amount Paid

Interest Paid I = 7.6% p.a.

Principal Paid

Balance After Payment $25,000.00

$25,000.00

$2200.00

$158.33

$2041.67

22,958.33

$2200.00

145.40

2054.60

20,903.74

$2200.00

132.39

2067.61

18,836.13

$2200.00

119.30

2080.70

16,755.42

$2200.00

106.12

2093.88

14,661.54

$2200.00

92.86

2107.14

12,554.40

$2200.00

79.51

2120.49

10,433.91

$2200.00

66.08

2133.92

8299.99

$2200.00

52.57

2147.43

6152.56

$2200.00

38.97

2161.03

3991.52

$2200.00

25.28

2174.72

1816.80

$1828.31

11.51

1816.80

0.00

$26,028.31

$1028.31

25,000.00

TOTALS

i. First two months interest = 158.33 + 145.40 = $303.73 ii. Principal repaid the end of two months = 2041.67 + 2054.60 = $4096.27 iii. Total interest over twelve months = $1028.31 iv. The final payment is $1816.80 + 11.51 = $1828.31

PART THREE Mathematics of Finance and Investment Chapter 9 Compound Interest—Future Value and Present Value Exercise 9.1 A. 1.

m  1; i  12%  0.12; n  5

m2; i 

7.4%  3.7%  0.037 ; n  8  2  16 2

m4; i 

5.5%  1.375%  0.01375 ; n  9  4  36 4

m  12 ; i 

m2; i 

11.5%  5.75%  0.0575 ; n  13.5  2  27 2

m4; i 

4.8%  1.2%  0.012 ; n  5.75  4  23 4

m  12 ; i  

m4; i 

10.75%  2.6875%  0.026875 ; n  3.75  4  15 4

m2; i 

12.25% 54  6.125%  0.06125 ; n  2  9 2 12

10.

m  12 ; i 

B. 1.

7%  0.583%  0.0583 ; n  4 12  48 12

8%  0.6%  0.006 ; n  12.5 12  150 12

8.1%  0.675%  0.00675 ; n  15.5 12  186 12

(a)

10%

(b)

(c)

10% / 4  2.5%

(d)

12 years  4  48

(e)

(1  0.025)48

(f )

3.271490

(a)

4.2%

(b)

(c)

4.2% / 2  2.1% yes

(d)

5.5 years  2  11

(e)

(1  0.021)11

(f )

1.256849

(a)

9.6%

(b)

(c)

9.6% /12  0.80%

(d)

7 years 12  84

(e)

(1  0.008)84

(f )

1.952921

(a)

16%

(b)

365

(c)

16% / 365  0.0438%

(d)

2 years  365  730

(e)

(1  0.000438)730

(f )

1.377031

(a)

7.25%

(b)

(c)

7.25% / 26 = 0.2788%

*Rounded

(d)

84 years  26  182 12

(e)

(1  0.002788)182

(f)

1.65996

Exercise 9.2 1.

3.5%  1.75%  0.0175; n  5  2  10 2 FV = 5000(1.0175)10  5000(1.189444)  $5947.22

PV = 5000; I / Y  3.5; C / Y  2; = 2; i 

I = 5947.22  5000  $947.22

(Set P / Y  2) 5000  PV 3.5 I/Y 10 N CPT FV 5947.22 2.

PV  1500; I/Y  3.45; C/Y  4;  2; i 

3.45%  0.8625%  0.008625; n  15  4  60 4

FV  1500(1.008625)60  1500(1.674109)  $2511.16 I  2511.16  1500  $1011.16

(Set P / Y  2) 2nd (CLR TVM) 5000  PV 3.5 I/Y 10 N CPT FV 5947.22 3.

PV = 500; I/Y  7; C/Y  4; i 

7%  1.75%  0.0175 4

October 31, 2008  July 31, 2029 contains 20 years, 9 months. 9  n  (20* 4)    4   80  3  83  12  FV = 500(1.0175)83  500(4.22043)  $2110.21

(Set P / Y = 4) 2nd (CLR TVM) 500  PV 7 I/Y 83 N CPT FV 2110.21 4.

 10  PV = 5000; I / Y = 7.75; C / Y = 2; i  3.875%  0.03875; n  2  5   11.6  12  FV  5000(1.03875)11.6  5000(1.55822)  7791.10 I  7791.10  5000  $2791.10

(Set P / Y  2) 5000  PV 7.75 I / Y 11.6 N CPT FV 7791.10

PV = 4000; I / Y = 3.83; C / Y = 1; i  3.83%  0.0383; n  4

8  4.6 12

FV  4000(1.0383)4.6  4000(1.191718)  $4766.87

(Set P / Y  1) 4000  PV 3.83 I / Y 4.6 N CPT FV 4766.87 PV = 4000; I/Y = 3.79; C / Y = 365; i = 3.79% = 0.0.0379; n = (4 8/12)(365)  1703.3 FV  4000(1.000104)1703.3 = 4000 (1.193461) = $4773.84 (Set P/Y = 365) 4000 +/– PV 3.79 I/Y 1703.3 N CPT FV 4773.84 The 3.79% compounded daily rate earns more interest. Interest is larger by $6.97.

PV = 8000; I / Y = 10.8; C / Y = 1; i  10.8%  0.108; n  7

5  7.416 12

FV  8000(1.108)7.416  8000(2.139620)  $17,116.96

(Set P / Y  1) 8000  PV 10.8 I / Y 7.416 N CPT FV 17,116.96 7.

PV  98.5; I / Y  3; C / Y  1; i  3%  0.03; n  33

FV  98.5(1.03)33  98.5(2.652335)  261.255021 Index at the beginning of 2024 would be 261.26

(Set P / Y  1) 98.5  PV 3 I / Y 33 N CPT FV 261.255021 8.

PV  2,500, 000; I / Y = 8; C / Y = 1; i  8%  0.08; n  5 FV  2,500, 000(1.08)5  2,500, 000(1.469328)  $3,673,320.19 Forecasted assets will amount to $3,673,320.19

(Set P / Y  1) 2,500,000  PV 8 I / Y 5 N CPT FV 3,673,320.19

Balance after 2.5 years: 3%  0.75%  0.0075; n  2.5  4  10 4 FV  2000(1.0075)10  2000(1.077583)  $2155.165091 PV  2000; I / Y  3; C / Y  4; i 

Balance after 6 years:

PV  2155.165091; I / Y  2.75; C / Y  12; i 

2.75%  0.2292%  0.002292; 12

n  3.5  12  42 FV  2155.165091(1.002292) 42  2155.16509(1.100913)  $2372.65 After six years the account is worth $2372.65

(Set P / Y  4) 2000  PV 3 I / Y 10 N CPT FV 2155.165091 (Set P / Y  12) 2155.165091  PV 2.75 I / Y 42 N CPT FV 2372.65 10. Balance after 3 years: 4.5%  0.375%  0.00375; n  3  12  36 12 FV = 2500(1.00375)36  2500(1.144248)  $2860.619581 PV = 2500; I / Y  4.5; C / Y  12; i 

Balance 1.5 years later:

5%  1.25%  0.0125; n  1.5  4  6 4 FV  2860.619581(1.0125) 6  2860.619581(1.077383)  3081.98

PV  2860.619581; I / Y  5; C / Y  4; i 

The accumulated value 1.5 years after the change is $3081.98

(Set P / Y  12) 2500  PV 4.5 I / Y 36 N CPT FV 2860.619581 (Set P / Y  4) 2860.619581  PV 5 I/Y 6 N CPT FV 3081.98

11. Debt value August 1, 2023:

PV  800; I / Y  10; C / Y  2; i 

10%  5%  0.05; 2

February 1, 2021  August 1, 2023, contains 2 years, 6 months: n  2.5  2  5 FV  800(1.05)5 = 800(1.276282) = $1021.02525

Debt value November 1, 2026:

PV  1021.02525; I / Y  11; C / Y  4; i 

11%  2.75%  0.0275; 4

August 1, 2023  November 1, 2026, contains 3 years, 3 months: n  3.25  4  13 FV  1021.02525(1.0275)13  1021.02525(1.422865)  $1452.78 The accumulated value of the debt on November 1, 2026, is $1452.78

(Set P / Y  2) 800  PV 10 I / Y 5 N CPT FV 1021.02525

(Set P / Y  4) 1021.02525  PV 11 I / Y 13 N CPT FV 1452.78 12. Balance July 1, 2026:

PV  1300; I / Y  8.5; C / Y  12; i 

8.5%  0.7083%  0.007083; 12

March 1, 2024  July 1, 2026, contains 2 years, 4 months: n  28 FV  1300(1.007083) 28  1300(1.218517)  $1584.071556

Balance April 1, 2029:

PV  1584.071556; I / Y  8; C / Y  4; i 

8%  2%  0.02; 4

July 1, 2026  April 1, 2029, contains 2 years, 9 months: n  2.75  4  11 FV  1584.071556(1.02)11 = 1584.071556(1.243374) = $1969.59 The balance on April 1, 2029, is $1969.59

(Set P / Y  12) 1300  PV 8.5 I / Y 28 N CPT FV 1584.071556 (Set P / Y  4) 1584.071556  PV 8 I / Y 11 N CPT FV 1969.59

13.

I / Y  6; C / Y  12; i 

6%  0.5%  0.005; 12

Balance July 1, 2022: PV  $1000; n  19 FV  1000(1.005)19  1000(1.099399)  $1099.398584 Balance November 1, 2024: PV  1099.398584 + 1000 = 2099.398584; n  28 FV  2099.398584(1.005)28  2099.398584(1.149873)  $2414.040929 Balance January 1, 2028: PV  2414.040929 + 1000 = 3414.040929; n  38 FV  3414.040929(1.005)38  3414.040929(1.208677)  $4126.47 The balance in the RRSP account is $4126.47 on January 1, 2028.

(Set P / Y  12) 1000  PV 6 I / Y 19 N CPT FV 1099.398584 2099.398584  PV 28 N CPT FV 2414.040929 3414.040929  PV 38 N CPT FV 4126.47 14. I / Y  7.5; C / Y  4; i 

7.5%  1.875%  0.01875 4

Balance on December 1, 2016: PV = 2000; n  2.75  4  11 FV = 2000(1.01875)11  2000(1.226715)  $2453.430923 Balance on September 1, 2020: PV  2453.430923 + 1800 = 4253.430923; n  1.75  4  7 FV  4253.430923(1.01875)7  4253.430923(1.138868)  $4844.095945 Balance on December 1, 2027: PV  4844.095945 + 1700 = 6544.095945; n  7.25  4  29 FV  6544.095945(1.01875)29  6544.095945(1.713804)  $11, 215.29 The accumulated value of the RRSP account on December 1, 2027 is $11, 215.29

(Set P / Y  4) 2000  PV 7.5 I / Y 11 N CPT FV 2453.430923 4253.430923  PV 7 N CPT FV 4844.095945

6544.095945  PV 29 N CPT FV 11215.29 15.

I / Y  10; C / Y  4; i 

10%  2.5%  0.025 4

Balance after 9 months: PV = 4000; n  3 FV = 4000(1.025)3  4000(1.076891)  $4307.5625 Balance after 18 months: PV = 4307.5625  1500  2807.5625; n  3 FV = 2807.5625(1.025)3  2807.5625(1.076891)  $3023.437735 Balance after 27 months: PV = 3023.437735  2000  1023.437735; n  3 FV = 1023.437735(1.025)3  1023.437735(1.076891)  $1102.13 The size of the final payment is $1102.13

(Set P / Y  4) 4000  PV 10 I / Y 3 N CPT FV 4307.5625 2807.5625  PV 3 N CPT FV 3023.437735 1023.437735  PV 3 N CPT FV 1102.13 16.

I / Y  9; C / Y  12; i 

9%  0.75%  0.0075 12

Balance after 4 months: PV  6000; n  4 FV  6000(1.0075)4  6000(1.030339)  $6182.035144 Balance after 9 months: PV  6182.035144  2000  4182.035144; n  5 FV  4182.035144(1.0075)5  4182.035144(1.038067)  $4341.231566 Balance after 1 year: PV  4341.231566  3000  1341.231566; n  3 FV  1341.231566(1.0075)3  1341.231566(1.022669)  $1371.636175

Balance 6 months later: PV  1371.636175  4000  5371.636175; n  6 FV  5371.636175(1.0075)6  5371.636175(1.045852)  $5617.94 The mortgage was for $5617.94

(Set P / Y  12) 6000  PV 9 I / Y 4 N CPT FV 6182.035144 4182.035144  PV 5 N CPT FV 4341.231566 1341.231566  PV 3 N CPT FV 1371.636175 5371.636175  PV 6 N CPT FV 5617.94 17. Balance after 1 year: 9%  2.25%  0.0225; n  1 4  4 4 FV  3000(1.0225) 4  3000(1.093083)  $3279.249956

PV  3000; I / Y  9; C / Y  4; i 

Balance after 2 years: 10%  5%  0.05; n  1  2  2 2 FV  3279.249956(1.05)2  3279.249956(1.1025)  $3615.373077 PV  3279.249956; I / Y  10; C / Y  2; i 

Balance after 4 years: PV  3615.373077  1500  2115.373077; i  0.05; n  2  2  4 FV  2115.373077(1.05)4  2115.373077(1.215506)  $2571.249196 Balance after 7 years: PV  2571.249196  1500  1071.249196; m  12; i 

10%  0.83%  0.0083; 12

n  3 12  36 FV  1071.249196(1.0083)36  1071.249196(1.348182)  $1444.24 The final payment is $1444.24

(Set P / Y  4) 3000  PV 9 I / Y 4 N CPT FV 3279.249956 3279.249956  PV 10 I/Y 2 N CPT FV 3615.373077 2115.373077  PV 4 N CPT FV 2571.249196

1071.249196  PV 36 N CPT FV 1444.24

18. Balance after 18 months: 11% 18  5.5%  0.055; n   2  3 2 12 3 FV  12, 000(1.055)  12, 000(1.74241)  $14, 090.8965 PV  12, 000; I / Y  11; C / Y  2; i 

Balance after 2 years: PV  14, 090.8965  5000  9090.8965; i  0.055; n  1 FV  9090.8985(1.055)  $9590.895808

Balance after 30 months: 12%  1%  0.01; n  6 12 FV  9590.895808(1.01)6  9590.895808(1.061520) = $10,180.92916 PV  9590.895808; I / Y  12; C / Y  12; i 

Balance after 5 years: PV  10,180.92916  4000  6180.93; i  0.01; n  30 FV  6180.92916(1.01)30  6180.92916(1.347849)  $8330.96 The final payment is $8330.96

(Set P / Y  2) 12,000  PV 11 I / Y 3 N CPT FV 14,090.8965 9090.8965  PV 1 N CPT FV 9590.895808 9590.895808  PV 12 I/Y 6 N CPT FV 10,180.92916

6180.92916  PV 30 N CPT FV 8330.96 19. Balance after 2 months: 7%  0.583%  0.00583; n  2 12 FV  15, 000(1.00583)2  15, 000(1.011701)  $15,175.51042 PV  15, 000; I / Y  7; C / Y  12; i 

Balance after 7 months: PV  15,175.51042  2000  13,175.51042; I / Y  7; C / Y  12; 7% i  0.583%  0.00583; n  5 12 FV  13,175.51042(1.00583)5  13,175.51042(1.029509)  $13,564.3057

Balance after 12 months: PV  13,564.3057  5000  8564.3057; i  0.005833; n  5 FV  8564.3057(1.00583)5  8564.3057(1.029509)  $8817.029242

Balance after 16 months: PV  8817.029242  4000  12,817.029242; i  0.00625; n  4 FV  12,817.029242(1.00625) 4  12,817.029242(1.025235)  $13,140.4715 Balance after 24 months: PV  13,140.4715  3000  10,140.43; i  0.00625; n  8 FV  10,140.4715(1.00625)8  10,140.4715(1.051108)  $10, 658.73 The final payment is $10,658.73

(Set P / Y  12) 15,000  PV 7 I / Y 2 N CPT FV 15,175.51042 13,175.51042  PV 5 N CPT FV 13,564.3057 8564.3057  PV 5 N CPT FV 8817.029242 12,817.029242  PV 7.5 I/Y 4 N CPT FV 13,140.4715 10,140.4715  PV 8 N CPT FV 10,658.73 20. Balance after 6 months:

Balance after 6 months (2 quarters): 8% PV  9000; I / Y  8; C / Y  4; i   2.0%  0.02; n  2 4 FV  9000(1.02) 2  9000(1.0404)  $9363.60 Balance after 21 months (7 quarters): 8%  2.0%  0.02; n  5 4 FV  6863.60(1.02)5  6863.60(1.104081)  $7577.969001 PV  9363.60  2500  6863.60; m  4; i 

Balance after 4 years: 7.75%  0.64583%  0.006458; n  27 12 FV  5077.969001(1.006458) 27  5077.969001(1.189835)  $6041.94 PV  7577.969001  2500  5077.969001; m  12; i 

The final payment is $6041.94

(Set P / Y  4) 9000  PV 8 I / Y 2 N CPT FV 9363.60 6863.60  PV 5 N CPT FV 7577.969001 (Set P / Y  12) 5077.969001  PV 7.75 I / Y 27 N CPT FV 6041.94

21. After 2 years: PV = 3500; i = 0.104/12; n = 24 FV = 3500(1 + 0.104/12) 24 = 4305.39 After 4 years: PV = 4035.39 – 1000 = 3305.39; n = 24 FV = 3305.39(1 + 0.104/12) 24 = 4065.99 After 5 years: PV = 4065.99 – 750 = 3315.99; n = 12 FV = 3315.99(1 + 0.104/12)12 = $3677.78 You owe $3677.78 at the end of five years. 22. Balance after 5 months: PV = 6000; n = 5 FV = 6000 (1 + 0.0615/12)5 = 6155.33 Balance after 10 months: PV = 6155.33 – 2000 = 4155.33 FV = 4155.33 (1 + 0.0615/12)5 = 4262.91 Balance after 1 year: PV = 4262.91 – 1500 = 2762.91 FV = 2762.91 (1 + 0.0615/12)2 = 2791.30 Balance after 20 months: PV = 2791.30 + 3200 = 5991.30 FV = 5991.30 (1 + 0.0615/12)8 = 6241.39 The mortgage was for $6241.39. Exercise 9.3 1.

FV  1600; I / Y  4; C/Y  2; i 

4%  2%  0.02; n  4.5  2  9 2

PV  1600(1.02)9  1600(0.836755)  $1338.81 Compound discount  1600  1338.81  $261.19

(Set P / Y  2) 1600 FV 4 I / Y 9 N CPT PV 1338.81 2.

FV  2500; I / Y  6; C / Y  4; i 

6%  1.5%  0.015; n  6.25  4  25 4

PV  2500(1.015)25  2500(0.689206)  $1723.01 Discount  2500  1723.01  $776.99

(Set P / Y  4) 2500 FV 6 I / Y 25 N CPT PV 1723.01 3.

FV  1250; I / Y  10; C/Y  4; i 

10%  2.5%  0.025; n  5  4  20 4

PV  1250(1.025) 20  1250(0.610271)  $762.84

(Set P / Y  4) 1250 FV 10 I / Y 20 N CPT PV  762.84 4.

FV  2000; I / Y  9; C / Y  12; i 

9%  0.75%  0.0075; n  7 12  84 12

PV  2000(1.0075) 84  2000(0.533845)  $1067.69

(Set P / Y  12) 2000 FV 9 I / Y 84 N CPT PV 1067.69 5.

FV  5000; I / Y  7; C/Y  4; i 

7%  1.75%  0.0175; 4

February 1, 2022  November 1, 2028, contains 6 years, 9 months: n  6.75  4  27 PV  5000(1.0175)27  5000(0.625995)  $3129.97

(Set P / Y  4) 5000 FV 7 I / Y 27 N CPT PV  3129.97 6.

FV  3000; I / Y  7.75; C/Y  365; i  0.021233%  0.000212; n = 5  365  1825 PV  3000(1.000212)1825  3000(0.678780)  $2036.34

(Set P / Y  365) 3000 FV 7.75 I/Y 1825 N CPT PV  2036.34 7.

1  8 FV  3000; I / Y  9; C / Y  2; i  4.5%  0.045; n  2  8   17  17.3 3  12  PV  3000(1.045)17.3  3000(0.466284)  $1398.85

(Set P / Y  2) 3000 FV 9 I / Y 17.3 N CPT PV 1398.85 8.

1  4 FV  1600; I / Y  5; C / Y  4; i  1.25%  0.0125; n  4  6   25  25.3 3  12  PV  1600(1.0125)25.3  1600(0.730005)  $1168.01

(Set P / Y  4) 1600 FV 5 I / Y 25.3 N CPT PV 1168.01

FV  9000; I / Y  8.25; C / Y  26; i  8.25% / 26  0.317308%; n  4  0.0825  PV  9000 1   26  

11 (26)  127.83 12

127.83

 9000 (1.003173) 127.83  9000 (0.666988)  $6002.89 (Set P / Y = 26) 9000 FV 8.25 I / Y 127.83 N CPT PV  6002.89

10.

FV  5000; I / Y  2.75%; C / Y  52; i  2.75% / 52  0.052885%; n  10  0.0275  PV  5000 1   52  

528.6

 5000 (1.000529)528.6  5000 (0.756155)  $3780.77

Set P / Y  52  5000 FV 2.75 I / Y 528.6 N CPT PV  3780.77

2 (52)  528.6 12

11.

FV  60, 000; I / Y  6; C / Y  1; i  6.0%  0.06; n  5

PV  60, 000(1.06) 5  60, 000(0.747258)  $44,835.49 Compare $60,000  44,835.49  $104,835.49 with $100,000

The two payments are worth $4835.49 more than one payment now. Choose the two payments. (Set P / Y  1) 60,000 FV 6 I / Y 5 N CPT PV  44,835.49 12.

FV  35, 000; I / Y  4.25; C / Y  1; i  4.25%  0.0425; n  3 PV  35, 000(1.0425) 3  35, 000(0.882616)  $30,891.56

Compare $20, 000  30,891.56  $50,891.56 with $50, 000 The two payments are worth $891.56 more than one payment now. Choose the two payments.

(Set P / Y  1) 35,000 FV 4.25 I / Y 3 N CPT PV  30,891.56 13.

FV  2200; I / Y  8; C / Y  12; i  8% /12  0.6%  0.006; n  18 PV  2200(1.006)18  2200(0.887274)  $1952

Compare $100  1952  $2052 with $2000 In terms of today’s dollar, choosing to pay $2000 now is better by $52. (Set P / Y  12) 2200 FV 8 I / Y 18 N CPT PV 1952 14.

FV  150; I / Y  7.5; C / Y  12; i  7.5% /12  0.625%  0.00625; n  12 PV  150(1.00625) 12  150(0.927960)  $139.19

FV  170; I / Y  7.5; C / Y  12; i  7.5%/12  0.625%  0.00625; n  15 PV  170(1.00625)15  170(0.910776)  $154.83

Compare $130  139.19  154.83  $424.02 with $350 The three payments are worth $74.02 more than one payment now. (Set P / Y  12) 150 FV 7.5 I / Y 12 N CPT PV 139.19

170 FV 7.5 I / Y 15 N CPT PV 154.83 15.

 0.04  PV1  30, 000 1   4    $24,586.33

20

 0.0395  PV1  24,586.33 1   12  

24

 $22, 721.70 16.

 0.0495  PV1  15, 000.67 1   365   11,146.38

2190

 0.05  PV2  (11,146.38  5000) 1   2    0.05   6146.38 1   2  

6

 $5300

Exercise 9.4

6% 54  1.5%  0.015; n   4  18 4 12 18 PV  6000(1.015)  6000(0.764912)  $4589.47 FV  6000; I / Y  6; C / Y  4; i 

(Set P / Y  4) 6000 FV 6 I / Y 18 N CPT PV  4589.47 2.

Discount period 2023-03-01 to 2027-08-01 contains 4 years, 5 months. 7.5%  0.625%  0.00625; n  4 12  5  53 12 PV  4200(1.00625) 53  4200(0.718766)  $3018.82 FV  4200; I / Y  7.5; C / Y  12; i 

(Set P / Y  12) 4200 FV 7.5 I / Y 53 N CPT PV  3018.82 3.

Discount period 2023-03-31 to 2026-09-30 contains 3 years, 6 months. 8.5%  4.25%  0.0425; n  3.5  2  7 2 PV  1800(1.0425) 7  1800(0.747253)  $1345.06 FV  1800; I / Y  8.5; C / Y  2; i 

(Set P / Y  2) 1800 FV 8.5 I / Y 7 N CPT PV 1345.06 4.

Discount period  15  6  9 years 9%  2.25%  0.0225; n  9  4  36 4 PV  7500(1.0225) 36  7500(0.44887)  $3366.53 FV  7500; I / Y  9; C / Y  4; i 

(Set P / Y  4) 7500 FV 9 I / Y 36 N CPT PV  3366.53 5.

Maturity value: 8%  4%  0.04; n  5  2  10 2 FV  3000(1.04)10  3000(1.480244)  $4440.73 PV  3000; I / Y  8; C / Y  2; i 

Proceeds: 9% 21  2.25%  0.0225; n   4  7 4 12 7 PV  4440.73(1.0225)  4440.73(0.855769)  $3800.24 FV  4440.73; I / Y  9; C / Y  4; i 

(Set P / Y  2) 3000  PV 8 I / Y 10 N CPT FV 4440.73 (Set P / Y  4) 4440.73 FV 9 I / Y 7 N CPT PV  3800.24

Maturity value: 8%  2.0%  0.02; n  7  4  28 4 FV  5000(1.02) 28  5000(1.741024)  $8705.12

PV  5000; I / Y  8; C / Y  4; i 

Proceeds: Discount period  7  2.5  4.5 years

6%  0.5%  0.005; n  4.5  12  54 12 PV  8705.12(1.005) 54  8705.12(0.763893)  $6649.78 FV  8705.12; I / Y  6; C / Y  12; i 

(Set P / Y  4) 5000  PV 8 I / Y 28 N CPT FV 8705.12 (Set P / Y  12) 8705.12 FV 6 I / Y 54 N CPT PV  6649.78

Maturity value: Due date 2027-06-01 10%  2.5%  0.025; n  6  4  24 4 FV  900(1.025) 24  900(1.808726)  $1627.85

PV  900; I / Y  10; C / Y  4; i 

Proceeds: Discount period 2026-12-01 to 2027-06-01 contains 6 months. 8.5% FV  1627.85; I / Y  8.5; C / Y  2; i   4.25%  0.0425; n  1 2 PV  1627.85(1.0425)1  1627.85(0.959233)  $1561.49

(Set P / Y  4) 900  PV 10 I / Y 24 N CPT FV 1627.85 (Set P / Y  2) 1627.85 FV 8.5 I / Y 1 N CPT PV  1561.49 8.

Maturity value: Due date 2031-04-01 7%  3.5%  0.035; n  10  2  20 2 FV  1300(1.035) 20  1300(1.989789)  $2586.73 PV  1300; I / Y  7; C / Y  2; i 

Proceeds: Discount period 2028-04-01 to 2031-04-01 contains 3 years. 9% FV  2586.73; I / Y  9; C / Y  4; i   2.25%  0.0225; n  3  4  12 4 PV  2586.73(1.0225)12  2586.73(0.765668)  $1980.58

(Set P / Y  2) 1300  PV 7 I / Y 20 N CPT FV 2586.73

(Set P / Y  4) 2586.73 FV 9 I / Y 12 N CPT PV  1980.58 9.

FV  10,000; I / Y  9; C / Y  12; i  0.75%  0.0075; n  22.5 Proceeds  10, 000(1.0075) 22.5  10, 000(0.845252)  $8452.52

(Set P / Y  12) 10,000 FV 9 I / Y 22.5 N CPT PV  8452.52 10. FV  7000; I / Y  8; C / Y  4; i  2%  0.02; n  3  4 

5 5 2  4  12   13  13.6 12 3 3

PV  7000(1.02) 13.6  7000(0.762894)  $5340.26

(Set P / Y  4) 7000 FV 8 I / Y 13.6 N CPT PV  5340.26

11.

FV  3800; I/Y  7.5; C/Y  1; i  7.5%  0.075; n  6

8  6.6 12

PV  3800(1.075) 6.6  3800(0.617462)  $2346.36

(Set P / Y  1) 3800 FV 7.5 I / Y 6.6 N CPT PV  2346.36 12.

 8 FV  5500; I / Y  4.5; C / Y  2; i  2.25%  0.0225; n  2  7   15.3  12  PV  5500(1.0225)15.3  5500(0.710934)  $3910.14

(Set P / Y  2) 5500 FV 4.5 I / Y 15.3 N CPT PV  3910.14 13.

Discount period  48  32  16 months FV  3750; I / Y  5.5; C / Y  2; i  2.75%  0.0275; n 

16 16 2  2   2  2.6 12 6 3

PV  3750(1.0275) 2.6  3750(0.930212)  $3488.29

(Set P/Y  2) 3750 FV 5.5 I/Y 2.6 N CPT PV  3488.29 14.

8 2  8 FV  5200; I / Y  9; C / Y  4; i  2.25%  0.0225; n  4  3   12   14  14.6 3 3  12  PV  5200(1.0225)14.6  5200(0.721558)  $3752.10 (Set P / Y  4) 5200 FV 9 I / Y 14.6 N CPT PV  3752.10 15. Due date: 2031-08-01 (Due date is eight years after date of issue). Discount period: 2027-04-01 to 2031-08-01: 4 years, 4 months

FV  4500; I / Y  6.5; C / Y  1; i  0.065; n  4

4  4.3 12

Proceeds  4500(1.065) 4.3  4500(0.761176)  3425.29 Discount  4500  3425.29  $1074.71

(Set P / Y  1) 4500 FV 6.5 I/Y 4.3 N CPT PV  3425.29

16. Due date: 2028-09-30 Discount period: 2026-07-31 to 2028-09-30: 2 years, 2 months  2 FV  2800; I / Y  8; C / Y  4; i  0.02; n  4  2   8.6  12 

Proceeds  2800(1.02) 8.6  2800(0.842297)  2358.43 Discount  2800  2358.43  $441.57

(Set P / Y  4) 2800 FV 8 I / Y 8.6 N CPT PV  2358.43 17. Due date: 2028-12-01 PV  1750; I / Y  6.5; C / Y=1; i  6.5%  0.065; n  6 Maturity value  1750(1.065)6  1750(1.459142)  $2553.50 Discount period: 2025-03-01 to 2028-12-01: 3 years, 9 months  9 FV  2553.50; I / Y  7; C / Y  2; i  3.5%  0.035; n  2  3   7.5  12  PV  2553.50(1.035) 7.5  2553.50(0.772587)  $1972.80

(Set P / Y  1) 1750  PV 6.5 I / Y 6 N CPT FV 2553.50 (Set P / Y  2) 2553.50 FV 7 I / Y 7.5 N CPT PV  1972.80 18.

PV  1200; I / Y  6; C / Y  12; i  0.5%  0.005; n  120 Maturity value  1200(1.005)120  1200(1.819397)  $2183.28

 2 FV  2183.28; I / Y  8; C / Y  4; i  2%  0.02; n  4  6   24.6  12  PV  2183.28(1.02) 24.6  2183.28(0.613568)  $1339.59 (Set P / Y  12) 1200  PV 6 I / Y 120 N CPT FV 2183.28

(Set P / Y  4) 2183.28 FV 8 I / Y 24.6 N CPT PV  1339.59

19. PV  2650; I / Y  9; C / Y  4; i  2.25%  0.0225; n  28 Maturity value  2650(1.0225)28  2650(1.864545)  $4941.04

 7 FV  4941.04; I / Y  8; C / Y  2; i  0.04; n  2  4   9.16  12 

Proceeds  4941.04(1.04)9.16  4941.04(0.698009)  $3448.89 Discount  4941.04  3448.89  $1492.15

(Set P / Y  4) 2650  PV 9 I / Y 28 N CPT FV 4941.04 (Set P / Y  2) 4941.04 FV 8 I / Y 9.16 N CPT PV  3448.89 20. Due date: 2032-06-15 (Due date is 10 years after date of issue). PV  4000; I / Y  8; C/Y  4; i  2%  0.02; n  40 Maturity value  4000(1.02) 40  4000(2.208040)  $8832.16 Discount period: 2027-04-15 to 2032-06-15: 5 years, 2 months  2 FV  8832.16; I / Y  10; C / Y  4; i  0.025; n  4  5   20.6  12 

Proceeds  $8832.16(1.025)20.6  8832.16(0.600307)  $5302.01 Discount  8832.16  5302.01  $3530.15

(Set P / Y  4) 4000  PV 8 I / Y 40 N CPT FV 8832.16 8832.16 FV 10 I / Y 20.6 N CPT PV  5302.01 Business Math News Box 1.

$10, 000 (1 0.20)  10, 000(0.80)  $8000

The value of the portfolio drops 20%, by $2000, down to $8000. 2.

$6000 (1  0.20)  4000  6000 (0.80)  4000  $8800.

The value of the portfolio decreases from $10 000 to $8800. This is a 12% drop overall. 3.

FV  $10,000(1.0375)3  $11,167.71

(a)

FV  $10,000(1.0325) (1.04) (1.055)  $11,328.59

(b)

FV  $10,000(1.035) (1.0375) (1.045) (1.0475) (1.05)  $12,342.07

Three-year GIC: 10, 000(1.042458)3  10, 000(1.132859)  $11,328.59

Five-year GIC: 10, 000(1.042984)5  10, 000(1.234208)  $12,342.08

The effective yield earns the same amount of interest on $10,000 at maturity as the escalating rate for each type of GIC calculated in Question #4. (Slight difference attributed to rounding). Exercise 9.5 1.

FV  4000; I / Y  7; C / Y  1; i  7%  0.07

(a)

E = 4000(1.07)- 5 = 4000(0.712986) = $2851.94

(Set P / Y = 1) 4000 FV 7 I / Y 5 N CPT PV - 2851.94 (b)

E = 4000(1.07)- 3 = 4000(0.816298) = $3265.19

4000 FV 3 N CPT PV - 3265.19 (c)

E = $4000

(d)

E = 4000(1.07)5 = 4000(1.402552) = $5610.21

4000  PV 5 N CPT FV 5610.21 2.

(a)

E = 5500(1.021)- 9 = 5500(0.829408) = $4561.75

(Set P / Y = 4) 5500 FV 8.4 I / Y 9 N CPT PV - 4561.75 (b)

E = 5500(1.021)- 4 = 5500(0.920231) = $5061.27

5500 FV 4 N CPT PV - 5061.27 (c)

E = $5500

(d)

E = 5500(1.021)3 = 5500(1.064332) = $5853.83

5500 ± PV 3 N CPT FV 5853.83

I / Y = 10; C / Y = 2; i = 10%/2 = 0.05 Focal date is four years from now. E1 = 2000(1.05)8 = 2000(1.477455) = $2954.91 E2 = 2000(1.05)2 = 2000(1.1025) = $2205 E3 = 2000(1.05)–4 = 2000(0.822702) = $1645.40 x = E1 + E2 + E3 = 2954.91 + 2205 + 1645.40 = $6805.31 The single payment four years from now is $6805.31

(Set P / Y  2)2000  PV 10 I / Y 8 N CPT FV 2954.91 2000  PV 2 N CPT FV 2205 2000 FV 4 N CPT PV  1645.40

I/Y = 7.5; C / Y = 12; i = 7.5%/12 = 0.00625 Focal date is 2.5 years from now. E1 = 600(1.00625)18 = 600(1.118681) = $671.21 E2 = 800(1.00625)–6 = 800(0.963307) = $770.65 E3 = 1200(1.00625)–42 = 1200(0.769755) = $923.71 x = E1 + E2 + E3 = 671.21 + 770.65 + 923.71 = $2365.57 The single payment 2.5 years from now is $2365.57

(Set P / Y  12) 600  PV 7.5 I / Y 18 N CPT FV 671.21 800 FV 6 N CPT PV  770.65 1200 FV 42 N CPT PV  923.71

I / Y = 6; C / Y = 12; i = 6% / 12 = 0.005 Focal date is 15 months from now.

E1  400(1.005)15  400(1.077683)  $431.07 E 2  721.28(1.005)7  721.28(1.035529)  $746.91

E3  500(1.005)9  500(1.045911)  $522.96 E1  E 2  E3  x 431.07  746.91  522.96  x x  655.02 The final payment is $655.02

(Set P / Y  12) 400  PV 6 I / Y 15 N CPT FV 431.07 721.28  PV 6 I / Y 7 N CPT FV 746.91 500  PV 9 N CPT FV 522.96

I / Y = 10.8; C / Y = 4; i = 10.8% / 4 = 0.027 Focal date is 18 months from now. E1 = 1200(1.027)10 = 1200(1.305282) = $1566.34 E2 = 1000(1.027)8 = 1000(1.237552) = $1237.55 E3 = 800(1.027)6 = 800(1.173337) = $938.67 E4 = 1000(1.027)3 = 1000(1.083207) = $1083.21

E1  E 2  E3  E 4  x 1566.34  1237.55  938.67  1083.21  x x  782.01 The size of the final payment is $782.01

(Set P / Y  4)1200  PV 10.8 I / Y 10 N CPT FV 1566.34 1000  PV 8 N CPT FV 1237.55 800  PV 6 N CPT FV 938.67 1000  PV 3 N CPT FV 1083.21

I / Y  7.5; C / Y  4; i 

7.5%  1.875%  0.01875 4

E1 = 1400(1.065)3 = 1400(1.207950) = $1691.13 E2 = 1691.13(1.01875)–4 = 1691.13(1.928388) = $1570.02 E3 = 800(1.01875)6 = 800(1.117907) = $894.33

E 2  E3  x 1570.02  894.33  x x  2464.35 The replacement payment is $2464.35

(Set P / Y  1) 1400  PV 6.5 I / Y 3 N CPT FV 1691.13 (Set P / Y  4) 1691.13 FV 7.5 I / Y 4 N CPT PV  1570.02 800  PV 6 N CPT FV 894.33

I/Y = 10.5; C / Y = 1; i = 10.5% = 0.105 E1 = 2000(1.105)7 = 2000(2.011574) = $4023.15 E2 = 2000(1.105)3 = 2000(1.349233) = $2698.47 E3 = 2000(1.105)5 = 2000(1.647447) = $3294.89

E1  E 2  E3  x 4023.15  2698.47  3294.89  x x  3426.73 The second payment is $3426.73

(Set P / Y  1) 2000  PV 10.5 I / Y 7 N CPT FV 4023.15 2000  PV 3 N CPT FV 2698.47 2000  PV 5 N CPT FV 3294.89

I / Y  7; C / Y  4; i 

7%  1.75%  0.0175 4

E1 = 1500(1.0175)13 = 1500(1.252990) = $1879.48 E2 = 1900(1.0175)8 = 1900(1.148882) = $2182.88 E3 = 2000(1.0175)3 = 2000(1.053424) = $2106.85

E1  E 2  E3  x 1879.48  2182.88  2106.85  x x  1955.51 The replacement payment is $1955.51

(Set P / Y  4)1500  PV 7 I / Y 13 N CPT FV 1879.48 1900  PV 8 N CPT FV 2182.88 2000  PV 3 N CPT FV 2106.85

10. Let the size of the equal payments be $x and the focal date be now. I / Y = 7.2; C / Y = 12; i = 7.2%/12 = 0.006 3000 = x(1.006)–12 + x(1.006)–36 + x(1.006)–60 3000 = 0.930731x + 0.806256x + 0.698427x 3000 = 2.435414x x = 1231.82 The size of the equal payments is $1231.82

(Set P / Y  12) 1 FV 7.2 I / Y 12 N CPT PV  0.930731 1 FV 36 N CPT PV  0.806256 1 FV 60 N CPT PV  0.698427

11. Let the size of the equal payments $x and the focal date be now. I / Y = 8; C / Y 4; i = 8% / 4 = 2% = 0.02 7500 = x(1.02)–4 + x(1.02)–12 7500 = 0.923845x + 0.788493x 7500 = 1.712338x x = 4379.98 The size of the equal payments is $4379.98

(Set P / Y  4) 1 FV 8 I / Y 4 N CPT PV  0.923845 1 FV 12 N CPT PV  0.788493

12. Let the size of the equal payments $x and the focal date be four years from now. I/Y = 10; C / Y 1; i = 10% / 1 = 10% = 0.10 3000 = x(1.10)3 + x(1.10)2 + x(1.10)1 + x 3000 = 1.331x + 1.21x + 1.1x + x 3000 = 4.641x x = 646.41 The size of the equal payments is $646.41

(Set P / Y  1)1  PV 10 I / Y 3 N CPT FV 1.331 1  PV 2 N CPT FV 1.21 13. Let the size of the equal payments $x and the focal date be five years from now. I / Y = 5.16; C / Y 12; i = 5.16% / 12 = 0.43% = 0.0043 9000 = x(1.0043)40 + x(1.0043)30 + x 9000 = 1.18724x + 1.137376x + x 9000 = 3.324615x x = 2707.08 The size of the equal payments is $2707.08

(Set P / Y  12)1  PV 5.16 I / Y 40 N CPT FV 1.18724 1  PV 30 N CPT FV 1.137376 14. I / Y = 12; C / Y = 2; i = 12% / 2 = 0.06 E1 = 800(1.06)12 = 800(2.012196) = $1609.76 E2 = 1000(1.06)–2 = 1000(0.889996) = $890 E3 = $x E4 = x(1.06)– 8 = x(0.627412) E1  E 2  E 3  E 4 1609.76  890  x  0.627412 x 2499.76  1.627412 x x  1536.03 The size of the two equal payments is $1536.03

(Set P / Y  2)800  PV 12 I/Y 12 N CPT FV 1609.76 1000 FV 2 N CPT PV  890 1 FV 8 N CPT PV  0.627412

15.

I / Y  6.9; C / Y  12; i 

6.9%  0.575%  0.00575 12

E1 = 3000(1.00575)12 = 3000(1.071224) = $3213.67 E2 = 2500(1.00575)–48 = 2500(0.759413) = $1898.53 E3 = $x E4 = x(1.00575)–72 = 0.661785x

E1  E 2  E 3  E 4 3213.67  1898.53  x  0.661785 x 5112.20  1.661785 x x  3076.33 The size of the two equal payments is $3076.33

(Set P / Y  12) 3000  PV 6.9 I / Y 12 N CPT FV 3213.67 2500 FV 48 N CPT PV  1898.53 1 FV 72 N CPT PV  0.661785

16.

I / Y  9; C / Y  12; i 

9%  0.75%  0.0075 12

E1 = 500(1.03)2 = 500(1.060900) = $530.45 E2 = 530.45(1.0075)3 = 530.45(1.022669) = $542.47 E3 = 800(1.05)3 = 800(1.157625) = $926.10 E4 = 926.10(1.0075)–9 = 926.10(0.934963) = $865.87

E2  E4  x 542.47  865.87  x x  1408.34 The remaining payment is $1408.34

(Set P / Y  4) 500  PV 12 I / Y 2 N CPT FV 530.45 (Set P / Y  12) 530.45  PV 9 I / Y 3 N CPT FV 542.47 (Set P / Y  2) 800  PV 10 I / Y 3 N CPT FV 926.10 (Set P / Y  12) 926.10 FV 9 I / Y 9 N CPT PV  865.87

17. I / Y  9; C / Y  12; i  11%/4  2.75%  0.0275; i 

9%  0.75%  0.0075 12

E1 = 900(1.0275)1 = 900(1.0275) = $924.75 E2 = 924.75(1.0075)–3 = 924.75(0.977833) = $904.25 E3 = 800(1.0275)10 = 800(1.311651) = $1049.32 E4 = 1049.32(1.0075)–30 = 1049.32(0.799187) = $838.60 E5 = $x E6 = x(1.0075)–36 = 0.764149x

E 2  E 4  E5  E 6 904.25  838.60  x  0.764149 x 1742.85  1.764149 x x  987.93 The size of the two equal payments is $978.93

(Set P / Y  4) 900  PV 11 I / Y 1 N CPT FV 924.75 (Set P / Y  12) 924.75 FV 9 I / Y 3 N CPT PV  904.25 (Set P / Y  4) 800  PV 11 I / Y 10 N CPT FV 1049.32 (Set P / Y  12) 1049.32 FV 9 I / Y 30 N CPT PV  838.60 1 FV 36 N CPT PV  0.764149 18.

I / Y  11; C / Y  4; i 

11%  2.75%  0.0275 4

E1 = 1400(1.0275)6 = 1400(1.176768) = $1647.48 E2 = 1600(1.115)5 = 1600(1.723353) = 2757.37 E3 = 2757.37(1.0275)–14 = 2757.37(0.683997) = $1886.03 E4 = $x E5 = x(1.0275)–10 = 0.762398x

E1  E3  E 4  E5 1647.48  1886.03  x  0.762398 x 3533.51  1.762398 x x  2004.94

The size of the two equal payments is $2004.94

(Set P / Y  4) 1400  PV 11 I / Y 6 N CPT FV 1647.48 (Set P / Y  1) 1600  PV 11.5 I / Y 5 N CPT FV 2757.37 (Set P / Y  2) 2757.37 FV 11 I / Y 14 N CPT PV 1886.03 (Set P / Y  4) 1 FV 11 I / Y 10 N CPT PV  0.762398

19. Total payment during next (third) payment = 450 (1 + 0.04/12)2 + 450 (1 + 0.04/12) 1 + 450 = 453.01 + 451.50 + 450 = 1354.51. Her next payment should be $1345.51. 20.

Use 1 year as the focal date: FV1  PV1  x  PV2 52

104

 0.029   0.029   0.029  1500 1   x  2 x 1    1500 1    52  52  52     1500 (1.029416)  1500(0.943665) 1544.12  1415.50  x  2 x (0.890504) 2959.62  x  1.781008 x 2959.62  2.781008 x

208

x  $1064.23 2 x  $2128.46 The first payment is $1064.23. The second payment is $2128.46. Review Exercise 1. (a)

FV  2000(1.0175)10  2000(1.189444)  $2378.89

(Set P / Y  2) 2000  PV 3.5 I / Y 10 N CPT FV 2378.89

(b)

Interest  2378.89  2000  $378.89

(a)

FV  5000(1.0225)3  5000(1.069030)  $5345.15

(Set P / Y  1) 5000  PV 2.25 I / Y 3 N CPT FV 5345.15 (b)

FV  5000(1.005)12  5000(1.061678)  $5308.39

(Set P / Y  4) 5000  PV 2 I / Y 12 N CPT FV 5308.39 3.

(a)

FV  1800(1.00925) 62  1800(1.769795)  $3185.63

Interest  3185.63  1800  $1385.63

(Set P / Y  4) 1800  PV 3.7 I / Y 62 N CPT FV 3185.63 (b)

FV  1250(1.00216)180  1250(1.476358)  $1845.45 Interest  1845.45  1250  $595.45

(Set P / Y  12)1250  PV 2.6 I / Y 180 N CPT FV 1845.45 4.

FV  6000(1.03)13.16  6000(1.475786)  $8854.72 Interest  8854.72  6000  $2854.72

(Set P / Y  2) 6000  PV 6 I / Y 13.16 N CPT FV 8854.72 5.

PV = 5200(1.01625) 13.3  5200(0.806602)  $4194.33

(Set P / Y  4) 5200 FV 6.5 I / Y 13.3 N CPT PV  4194.33 6.

(a)

PV = 3600(1.04) 18  3600(0.493628)  $1777.06 Discount = 3600  1777.06  $1822.94

(Set P / Y  2) 3600 FV 8 I / Y 18 N CPT PV  1777.06 (b)

PV  9000(1.017)20  9000(0.713807)  $6424.27 Discount  9000  6424.27  $2575.73

(Set P / Y  4) 9000 FV 6.8 I / Y 20 N CPT PV  6424.27

PV  4000(1.022) 30  4000(0.520563)  $2082.25

Discount  4000  2082.25  $1917.75

(Set P / Y  4) 4000 FV 8.8 I / Y 30 N CPT PV  2082.25 8.

(a)

E  3000(1.005) 18  3000(0.914136)  $2742.41

(Set P / Y  12) 3000 FV 6 I / Y 18 N CPT PV  2742.41 (b)

E  3000(1.005)6  3000(0.970518)  $2911.55

3000 FV 6 N CPT PV  2911.55 (c)

E  3000(1.005)18  3000(1.093929)  $3281.79

3000  PV 18 N CPT PV 3281.79 9.

Discount period 2024-09-30 to 2028-06-30 contains 3 years, 9 months PV = 1500(1.025) 15  1500(0.690466)  $1035.70

(Set P / Y  4) 1500 FV 10 I / Y 15 N CPT PV  1035.70 10.

FV  75,000; I / Y  6.5; C / Y  2; i = 0.0325; n 

42 2  7 12

PV  75,000(1.0325)7  75,000(0.799410)  $59,955.75

(Set P / Y  2) 75,000 FV 6.5 I / Y 7 N CPT PV  59,955.75 11. Due date May 1, 2031 Maturity value  1750(1.02)20  1750(1.485947)  $2600.41

Discount period 2027-08-01 to 2031-05-01 contains 3 years, 9 months Proceeds  2600.41(1.015) 15  2600.41(0.799852)  $2079.94

(Set P / Y  2) 1750  PV 4 I / Y 20 N CPT FV 2600.41 (Set P / Y  4) 2600.41 FV 6 I / Y 15 N CPT PV  2079.94 12.

Maturity value  10, 000(1.02)28  10, 000(1.741024)  $17, 410.24

Proceeds  17, 410.24(1.035)6  17, 410.24(0.813501)  $14,163.24

(Set P / Y  4) 10,000  PV 8 I / Y 28 N CPT FV 17, 410.24 (Set P / Y  2)17, 410.24 FV 7 I / Y 6 N CPT PV  14,163.24 13. Due date 2033-06-01 Maturity value  40, 000(1.06)30  40, 000(5.743491)  $229, 739.65

Discount period 2026-09-01 to 2033-06-01 contains 6 years, 9 months

Proceeds  229,739.65(1.0275)27  229,739.65(0.480718)  $110, 440.03

(Set P / Y  2) 40,000  PV 12 I / Y 30 N CPT FV 229,739.65 (Set P / Y  4) 229,739.65 FV 11 I / Y 27 N CPT PV 110, 440.03 14.

Maturity value  16,500(1.06)30  16,500(5.743491)  $94, 767.60

Discount period contains 11 years, 8 months

Proceeds  94,767.60(1.0075)140  94,767.60(0.351311)  $33, 292.94

(Set P / Y  2)16,500  PV 12 I / Y 30 N CPT FV 94,767.60 (Set P/Y  12) 94,767.60 FV 9 I/Y 140 N CPT PV  33, 292.94 15. Accumulated value after two years = 20,000 (1.05)4 = 20,000 (1.215506) = $24,310.125 Balance after payment = 24,310.125 – 8000 = $16 310.125 Accumulated value after 3.5 years = 16,310.125(1.05)3 = 16,310.125(1.157625) = $18,881.00845 Balance after payment = 18,881.00845 – 10,000 = $8881.00845 Accumulated value one year later = 8881.00845(1.05) 2 = 8881.00845(1.1025) = $9791.31

(Set P / Y  2) 20, 000  PV 10 I / Y 4 N CPT FV 24,310.125 16,310.125  PV 3 N CPT FV 18,881.00845 8881.00845  PV 2 N CPT FV 9791.31

16. Period 2022-03-01 to 2024-09-01 contains 2 years, 6 months: n = 10 Balance September 1, 2024 = 1750(1.0075)10 = 1750(1.077583) = $1885.769455 Period 2024-09-01 to 2026-06-01 contains 1 year, 9 months: n = 21 Balance June 1, 2026  1885.769455(1.003) 21  1885.769455(1.072383)  $2022.267746

Period 2026-06-01 to 2028-12-01 contains 2 years, 6 months: n = 5 The value of the RRSP deposit on December 1, 2028 = 2022.267746(1.0225)5  2022.267746(1.117678)  $2260.24

(Set P / Y  4)1750  PV 3 I / Y 10 N CPT FV 769 455 (Set P / Y  12)1885.769455  PV 4 I / Y 21 N CPT FV 267 746 (Set P / Y  2) 2022.267746  PV 4.5 I / Y 5 N CPT FV 2260.24

17. Balance after 2.5 years FV1 = 2500(1.0125)10 = 2500(1.132271) = $2830.677074 Balance two years later FV2  2830.677074(1.005) 24  2830.68(1.127160)  $3190.63

(Set P / Y  4) 2500  PV 5 I / Y 10 N CPT FV 2830.677074 (Set P / Y  12) 2830.677074  PV 6 I / Y 24 N CPT FV 3190.63 18. Balance one year after first deposit = 2000(1.015)4 = 2000(1.061364) = $2122.727101 Balance two years after first deposit = (2122.727101 + 2000)(1.015)4 = 4122.727101(1.061364) = $4375.712274 Balance four years after first deposit  (4375.712274  2000)(1.015)8  6375.712274(1.126493)  $7182.19

(Set P / Y  4) 2000  PV 6 I / Y 4 N CPT FV 2122.727101 4122.727101  PV 4 N CPT FV 4375.712274 6375.712274  PV 8 N CPT FV 7182.19 19. Balance February 1, 2017 = 2500(1.0125)4 = 2500(1.050945) = $2627.363342 Balance February 1, 2022 = (2627.363342 + 2000) (1.0125)20 = 4627.363342(1.282037) = $5932.452089 Balance August 1, 2026  (5932.452089  1500)(1.0125)18  (7432.452089)(1.250577)  $9294.86

(Set P / Y  4) 2500  PV 5 I / Y 4 N CPT FV 2627.363342 4627.363342  PV 20 N CPT FV 5932.452089 7432.452089  PV 18 N CPT FV 9294.86

20. Balance after 15 months = 8000(1.02)5 – 3000 = 8000(1.104081) – 3000 = 8832.646426 – 3000 = $5832.646426 Balance after 24 months = 5832.646426(1.02)3 = 5832.646426(1.061208) = $6189.651048 Balance after 30 months = 6189.651048(1.0225)2 – 4000 = 6189.651048(1.045506) – 4000 = $6471.318856 – 4000 = $2471.318856 Balance after four years  2471.318856(1.0225)6  2471.32(1.142825)  $2824.29

(Set P / Y  4) 8000  PV 8 I / Y 5 N CPT FV 8832.646426 5832.646426  PV 3 N CPT FV 6189.651048 6189.651048  PV 9 I / Y 2 N CPT FV 6471.318856 2471.318856  PV 6 N CPT FV 2824.29 21. Let the single payment be $x and the focal date be two years from now.

x  1000(1.024)6  1200(1.024) 2  1500(1.024) 2  1000(1.152922)  1200(1.048576)  1500(0.953674)  1152.92  1258.29  1430.51  $3841.72

(Set P / Y  4)1000  PV 9.6 I / Y 6 N CPT FV 1152.92 1200  PV 2 N CPT FV 1258.29 1500 FV 2 N CPT PV  1430.51

22. Let the second payment be $x and the focal date be 15 months from now. Accumulated value of $10 000 one year from now

10, 000, (1.05) 2  10, 000(1.1025)  $11, 025 11, 025(1.0075)3  6000(1.0075)9  x 11, 025(1.022669)  6000(1.069561)  x 11, 274.93  6417.37  x x  $4857.56

(Set P / Y  2)10,000  PV 10 I / Y 2 N CPT FV 11,025 (Set P / Y  12)11,025  PV 9 I / Y 3 N CPT FV 11, 274.93 6000  PV 9 N CPT FV 6417.37 23. Let the size of the equal payments be $x and the focal date be now. Accumulated value of $3000 due in two years

 3000(1.055)4  3000(1.238825)  $3716.47 Accumulated value of $2500 due in 15 months  2500(1.0225)5  2500(1.117678)  $2794.19 3716.47(1.007) 24  2794.19(1.007) 15  x  x(1.007) 18 3716.47(0.845849)  2794.19(0.900654)  x  0.882002 x 3143.57  2516.60  1.882002 x x  $3007.53

(Set P / Y  2) 3000  PV 11 I / Y 4 N CPT FV 3716.47 (Set P / Y  4) 2500  PV 9 I / Y 5 N CPT FV 2794.19 (Set P / Y  12)3716.47 FV 8.4 I / Y 24 N CPT PV  3143.57 2794.19 FV 15 N CPT PV  2516.60 1 FV 18 N CPT PV  0.882002 24. Let the single payment be $x and the focal date be two years from now.

400(1.02)8  500(1.02)2  900(1.02)4  x 400(1.171659)  500(1.0404)  900(0.923845)  x 468.66  520.20  831.46  x x  $1820.32

(Set P / Y  4) 400  PV 8 I / Y 8 N CPT FV 468.66 500  PV 2 N CPT FV 520.20

900 FV 4 N CPT PV  831.46 25. Let the size of the equal payments be $x and the focal date be one year from now.

2600(1.048)4  2400(1.048)2  x  x (1.048)6 2600(1.206272)  2400(0.910495)  x  0.754801x 3136.31  2185.19  1.754801x 5321.50  1.754801x x  $3032.54

(Set P / Y  2) 2600  PV 9.6 I / Y 4 N CPT FV 3136.31 2400 FV 2 N CPT PV  2185.19 1 FV 6 N CPT PV  0.754801 26. Let the size of the equal payments be $x and the focal date be now.

7000(1.01) 24  x  x(1.01) 24  x(1.01) 36 7000(1.269735)  x  0.787566 x  0.698925 x 8888.14  2.486491x x  $3574.57

(Set P / Y  12) 7000  PV 12 I / Y 24 N CPT FV 8888.14 1 FV 24 N CPT PV  0.787566 1 FV 36 N CPT PV  0.698925 Self-Test 1.

FV  3300; I / Y  4; C / Y  4; i  1%; n  44

PV  3300(1.01)44  3300  0.64544   $2129.97

(Set P / Y  4) 3300 FV 4 I / Y 44 N CPT PV  2129.97 2.

PV  1300; I / Y  7.5; C / Y  12; i  0.625%; n  84

FV  1300(1.00625)84  1300(1.687699)  $2194.01 Compound interest  2194.01  1300  $894.01

(Set P / Y  12) 1300  PV 7.5 I / Y 84 N CPT FV 2194.01 3.

PV  1400; I / Y  7.75; C / Y = 1; i  7.75%; n  71/12  5.916 FV  1400(1.0775)5.916  1400(1.555257)  $2177.36

(Set P / Y  1)1400  PV 7.75 I / Y 5.916 N CPT FV 2177.36 4.

FV  5900; I / Y  7.5; C / Y  2; i  3.75%; n  30 PV  5900(1.0375)30  5900(0.331403)  $1955.28

(Set P / Y  2)5900 FV 7.5 I / Y 30 N CPT PV 1955.28 5.

FV  8800; I / Y  9.6; C / Y  12; i  0.8%; n  90

PV  8800(1.008)90  8800(0.488149)  $4295.71 Compound discount  8800  4295.71  $4504.29

(Set P / Y  12) 8800 FV 9.6 I / Y 90 N CPT PV  4295.71 6.

Maturity value on January 1, 2030: Interest period: July 1, 2021 to January 1, 2030: 8.5 years PV  8000; I/Y  7; C / Y = 4; i  1.75%; n  34 FV  8000(1.0175)34  8000(1.803725)  $14, 429.80

Proceeds on July 1, 2025 Discount period: July 1, 2025 to January 1, 2030: 4.5 years FV  14,429.80; I / Y  8; C / Y  2; i  4%; n  9 PV  14,429.80(1.04) 9  14,429.80(0.702587)  $10,138.19

(Set P / Y  4) 8000  PV 7 I / Y 34 N CPT FV 14, 429.80

(Set P / Y  2)14, 429.80 FV 8 I / Y 9 N CPT PV  10,138.19 7.

Maturity value = $1100 Discount period contains 3 years, 7 months Proceeds  1100(1.075) 3.583  1100(0.771708)  $848.88

(Set P / Y  1)1100 FV 7.5 I / Y 3.583 N CPT PV  848.88 8.

Maturity value  8200(1.01125)132  8200(4.378512)  $35,903.80

Discount period contains 3 years, 10 months

Proceeds  35,903.80(1.0525)7.6  35,903.80(0.675508)  $24, 253.31

(Set P / Y  12)8200  PV 13.5 I / Y 132 N CPT FV 35,903.80 (Set P / Y  2)35,903.80 FV 10.5 I / Y 7.6 N CPT PV  24, 253.31 9.

Maturity value: PV  10, 200; I / Y = 11.6; C / Y = 2; i  5.8%; n  10

FV  10, 200(1.058)10  10, 200(1.757344)  $17,924.90 Proceeds: FV  17,924.90; I / Y  10; C / Y  4; i  2.5%; n  8 PV  17,924.90(1.025) 8  17,924.90(0.820747)  $14, 711.80

(Set P / Y  2)10, 200  PV 11.6 I / Y 10 N CPT FV 17,924.90 (Set P / Y  4)17,924.90 FV 10 I / Y 8 N CPT PV  14,711.80 10.

Balance after five years: PV = 3600; I/Y = 4.8; C/Y = 12; i  0.4%; n  60

FV  3600(1.004)60  3600(1.270641)  $4574.306587 Balance seven years later (after 12 years) PV  4574.306587; I / Y  6; C / Y  2; i  3%; n  14 FV  4574.306587(1.03)14  4574.306587(1.512590)  $6919.05

(Set P / Y  12)3600  PV 4.8 I / Y 60 N CPT FV 4574.306587 (Set P / Y  2) 4574.306587  PV 6 I / Y 14 N CPT FV 6919.05 11. Let the size of the single payment be $x and the focal date be one year from now.

x  4000(1.035) 2  4000(1.035) 8  3000 1.035 

10

 4000(1.071225)  4000(0.759412)  3000(0.708919)  4284.90  3037.65  2126.76  $9449.31

(Set P / Y  2) 4000  PV 7 I / Y 2 N CPT FV 4284.90 4000 FV 8 N CPT PV  3037.65 3000 FV 10 N CPT PV  2126.76 12. Maturity value of $600 due in nine months: PV  600; I / Y  10.5; C / Y  12; i  0.875%; n  9 FV  600(1.00875)9  600(1.081563)  $648.94

Let the size of the final payment be $x and the focal date be 24 months from now.

800 (1.02375)8  648.94(1.02375)5  800(1.02375)6  x 800 (1.206567)  648.94(1.124526)  800(1.151234)  x 965.25  729.75  920.99  x x  $774.01

(Set P / Y  12) 600  PV 10.5 I / Y 9 N CPT FV 648.94 (Set P / Y  4) 648.94  PV 9.5 I / Y 5 N CPT FV 729.75

800  PV 8 N CPT FV 965.25 800  PV 6 N CPT FV 920.99

13. Balance after two years: PV  5000; I / Y  10; C / Y  2; i  5%; n  4

FV  5000(1.05)4  5000(1.215506)  $6077.53125 Balance  6077.53125  2000  $4077.53125 Balance after three years: PV  4077.53125; I / Y  10; C/Y  2; i  5%; n  2 FV  4077.53125(1.05)2  4077.53125(1.1025)  $4495.478203 Balance  4495.478203  2500  $1995.478203 Balance after five years: PV  1995.48; I / Y  10; C / Y  2; i  5%; n  4 FV  1995.478203(1.05)4  1995.478203(1.215506)  $2425.52

(Set P / Y  2) 5000  PV 10 I / Y 4 N CPT FV 6077.53125 4077.53125  PV 2 N CPT FV 4495.478203 1995.478203  PV 4 N CPT FV 2425.52 14. Let the size of the three equal payments be $x and the focal date be now.

7000  x(1.0275)1  x(1.0275)5  x(1.0275)9 7000  0.973236 x  0.873154  0.783364 x 7000  $2661.85

(Set P / Y  4) 1 FV 11 I / Y 1 N CPT PV  0.973236 1 FV 5 N CPT PV  0.873154 1 FV 9 N CPT PV  0.783364

Challenge Problems 1.

Let the amount of money to be received by each child be $x.

PV  50,000; I / Y  7.75; C / Y  2; i 

7.75%  3.875%  0.03875; 2

n1  (21  19)(2)  4; n2  (21  16)(2)  10; n3  (21  13)(2)  16 50, 000  x(1.03875)4  x(1.03875)10  x(1.03875)16 50, 000  0.858926 x  0.683738 x  0.544281x 50, 000  2.086945 x x  23,958.46

Each child will receive $23,958.46

(a) Certificate A:

Value at the end of year 1  1000(1.04)2  $1081.60; Value at the end of year 2  1081.60(1.02) 4  $1170.758624; Value at the end of year 3  1170.758624(1.006)12  $1267.931013; Value at the end of year 4  1267.931013(1.000219)365  $1373.52 Certificate B:

Value at the end of year 1  1000(1.000219)365  $1083.277572; Value at the end of year 2  1083.277572(1.006)12  $1173.189076; Value at the end of year 3  1173.189076(1.02) 4  $1269.897586; Value at the end of year 4  1269.897586(1.04)2  $1373.52. (b) Value of the third certificate  1000(1.000192)1460  $1323.09 The value of the third certificate is $50.43 less than the value of Certificates A and B. Case Study 9 1.

(a)

Present value of cash payments for Plan A: PV of $750,000 one year from now  750, 000(1.02252 )  $717,356 PV of $750,000 two years from now  750, 000(1.02252 )(1.0252 )  682, 790 PV of $750,000 three years from now  750, 000(1.02252 )(1.0252 )(1.0252 )  649,889 PV of $750,000 four years from now  750, 000(1.02252 )(1.0252 )(1.0252 )(1.02752 )  615,567 Total present value of payments  $2,665,602 Since the present value of the cash payment is less than $2,800, 000, the cash requirements of Plan A can be met by the investment.

(b)

The difference between the investment and the present value of the cash payments  2,800,000  2,665,602  $134,398. The accumulated value of the difference after four years  134,398(1.02252 )(1.0252 )(1.0252 )(1.02752 )  163,748 The difference between the cash required and the cash available from the investment is $163,748.

(a)

Present value of cash payments for Plan B: PV of $300,000 now PV of $700,000 one year from now  750, 000(1.02252 ) PV of $700,000 two years from now  900, 000(1.02252 )(1.0252 )

 

$300,000 717,356



819,348

975, 000(1.02252 )(1.0252 )(1.0252 )(1.02752 )



800,237

Total present value of payments

 $2, 636,941

PV of $975,000 four years from now

Since the present value of the cash payments is less than $2,800, 000, the cash

requirements of Plan B can be met by the investment.

(b)

The difference between the investment and the present value of the cash payments  2,800,000  2,636,941  $1,630,59. The accumulated value of the difference after four years = 163, 059(1.02252 )(1.0252 )(1.0252 )(1.02752 ). The difference between the cash required and the cash available from the investment is

$198, 669. 3.

(a)

Present value of cash payments for Plan A: PV of $750,000 now

$750,000

PV of $750,000 one year from now  750, 000(1.02252 )

717,356

PV of $750,000 two year from now  750, 000(1.02252 )(1.0252 )

682,790

750, 000(1.02252 )(1.0252 )(1.0252 )(1.02752 )

615,567

Total present value of payments

= $2, 765, 713

PV of $750,000 four years from now =

Since the present value of the cash payments is less than $2,800, 000, the cash

requirements of Plan A can be met by the investment.

(b)

The difference between the investment and the present value of the cash payments  2,800,000  2,765,713  $34,287.

The accumulated value of the difference after five years = 34, 287(1.02252 )(1.0252 )(1.0252 )(1.02752 )  $41, 775. The difference between the cash required and the cash available from the investment is

$41,775. 4.

(a)

Present value of cash payments for Plan A = 0.049/4 = 0.01225 PV of $750,000 one year from now  750, 000(1.012254 ) $714,349



PV of $750,000 two years from now  750, 000(1.012258 ) 680,392



PV of $750,000 three years from now  750, 000(1.0122512 ) 648,049



PV of $750,000 four years from now  750, 000(1.0122516 ) 617,244



Total present value of payments 

$2,660,034 Since the present value of the cash payments is less than $2,800, 000, the cash

requirements of Plan A can be met by the investment.

(b)

Present value of cash payments for Plan B = 0.049/4 = 0.01225 PV of $300,000 now $300,000



PV of $700,000 one year from now  700, 000(1.012254 ) 666,725



PV of $900,000 two years from now  900, 000(1.012258 ) 816,470



PV of $975,000 four years from now  975, 000(1.0122516 ) 802,417



Total present value of payments 

$2,585, 612 Since the present value of the cash payments is less than $2,800, 000, the cash

requirements of Plan B can be met by the investment.

(c)

The treasurer should recommend Plan B , since it has the lower present value ($2,660,034  $2,585,612).

Chapter 10 Compound Interest—Further Topics

Exercise 10.1 1.

PV  400; FV  760 I / Y  7; C / Y  2; i  3.5%  0.035 n 760  400 1.035 ln( 760 400 ) ln(1.035) It takes about 18.66 half-years. 0.641854 n 0.03440 n  18.657769  Set P / Y = 2  400  PV 760 FV 7 I / Y CPT N 18.657769 n

PV  580; FV  600; I Y  4.5;C Y  12; i  0.375%  0.00375 n 600  580 1.00375  1.00375n  1.034483 n ln1.00375  ln1.034483 0.003743n  0.033902 n  9.057354  months  9.057354 Time in days   365  275.494517  275days. 12 Set P Y = 12  2nd  CLR TVM  580  PV 600 FV 4.5 I Y CPT N 9.057354

3. Let PV  1; FV  4; I / Y  8;C / Y  4; i  2.0%  0.02 1.02n  4 ln( 14 ) n 1 ln(1.02) Number of years   70.005578  17.501394  17.5 years 4 1.386294 n 0.019803 n  70.005578  quarters 

Set P / Y  4  1  PV 4 FV 8 I / Y CPT N 70.005578

PV  800; I  320; FV  1120; I / Y  6;C / Y  12; i  0.5%  0.005 800(1.005)n  1120 ln( 1120 800 ) 67.462544 ln(1.005) Number of years   5.621879  5.62 years 12 0.336472 n 0.004988 n  67.462544 months Set P / Y  12  800  PV 1120 FV 6 I / Y CPT N 67.462544 n

PV  2000; I  604.35; FV  2604.35; I / Y  8;C / Y  4; i  2%  0.02 2604.35  2000 1.02 

ln( 2604.35 2000 ) ln(1.02) Number of months  13.333379  3  40.000138  40 months 0.264036 n 0.019803 n  13.333379 quarters n

Set P / Y  4  2000  PV 2604.35 FV 8 P / Y CPT N 13.333379

PV  1000; I  157.63; FV  1157.63; I / Y  10;C / Y  2; i  5%  0.05 1157.63  1000 1.05 

548 days after May 1, 2027 is October 30, 2028

ln( 1157.63 1000 ) ln(1.05) 0.146375 n 0.048790 n  3.000089  half- years  n

n  3.000089 2  1.500044 years n  1.500044  365  547.516156 n  548days

Set P / Y  2  1000  PV 1157.63 FV 10 I / Y CPT N 3.000089

PV  600; I  150; FV  750; I / Y  9;C / Y  12; i  0.75%  0.0075 750  600 1.0075

908days after May 15, 2025 is November 9, 2027

750 600

ln( ) ln(1.0075) 0.223144 n 0.007472 n  29.863906  months  n

n  29.863906 12  2.488659 years n  2.488658  365  908.360485 n  908days

Set P / Y = 12  600  PV 750 FV 9 I / Y CPT N 29.863906

Let the focal date be now; I / Y  8; P / Y  4; i 

8%  2.0%  0.02 4

3000  1600(1.02) 4  1700(1.02) n 3000  1478.152682  1700(1.02) n 1521.847318  1700 1.02 

n

1700 ln( 1521.847318 ) ln(1.02) 0.110703 n 0.019803 n  5.590335 (quarters)

n

The second payment should be paid in 5.59 quarters.

Set P / Y = 4  1600 FV 8 I / Y 4 N CPT PV 1478.152682 1700 FV 1521.847318  PV CPT N 5.590335 (quarters)

9%  0.75%  0.0075 12 5000  2000(1.0075) 9  2500(1.0075)24  1200(1.0075)  n

Let the focal date be now; I / Y  9; P / Y  12; i 

5000  1869.926354  2089.578510  1200(1.0075) n 1040.495136  1200 1.0075 

n

1200 ln( 1040.495136 ) ln(1.0075) 0.142625 n 0.007472 n  19.087872 (months)

n

The final payment should be made in 19.09 months.

Set P / Y  12  2000 FV 9 I / Y 9 N CPT PV 1869.926354 Set P / Y  12  2500 FV 9 I / Y 24 N CPT PV 2089.578510 1200 FV 1040.495136  PV CPT N 19.087872 (months)

10.

10%  0.83%  0.0083 12 14, 000(1.0083) 8  15, 000(1.0083) 14  29, 000(1.0083)  n Let the focal date be now; I / Y  9; P / Y  12; i 

13,100.71629  13,354.68102  29, 000(1.0083)  n 26, 455.39731  29, 000(1.0083) n n

29,000 ln( 26,455.39731 )

ln(1.0083) 0.091836 n 0.008299 n  11.066131 (months) The single payment would be made in 11.07 months.

Set P / Y = 12 14,000 FV 9 I / Y 8 N CPT PV 13,100.71629 Set P / Y = 12 15,000 FV 9 I / Y 14 N CPT PV 13,354.68102 29,000 FV 26,455.39731  PV CPT N 11.066131 (months)

11.

8.7%  0.725%  0.00725 12 8000(1.00725)2  1600(1.00725)5  4500(1.00725)13  15, 000(1.00725) n Let the focal date be now; I / Y  8.7; P / Y  12; i 

7885.249415  1543.240465  4096.641168  15, 000(1.00725) n 13,525.13105  15, 000 1.00725

n

n

15,000 ln( 13,525.13105 )

ln(1.00725) 0.103501 n 0.007224 n  14.327645 (months) The one payment would be made in 14.33 months.

Set P / Y = 12 8000 FV 8.7 I / Y 2 N CPT PV 7885.249415 Set P / Y = 12 1600 FV 8.7 I / Y 5 N CPT PV 1543.240465 Set P / Y = 12  4500 FV 8.7 I / Y 13 N CPT PV 4096.641168 15,000 FV 13,525.13105  PV CPT N 14.327645 (months)

12.

Let the focal date be now; I / Y  9; C / Y  12; i  0.75%  0.0075 4000  5000 1.0075 

 15, 000 1.0075 

n

4000  5000  0.764149   6000  0.638700   15, 000 1.0075 

n

4000  3820.74  3832.20  15, 000 1.0075 

n

11, 652.94  15, 000 1.0075 

n

36

 6000 1.0075 

60

n

15,000 ln( 11,652.94 )

ln(1.0075) 0.252492 n 0.007472 n  33.791648 (months) The equated date is 33.79 months from now.

 Set P / Y = 12  5000 FV 9 I / Y 36 N CPT P V  3820.74 6000 FV 60 N CPT PV  3832.20 11652.94  PV 15000 FV CPT N 33.791648

13.

Let the focal date be now; I / Y  7; C / Y  4; i  1.75%  0.0175 1200 1.0175   1400 1.0175   1600 1.0175 

 4000 1.0175 

n

1139.142335  1261.599558  1345.165758  4000 1.0175 

n

3745.907652  4000 1.0175 

n

3

6

10

4000 ln( 3745.907652 ) ln(1.0175) 0.065630 n 0.017349 n  3.783029 (quarters) The alternative single payment would have to be made 3.78 quarters from now.

n

3.78 12 = 11.35 months from now 4  Set P / Y = 4 1200 FV 7 I / Y 3 N CPT P V 1139.14 Number of months =

1400 FV 6 N CPT PV 1261.60 1600 FV 910 N CPT PV 1345.17 3745.907652  PV 4000 FV CPT N 3.783029

14.

Let the focal date be now; I / Y  7.2; C / Y  2; i  3.6%  0.036 12, 000 1.036   12, 000 1.036   12, 000 1.036  2

4

8

 16, 000 1.036   10, 000 1.036   10, 000 1.036  3

5

n

11,180.51311  10, 416.98945  9042.805770  14,389.33476  8379.174270  10, 000 1.036  7871.799301  10, 000 1.036 

n

10,000 ln( 7871.799301 ) ln(1.036) 0.239298 n 0.035367 n  6.766123 (half-years)

n

The third payment is to be received in 6.77 half-years.

Set P / Y = 2  12 000 FV 7.2 I / Y 2 N CPT P V 11,180.51311 12, 000 FV 4 N CPT PV 10, 416.98945 12, 000 FV 8 N CPT PV  9042.805770 16, 000 FV 3 N CPT PV 14,389.33476 10, 000 FV 5 N CPT PV  8379.174270 7871.799301  PV 10,000 FV CPT N 6.766123

15.

Let the focal date be now; I / Y  6; C / Y  12; i  0.5%  0.005 200, 000 1.005  300, 000 1.005  500, 000 1.005 6

8

150, 000 1.005  860, 000 1.005 4

18

n

194,103.6156  288, 265.5611  457, 068.0799  147,037.1283  860, 000 1.005 792, 400.1284  860, 000 1.005 n

n

860,000 ln( 792,400.1284 )

ln(1.005) 0.081866 n 0.004988 n  16.414081 months The second payment is to be received in 16.41months.

Set P / Y  12  200, 000 FV 6 I / Y 6 N CPT P V 194,103.6156 300, 000 FV 8 N CPT PV  288, 265.5611 500, 000 FV 18 N CPT PV  457, 068.0799 150, 000 FV 4 N CPT PV 147, 037.1283 792,400.1284  PV 860,000 FV CPT N 16.414081 Exercise 10.2

n

PV  420; FV  1000; n  38; C / Y  4 1000  420 (1  i) 38 1

38 æ1000 ö ÷ i  çç ÷ ÷ 1 çè 420 ø 1

i  (2.380952) 38  1 i  1.023092  1 i  0.023092  2.3092% The nominal annual rate is 2.3092%  4  9.2366% compounded quarterly. (Set P/ Y = 4) 420  PV 1000 FV 38 N CPT I / Y 9.2366 2.

PV  2000; FV  2800; n  87; C / Y  12 2800  2000 (1  i )87 1

87 æ2800 ö ÷ i  çç 1 ÷ çè 2000 ÷ ø 1

i  (1.4) 87  1 i  1.003875  1 i  0.003875  0.3875% The nominal annual rate is 0.3875% 12  4.6500% compounded monthly. (Set P/ Y = 12) 2000  PV 2800 FV 87 N CPT I / Y 4.6500 3.

Let PV  1; FV  2 n  27;C / Y  4 (1  i ) 27  2 1

æ2 ö27 i  çç ÷ 1 ÷ çè1 ÷ ø i  1.026004  1 i  0.026004  2.6004% The nominal annual rate is 2.6004  4  10.4018% compounded quarterly. (Set P/ Y = 4) 1  PV 2 FV 27 N CPT I / Y 10.4018

n  110; C / Y  12 (1  i )110  3 1

æ3 ö110 i  çç ÷ 1 ÷ çè1 ÷ ø i  1.010037  1 i  0.010037  1.0037% The nominal annual rate is 1.0037 12  12.0449% compounded monthly. (Set P/ Y = 12) 1  PV 3 FV 110 N CPT I / Y 12.044910

Let PV  1; FV  3 n  12; C / Y  1 (1  i )12  3 1

æ3 ö12 i  çç ÷ ÷ ÷ 1 çè1 ø i  1.095873  1 i  0.095873  9.5873% The nominal annual rate is 9.5873% compounded annually. (Set P/ Y = 1) 1  PV 3 FV 12 N CPT I / Y 9.5873

PV  800; FV  952.75; n  60; C / Y  12 952.75  800 (1  i )60 1

60 æ952.75 ö ÷ i  çç 1 ÷ çè 800 ÷ ø 1

i  (1.190938) 60  1 i  1.002917 - 1 i  0.002917  0.2917% The nominal annual rate is 0.2917% 12  3.4999% compounded monthly. (Set P / Y = 12) 800  PV 952.75 FV 60 N CPT I / Y 3.4999

PV  4000; FV  6000; n  13; C / Y  2 6000  4000 (1 + i )13 1

æ6000 ÷ ö13 i  çç ÷ ÷ 1 çè 4000 ø 1

i  (1.50) 13  1 i  1.031681  1 i  0.031681  3.1681% The nominal annual rate is 3.1681%  2  6.3362% compounded semi-annually. (Set P / Y = 2) 4000  PV 6000 FV 13 N CPT I / Y 6.3362 8.

PV  1200; FV  1400; n  13; C / Y  4 1400  1200 (1 + i )13 1

æ1400 ö÷13 i  çç ÷ 1 çè1200 ø÷ &131  1 i  (1.16) i  1.011928  1 i  0.011928  1.1928% The nominal annual rate is 1.1928%  4  4.7713% compounded quarterly. (Set P / Y = 4) 1200  PV 1400 FV 13 N CPT I / Y 4.7713 9.

PV  600; FV  705.25; n  15; C / Y  12 705.25  600 (1 + i)15 1

15 æ705.25 ö ÷ i  çç ÷ ÷ 1 çè 600 ø 1 i  1.175416&15  1

i  1.010833  1 i  0.01833  1.0833% The nominal annual rate is 1.0833%  12  12.9997% compounded monthly. (Set P / Y = 12) 600  PV 705.25 FV 15 N CPT I / Y 12.999725

10.

PV  1000; FV  1250; n  12 /12  365  365; C / Y  365 1250  1000 (1 + i)365 1

æ1250 ÷ ö365 i  çç 1 ÷ çè1000 ÷ ø 1

i  1.25 365  1 i  1.000612  1 i  0.000612  0.061154% The nominal annual rate is 0.061154%  365  22.3212% compounded daily. (Set P / Y = 365) 1000  PV 1250 FV 365 N CPT I / Y 22.3212 Exercise 10.3 A. 1. (a)

i  0.095 / 2  0.0475; C / Y  2 f  (1  i ) m  1  (1.0475) 2  1  1.097256  1  0.097256  9.7256%

(b) i  0.105 / 4  0.02625; C / Y  4 f  (1  i ) m  1  (1.02625) 4  1  0.109207  10.9207%

(c)

i  0.05 /12  0.00416; C / Y  12 f  (1  i)m  1  (1.00416)12  1  0.051162  5.1162%

(d)

i  0.072 / 26  0.002769; C / Y  26 f  (1  i ) m  1  (1.002769) 26  1  0.074548  7.4548%

(e)

i  0.036 / 365  0.000099; C / Y  365 f  (1  i ) m  1  (1.000099)365  1  0.036654  3.6654%

(f)

i  0.082 / 52  0.001577; C / Y  52 f  (1  i ) m  1  (1.001577)52  1  0.085386  8.5386%

2. (a)

i2  1  i1  m2  1 2

 0.09  4  1   1 2   2

 (1.045) 4  1  0.022252 j  0.022252  4  0.089010  8.9010% m1

(b)

i2  1  i1  m2  1 4

 0.065 12  1   1 4   4

 (1.01625) 12  1  0.005388 j  0.005388 12  0.064651  6.4651% m1

(c)

i2  1  i1  m2  1 2

 0.075  365  1   1 2   2

 (1.0375) 365  1  0.000202 j  0.000202  365  0.073635  7.3635%

(d)

i2  1  i1  m2  1 4

 0.0425  2  1   1 4   4

 (1.010625) 2  1  0.021363 j  0.021363  2  0.042726  4.2726%

B. 1.

PV  100; FV  150; n  24; C / Y  4

150  100 (1  i ) 24 1

æ150 ö24 i  çç ÷ 1 ÷ çè100 ÷ ø i  (1.5) 24  1 i  1.017038  1  0.017038  1.7038% per quarter 1

The nominal rate is 1.7038%  4  6.8152% compounded quarterly.

f  (1  i ) m  1  1.0170384  1  1.069913  1  0.069913  6.9913%

Set P / Y  4 100  PV 150 FV 24 N CPT I / Y 6.8152 STO 1 2nd I Conv Nom  RCL 1 Enter C / Y  4 Enter CPT Eff  6.9913

Alternatively, if you recognize the effective annual rate implies you can simply use m  1 and n  6 1  6 in this case, then 1

 150  6 i  1  100   6.9913%

PV  450; FV  750; n  41; C / Y  12 750  450 (1  i ) 41 1

41 æ750 ö ÷ i  çç 1 ÷ çè 450 ÷ ø

i  (1.666667) 41  1 i  1.012537  1  0.012537  1.2537% per month The nominal rate is 1.2537% 12  15.0445% compounded monthly. 1

f  (1  i) m  1  1.01253712  1  1.161265  1  0.161265  16.1265%

Set P / Y  12 450  PV 750 FV 41 N CPT I / Y 15.0445 STO 1 2nd I Conv Nom  RCL 1 Enter C / Y  12 Enter CPT Eff  16.1265

Alternatively, 1

3.416& æ750 ö ÷ i = çç ÷ ÷ - 1 çè 450 ø = 16.1265%

PV  1100; FV  1350; n  9; C / Y  2 1350  1100 (1  i )9 1

9 æ1350 ö ÷ i  çç ÷ ÷ 1 çè1100 ø 1

i  (1.227273) 9  1 i  1.023016  1  0.023016  2.3016% per half year The nominal rate is 2.3016  2  4.6032% compounded semi-annually. f  1 + i m  1  1.0230162  1  1.046561  1  0.046561  4.6561%

Set P / Y  2 1100  PV 1350 FV 9 N CPT I / Y 4.6032 STO 1 2nd I Conv Nom  RCL 1 Enter C / Y  2 Enter CPT Eff  4.6561

Alternatively, 1

4.5 æ1350 ö ÷ i = çç - 1 ÷ çè1100 ÷ ø = 4.6561%

PV  2300; FV  2800; n  38; C / Y  12 2800  2300 (1  i )38 1

æ2800 ö÷38 i  çç ÷ 1 çè 2300 ø÷ 1

i  (1.217391) 38  1 i  1.005190  1  0.005190  0.5190% Nominal rate  0.5190%´ 12  6.2280% f  (1  i ) m  1  1.00519012  1  1.064089  1  0.064089  6.4089%

Set P / Y  12  2300  PV 2800 FV 38 N CPT I / Y 6.2280 STO 1 2nd I Conv Nom  RCL 1 Enter C / Y  12 Enter CPT Eff  6.4089 Alternatively, 1

æ2800 ÷ ö3.16& i = çç - 1 ÷ çè 2300 ÷ ø = 6.4089% 5.

f  9.25%  0.0925; C / Y  4 f  (1  i ) m  1 0.0925  (1  i ) 4  1 1

æ1.0925 ÷ ö4 i  çç 1 ÷ çè 1 ÷ ø i  1.0925 4  1 i  1.022364 - 1 i  0.022364  2.2364% 1

The nominal annual rate is 2.2364%  4  8.9454% compounded quarterly. 2nd I Conv Eff  9.25 Enter C/ Y  4 Enter CPT Nom  8.9454

f  6.37%  0.0637; C / Y  2 f  (1  i) m  1 0.0637  (1  i) 2  1 1

æ1.0637 ÷ ö2 i  çç 1 ÷ çè 1 ÷ ø i  (1.0637) 2  1 i  1.031358  1 i  0.031358  3.1358% 1

The nominal annual rate is 3.1358%  2  6.2717% compounded semi-annually. 2nd I Conv Eff  6.37 Enter C/ Y  2 Enter CPT Nom  6.2717 7.

f  6.4%  0.064; C / Y  12 f  (1  i ) m  1 1

12 æ1.064 ö ÷ i  çç ÷ ÷ 1 çè 1 ø

i  1.06412  1 i  1.005183  1 i  0.005183  0.5183% 1

The nominal annual rate is 0.5183% 12  6.2196% compounded monthly. 2nd I Conv Eff  6.4 Enter C/ Y  12 Enter CPT Nom  6.2196 8.

f  5.3%  0.053; C / Y  4 f  (1  i ) m  1 1

æ1.053 ÷ ö4 i  çç 1 ÷ çè 1 ÷ ø i  (1.053) 4  1 i  1.012995 i  0.012995  1.2995% 1

The nominal annual rate is 1.2995%  4  5.1978% compounded quarterly. 2nd I Conv Eff  5.3 Enter C/ Y  4 Enter CPT Nom  5.1978

Value of $1 in one year at 7.5% compounded semi-annually. FV1  1.03752  1.076406 Value of $1 in one year at a monthly rate i, FV2  (1  i )12 for FV1  FV2 1

æ1.076406 ÷ ö12 i  çç 1 ÷ ÷ çè ø 1 i  1.07640612  1 i  1.006155 i  0.006155  0.6155% 1

The nominal rate is 0.6155% 12  7.3854% compounded monthly. 2nd I Conv Nom  7.5 Enter C / Y  12 Enter CPT Eff  7.6406 STO 1 2nd I Conv Eff  RCL 1 Enter C/ Y  12 Enter CPT Nom  7.3854

10.

Value of $1 in one year at 6% compounded quarterly. FV1  1.0154  1.061364 Value for $1 in one year at a daily rate i, FV2  (1  i )365 for FV1  FV2 1

365 æ1.061364 ö ÷ i  çç ÷ ÷ 1 çè ø 1 1

i  (1.061364) 365  1 i  1.000163 i  0.000163  0.0163% The nominal annual rate is 0.016318%  365  5.9559% compounded daily. 2nd I Conv Nom  6 Enter C / Y  4 Enter CPT Eff  6.1364 STO 1 2nd I Conv Eff  RCL 1 Enter C / Y  365 Enter CPT Nom  5.9559

11.

PV  600; i  0.2916%; n  60; C / Y  12



(a) FV  600 1.002916

  600 1.190943  $714.57 60

(b) Interest earned  714.57  600  $114.57 (c) f  (1  i ) m  1  1.00291612  1  1.035567  1  0.035567  3.5567%

 Set P / Y  12  600  PV 60 N 3.5 I / Y CPT FV 714.565698 STO 1 2nd I Conv Nom  3.5 Enter C / Y  12 Enter CPT Eff  3.5567 OR:  Set P / Y  1 600  PV RCL 1 FV 5 N CPT I / Y 3.5567

12.

PV  5000; i  0.0275 / 365  0.000075; n  730; C / Y  365 (a) FV  5000 1.000075 

730

 5000 1.056538   $5282.69

(b) Interest earned  5282.69  5000  $282.69 (c) f  (1  i ) m  1  1.000075365  1  1.027881  1  0.027881  2.7881%

Set C / Y  365 5000  PV 730 N 2.75 I / Y CPT FV 5282.692129 STO 1 2nd I Conv Nom  2.75 Enter C / Y  365 Enter CPT Eff  2.7881 OR:  Set P / Y  1 5000  PV RCL 1 FV 2 N CPT I / Y 2.7881

13.

PV  1200; i  0.0425 / 4  0.010625; n  40; C / Y  4 (a) FV  1200 1.010625   1200 1.526165   $1831.40 40

(b) Interest earned  1831.40  1200  $631.40 (c) f  (1  i ) m  1  1.0106254  1  1.043182  1  0.043182  4.3182%

 Set C / Y  4  1200  PV 40 N 4.25 I / Y CPT FV 1831.398002 STO 1 2nd I Conv Nom  4.25 Enter C / Y  4 Enter CPT Eff  4.3182 OR:  Set P / Y  1 1200  PV RCL 1 FV 10 N CPT I / Y 4.3182

14.

PV  1750; i  0.05 / 2  0.025; n  16; C / Y  2 (a) FV  1750 1.025   1750 1.484506   $2597.88 16

(b) Interest earned  2597.88  1750  $847.88 m (c) f  (1  i)  1

 1.0252  1  1.050625  1  0.050625  5.0625%

 Set P / Y  2  1750  PV 16 N 5 I / Y CPT FV 2597.884836 STO 1 2nd I Conv Nom  5 Enter C / Y  2 Enter CPT Eff  5.0625 OR:  Set P / Y  1 1750  PV RCL 1 FV 8 N CPT I / Y 5.0625

Business Math News Box 1.

(a) Payments are made monthly at an annual interest rate of 29.9% compounded monthly. The periodic interest rate is 2.4917%. i  0.299 / 12  0.02491666  2.4917% Payment Number 1 2 3 4 5 6 Total

Amount Paid

Interest Paid (2.4917%)

Principal Repaid

$200.00 $200.00 $200.00 $200.00 $200.00 $200.00

$24.92 $20.55 $16.08 $11.50 $ 6.80 $ 1.99 $81.84

$175.08 $179.45 $183.92 $188.50 $193.20 $ 79.85 $1000.00

Outstanding Principal Balance $1000.00 $824.92 $645.47 $461.55 $273.05 $ 79.85 $ 0.00

He will make six payments (five payments of $200 and a sixth payment of $81.84 ). He will pay $81.84 in interest. (b) Payments are made monthly at an annual interest rate of 19.5% compounded monthly. The periodic interest rate is 1.6250%. I  0.195/12  0.01625  1.625%

Payment Number

Amount Paid

Interest Paid (1.625%)

Principal Repaid

1 2 3 4 5 6 Total

$200.00 $200.00 $200.00 $200.00 $200.00 $ 51.73

$16.25 $13.26 $10.23 $ 7.15 $ 4.01 $ 0.83 $51.73

$183.75 $186.74 $189.77 $192.85 $195.99 $ 50.90 $1000.00

Outstanding Principal Balance $1000.00 $816.25 $629.51 $439.74 $246.89 $ 50.90 $ 0.00

He will make six payments (five payments of $200 and a sixth payment of $51.73). He will pay $51.73 in interest. Difference in interest  $81.84  $51.73  $30.11

Therefore, he would have saved $30.11 in interest if he had used the CIBC credit card instead of his Best Buy credit card to make his purchase. 2.

Using Formula 10.3 to find the effective rate of interest, f  (1  i)m  1 f  (1 + .02)12 – 1 f  1.26824  1 f  026824  26.82% The effective rate of interest, f, for a credit card that is advertised as 2% per month is 26.82%.

Review Exercise 1.

PV  400; I  100; FV  500; n  48; C / Y  12 500  400 (1  i) 48 1

48 æ500 ö ÷ i  çç 1 ÷ çè 400 ÷ ø 1

i  (1.25) 48  1 i  1.004660 i  0.004660  0.4660% The nominal annual rate is 0.4660% 12  5.5920% compounded monthly. (Set P/ Y = 12) 400  PV 500 FV 48 N CPT I / Y 5.592

PV  300; I  80; FV  380; n  24; C / Y  4 380  300 (1  i ) 24 1

24 æ380 ö ÷ i  çç 1 ÷ çè300 ÷ ø &241  1 i  (1.26)

i  1.009898 i  0.009898  0.9898% The nominal annual rate is 0.9898%  4  3.9593% compounded quarterly. (Set P/ Y = 4) 300  PV 380 FV 24 N CPT I / Y 3.9593

Let the focal date be today; i  0.75%  0.0075; C / Y  12 500 1.0075   600  1300 1.0075 

n

500 1.045852   600  1300 1.0075 

n

522.93  600  1300 1.0075

n

1122.93  1300 1.0075 

n

1300 ln( 1122.93 ) ln(1.0075) 0.146423 n 0.007472 n  19.596177 (months)

n

The payment date is 19.60 months from now.

 Set P / Y  12  500  PV 9 I / Y 6 N CPT FV 522.93 522.93  600  1122.93 1122.93  PV 1300 FV CPT N 19.5962

Let the focal date be today; I/Y  7.25 C / Y  2; i  3.625%  0.03625 .

600(1.03625)0.6  400  1100(1.03625)  n 600(1.024023)  400  1100(1.03625)  n 614.41  400  1100(1.03625)  n 1014.41  1100(1.03625)  n 1100 ln  1014.41  ln(1.03625) 0.081003 n= 0.035608 n = 2.274827(semi-annual periods)

The payment date is 2.27 semi-annual periods from now. 2.27 Number of months =  12 = 13.62 months from now 2 (Set P / Y  2) 600  PV 7.25 I / Y 0.6 N CPT FV 614.41 614.41  400  1014.41 1014.41  PV 1100 F V CPT N 2.274827

Let PV  $1; FV  $3; i  5.0%  0.05; C / Y  2 3  1(1.05) n ln( 13 ) ln(1.05) 1.098612 n 0.048790 n  22.517085 n  22.52 half-years 22.52 Number of years = = 11.26 years 2 Set P / Y  2  1  PV 3 FV 10 I / Y CPT N 22.517085 n

Let PV  $1; FV  $2; i  0.6%  0.006; C / Y  12



2  1 1.006



ln( 12 ) ln(1.006) 0.693147 n 0.006645 n  104.318267 n

n  104.32 months

Set P / Y  12 1  PV 2 FV 8 I / Y CPT N 104.318267

Value of $1 in one year at i % monthly, FV1  (1  i )12 Value of $1 in one year at 6.2% effective, FV2  1.062 For FV1  FV2 (1  i )12  1.062 i  (1.062)12  1 i  1.005025 i  0.005025  0.5025% 1

The nominal rate is 0.5025% 12  6.0305% compounded monthly. 2nd I Conv Eff  6.2 Enter C / Y  12 Enter CPT Nom  6.0305

Value of $1 in one year at i % monthly, FV1  (1  i ) 4 Value of $1 in one year at 5.99% effective, FV2  1.0599 For FV1  FV2 , (1  i) 4  1.0599 i  1.0599 4  1 i  1.014650  1 i  0.014650  1.4650% 1

The nominal rate is 1.4650%  4  5.8600% compounded quarterly. 2nd I Conv Eff  5.99 Enter C/ Y  4 Enter CPT Nom  5.8600

(a)

PV  2500; FV  4000; n  32; C / Y  4 4000  2500 (1  i )32 1

i  (1.6) 32  1 i  1.014796  1 i  0.014796  1.4796% The nominal annual rate is 1.4796%  4  5.9184% compounded quarterly.

Set P / Y  4  2500  PV 4000 FV 32 N CPT I / Y 5.9184

(b)

Let PV  100; FV  200; n  10; C / Y  2 200 100(1 + i )10 1

10 æ200 ö ÷ i  çç ÷ ÷ 1 çè100 ø

i  1.071773  1 i  0.071773  7.1773% The nominal annual rate is 7.1773%  2  14.3547% compounded semi-annually. (Set P/ Y = 2) 100  PV 200 FV 10 N CPT I / Y 14.3547 (c)

f  9.2%  0.092; C / Y  12 1  f  (1 + i) m 1

æ1.092 ÷ ö12 i  çç 1 ÷ çè 1 ÷ ø i  (1.092) 12  1 i  1.007361  1 i  0.007361  0.7361% 1

The nominal annual rate is 0.7361% 12  8.8334% compounded monthly. 2nd I Conv Eff  9.2 Enter C/ Y  12 Enter CPT Nom  8.8334

(d)

1  i  1.024 1  i  1.082432 i  8.2432% The nominal annual rate is 8.2432% compounded annually. 2nd I Conv Nom  8 Enter C/ Y  4 Enter CPT Eff  8.2432

10. (a)

Let PV  1500; FV  1800; n  48; C / Y  12 1800  1500 (1 + i) 48 1

æ1800 ÷ ö48 i  çç 1 ÷ çè1500 ÷ ø 1

i  (1.2) 48  1 i  1.003806  1 i  0.003806  0.3806% The nominal annual rate is 0.3806% 12  4.5667% compounded monthly. (Set P/ Y = 12) 1500  PV 1800 FV 48 N CPT I / Y 4.5667 (b)

Let PV  100; FV  200; n  28; C / Y  12 1

200  100 (1 + i ) 28 1

28 æ200 ö ÷ i  çç ÷ ÷ 1 çè100 ø i  1.025064  1 i  0.025064  2.5064%

The nominal annual rate is 2.5064%  4  10.0257% compounded quarterly. (Set P/ Y = 4) 100  PV 200 FV 28 N CPT I / Y 10.0257

(c)

f  7.75%  0.0775; C / Y  12 1  f  (1 + i) m 1

12 æ1.0775 ö ÷ i  çç ÷ ÷ 1 çè 1 ø

i  (1.0775) 12  1 i  1.006240 i  0.006240  0.6240% 1

The nominal annual rate is 0.6240% 12  7.4876% compounded monthly. 2nd I Conv Eff  7.75 Enter C/ Y  12 Enter CPT Nom  7.4876

(d)

1  i  1.0154 1  i  1.061364 i  6.1364% The nominal annual rate is 6.1364% 2nd I Conv Nom  6 Enter C/ Y  4 Enter CPT Eff  6.1364

11. (a)

Using f  (1  i) m  1, f  1.0037512  1  1.045940  1  0.045940  4.5940% 2nd I Conv Nom  4.5 Enter C/ Y  12 Enter CPT Eff  4.5940

(b)

3000  2000 (1  i) 28 ; n  28; C / Y  4 (1  i ) 28  1.5 1

1  i  1.5 28 1  i  1.014586 f  1.0145864  1  1.059634  1  0.059634  5.9634%

Set P / Y  4  2000  PV 3000 FV 28 N CPT I / Y 5.834501 STO 1 2nd I Conv Nom  RCL 1 Enter C/ Y  4 Enter CPT Eff  5.9634

12. (a)

f  1.00512  1  1.061678  1  0.061678  6.1678% 2nd I Conv Nom  6 Enter C/ Y  12 Enter CPT Eff  6.1678

(b)

2000  1100 (1 + i )84 ; n  84; C / Y  12 && (1 + i )84  1.81 1

&&ö84 æ1.81 ÷ ç ÷ i  çç 1 ÷ ÷ è 1 ø i  1.007142 f  1.00714212  1  1.089158  1  0.089158  8.9158% (Set P / Y = 12) 1100  PV 2000 FV 84 N CPT I / Y 8.570993 STO 1 2nd I Conv Nom  RCL 1 Enter C/ Y  12 Enter CPT Eff  8.9158 13.

1  i   1.02125 12

i  1.0212512  1 i  1.007034  1 i  0.007034  0.7034% 4

The nominal annual rate is 0.7034% 12  8.4405% compounded monthly. 2nd I Conv Nom  8.5 Enter C / Y  4 Enter CPT Eff  8.774796 ST O 1 2nd I Conv Eff  RCL 1 Enter C/ Y  12 Enter CPT Nom  8.4405

14.

1  i   1.05 4

i  1.05 4  1 i  1.012272  1 i  0.012272  1.2272% 1

The nominal annual rate is 1.2272%  4  4.9089% compounded quarterly. 2nd I Conv Eff  5 Enter C/ Y  4 Enter CPT Nom  4.9089

15. (a)

i  4.00%  0.04;C / Y  1 f  (1  i ) m  1  1.04   1  1.04  1  0.04  4% 1

2nd I Conv Nom  4 Enter C / Y  1 Enter CPT Eff  4.0 (b)

i  1.875%  0.01875; C / Y  2 f  (1  i)m  1  1.01875   1  1.037852  1  0.037852  3.7852% 2

2nd I Conv Nom  3.75 Enter C / Y  2 Enter CPT Eff  3.7852 (c)

i  0.875%  0.00875; C / Y  4 f  (1  i)m  1  1.00875   1  1.035462  1  0.035462  3.5462% 4

2nd I Conv Nom  3.5 Enter C / Y  4 Enter CPT Eff  3.4562 (d)

i  0.2708%  0.002708; C / Y  12 f  (1  i )m  1  1.002708   1  1.032989  1  0.032989  3.2989% 12

2nd I Conv Nom  3.25 Enter C / Y  12 Enter CPT Eff  3.2989 He should choose 4% compounded annually. 16. (a)

9517.39  7500(1.015) n ; C / Y  2 ln( 9517.39 7500 ) ln(1.015) 0.238218 n 0.014889 n  15.999989 n  16 (half-years) n

It will take 16 half-years (or 8 years).

Set P / Y  2  7500  PV 9517.39 FV 3 I / Y CPT N 15.999989

(b)

Let PV  $1; FV  $3; i  2.25%  0.0225; C / Y  4 3  11.0225 

ln( 13 ) ln(1.0225) 1.098612 n 0.02225 n  49.374482 (quarters) n

Money will triple in 49.37 quarters.

Set P / Y  4  1  PV 3 FV 9 I / Y CPT N 49.374482 (c)

PV  10, 000; FV  13, 684; i  0.875%  0.00875; C / Y  12 13, 684  10, 000 (1.00875) n n

13,684 ln( 10,000) )

ln(1.00875) 0.313642 n 0.008712 n  36.001413 (months) It will take 3 years

Set P / Y  12  10, 000  PV 13, 684 FV 10.5 I / Y CPT N 36.001413 17.

Let the focal date be now; i  2%  0.02 2000 1.02   1500 1.02   1800  1700 1.02 

n

2000  0.942322   1500  0.923845   1800  1700 1.02 

n

1884.64  1385.77  1800  1700 1.02 

n

3

4

1470.41  1700 1.02 

n

1700 ln( 1470.41 ) ln(1.02) 0.145087 n 0.019803 n  7.326653 (quarters) The second payment is to be made 7.33 quarters from now. 7.33 Number of months = 12 = 22 months from now 4  Set P / Y  4  2000 FV 8 I / Y 3 N CPT PV 1884.64

n

1500 FV 4 N CPT PV 1385.77 1884.64  1385.77  1800  1470.41 1470.41  PV 1700 FV CPT N 7.326653

18.

Let the focal date be 4 years from now. I Y  7%; C / Y  12, i  7% /12  0.583%  0.00583



3885.35 1.00583

  4000 n

4000 ln( 3885.35 ) ln(1.00583) 0.029081 n 0.005816 n  5.000175  months 

n

The debt was paid five months before the due date.

Set P / Y  12  3885.35  PV 4000 FV 7 I / Y CPT N 5.000175 19.

Let the focal date be now; i  0.5%  0.005; C / Y  12. 2000  2500 1.005   4000 1.005 

 9000 1.005 

n

2000  2500  0.970518   4000  0.941905   9000 1.005 

n

2000  2426.30  3767.62  9000 1.005 

n

8193.92  9000 1.005 

n

6

12

9000 ln( 8193.92 ) ln(1.005) 0.093832 n 0.004988 n  18.813309 (months)

n

The date of discharge is 18.81 months from now.

 Set P / Y  12  2500 FV 6 I / Y 6 N CPT PV  2426.30 4000 FV 12 N CPT PV  3767.62 2000  2426.30  3767.62  8193.92 8193.92  PV 9000 FV CPT N 18.813309

20.

Let the focal date be now; i  5.5%  0.055; C / Y  2 5000 1.055   6000 1.055 

 5000  6000 1.055 

n

5000  0.947867   6000  0.874720   5000  6000 1.055 

n

4739.34  5248.32  5000  6000 1.055 

n

1

2.5

4987.66  6000 1.055 

n

6000 ln( 4987.66 ) ln(1.055) 0.184793 n 0.05354 n  3.451437 (half-years)

n

The payment of $6000 should be made 3.45 half-years from now.

 Set P / Y  2  5000 FV 11 I / Y 1 N CPT PV  4739.34 6000 FV 2.5 N CPT PV  5248.32 54739.34  5248.32  5000  4987.66 4987.66  PV 6000 FV CPT N 3.451437 21.

Let the focal date be now; I / Y  8%;C / Y  4, i  8% / 4  2%  0.02 12, 000  7500 1.02   7500 1.02  8

n

12, 000  7500  0.853490   7500 1.02  12, 000  6401.18  7500 1.02  5598.82  7500 1.02 

n

7500 ln( 5598.82 ) ln(1.02) 0.292348 n 0.019803 n  14.762794 (quarters) The second payment should be made in 14.76 quarters. 14.76 quarters Number of years = = 3.69 years 4 Set P / Y  4  7500 FV 8 I / Y 8 N CPT PV  6401.18

n

5598.82  PV 7500 FV CPT N 14.762794

Let the focal date be now; I / Y  10; C / Y  4, i  10% / 4  2.5%  0.025

22.

3000 1.025

10

 5000 1.025

20

 4000 1.025

17

 6000 1.025

n

3000  0.781198  5000  0.610271  4000  0.657195  6000 1.025 2343.59  3051.36  2628.78  6000 1.025

n

5394.95  2628.78  6000 1.025

n

2766.17  6000 1.025

n

6000 ln( 2766.17 ) ln(1.025) 0.774297 n 0.024693 n  31.356923 (quarters) The second payment should be made in 31.36 quarters. 31.26 quarters Number of years = = 7.84 years 4 Set P / Y  4  3000 FV 10 I / Y 10 N CPT PV  2343.59

n

5000 F V 20 N CPT PV  3051.36 4000 F V 17 N CPT PV  2628.78 2766.17  P V 6000 FV CPT N 31.356923 Self-Test 1.

PV  11, 000; FV  12,950; I / Y  5%; C / Y  2; i  5% / 2  2.5%  0.025 12,950  11, 000 1.025  n

ln( 12,950 11,000 )

ln(1.025) 0.163201 n 0.024693 n  6.60919  half-years  The money was invested for 0.5  6.60919  3.3 years.

Set P / Y  2  11000  PV 12950 FV 5 I / Y CPT N 6.60919

2177.36  1400 1  i 

1  i 

5.916

; n  5.916; C / Y  1

 1.555257 1

i  1.555257 5.916  1 i  1.077500  1 i  1.077500  1  0.077500  7.75%

Set P / Y  1 1400  PV 2177.36 FV 5.916 N CPT I / Y 7.75 3.

i  0.45%; C / Y  12 f  1.004512  1  1.055357  1  5.5357% 2nd I Conv Nom  5.4 Enter C/ Y  12 Enter CPT Eff  5.5357

Let the focal date be now; i  0.85%

1000 1.0085   400 1.0085   1746.56 1.0085 

n

1106.91  380.19  1745.56 1.0085 

n

1487.10  1746.56 1.0085 

n

6

ln( 1746.56 1487.10 ) ln(1.0085) 0.160820 n 0.008464 n  19.000323 n

Payment should be made 19 months from now.

Set P / Y  12  1000  PV 10.2 I / Y 12 N CPT F V 1106.91 400 FV 6 N CPT PV  380.19 1106.91  380.19  1487.10 1487.10  PV 1746.56 FV CPT N 19.000323 5.

PV  6900; FV  6900  3000  9900; n  10;C / Y  2 9900  6900 1  i 

1  i   1.434783 10

i  (1.434783) 10  1 i  1.036761  1 i  0.036761  3.6761% The nominal annual rate is 3.6761%  2  7.3522% compounded semi-annually.

Let PV  1; then FV  2; i  1.8% 2  11.018

ln( 12 ) ln(1.018) 0.693147 n 0.017840 n  38.853720 (quarters) 38.853720 n  9.71 years 4 (Set P Y  4) 1  PV 2 FV 7.2 I Y CPT N 38.853720 n

f  10.25%; C / Y  2 1.1025  1  i 

2 1

i  (1.1025) 2  1 i  1.05  1 i  0.05  5% The nominal annual rate is 5%  2  10% compounded semi-annually. 2nd I Conv Eff  10.25 Enter C / Y  2 Enter CPT Nom  10

Let the focal date be now; I / Y  9; C / Y  12, i  9% / 12  0.75%  0.0075 1000 1.075   3000 1.075   4000 1.075  2

8

14

 2000  6000 1.0075 

n

1000  0.865333  3000  0.560027   4000  0.363313  2000  6000 1.0075  865.33  1682.11  1453.25  2000  6000 1.0075  2000.69  6000 1.0075 

n

6000 ln( 2000.69 ) ln(1.0075) 1.098268 n 0.007472 n  146.98452 (months)

n

The second payment should be made in 146.98 months.

Set P / Y  12  1000 FV 9 I / Y 2 N CPT PV  865.33 3000 F V 8 N CPT PV  1682.11 4000 FV 14 N CPT PV  1453.25 2000.69  P V 6000 FV CPT N 146.98452

n

At 3.95% compounded semi-annually, the effective rate will be 1.01975  1  3.98901%. 2

At 3.92% compounded quarterly, the effective rate will be 1.0098  1  3.978001%. 4

At 3.9% compounded monthly, the effective rate will be 1.00325  1  3.970473%. 12

He will maximize his return at 3.95% compounded semi-annually. 2nd I Conv Nom  3.95 Enter C / Y  2 Enter CPT Eff  3.98901 2nd I Conv Nom  3.92 Enter C / Y  4 Enter CPT Eff  3.978001 2nd I Conv Nom  3.9 Enter C / Y  12 Enter CPT Eff  3.970473

Challenge Problems 1.

Let the number of semi-annual compounding periods be n. The future value of Olga’s investment on that date  800 1  0.045 

The future value of Ursula’s investment on that date

 600 1  0.035 

800 1.045   2  600 1.035  n

2 1.045   3 1.035n  n

ln 2  n  ln 1.045   ln 3  n  ln 1.035  0.693147  0.044017 n  1.098612  0.034401n 0.009616n  0.405465 n  42.168046 half-years n  42.168046(365 / 2) days  7695.67 days The future value of Olga’s investment will be twice the future value of Ursula’s investment in 7696 days. 2.

Let the amount invested be $1. The maturity value of the investment  11.01875  1.005416 

1.000164   1.077136 1.066972 1.061838 4

365

 1.220335

The equivalent future value compounded semi-annually is 1  i  . 6

1  i 6  1.220335 1

1  i  1.220335 6 1  i  1.033744 i  0.033744 The rate compounded semi-annually will be 6.7489%.

Case Study 1.

(a)

Effective rates of interest for the 48-month option: Car dealer: 4.5% compounded monthly 12

12  0.045  f  1    1  1.00375   1  0.04594 12  

 4.59%

Credit union: 5.3% compounded quarterly 4

 0.053  4 f  1    1  1.01325  1  0.054063 4    5.41% Bank: 5.1% compounded semi-annually 2

 0.051  2 f  1    1  1.0255  1  0.051650 2    5.17% (b)

Accumulated value of least expensive option (dealer): S  10,150 1.00375   10,150 1.196814   $12,147.67 48

Accumulated value of most expensive option (credit union): S  10,150 1.01325   10,150 1.234428   $12,529.45 16

Interest savings  12,529.45  12,147.67  $381.78 2.

(a)

Effective rates of interest for the 36-months option: Credit union: 5% compounded quarterly 4

 0.05  4 f  1    1  1.0125  1  0.050945 4    5.09% Bank: 4.80% compounded semi-annually 2

 0.048  2 f  1    1  1.024  1  0.048576 2    4.86% (b)

Accumulated value of least expensive option (bank): S  10,150 1.024   10,150 1.152922   $11, 702.15 6

Accumulated value of most expensive option (credit union): S  10,150 1.0125   10,150 1.160755   $11, 781.66 12

Interest savings  11, 781.66  11, 702.15  $79.51 3.

Chapter 11

Ordinary Simple Annuities

Exercise 11.2 1.

PMT  200; i  1.25%; n  48 1.012548  1  FVn  200    200 (65.228388)  $13, 045.68  0.0125  (Set P / Y, C / Y  4) 0 PV 200  PMT 5 I / Y 48 N CPT FV 13, 045.68

PMT  60; i  0.4%; n  72 1.00472  1  FVn  60    60 (83.247831)  $4994.87 0.004   (Set P / Y, C / Y  12) 0 PV 60  PMT 4.8 I / Y 72 N CPT FV 4994.87

5.4%  0.45%  0.0045; n  16 12 1.004516  1  FVn  240    240 (16.551508)  $3972.36  0.0045  PMT  240; i 

(Set P / Y, C / Y  12) 0 PV 240  PMT 5.4 I / Y 16 N CPT FV 3972.36 4.

19.8%  1.65%  0.0165; n  19 12 1.016519  1  FVn  600    600 (22.103618)  $13, 262.17  0.0165  PMT  600; i 

(Set P / Y, C / Y  12) 0 PV 600  PMT 19.8 I / Y 19 N CPT FV 13, 262.17069

6.6%  0.55%  0.0055; n  6 12 1.00556  1 FVn  151.72    151.72 (6.083108)  $922.93  0.0055  PMT  151.72; i 

(Set P / Y, C / Y  12) 0 PV 151.72  PMT 6.6 I / Y 6 N CPT FV 922.929070 Amount owed  $992.93

PMT  1500; i  3.5%; n  30 1.03530  1  FVn  1500    1500 (51.622677)  $77, 434.02  0.035  Interest  77, 434.02  1500(30)  77, 434.02  45, 000  $32, 434.02 (Set P / Y, C / Y  2) 0 PV 1500  PMT 7 I / Y 30 N CPT FV 77, 434.02

2.4%  0.2%  0.002; n  96 12 1.00296  1 FVn  120    120 (105.719104)  $12, 686.29  0.002  PMT  120; i 

Interest  12, 686.29  120(96)  12, 686.29  11,520  $1166.29 (Set P / Y, C / Y  12) 0 PV 120  PMT 2.4 I / Y 96 N CPT FV 12, 686.29249

PMT  15; i  0.375%; n  192 1.00375192  1  FVn  15    15(280.444934)  $4206.67  0.00375  Interest  4206.67  15(192)  4206.67  2880  $1326.67 (Set P / Y, C / Y  12) 0 PV 15  PMT 4.5 I / Y 192 N CPT FV 4206.67

(a) After 15 years: PMT  250; i  0.3%; n  180

1.003180  1  FVn  250    250(238.206744)  $59,551.6859  0.003  Amount 25 years from now:

PV  $59,551.6859; i  0.3%; n  10 12  120 FV  $59,551.6859(1.003)120  $59,551.6859(1.432557)  $85,311.19 (b) Contribution  250(180)  $45,000 (c) Interest  85,311.19  45,000  $40,311.19

(Set P / Y, C / Y  12) 0 PV 250  PMT 3.6 I / Y 180 N CPT FV 59,551.6859 STO 1 RCL 1  PV 0 PMT 120 N CPT FV 85,311.19459

10. (a) After ten years: PMT  2000; i  3.8%; n  10

1.03810  1 FV1  2000    2000(11.895346)  $23,790.69116  0.038  Five years later: PV  23, 790.69; i  3.8%; n  5

FV2  23,790.69116(1.038)5  23,790.69116(1.204999)  $28,667.76

(Set P / Y, C/Y  1) 0 PV 2000  PMT 3.8 I / Y 10 N CPT FV 23,790.69116 STO 1 RCL 1  PV 0 PMT 5 N CPT FV 28, 667.76 (b) Contribution  10(2000)  $20,000 (c) Interest  28, 667.76  20, 000  $8667.76 11. (a) After five years: PMT  250; i  0.3750%; n  60

1.0037560  1  FVn  250    250(67.145552)  $16,786.38803  0.00375  Amount four years later: PV  16,786.38803; i  0.375%; n  48 FV  16,786.38803(1.00375) 48  16,786.38803(1.196814)  $20, 090.19

(b) Contribution  60(250)  $15,000 (c) Interest  20, 090.19  15, 000  $5090.19

(Set P / Y, C / Y  12) 0 PV 250  PMT 4.5 I / Y 60 N CPT FV 16, 786.38803 STO 1 RCL 1  PV 0 PMT 48 N CPT FV 20, 090.19 12. (a) After three years: PMT  560; i  1.3%; n  12

1.01312  1  FVn  560    560(12.89629)  $7221.92267  0.013  Amount seven years later:

PV  7221.92267; i  1.3%; n  28 FV  7221.92267 1.013  7221.92267(1.435703)  $10,368.53 28

(b) Contribution  12(560)  $6720 (c) Interest  10,368.53  6720  $3648.53

(Set P / Y, C / Y  4) 0 PV 560  PMT 5.2 I / Y 12 N CPT FV 7221.92267 STO 1 RCL 1  PV 0 PMT 28 N CPT FV 10,368.53

13.

After two years: PMT  92; i  0.33583%; n  24

1.003358324  1  FV1  92    92 (24.950135)  2295.41  0.0033583 

Two years later: PV  2295.41; i  0.4416%; n  24



FV2  2295.41 1.004416

  2295.41(1.1115624)  $2551.49 24

(Set P / Y, C / Y  12) 0 PV 92  PMT 4.03 I / Y 24 N CPT FV 2295.41 STO 1 (Set P/Y, C / Y  12) RCL 1  PV 0 PMT 5.3 I / Y 24 N CPT FV 2551.49

14. After six years: PMT  300; i  2.5%; n  12 1.02512  1  FVn  300    300(13.795553)  $4138.665891 0.025  

Ten years later: PV  4138.665891; i  1.5%; n  40 FV  4138.665891(1.015) 40  4138.665891(1.814018)  $7507.62

(Set P / Y, C / Y  2) 0 PV 300  PMT 5 I / Y 12 N CPT FV 4138.665891 STO 1 (Set P / Y, C / Y  4) RCL 1  PV 0 PMT 6 I/Y 40 N CPT FV 7507.62

15.

 1 0.0175  1  12   = 200(131.036137) = $26,207.23 with $200 payments for 10 FV = 200   0.0175   12   years 120

 1 0.0175   1  12    200 (62.655564)  $12,531.11 FV1  200   0.0175   12   60

 0.175  FV2  12,531.111    12,531.11(1.091373)  $13, 676.11 12   60  1 0.0175  1  12   FV3  400  0.0175   400 (62.655564)  $25, 062.23   12   Total amount in savings account  $13, 676.11  25, 062.23  $38, 738.34 Mariana has $38,738.34 in her savings account after 10 years. Doubling the payment doubled the amount contributed for the last 5 years and increased her account by an extra $12,531.11.

16. (a)

 (1  0.10)15 1  FV = 2400    2400 (31.772482)  $76, 253.96 0.10  

(b)

 1 0.10   1  4    600 (135.99159)  $81,594.95 FV = 600   0.10   4  

(c)

 1 0.10   1  FV = 200   120.10   200 (414.470346)  $82,894.07   12  

180

40  1 0.035   1   0.035  4   17. FV = 10, 000 1    1200  0.035  4     4    10, 000 (1.416909)  1200 (47.646724)  14,169.09  57,176.07 40

 $71,345.16 Interest  71,345.16  (10, 000  1200 (40))  71,345.16  58,000  $13,345.16 18.

After the first two years: PMT = 25; i =

0.04 & n = 24 = 0.003; 12

 (1.003) 24  1  FV1  25    25(24.942888)  $623.572194 0.003  

Four years later: n = 48 &48 = 623.572194 (1.173199) FV2 = 623.572194 (1.003) = $731.57

Between years 2 and 6: PMT = $200; n = 48  (1.003) 48  1  FV3  200    200 (51.959601)  $10,391.92 0.003  

Amount in the account after 6 years: $731.57 + 10,391.92 = $11,123.49

Total contributions = $25(24) + 200(48) = 600 + 9600 = $10, 200 Interest = $11,123.49 - 10, 200 = $923.49

19.

After the first five years: PMT = 1500; i =

0.036 = 0.009 4

 (1.009) 20  1  FV1  1500    1500 (21.805976)  $32, 708.96409  0.009 

Two years later: FV2 = 32, 708.96409(1.009)8 = 32, 708.96409(1.074309) = $35,139.54

Between years 5 and 7:  (1.009)8  1  FV3  1700    1700 (8.256587)  $14, 036.20  0.009 

Total amount in the account after 7 years:

$35,139.54 + 14,036.20 = $49,175.74 Total amount one year after the last deposit = 49,175.74 (1.009)4 = $50,970.11

Exercise 11.3 1.

PMT  375; i  3.5%; n  30 1  1.03530  PVn  375    375(18.392045)  $6897.02  0.035 

(Set P / Y, C / Y  2) 0 FV 375  PMT 7 I / Y 30 N CPT PV 6897.02 2.

PMT  60; i  0.375%; n  108 1  1.00375108  PVn  60    60(88.671407)  $5320.28  0.00375 

(Set P / Y, C / Y  12) 0 FV 60  PMT 4.5 I / Y 108 N CPT PV 5320.28

(a) PMT  600; i  1.9%; n  20

1  1.01920  PVn  600    600(16.510333)  $9906.20  0.019  (b) Interest  600(20)  9906.20  12, 000  9906.20  $2093.80

(Set P / Y, C / Y  4) 0 FV 600  PMT 7.6 I / Y 20 N CPT PV 9906.20 4.

(a) PMT  252.17; i  0.7%; n  36

1  1.007 36  PVn  252.17    252.17(31.724659)  $8000.01  0.007  (b) Interest  36(252.17)  8000.01  9078.12  8000.01  $1078.11

(Set P / Y, C / Y  12) 0 FV 252.17  PMT 8.4 I / Y 36 N CPT PV 8000.01

(a) PMT  69.33; i  0.9%; n  36

1  1.00936  PVn  69.33    69.33(30.633420)  $2123.82  0.009  Cash price  400  2123.82  $2523.82 (b) Cost of financing  36(69.33)  2123.82  2495.88  2123.82  $372.06

(Set P / Y, C / Y  12) 0 FV 69.33  PMT 10.8 I / Y 36 N CPT PV 2123.82

(a) PMT  458; i  0.76%; n  60

1  1.007660  PV  458    458(47.948854)  $21,960.58  0.0076 

Purchase price  2000  21,960.58  $23,960.58

(Set P / Y, C / Y  12) 0 FV 458  PMT 9.2 I / Y 60 N CPT PV 21,960.58 (b) Interest  60(458)  21,960.58  27, 480  21,960.58  $5519.42 7.

(a) PMT  343; i  0.3483%; n  96

1  1.00348396  PV  343    343(81.478437)  $27,947.10 0.003483  

(Set P / Y, C / Y  12) 0 FV 343  PMT 4.18 I / Y 96 N CPT PV 27,947.10 (b) Interest  96(343)  27,947.10  32,928  27,947.10  $4980.90 8.

(a) PMT  135; i  0.73%; n  60

1  1.007360  PV  135    135(48.399376)  $6533.92  0.0073  The cash price was $6533.92

(Set P / Y, C / Y  12) 0 FV 135  PMT 8.8 I / Y 60 N CPT PV 6533.92 (b) Cost of financing  60(135)  6533.92  8100  6533.92  $1566.08 9.

Wayne’s price: PMT  74.; i  0.7%; n  24

1  1.007 24  PV  74    74(22.021609)  $1629.60  0.007  (Set P / Y, C / Y  12) 0 FV 74  PMT 8.4 I / Y 24 N CPT PV 1629.60 Roberto’s price: PMT  244; i  1.6%, n  8 1  1.0168  PV  244    244(7.453418)  $1818.63  0.016  Wayne paid less.

(Set P / Y, C / Y  12) 0 FV 74  PMT 8.4 I / Y 24 N CPT PV 1629.60 (Set P / Y, C / Y  4) 0 FV 244  PMT 6.4 I / Y 60 N CPT PV 1818.63

10.

PMT  97; i  1.075%; n  36 1  1.0107536  PV  97    97(29.721386)  $2882.97  0.01075  The cash price is $2882.97

(Set P / Y, C / Y  12) 0 FV 97  PMT 12.9 I / Y 60 N CPT PV 2882.97 11. (a) PMT  234.60; i  0.6%; n  72

1  1.00672  PVn  234.60    234.60(58.325343)  $13, 683.13  0.006 

(Set P / Y, C / Y  12) 0 FV 234.60  PMT 7.2 I / Y 72 N CPT PV 13,683.13 (b) PMT  234.60; i  0.6%; n  6

1.0066  1 FVn  234.60    234.60(6.090723)  $1428.88  0.006 

0 PV 234.60  PMT 6 N CPT FV 1428.88 (c) Payout amount  Balance owing  After six months

 13, 683.13 (1  0.006)6  13, 683.13 (1.006)6  13, 683.13 (1.036544)  $14,183.17

(Set P / Y, C / Y  12) 13,683.13 FV 0  PMT 7.2 I / Y 6 N CPT FV 14,183.17 (d) Interest  14,183.17  13, 683.13  $500.04 (e) Payment required after six months Normal payment  6  234.60

Additional interest

$1428.88 1407.60 $21.28

12. (a) PMT  642.79; i  0.75%; n  36

1  1.007536  PVn  642.79    642.79(31.446805)  $20, 213.69 0.0075  

(Set P / Y, C / Y  12) 0 FV 642.79  PMT 9 I / Y 36 N CPT PV 20, 213.69 (b) PMT  642.79; i  0.75%; n  9

1.00759  1  FVn  642.79    642.79(9.274779)  $5961.73 0.0075  

(Set P / Y, C / Y  12) 0 PV 642.79  PMT 9 I / Y 9 N CPT FV 5961.73

 20, 213.69 (1  0.0075)9  20, 213.69 (1.0075)9  20, 213.69 (1.069561)  $21, 619.77

(Set P / Y, C / Y  12) 20, 213.69 PV 0  PMT 9 I / Y 9 N CPT FV 21,619.77 (d) Interest  21, 619.77  20, 213.69  $1406.08 (e) Payment needed after nine months

Normal payments  9  642.79

$5961.73 5785.11

Additional interest

$176.62

1  1.0225 40  13. PVn = 2500   = 2500(26.193522) = $65,483.81  0.0225 

Interest = 2500 (40) – 65,483.81 = 100,000 – 65483.81 = $34,516.19 1  1.0125 40  14. (a) PVn = 1000    0.0125 

= 1000 (31.326933) = $31,326.93 (b) Payout = 40(1000) = $40,000 (c) Interest = 40,000 – 31,326.93= $8673.07

1  1.0025120  15. PVn = 5000   = 5000(103,561753) = $517,808.77  0.0025 

1  1.0025180  PVn = 7000   = 7000(144.805471) = $1,013,638.30  0.0025 

PV = 1,013,638.30 (1.0025)–120 = $751,202.90 $517,808.77 + $751,202.90 = $1,269,011.67

120  11 0.065   0.065   12   PV  200  0.065   2000 1   12     12    200 (88.0685)  2000 (0.522962)  17, 613.70  1045.92 120

16.

 $18, 659.62 17.

1 1 0.055  12  PVLOAN  350   0.055  12 

 48

   350 (42.998777)  $15, 049.57  

Balance of the loan after 2 years: Accumulated value of original principal after 2 years 24

æ 0.055 ö ÷ FV1 = 15, 049.57 çç1 + ÷ ÷ = 15, 049.57 (1.115998) = $16, 795.28 çè 12 ø Accumulated value of the first 24 (2 years) payments

 1 0.055   1  12    350 (25.30856)  $8858 FV2  350   0.055   12   24

Outstanding balance of the loan contract after 2 years $16, 795.28 - 8858 = $7937.28

(a) $7937.28

æ ö (b) FV3 = 7937.28çç1 + 0.055 ÷ ÷ ÷ = 7937.28(1.115998) = $8857.99 çè 12 ø - 24

æ 0.05 ö ÷ PV2 = 8857.99 çç1 + ÷ ÷ çè 12 ø

= 8857.99 (0.905025) = $8016.71

- 24

æ 0.06 ÷ ö (c) PV3 = 8857.99 çç1 + ÷ ÷ çè 12 ø (d)

= 8857.99(0.887186) = $7858.68

Selling price of a loan contract decreases if the interest rate on the contract increases. (A higher interest rate means you’d have to set less aside today in order to earn a specified future amount).

18.

For the first 5 years: PMT = 360; i =

0.0498 = 0.00415; n = 60 12

1  (1.00415) 60  PV1  360    360 (53.016444)  $19, 085.92 0.00415  

From years 5 to 7 : PMT = 425; n = 24 1  (1.00415) 24  PV2  425    425(22.798553)  $9689.384906 0.00415  

Present value of $425 payments:

PV3 = 9689.384906(1.00415)- 60 = 9689.384906(0.779982) = $7557.54

Present value of total payments = $19, 085.92 + 7557.54 = $26, 643.46 Total payments = $360(60) + 425(24) = 21, 600 + 10, 200 = $31,800 Interest paid = $31,800 - 26, 643.46 = $5156.54 19.

(a)

For the first 10 years : PMT = 300; i =

0.035 = 0.000673; n = 520 52

1  (1.000673) 520  PV1  300    300 (438.625851)  $131,587.76 0.000673  

For the next 7 years : PMT = 415; n = 364

1  (1.000673) 364  PV2  415    415(322.743129)  $133,938.3987 0.000673  

Present value of the $415 payments: - 520

PV3 = 133,938.3987 (1.000673)

= 133,938.3987 (0.704771) = $94,395.91

Purchase price = $50, 000 + 131, 587.76 + 94, 395.91 = $275,983.67 (b)

Interest = Total payments - Purchase price = ($300 (520) + 425(364)) - 275, 983.67 = 310, 700 - 275,983.67 = $34, 716.33

20.

For the first 5 years : PMT = 500; i =

0.04 = 0.01; n = 20 4

1  (1.01)20  PV1  500    500(18.045553)  $9022.78 0.01  

For the next 5 years: i =

0.0375 = 0.009375 4

1  (1.009375)20  PV2  500    500(18.159621)  $9079.810604 0.009375  

Present value of payments from year 5 to 10:

PV3 = 9079.810604(1.01)- 20 = 9079.810604(0.819544) = $7441.31 Purchase price of the annuity = $9022.78 + 7441.31 = $16, 464.09

21.

Present value of the annuity: PMT = 2500; i = when he starts college

0.03 = 0.015; n = 6 PMTS 2

1  (1.015) 6  PV1  2500    2500 (5.697187)  $14, 242.96791  0.015  0.0285 For the first 7 years: i   0.01425 2 Deposit required today  14, 242.96791(1.015) 16 (1.01425) 14  14, 242.96791(0.788031)(0.820294)  $9206.90

Exercise 11.4 1.

FVn  20, 000; i  1.5%; n  60  20, 000  0.015  PMT    60  (1.015)  1   300  PMT   1.443220  PMT  $207.87

(Set P / Y, C / Y  4) 0 PV 20,000 FV 6 I / Y 60 N CPT PMT  207.87

PVn  8000; i  0.7%; n  60  8000  0.007  PMT   60  1  (1.007)  56 PMT  0.341991 PMT  $163.75

(Set P / Y, C / Y  12) 0 FV 8000  PV 8.4 I / Y 60 N CPT PMT 163.75 3.

PVn  7500; i  4.8%; n  20  7500  0.048  PMT   20  1  (1.048)  360 PMT  0.608462 PMT  $591.66

(Set P / Y, C / Y  2) 0 FV 7500  PV 9.6 I / Y 20 N CPT PMT 591.66 4.

PVn  320, 000; i  0.5416%; n  144  320, 000  0.005416  PMT   144   1  (1.005416)  1733.3 0.540628 PMT  $3206.15 PMT 

(Set P / Y, C / Y  12) 0 FV 320,000  PV 6.5 I / Y 144 N CPT PMT 3206.15

FVn  11, 000; i  1.0%; n  36 11, 000  0.01  PMT    36  (1.01)  1  110 PMT  0.430769 PMT  $255.36

(Set P / Y, C / Y  4) 0 PV 11,000 FV 4 I / Y 36 N CPT PMT  255.36

FVn  18, 000; i  2.5%; n  30 18, 000  0.025  PMT    30  (1.025)  1  450 PMT  1.097568 PMT  $410

(Set P / Y, C / Y  2) 0 PV 18,000 FV 5 I / Y 30 N CPT PMT  410 7.

FVn  3500; i  0.9375%; n  28  3500  0.009375  PMT    28  (1.009375)  1  32.8125 PMT  0.298588 PMT  $109.89

(Set P / Y, C / Y  4) 0 PV 3500 FV 3.75 I / Y 28 N CPT PMT 109.89 8.

FVn  10, 000; i  1.5%; n  16 10, 000  0.015  PMT    16  (1.015)  1  150 PMT  0.268986 PMT  $557.65

(Set P / Y, C / Y  2) 0 PV 10,000 FV 3 I / Y 16 N CPT PMT  557.65 9.

PVn  7200; i  1.625%; n  12  7200  0.01625  PMT  PMT  12  1  (1.01625)  117 PMT  0.175875 PMT  $665.25

(Set P / Y, C / Y  4) 0 FV 7200  PV 6.5 I / Y 12 N CPT PMT 665.25

10.

Amount financed: $15, 300  (0.15) 15,300  $15, 300  2295  $13, 005 PVn  13, 005; i  0.6%; n  48 13, 005  0.006  PMT   48   1  (1.006)  86.70 PMT  0.273079 PMT  $317.49

(Set P / Y, C / Y  12) 0 FV 13,005  PV 8 I / Y 48 N CPT PMT 317.49 11.

Amount financed: $16, 500  2000  $14, 500 PVn  14, 500; i  0.625%; n  60 14,500  0.00625  PMT   60   1  (1.00625)  90.625 PMT  0.311908 PMT  $290.55

(Set P / Y, C / Y  12) 0 FV 14,500  PV 7.5 I / Y 60 N CPT PMT 290.55 12.

Amount financed: $10, 600  (0.1)10, 600  $10, 600  1060  $9540 PVn  9540; i  0.6%; n  48  9540  0.006  PMT   48  1  (1.006)  57.24 PMT  0.249593 PMT  $229.33

(Set P / Y, C / Y  12) 0 FV 9540  PV 7.2 I / Y 48 N CPT PMT 229.33 13.

Amount financed: $250,000  (0.20)250,000  $250,000  50, 000  $200,000 PVn  200,000; i  2.5%; n  100  200, 000  0.025  PMT  PMT  100   1  (1.025)  5000 PMT  0.915353 PMT  $5462.38

(Set P / Y, C / Y  4) 0 FV 200,000  PV 10 I / Y 100 N CPT PMT 5462.38

14.

Amount financed: $949.99  75.12  $874.87 PVn  874.87; i  0.9%; n  15  874.87  0.009  PMT   15   1  (1.009)  7.873830 PMT  0.125756 PMT  $62.61

(Set P / Y, C / Y  12) 0 FV 874.87  PV 10.8 I / Y 15 N CPT PMT 62.61 15. Amount after 15 years: PMT  1200; i  3.75%; n  30

1.037530  1  FVn  1200    0.0375   1200(53.799237)  $64,559.08

(Set P / Y, C / Y  2) 0 PV 1200  PMT 7.5 I / Y 30 N CPT FV 64,559.08 RRIF withdrawals: PVn  64,559.08; i  3.75%; n  40

 64,559.08  0.0375  PMT    40  1  (1.0375)  2420.9655 PMT  0.770662 PMT  $3141.41

0 FV 64,559.08  PV 40 N CPT PMT 3141.41

16. Amount after 21 years: PMT  60; i  1.1875%; n  84

1.01187584  1  FVn  60    0.011875   60(142.788050)  $8567.28

(Set P / Y, C / Y  4) 0 PV 60  PMT 4.75 I / Y 84 N CPT FV 8567.28 Withdrawals: PVn  8567.28; i  1.1875%; n  16

 8567.28  0.011875  PMT   16   1  (1.011875)  101.736450 PMT  0.172115 PMT  $591.10

0 FV 8567.28  PV 16 N CPT PMT 591.10 17. Amount after 18 years:

PMT  200; i  0.75%; n  72 1.007572  1 FV  200    200(95.007028)  $19, 001.41  0.0075 

(Set P / Y, C / Y  4) 0 PV 200  PMT 3 I / Y 72 N CPT FV 19,001.41 Monthly payments received: PV  19, 001.41; i  0.325%; n  48

19, 001.41 0.00325  PMT    48  1  (1.00325)  61.754583 PMT  0.144224 PMT  428.18 PMT  $428.18

(Set P / Y, C / Y  12) 0 FV 19,001.41  PV 3.9 I / Y 48 N CPT PMT 428.18

18. Amount after five years: PV  350; i  0.8%; n  20 1.00820  1  FV  350    350(21.595505)  $7558.43  0.008 

(Set P / Y, C / Y  4) 0 PV 350  PMT 3.2 I / Y 20 N CPT FV 7558.43 Size of each withdrawal:

PV  7558.43; i  0.8%; n  36

 7558.43  0.008  PMT   36   1  (1.008)  60.467440 PMT  0.249379 PMT  242.47 PMT  $242.47

0 FV 7558.43  PV 36 N CPT PMT 242.47 19. Amount after 14 years:

PV  1500; i  1.4%; n  28 1.01428  1  FV  1500    1500(33.994267)  $50,991.40  0.014 

(Set P / Y, C / Y  2) 0 PV 1500  PMT 2.8 I / Y 28 N CPT FV 50,991.40 Size of each withdrawal: PV  50,991.40; i  1.4%; n  40

 50,991.40  0.014  PMT    40  1  (1.014)  713.8796 PMT  0.426568 PMT  1673.54 PMT  $1673.54

0 FV 50,991.40  PV 40 N CPT PMT 1673.54

20. Amount after eight years:

PV  10, 000; i  4.15%; n  8 1.04158  1  FV  10, 000    10, 000(9.263619)  $92, 636.19  0.0415 

(Set P / Y, C / Y  1) 0 PV 10,000  PMT 4.15 I / Y 8 N CPT FV 92,636.19

Size of each withdrawal: PV  92,636.19; i  4.15%; n  10

 92,636.19  0.0415  PMT    10  1  (1.0415)  3844.401885 PMT  0.334103 PMT  11,506.65 PMT  $11,506.65

0 FV 92,636.19  PV 10 N CPT PMT 11,506.65

Exercise 11.5 1.

FVn  4500; PMT  50; i  0.50%

1.005n  1  4500  50    0.005   0.005 ln ( 4500 50 )  1 n ln(1.005) ln1.45 n ln1.005 0.371564 n 0.004987 n  74.498339 months n  75 months

(Set P / Y, C / Y  12) 0 PV 4500 FV 50  PMT 6 I / Y CPT N 74.498339

FVn  5000; PMT  60; i  0.50%

1.005n  1  5000  60    0.005   0.005 ln[( 5000 60 )  1] n ln(1.005) 0.348307 n 0.004988 n  69.835347 months n  70 months

(Set P / Y, C / Y  12) 0 PV 5000 FV 60  PMT 6 I / Y CPT N 69.835347 3.

FVn  20, 000; PMT  0; i  3.25%; n  8 PV  20, 000(1.0325)8  15, 484.9395 FVn  15, 484.9395; PMT  646.56; i  3.25% 1.0325n  1 15, 484.9395  646.56    0.0325   .0325 ln[( 15,484.9395 ) 1 646.56 n ln(1.0325) ln(1.778366) n ln(1.0325) 0.575695 n 0.31983 n  18.000010 semi-annual periods The number of deposits is 18.

(Set P / Y, C / Y  2) 20, 000 FV 0 PMT 6.5 I / Y 8 N CPT PV 15, 484.9395 0 PV 15, 484.9395 FV 646.56  PMT CPT N 18.000010

PMT  1000; FV  36, 000; i  0.53%  0.0053

1.0053n  1  36, 000  1000    0.0053   0.0053  ln[( 36,0001000 )  1 n  ln(1.0053)   ln(1.192) n ln(1.0053) 0.175633 n 0.005319 n  33.018845 It will have to be deposited for 2 years, 10 months.

(Set P / Y, C / Y  12) 0 PV 1000  PMT 36,000 FV 6.4 I / Y CPT N 33.018845 5.

PMT  96; FV  3600; i  0.3%

1.003n  1  3600  96    0.003   0.003 ln[( 3600 96 )  1] n ln(1.003) ln(1.1125) n ln(1.003) 0.106610 n 0.002996 n  35.584918 It will take 3 years.

(Set P / Y, C / Y  12) 0 PV 96  PMT 3600 FV 3.6 I / Y CPT N 35.584918 6.

PMT  400; FV  5000; i  0.425%

1.00425n  1  5000  400    0.00425  0.00425 ln[( 5000 400 )  1] n ln(1.00425) ln(1.053125) n ln(1.00425) 0.051762 n 0.004241 n  12.205141 (quarters) It will take 3 years, 3 months.

(Set P / Y, C / Y  4) 0 PV 400  PMT 5000 FV 1.7 I / Y CPT N 12.205141 7.

PVn  8000; PMT  300; i  0.3% 1.1003 n  8000  300    0.003  n

 0.003 ln[1  ( 8000300 )]  ln(1.003)

ln(0.91)  ln(1.003) 0.093090 n 0.003328 n  27.973646 months n

It will take 2 years, 4 months.

(Set P / Y, C / Y  12) 0 FV 800  PV 300 PMT 4 I / Y CPT N 27.973646 8.

PVn  265, 000; PMT  15, 600; i  3.5% 1  1.035 n  265, 000  15, 600    0.035  n

 0.035 ln[1  ( 265,000 )] 15,600

 ln(1.035) ln(0.405449) n  ln(1.035) 0.902761 n 0.034401 n  26.241960 semi-annual periods The term of the mortgage is 13 years, 6 months.

(Set P / Y, C / Y  2) 0 FV 265,000  PV 15,600 PMT 7 I / Y CPT N 26.241960

PVn  12,000; PMT  292.96; i  0.6%

1  1.006 n  12, 000  292.96    0.006   0.006 ln[1  ( 12,000 )] 292.96 n  ln(1.006) ln(0.726925) n  ln(1.006) 0.318932 n 0.006645 n  47.999048 months

The term of the loan is 4 years.

(Set P / Y, C / Y  12)0 FV 12,000  PV 292.96 PMT 8 I / Y CPT N 47.999048 10.

PVn  4000; PMT  500; i  1.0% 1  1.01 n  4000  500    0.01   0.01 ln[1  ( 4000500 )] n  ln(1.01) ln(0.92) n  ln(1.01) 0.083382 n 0.009950 n  8.379782 quarters

It will take 2 years, 3 months.

(Set P / Y, C / Y  4) 0 FV 4000  PV 500 PMT 4 I / Y CPT N 8.379782 11.

PVn  12, 000; PMT  1100; i  1.95%

(Set P / Y, C / Y  2) 0 FV 12,000  PV 1100 PMT 3.9 I / Y CPT N 12.384875

Payments can be withdrawn for 6 years, 6 months. 12.

PVn  7000; PMT  800; i  0.3816% (Set P / Y, C / Y  12) 0 FV 7000  PV 800 PMT 4.58 I / Y CPT N 8.916416

It will take 9 months. 13.

PVn  5741; PMT  650; i  0.42916% (Set P / Y, C / Y  12) 0 FV 5741  PV 650 PMT 5.15 I / Y CPT N 9.023365

The money will last 10 months. 14.

PVn  6000; PMT  730; i  0.975%

(Set P / Y, C / Y  4) 0 FV 6000  PV 730 PMT 3.9 I / Y CPT N 8.608930 Payments can be withdrawn for 2 years, 3 months.

15.

  22,000 0.048  ln 1   195 26       n  0.048  ln 1  26  ln (0.791716)  ln (1.001846) 0.233553  0.001844  126.624381 

 127 payments nfinal  126.624381  126  0.624381  1 1 0.048  26  PV  195   0.048  26   195(0.623446)  121.572021

0.624381

   

æ 0.048 ÷ ö = $121.80 Final payment is 121.572021çç1 + ÷ ÷ çè 26 ø

16.

 1,000,000 0.04   ln  1440 12  1     n  .04 ln 1  12  ln (3.314815) ln (1.003) 1.198402  0.003328  360.119396 

 361months

(30 years, 1 month)

Accumulated value of 360 payments:  (1.003)360  1  FV1  1440    $999, 431.1421  0.003 

Accumulated value at the end of 361 months: & = $1, 002, 762.58 FV2 = 999, 431.1421(1.003)

Therefore, you only need to contribute 360 payments, not 361, because the balance in the account after 360 payments will earn enough interest in the last month to exceed $1,000,000. Your last contribution is $1440. Business Math News Box

1  1  0.211   12   $5000  PMT   0.211   12   5000  PMT  26.506616  36

PMT  $188.63

INTEREST = (36 ´ $188.63) - 5000 = 6790.68 - 5000 = $1790.68

Calculating the number of monthly payments:

n

 5000 0.211  ln 1 50012     



 ln 1 0.211 12





ln (0.824167) 0.193383   ln (1.017583) 0.017431  11.094467  12 payments

You will make 11 payments of $500, and a smaller 12th payment. Use the PV annuity formula to calculate the PV of the 11 full monthly payments:

1  1  0.211 11  12   500(9.922756)  $4961.38 PV  500  0.211   12 Therefore, the PV of the 12th smaller payment is:

$5000 - 4961.38 = $38.62 Substitute $38.62 into the PV lump sum formula with n = 12 - 12

æ 0.211ö ÷ $38.62 = FV çç1 + ÷ çè ø 12 ÷ 38.62 = FV (0.811260) FV = $47.60

The smaller 12th payment is $47.60.

INTEREST   (11  $500)  47.60  5000  $547.60 Difference in interest  $1790.68  547.60  $1243.08 You would pay $1243.08 less interest if you increased your monthly payment to $500. 4.

It would take you 12 months (1 year) to pay off the debt.

Exercise 11.6 1.

FVn  12, 239.76; PMT  350; n  24

(Set P / Y, C / Y  4) 0 PV 12, 239.76 FV 350  PMT 24 N CPT I / Y 12.5000

Nominal rate is 12.5000% compounded quarterly. 2.

FVn  33, 600; PMT  2500; n  10

(Set P / Y, C / Y  1) 0 PV 33,600 FV 2500  PMT 10 N CPT I / Y 6.4142 Nominal rate is 6.4142% compounded annually. 3.

FVn  35, 000; PMT  250; n  120

(Set P / Y, C / Y  12) 0 PV 35,000 FV 250  PMT 120 N CPT I / Y 3.0356

Nominal rate is 3.0356% compounded monthly. 4.

FVn  45, 000; PMT  400; n  96

(Set P / Y, C / Y  12) 0 PV 45,000 FV 400  PMT 96 N CPT I / Y 3.9106 Nominal rate is 3.9106% compounded monthly. 5.

PVn  6000; PMT  144.23; n  48

(Set P / Y, C / Y  12) 0 FV 6000  PV 144.23 PMT 48 N CPT I / Y 7.1983

Nominal rate is 7.1983% compounded monthly. 6.

PVn  119,875.67; PMT  1800; n  108

(Set P / Y, C / Y  12) 0 FV 119,875.67  PV 1800 PMT 108 N CPT I / Y 11.7000

Nominal rate is 11.7000% compounded monthly. 7.

PVn  21,500; PMT  1000; n  28

(Set P / Y, C / Y  4) 0 FV 21,500  PV 1000 PMT 28 N CPT I / Y 7.6850 Nominal rate is 7.6850% compounded quarterly.

Amount financed: $11, 400  (0.10)11, 400  11, 400  1140  $10, 260 PVn  10, 260; PMT  286.21; n  42

(Set P / Y, C / Y  12) 0 FV 10, 260  PV 286.21 PMT 42 N CPT I / Y 9.1089

Nominal rate is 9.1089% compounded monthly. 9.

Amount financed: $50,000  (0.20)50,000  50,000  10,000  $40,000 PVn  40,000; PMT  1000; n  100

(Set P / Y, C / Y  4) 0 FV 40,000  PV 1000 PMT 100 N CPT I / Y 8.8899

Nominal rate is 8.8899% compounded quarterly. 10.

Amount financed: $35,000  (0.10)35,000  35,000  3500  $31,500 PVn  31,500; PMT  2100; n  24

(Set P / Y, C / Y  2) 0 FV 31,500  PV 2100 PMT 24 N CPT I / Y 8.3203 Nominal rate is 8.3203% compounded semi-annually. 11. Note: The calculations for this question were done using Excel's RATE function. Answer: Nper 18 PMT –725 PV 9000 Result 0.042421 f = (1 + 0.042421)2 – 1 = 8.664154% 12. Answer: PV = $300,000; PMT = $20,000; i = 4.25% / 4 = 1.0625% n = ln [1 – (PV i/PMT)] / ln(1 + i) n = ln [1 – (300,000  0.010625 / 20,000)] / – ln(1.010625) n = ln (0.840625) / – ln(1.010625) n = – 0.173610 / – 0.010569

n = 16.426380 (quarters) n = 17 quarters n = 4 years 3 months 13. Answer: PMT = 5000(0.035) = $175 PP = 5000(1 + 0.025)–30 + 175[1 – (1 + 0.025)–30 / 0.025 ] PP = 2383.71 + 3662.80 PP = $6046.51 Review Exercise 1.

(a) PMT = 360; i  1.75%; n  48

1.017548  1  FVn  360    360(74.262784)  $26, 734.60  0.0175  (b) Amount deposited  48(360)  $17, 280 (c) Interest = 26, 734.60 - 17, 280 = $9454.60

(Set P / Y, C / Y  4) 0 PV 360  PMT 7 I / Y 48 N CPT FV 26,734.60 2.

(a) PMT  1000; i  0.325%; n  114

1  1.00325114  PV  1000    1000(95.137251)  $95,137.25  0.00325 

(Set P / Y, C / Y  12) 0 FV 1000  PMT 3.9 I / Y 114 N CPT PV 95,137.25 (b) Amount paid out  1000(114)  $114,000 (c) Interest earned  114,000  95,173.25  $18,862.75 3.

FVn  10, 000; PMT  300; i  2.25%

1.0225n  1 10, 000  300    0.0225  0.0225 ln[( 10,000300 )  1] n 0.0225 ln(1.75) n ln(1.0225) 0.559616 n 0.022251 n  25.150583 semi-annual periods n  13 years

(Set P / Y, C / Y  2) 0 PV 300  PMT 10,000 FV 4.5 I / Y CPT N 25.150583

FV  10,000; i  2.4%; n  5 1.0245  1  10, 000  PMT    0.024 

10, 000  0.024  PMT    5  (1.024)  1  PMT  $1906.28

(Set P / Y, C / Y  2) 0 PV 5 N 10,000 FV 4.8 I / Y CPT PMT 1906.28 5.

PV  $30,000; i  0.5175; n  144

 30, 000  0.005175  PMT   144   1  (1.005175)  155.25 PMT  0.524448 PMT  $296.03

(Set P / Y, C / Y  12) 30,000  PV 144 N 0 FV 6.21 I / Y CPT PMT 296.025703

At the end of four years: PMT  1500; i  2.13%; n  8 1.02138  1  FV  1500    1500(8.622495)  $12,933.74  0.0213 

Six years after last payment: P  12,933.74; i  2.13%; n  12

FV  12,933.74(1.0213)12  12,933.74(1.287775)  $16,655.75

(Set P / Y, C / Y  2) 0 PV 1500  PMT 4.26 I / Y 8 N CPT FV 12,933.74 0 PMT 12,933.74  PV 12 N CPT FV 16, 655.75 7.

PMT  124; i  0.75%; n  30 1  1.007530  PVn  124    124(26.775080)  $3320.11  0.0075 

The purchase price was 300  3320.11  $3620.11

(Set P / Y, C / Y  12) 0 FV 124  PMT 9 I / Y 30 N CPT PV 3320.11

PVn  11,500; PMT  1450; i  5.25% 1  1.0525 n  11,500  1450    0.0525   0.0525 ln[1  ( 11,5001450 )]  ln(1.0525) ln(0.583621) n  ln(1.0525) 0.538504 n 0.051168 n  10.524175 half-years

n

The term of the contract is 5 years, 6 months.

(Set P / Y, C / Y  2) 0 FV 11,500  PV 1450 PMT 10.5 I / Y CPT N 10.524175

FVn  106, 000; PMT  1100; n  60

(Set P / Y, C / Y  4) 0 PV 1100  PMT 106,000 FV 60 N CPT I / Y 6.0185 Nominal rate is 6.0185% compounded quarterly. 10.

PVn  5600; PMT  121.85; n  54

(Set P / Y, C / Y  12) 0 FV 5600  PV 121.85 PMT 54 N CPT I / Y 7.2505 Nominal rate is 7.2505% compounded monthly. 11. (a) Amount after ten years:

PMT  250; i  1%; n  40 1.0140  1 FVn  250    250(48.886373)  $12, 221.59334  $12, 221.59  0.01  (b) Amount three years later:

*Rounded

PV  12, 221.59334; i  0.416%; n  36 FV  12, 221.59334(1.00416)36  12, 221.59334(1.161472)  $14,195.04129  $14,195.04

*Rou

PV  14,195.04; i  0.416%; n  120 PMT 

14,195.04  0.00416 59.146   $150.56 120 1  (1.00416) 0.392839

(Set P / Y, C / Y  4) 0 PV 250  PMT 4 I / Y 40 N CPT FV 12, 221.59334 (Set P / Y, C / Y  12) 0 PMT 12, 221.59334  PV 5 I / Y 36 N CPT FV 14,195.04129 14,195.04129  PV 0 FV 120 N CPT PMT 150.56

12. Amount needed at retirement:

PMT  500; i  0.5%; n  240 1  1.005240  PVn  500    500(139.580772)  $69, 790.39  0.005  Amount needed 12 years earlier: FV  69, 790.39; i  0.5%; n  144 PV  69, 790.39(1.005144 )  69, 790.39(0.487626)  $34, 031.63

(Set P / Y, C / Y  12) 0 FV 500  PMT 6 I / Y 240 N CPT PV 69, 790.39 0 PMT 69, 790.39 FV 144 N CPT PV  34, 031.63 13. At the end of four years: PV  10, 000; i  1.75%; n  8 FV  10, 000(1.0175)8  10, 000(1.148882)  $11, 488.81783

Annuity term:

PV  11, 488.82; PMT  932; i  1.75% 1  1.0175 n  11, 488.82  932    0.0175  ln[1  ( 11,488.82932 0.0175 )]  ln(1.0175) ln(0.784276) n  ln(1.0175) 0.242994 n 0.017349 n  14.0065 (half-years) n

n  7 years.

(Set P / Y, C / Y  2) 0 PMT 10, 000  PV 3.5 I / Y 8 N CPT FV 11, 488.82 11, 488.82 PV 932  PMT 0 FV CPT N 14.0065 14. At the end of 13 years: PMT  125; n  156; i  0.39 1.0039156  1  FV  125    125(214.1836)  $26, 772.95  0.0039 

Interest rate on annuity: PV  26,772.95; PMT  890; C / Y  4; n  36

(Set P / Y, C / Y  4) 26, 772.95 PV 890  PMT 0 FV 36 N CPT I / Y 4.019675 The nominal annual rate is 4.0197% compounded quarterly. 15. (a) FV  7200; PMT  $30; n  168

(Set P / Y, C / Y  12) 0 PV 30  PMT 7200 FV 168 N CPT I / Y 4.860588

The nominal annual rate is 4.8606% compounded monthly. (b) PV  7200; PMT  135; i  0.405049

1  1.00405049 n  7200  135    0.00405049 

ln[1  ( 7200  0.00405049 )] 135  ln(1.00405049) ln(0.783974) n  ln(1.00405049) 0.243379 n 0.004042 n = 60.212519 (months) n

The term is 5 years, 1 month.

(Set P / Y, C / Y  12) 7200 PV 135  PMT 0 FV 4.860588 I / Y CPT N 60.2122519

Self-Test PMT  2400; i  1.375%; n  32 1.

1  1.0137532  PVn  2400    2400(25.747647)  $61, 794.35  0.01375  FV  61, 794.35; i  1.375%; n  40 PV  61, 794.35 1.01375 

40

 61, 794.35(0.579116)  $35, 786.08

(Set P / Y, C / Y  4) 0 FV 2400  PMT 5.5 I / Y 32 N CPT PV 61, 794.35 0 PMT 61, 794.35 FV 40 N CPT PV  35, 786.08 2.

PMT  4800; i  1.5%; n  24

1  1.01524  PVn  4800    4800(20.030405)  $96,145.95  0.015  FVn  96,145.95; i  0.625%; n  120  96,145.95  0.00625  PMT    120  (1.00625)  1  600.912188 PMT  1.112065 PMT  $540.36

(Set P / Y, C / Y  4) 0 FV 4800  PMT 6 I / Y 24 N CPT PV 96,145.95 (Set P / Y, C / Y  12) 0 PV 96,145.95 FV 7.5 I / Y 120 N CPT PMT 540.36 3.

PVn  48, 000; PMT  4000; n  20

(Set P / Y, C / Y  2) 0 FV 48,000  PV 4000 PMT 20 N CPT I / Y 10.9002 Nominal rate is 10.9002% compounded semi-annually. 4.

PVn  14, 400; PMT  600; i  2.625% 1  1.02625 n  14, 400  600    0.02625   0.02625 ln[1  ( 14,400600 )] n  ln(1.02625) ln(0.37) n  ln(1.02625) 0.994252 n 0.025911 n  38.371256 quarterly payments Copyright © 2025 Pearson Canada Inc.

There will be 39 payments but the 39th payment will be a smaller amount than the other payments.

(Set P / Y, C / Y  4) 0 FV 14, 400  PV 600 PMT 10.5 I / Y CPT N 38.371256 5.

PMT  574; i  0.5%; n  72

After 6 years: 1.00572  1  FV  574    574(86.408856)  $49,598.68  0.005 

Term of the annuity:

PV  49,598.68; PMT  3600; i  0.01475 1  1.01475 n  49,598.68  3600    0.01475   0.01475  ln 1  49,598.68 3600  n  ln(1.01475) ln(0.796783) n  ln(1.01475) 0.227173 n 0.014642 n  15.515156 quarters





It will take 47 months.

(Set P / Y, C / Y  12) 574  PMT 0 PV 6 I / Y 72 N CPT FV 49,598.68 (Set P / Y, C / Y  4) 49,598.68 PV 3600  PMT 0 FV 5.9 I / Y CPT N 15.515156 6.

PMT  450; i  1.0%; n  84 1  1.0184  PVn  450    450(56.648453)  $25, 491.80 0.01  

Amount paid  450  84  $37,800 Interest paid  37,800  25, 491.80  $12,308.20

(Set P / Y, C / Y  12) 0 FV 450  PMT 12 I / Y 84 N CPT PV 25, 491.80

Accumulated value of the deposits for first five years:

PMT  540; i  1.25%; n  20 1.012520  1  FVn  540    540(22.562979)  $12,184.00841  0.0125  PV  12,184.00841; i  1.375%; n  32 FV  12,184.00841(1.01375)32  12,184.00841(1.548060)  $18,861.5744 Accumulated value of the deposits for remaining eight years:

PMT  540; i  1.375%; n  32 1.0137532  1  FVn  540    540(39.858899)  $21,523.80557  0.01375  Balance  18,861.5744  21,523.80557  $40,385.38 (Set P / Y, C / Y  4) 0 PV 540  PMT 5 I / Y 20 N CPT FV 12,184.00841 0 PMT 12,184.00841  PV 5.5 I / Y 32 N CPT FV 18,861.5744 0 PV 540  PMT 32 N CPT FV 21,523.80557

PMT  3200; i  3.25%; n  8 1.03258  1 FVn  3200    3200(8.971617)  $28, 709.17  0.0325  Total paid  3200  8  $25, 600 Interest  28, 709.17  25, 600  $3109.17

(Set P / Y, C / Y  2) 0 PV 3200  PMT 6.5 I / Y 8 N CPT FV 28,709.17 9.

FVn  67, 200; i  3.25%; n  16

 67, 200  0.0325  PMT    16  (1.0325)  1  2184 PMT  0.668173 PMT  $3268.62

(Set P / Y, C / Y  2) 0 PV 67, 200 FV 16 N 6.5 I / Y CPT PMT  3268.62

Challenge Problems

Accumulated value of the deposits of $500 on December 1, 2027

1.01259  1  FVn  500  (1.01253 )(1.016253 )   0.0125   500(9.463374)(1.037971)(1.049547)  5154.69 Accumulated value of the withdrawals of $300 on December 1, 2027

1.01253  1  1.016253  1  3 FVn  300  (1.01625)  300   0.01625   0.0125     300(3.037656)(1.049546)  300(3.049014)  956.45  914.70  1871.15 Balance in the account on December 1, 2027  5154.69  1871.15  $3283.54

PMT = 400; n = 19(12) = 228; i =

= 0.5% = 0.005; I/Y = 6; P/Y = C/Y = 12

FV = 400[ (1+ 0.06/12)228 – 1 / 0.06/12 ] = $169,431.94 FV = 24,500(1 + 0.06/12)228 = $76,388.53 Total = 169,431.94 + 76,388.53 = $245,820.47 FV = 245,820.47(1 + 0.06/12)60 = $331,574.98 Case Study 1.

(a) Cash value of TV A:

PMT  230; i  0.012916; n  12 1  0.01291612  PVn  230    230(11.050406)  $2541.59  0.012916  Cash value  Down payment  PVn  500  2541.59  $3041.59 (b) Cash value of TV B:

PMT  260; i  0.012916; n  18 1  1.01291618  PVn  260    260(15.969248)  $4152  0.012916  Cash value = Down payment  PVn  100  4152  $4252 2.

(a) Overhead assigned to TV A  15% of $1950  $292.50 (b) Overhead assigned to TV B  15% of $2160  $324

(a) Profit on TV A  Cash value  (Overhead  Cost)  3041.59  (292.50  1950)  $799.09 Profit of TV A as a percent of cost 

799.09 (100%)  40.98% 1950

(b) Profit on TV B  Cash value  (Overhead  Cost)  4252  (324  2160)  $1768 Profit of TV B as a percent of cost 

1768 (100%)  81.85% 2160

(c) On the basis of the relatively higher profit on TV B, Suzanne should recommend that TV B be more heavily promoted. 4.

Since the special plans stay the same, the cash values of TV A and TV B remain at $3041.59 and $4252 respectively.

TV A: New cost = 91% of $1950 = $1774.50 New overhead  15% of $1774.50  $266.18 New profit  3041.59  (266.18  1774.50)  $1000.91

Profit on TV A as a percent of cost 

1000.91 (100%)  56.41% 1774.50

TV B: New cost = 94% of $2160  $2030.40

New overhead = 15% of $2030.40 = $304.56 New profit = 4252  (304.56 + 2030.40) = $1917.04

Profit on TV B as a percent of cost 

1917.04 (100%)  94.42% 2030.40

On the basis of the relatively higher profit on TV B, Suzanne should continue to recommend that TV B be more heavily promoted.

Chapter 12 Ordinary General Annuities Exercise 12.1 1.

& n  12; c  12  3 PMT  300; i  0.3416%; 4 p  1.003416&3  1  1.010285  1  1.0285% 1.01028512  1  FV  300    300(12.702634)  $3810.79  0.010285 

 Set P/Y  4; C/Y  12  0 PV 300  PMT 4.1 I/Y 12 N CPT FV 3810.79 2.

PMT  1500; i  1.5%; n  9; c 

4 2 2

p  1.0152  1  1.030225  1  3.0225% 1.0302259  1 FV  1500    1500(10.168425)  $15, 252.64  0.030225 

Set P/Y  2;C/Y  4  0 PV 1500  PMT 6 I/Y 9 N CPT FV 15, 252.64 3.

PMT  25; i  1.25%; n  120; c 

4  0.3& 12

p  1.01250.3  1  1.004149  1  0.414943% 1.004149120  1  FV  25    25(155.110514)  $3877.76  0.004149 

Set P/Y  12;C/Y  4  0 PV  25  PMT 120 N 5 I/Y CPT FV 3877.76 4.

PMT  270; i  1.55%; n  60; c 

4  0.3& 12

p  1.01550.3  1  1.0051402  1  0.51402% 1.005140260  1  FV  270    270(70.072529)  $18, 919.58  0.0051402  He will have saved $18, 949.58 and be over by 18, 919.59  18, 500  $419.58

Set P/Y  12;C/Y  4  0 PV 270  PMT 6.2 I/Y 60 N CPT FV 18, 919.58

(a)

PMT  1000; i  1.5%; n  10; c 

4 4 1

p  1.0154  1  1.061364  1  6.1364% 1.06136410  1 FV  1000    1000(13.265504)  $13, 265.50  0.061364 

Set P/Y  1; C/Y  4  0 PV 1000  PMT 6 I/Y 10 N CPT FV 13, 265.50

(b)

Interest  13, 265.50  10(1000)  13, 265.50  10, 000  $3265.50

(a)

& n  520; c  12  0.230769 PMT  20; i  0.3%; 52 p  1.003&0.230769  1  1.000768  1  0.076825% 1.000768520  1 FV  20    20(638.899983)  $12, 778  0.000768 

Set P/Y  52; C/Y  12  0 PV 20  PMT 4 I/Y 520 N CPT FV 12, 778 (b) 7.

Interest  12, 778  20  52 10  12, 778  10, 400  $2378

Amount after 10 years:

PMT  500; i  2.25%; n  40; c 

2 1  4 2

p  1.0225 2  1  1.011187  1  1.1187% 1.011187 40  1  FV  500    500(50.101736)  $25, 050.86786  0.011187  Amount five years later: PV  25, 050.86786; i  2.25%; n  10 FV  25, 050.86786 1.022510   25, 050.86786(1.249203)  $31, 293.63

Set P/Y  4;C/Y  2  0 PV 500  PMT 4.5 I/Y 40 N CPT FV 25, 050.86786 Set P/Y  2;C/Y  2  0 PMT 25, 050.86786  PV 4.5 I/Y 10 N CPT FV 31, 293.63 8.

Amount after 15 years:

12 6 2 p  1.00256  1  1.015094  1  1.5094%

PMT  950; i  0.25%; n  30; c 

1.01509430  1  FV  950    950(37.593040)  $35, 713.38846  0.015094 

Amount 10 years later: PV  35, 713.38846; i  0.25%; n  120 FV  35, 713.38846 1.0025120   35, 713.38846(1.349354)  $48,189.99

Set P/Y  2;C/Y  12  0 PV 950  PMT 3 I/Y 30 N CPT FV 35, 713.38846 Set P/Y  12;C/Y  12  0 PMT 35, 713.38846  PV 3 I/Y 120 N CPT FV 48,189.99 9.

PMT  30; i  0.4308%; n  208; c 

12 3  52 13

p  1.00430813  1  1.000993  1  0.0993% 1.000993208  1  FV  30    30(230.902037)  $6927.06  0.000993 

Set P/Y  52;C/Y  12  0 PV 30  PMT 5.17 I/Y 208 N CPT FV 6927.06 10.

PMT  18; i  1.0%; n  24; c 

4 1  12 3

p  1.013  1  1.003322  1  0.3322% 1.00332224  1  FV  18    18(24.939685)  $488.91  0.003322 

Set P/Y  12;C/Y  4  0 PV 18  PMT 4 I/Y 24 N CPT FV 448.91 11.

PMT  1000; i  3.24% /12  0.0027  0.27%; n  4; c 

12  12 1

p  1.002712  1  1.032885  1  0.032885  3.2885% 1.0328854  1  FV  1000    1000(4.201674)  $4201.67  0.032885 

Set P/Y  1;C/Y  12  0 PV 1000  PMT 3.24 I/Y 4 N CPT FV 4201.674368 12.

PMT  4000; i  1.95% /1  0.0195  1.95%; n  5(4)  20; c 

1  0.25 4

p  1.01950.25  1  1.004840  1  0.004840  0.4840% 1.00484020  1 FV  4000    4000(20.946813)  $83, 787.25  0.004840 

Set P/Y  4;C/Y  1 0 PV 4000  PMT 1.95 I/Y 20 N CPT FV 83, 787.25248

13. PMT = $323.80; i = 5.45%; c = 1/12 p = (1.0545)1/12 – 1 = 0.004432 = 0.4432% FV13 = 323.80(1.004432)2 = $326.68 FV14 = 323.80(1.004432)1 = $325.24 15th payment

$323.80 $975.72

14. FV = $20,000; PMT = $500; i = 3.12%; n = (2)(12) = 24; c = 1/12 p = (1.0312)1/12 – 1 = 0.002564 = 0.256354% 1.00256424  1 FV = 500    0.002564 

FV = 500(24.721020) FV = $12,360.51 $20,000 – 12,360.51 = $7639.49 m1 m2

 0.045 12 15. i2  (1  i1 )  1  1    1  0.003675 1    (1.003675)240  1  FV1  100    100 (384.159795)  $38, 415.98  0.003675   (1.045)20  1  FV2  1200    1200 (31.371423)  $37, 645.71 0.045   Difference  $38, 415.98  37, 645.71  $770.27 16.

For the first 5 years: PMT = 100; i =

0.06 = 0.005; n = 60 12

 (1.005)60  1 FV1  100    100(69.770031)  $6977.003051  0.005  For the next 10 years: n = 120

FV2 = $6977.003051(1.005)120 = 6977.003051(1.819397) = $12,693.94

For the last 10 years: PMT = 100; n = 26´ 10 = 260 12

12  0.06  26 26 i2  1   1  (1.005)  1  1.002305  1  0.002305  12    (1.002305)260  1  FV3  100    100 (355.549319)  $35,554.93  0.002305  After 10 years of working full time, the balance in the account will be $12,693.94 + 35,554.93 = $48,248.87

Exercise 12.2 1.

PMT  250; i  0.25%; n  48; c 

12 3 4

p  1.00253  1  1.007519  1  0.7519% 1  1.00751948  PV  250    250(40.167520)  $10, 041.88  0.007519 

Set P/Y  4;C/Y  12  0 FV 250  PMT 3 I/Y 48 N CPT PV 10, 041.88 2.

PMT  1560; i  1.5%; n  9; c 

4 4 1

p  1.0154  1  1.061364  1  6.1364% 1  1.0613649  PV  1560    1560(6.761510)  $10, 547.96  0.061364 

Set P/Y  1;C/Y  4  0 FV 1560  PMT 6 I/Y 9 N CPT PV 10, 547.96 3.

PMT  825; i  3.5%; n  64; c 

2 1  4 2

p  1.035 2  1  1.017350  1  1.7350% 1

1  1.01735064  PV  825    825(38.468566)  $31, 736.57  0.017350 

Set P/Y  4;C/Y  2  0 FV 825  PMT 7 I/Y 64 N CPT PV 31, 736.57

PMT  175; i  1.25%; n  240; c 

4 1  12 3

p  1.0125 3  1  1.004149  1  0.4149% 1  1.004149240  PV  175    175(151.788066)  $26, 562.91  0.004149 

Set P/Y  12;C/Y  4  0 FV 175  PMT 5 I/Y 240 N CPT PV 26, 562.91 5.

(a)

PMT  6000; i  0.5%; n  12; c 

12 6 2

p  1.0056  1  1.030378  1  3.0378% 1  1.03037812  PV  6000    6000(9.931610)  $59,589.66  0.030378  Purchase price  59,589.66 + 10,000  $69,589.66

Set P/Y  2;C/Y  12  0 FV 6000  PMT 6 I/Y 12 N CPT PV 59,589.66 (b) 6.

Interest  12(6000)  59,589.66  72,000  59,589.66  $12,410.34

(a)

PMT  565; i  2.25%; n  48; c 

4 1  12 3

1 3

p  1.0225  1  1.007444  1  0.7444% 1  1.00744448  PVA  565    5650  40.235947   $22,733.31  0.007444  Purchase price  22,733.31 + 5000  $27,733.31

Set P/Y  12; C/Y  4  0 FV 565  PMT 9 I/Y 48 N CPT PV 22, 733.31 (b)

Interest  48  565  22, 733.31  27,120  22, 733.31  $4386.69

PMT  1715.59; i  2.8%; n  300; c 

2 1  12 6

p  1.028 6  1  1.004613  1  0.4613% 1  1.004613300  PV  1715.59    1715.59 162.278372   $278,403.15  0.004613 

Set P/Y  12; C/Y  2  0 FV 1715.59  PMT 5.6 I/Y 300 N CPT PV 278,403.15

8. 1  0.25 4 p  1.0730.25  1  1.017771  1  1.7771%

PMT  924.37; i  7.3%; n  28; c 

1  1.01777128  PV  924.37    924.37  21.908941  $20, 251.97  0.017771 

Set P/Y  4; C/Y  1 0 FV 924.37  PMT 7.3 I/Y 28 N CPT PV 20, 251.97 9.

12 6 2 P  1.003836  1  1.023222  1  2.3222% PMT  1800; i  0.383%; n  40; c 

1  1.02322240  PV  1800    1800  25.871637   $46, 568.95  0.023222 

Set P/Y  2; C/Y  12  0 FV 1800  PMT 4.6 I/Y 40 N CPT PV 46,568.95 10.

PMT  178; i  1.3%; n  96; c 

4 1  12 3

p  1.0133  1  1.004315  1  0.4315% 1  1.00431596  PV  178    178  78.464244   $13, 966.64  0.004315 

Set P/Y  12; C/Y  4  0 FV 178  PMT 5.2 I/Y 96 N CPT PV 13,966.64 11. Amount now if quarterly payment is chosen: PMT  145; i  1.975%; n  40; c 

2 1  4 2

p  1.01975 2  1  1.009827  1  0.9827% 1

1  1.00982740  PV  145    145  32.942971  $4776.73  0.009827  Comparing the two offers: $5000  4776.73  233.27 The lump-sum cash offer is $223.27 higher than the quarterly payments offer.

Set P/Y  4; C/Y  2  0 FV 145  PMT 3.95 I/Y 40 N CPT PV 4776.73

12.

Amount now if monthly payment is chosen:

PMT  200, 000; i  1.425%; n  24; c 

4 1  12 3

p  1.01425 3  1  1.004728  1  0.4728% 1

1  1.00472824  PV  200, 000    200, 000  22.638020   $4, 527, 604.03  0.004728  Comparing the two offers: $4,527, 604.03  4, 000, 000  527, 604.03 The monthly payments offer is 527, 604.03 higher than the lump-sum offer.

 Set P/Y = 12;C/Y = 4  0 FV 200, 000  PMT 5.7 I/Y 24 N CPT PV 4,527, 604.03 13. Answer: PMT = 2000, i = 0.05/2 = 0.025, n = 6 × 4 = 24, c = 2/4 Balance at 4 years (annuity starts): p = (1.025)2/4 – 1 = 0.012423 1  1.01242324  PVg = 2000    0.012423 

= 2000(20.642960) = $41,285.92 Balance now: PV = 41,285.92(1.025)8 = $33,885.28 Exercise 12.3 1.

FV  12, 000; i  0.5%; n  60; c 

12 3 4

p  1.0053  1  1.015075  1  1.5075% 1.01507560  1  12, 000  PMT    0.015075  12, 000  96.456485 PMT PMT  $124.41

Set P/Y  4;C/Y  12  0 PV 12, 000 FV 6 I/Y 60 N CPT PMT 124.41

PV  6000; i  3.5%; n  60; c 

2 1  12 6

1 6

p  1.035  1  1.005750  1  0.5750% 1  1.00575060  6000  PMT    0.005750  6000  50.622467 PMT PMT  $118.52

Set P/Y  12;C/Y  2  0 FV 6000  PV 7 I/Y 60 N CPT PMT 118.52 3.

PV  9500; i  1.85%; n  16; c 

4 2 2

p  1.01852  1  1.037342  1  3.7342% 1  1.03734216  9500  PMT    0.037342  9500  11.884114 PMT PMT  $799.39

Set P/Y  2;C/Y  4  0 FV 9500  PV 7.4 I/Y 16 N CPT PMT 799.39 4.

FV  16, 000; i  0.35%; n  18; c 

12  12 1

p  1.003512  1  1.042818  1  4.2818% 1.04281818  1  16, 000  PMT    0.042818  16, 000  26.31909 PMT PMT  $607.92

Set P/Y  1;C/Y  12  0 PV 16, 000 FV 4.2 I/Y 18 N CPT PMT  607.92 5.

PV  330, 000; i  2.55%; n  216; c 

2  0.16 12

p  1.02550.16  1  1.004206  1  0.4206% 1  1.004206216  330, 000  PMT    0.00420622  330, 000  PMT (141.732441) PMT  $2328.33

Set P/Y  12;C/Y  2  330, 000 PV 0 FV 5.1 I/Y 216 N CPT PMT  2328.33

FV  12, 000; i  0.5%; n  40; c 

12 3 4

p  1.0053  1  1.015075  1  1.5075% 1.01507540  1  12, 000  PMT    0.015075  12, 000  54.354225 PMT PMT  $220.77

Set P/Y  4;C/Y  12  0 PV 12, 000 FV 6 I/Y 40 N CPT PMT  220.77 7.

FV  20, 000; i  3.5%; n  80; c 

2 1  4 2

p  1.035 2  1  1.017349  1  1.7349% 1

1.01734980  1  20, 000  PMT    0.017349  20, 000  170.567460 PMT PMT  $117.26

Set P/Y  4;C/Y  2  0 PV 20, 000 FV 7 I/Y 80 N CPT PMT 117.26 8.

PV  0.85  27,900   23,715; i  4%; n  48; c 

2 1  12 6

p  1.04 6  1  1.006558  1  0.6558% 1  1.00655848  23,715  PMT    0.006558  23, 715  41.064609 PMT PMT  $577.50

Set P/Y  12;C/Y  2  0 FV 23,715 PV 8 I/Y 48 N CPT PMT  577.50 9.

FV  10, 000.28; i  3.7%; n  180; c 

1  0.083 12

p = 1.037.083 - 1 = 1.003032 - 1 = 0.3032% é1.003032180 - 1ù ú 10, 000.28 = PMT ê ê 0.003032 ú ë û 10, 000.28 = PMT (238.955349) PMT = $41.85

(Set P/Y = 12;C/Y = 1)0 PV 10, 000.28 FV 3.7 I/Y 180 N CPT PMT - 41.85

10.

& n  28; c  12  3 PV  120, 000; i  0.583%; 4 3 & p  1.00583  1  1.017602  1  1.7602% 1  1.01760228  120, 000  PMT    0.017602  120, 000  21.957428 PMT PMT  $5465.12

(Set P/Y  4;C/Y  12) 0 FV 120, 000 PV 7 I/Y 28 N CPT PMT  5465.12 2 11. FV  800,000; i  2.22%; n  8; c   2 1 p  1.02222  1  1.044893  1  0.044893  4.44893% 1.0448938  1  800, 000  PMT    0.044893  800, 000  PMT (9.376426) PMT  $85,320.36

Set P/Y  1; C/Y  2  0 PV 800, 000 FV 4.44 I/Y 8 N CPT PMT  85,320.35561 12.

PV  200; i  6.6%; n  40; c 

1  0.25 4

p  1.0660.25  1  1.016107  1  1.6107% 1  1.016107 40  200 = PMT    0.016107  200  PMT  29.320177  PMT  $6.82

(Set P/Y  4;C/Y  1) 0 FV 200  PV 6.6 I/Y 40 N CPT PMT 6.821241 13. Purchase Price = $35,100 + $365 + 125 = $35,590 Down payment = 25% of $35,590 = $8897.50 Amount of the Loan: 35,590 – 8897.50 = $26,692.50 n = (4.25)(12) = 51 payments; i = 0.0219 c =

4 12

p = (1 + 0.0219)4/12 – 1 = 0.007247 = 0.7247% 1  1.007247 51  26,692.50 = PMT    0.007247  26,692.50 = PMT (42.509044) PMT = $627.93

(Set P/Y  12;C/Y  4) 0 FV 26,692.50 PV 8.76 I/Y 51 N CPT PMT  627.93 14. PV = $115,500; i = 6.3%; c = 1/2 p = (1 + 0.063)1/2 – 1 = 0.031019 = 3.101891% n = (2)(10 – 2.5 years) = (2)(7.5 years) = 15 payments 1  1.031019 15  115,000 = PMT    0.031019  115,000 = PMT (11.850428)

PMT = $9704.29

(Set P/Y  2;C/Y  1) 0 FV 115,000 PV 6.3 I/Y 15 N CPT PMT  9704.29 15. $39,000 = (PV of the loan payment annuity) + (PV of the $17,500 balance) For PVannuity 2 n = 36; i = 0.037; c  12 2/12 p = (1 + 0.037) – 1 = 0.006074 = 0.607369% 1  1.006074 36  PVann = PMT    0.006074  PVann = PMT(32.248554) For PV balance n = 6; i = 0.074/2 PVbal = 17,500 ( 1 + 0.074/2) –6 = $14,072.31 $39,000 – 14,072.31 = $24,927.69 24,927.69 = PMT(32.248554) PMT = $772.99

(Set P/Y  12;C/Y  2) 0 FV 24,927.69  PV 7.4 I/Y 36 N CPT PMT 772.986296 Exercise 12.4 1.

FV  5000; PMT  100; i  1.5%; c 

4 1  12 3

1 3

p  1.015  1  1.004975  1  0.4975% 1.004975n  1  5000  100    0.004975  1.004975n  1 50  0.004975 n 1.0049756  1.248760 n ln1.004975  ln10.248760 0.004963n  0.222151 n  44.762662 months  3 years, 9 months

(Set P/Y  12; C/Y  4) 0 PV 100  PMT 5000 FV 6 I/Y CPT N 44.762662

FV  15, 000; PMT  90; i  1%; c 

4 1  12 3

p  1.013  1  1.003322  1  0.3322% 1.003322n  1 15, 000  90    0.003322  1.003322n  1 166.6& 0.003322 n 1.003322  1.553717 n ln1.003322  ln1.553717 0.003317n  0.440650 n  132.854320 months  11 years, 1 month

(Set P/Y  12;C/Y  4) 0 PV 90  PMT 15,000 FV 4 I/Y CPT N 132.854320

PV  3000; PMT  90; i  5.25%; c 

2 1  12 6

p  1.0525 6  1  1.008565  1  0.8565% 1

1  1.008565 n  3000  90    0.008565  1  1.008565 n 33.3& 0.008565 n 1.008565  0.714516 n ln1.008565  ln 0.714516 0.008528n  0.336150 n  39.416954 months  3 years, 4 months

(Set P/Y  12; C/Y  2) 0 FV 3000  PV 90 PMT 10.5 I/Y CPT N 39.416954

PV  28, 000;PMT  3400; i  4.04%; c 

2  0.5 4

p  1.04040.5  1  1.02  1  2% 1  1.02 n  28, 000  3400    0.02  1  1.02 n  8.235294     0.02  0.164706  1  1.02 n 1.02 n  0.835294  n ln1.02  ln 0.835294  0.019803n  0.179972 n  9.088258 quarters  2 years, 6 months

(Set P/Y  4;C/Y  2)28,000 PV 3400  PMT 0 FV 8.08 I/Y CPT N 9.088258

FV  120, 000; PMT  500; i  1.35%; c 

4 1  12 3

p  1.0135 3  1  1.004480  1  0.4480% 1

1.004480n  1  120, 000  500    0.004480  1.004480n  1 240  0.004480 n 1.004481  2.075176 n ln1.004480  ln 2.075176 0.004470n  0.730046 n  163.325073 months  13 years, 8 months

(Set P/Y  12;C/Y  4) 0 PV 500  PMT 120,000 FV 5.4 I/Y CPT N 163.325073

FV  100, 000; PMT  2000; i  1.5%; c 

4 4 1

p  1.0154  1  1.061364  1  6.1364% 1.061364n  1  100, 000  2000    0.061364  1.061364n  1 50  0.061364 n 1.061364  4.068178 n ln1.061364  ln 4.068178 0.059554n  1.403195 n  23.561549 years  24 years

(Set P/Y  1; C/Y  4) 0 PV 2000  PMT 100, 000 FV 6 I/Y CPT N 23.561549

PV  16, 000; PMT  800; i  3.4%; c 

2 1  12 6

p  1.034 6  1  1.005588  1  0.5588% 1

1  1.005588 n  16, 000  800    0.005588  1  1.005588 n 20  0.005588 n 1.005588  0.888240 n ln1.005588  ln 0.888240 0.005572n  0.118514 n  21.267743 months  1 year, 10 months

(Set P/Y  12;C/Y  2) 0 FV 16, 000  PV 800 PMT 6.8 I/Y CPT N 21.267743

PV  240,000; PMT  1502; i  2.875%; c 

2 1  12 6

p  1.02875 6  1  1.004735  1  0.4735% 1  1.004735 n  240,000  1502    0.004735  1  1.004735 n 159.786951  0.004735 n 1.004735  0.243368 n ln1.004735  ln 0.243368  0.004724n  1.413180 n  299.144104 months  25 years

(Set P/Y  12;C/Y  2) 0 FV 240,000 PV 1502  PMT 5.75 I/Y CPT N 299.144104

FV  20, 000; PMT  370.37; i  3.25%; c 

2 1  4 2

p  1.0325 2  1  1.016120  1  1.6120% 1

1.016120n  1  20, 000  370.37    0.016120  1.016120n  1 54.000054  0.016120 n 1.016120  1.870485 n ln1.016120  ln1.870485 0.015992 n  0.626198 n  39.158098 quarters  40 deposits

(Set P/Y  4;C/Y  2) 0 PV 370.37  PMT 20,000 FV 6.5 I/Y CPT N 39.158098

10.

PV  100, 000; PMT  4500; i  0.5625%; c 

12 3 4

p  1.0056253  1  1.016970  1  1.6970% 1  1.016970 n  100, 000  4500    0.016970  1  1.016970 n 22.2& 0.016970 n 1.016970  1  0.377113 n ln1.016970  ln 0.622887 0.016828n  0.473391 n  28.131606 quarters  7 years, 3 months

(Set P/Y  4;C/Y  12) 0 FV 100,000  PV 4500 PMT 6.75 I/Y CPT N 28.131606

11.

PV  35, 000; PMT  6000; i  8.0% / 4  2%; c 

4 4 1

p  1.024  1  1.082432  1  0.082432  8.2432% ln[1  ( 35,000(0.082432) )] 6000 n  ln(1  0.082432) ln[1  0.480854] n  ln(1.082432) 0.655571 n 0.079211 n  8.276309 years  9 payments

(Set P/Y  1;C/Y  4) 0 FV 35, 000  PV 6000 PMT 8 I/Y CPT N 8.276309

12.

FV  50, 000; PMT  10,000; i  4.0% / 4  1%; c 

4 2 2

p  1.012  1  1.0201  1  0.0201  2.01% n

ln[( 50,000(0.0201) )  1] 10,000

ln(1  0.0201) ln[1.1005] n ln(1.0201) 0.095765 n 0.019901 n  4.81 semi-annual periods  2 years, 6 months

(Set P/Y  2; C/Y  4) 0 PV 50, 000 FV 10, 000  PMT 4 I/Y CPT N 4.812133 13. Answer: FV  $75,000; PMT  $250; i  6%; c 

p  (1.06)1/12  1  0.004868  75, 000  0.004867   ln    1 250     n ln (1.004868) ln 2.460265 n ln 1.004868 n  0.900269/ 0.004856 n  185.402987(months) n  186 months 186/12 = 15.5 years = 15 years, 6 months Business Math News Box

1 12

Total monthly lease payment = Depreciation fee + Finance fee + Sales tax Depreciation fee = (Selling price - Residual value) ¸ Term (in months) = (30,000 - 17,000) ¸ 36 = $361.11 æAPR ÷ ö Finance fee = (Selling price + Residual value) ´ çç ÷ çè 2400 ÷ ø æ 1.9 ö÷ = (30, 000 + 17, 000)´ çç ÷ çè 2400 ø÷ = 47, 000´ (0.000792) = $37.21 Sales tax = (Depreciation fee + Finance fee) ´ Sales tax rate in Ontario = (361.11 + 37.21)´ 0.13 = (398.32)´ 0.13 = $51.78 Total monthly lease payment = $361.11 + 37.21 + 51.78 = $450.10

Amount to finance = Selling price + Sales tax = (30, 000)´ (1.13) = $33,900 $33,900 ¸ 36 payments = $941.67 monthly payment

At the end of the three-year lease, you’d be left with nothing. If you buy the car, you have a vehicle that you can sell for $17,000 or you can keep driving it for several more years, paying only maintenance and insurance costs.

Exercise 12.5 1.

FV  11, 600; PMT  253; n  42; c  4  1 12 3 (Set P/Y  12; C/Y  4) 0 PV 11, 600 FV 253  PMT 42 N CPT I/Y 5.0895 The nominal annual rate is 5.0895% compounded quarterly.

FV  7720; PMT  416; n  16; c  1 2 (Set P/Y  2;C/Y  1) 0 PV 7720 FV 416  PMT 16 N CPT I/Y 3.9247 The nominal annual rate is 3.9247% compounded annually.

PV  6000; PMT  144.23; n  48; c  2  1 12 6 (Set P/Y  12; C/Y  2) 0 FV 6000  PV 144.23 PMT 48 N CPT I/Y 7.3071 The nominal annual rate is 7.3071% compounded semi-annually. 4.

PV  22,800   0.10  22,800  20,520; PMT  572.42; n  42; c  1 12 (Set P/Y  12; C/Y  1) 0 FV 20,500 PV 572.42  PMT 42 N CPT I/Y 9.5618 The nominal annual rate is 9.562% compounded annually.

PV  300, 000   0.20  300, 000  240, 000; PMT  400; n  1300; c  12 52 (Set P/Y  52; C/Y  12) 0 FV 240, 000 PV 400  PMT 1300 N CPT I/Y 7.266998 The nominal annual rate is 7.2670% compounded monthly.

PV  250, 000; PMT  22, 000; n  15; c  1 2 (Set P/Y  2;C/Y  1) 0 FV 250, 000  PV 22, 000 PMT 15 N CPT I/Y 7.5154 The nominal annual rate is 7.5154% compounded annually.

FV  20, 000; PMT  400; n  32; c  12  3 4 (Set P/Y  4; C/Y  12) 0 PV 20, 000 FV 400  PMT 32 N CPT I/Y 10.7784 The nominal annual rate is 10.7784% compounded monthly. 8.

PV  21, 500; PMT  2000; n  14; c  4  2 2 (Set P/Y  2; C/Y  4) 0 FV 21, 500  PV 2000 PMT 14 N CPT I/Y 7.4025 The nominal annual rate is 7.4025% compounded quarterly.

FV  4850; PMT  120; n  36; c  4  1 12 3 (Set P/Y  12; C/Y  4) 0 PV 4850 FV 120  PMT 36 N CPT I/Y 7.8560 The nominal annual rate is 7.856% compounded quarterly.

10.

FV  3500; PMT  420; n  8; c  1 4 (Set P/Y  4; C/Y  1) 0 PV 3500 FV 420  PMT 8 N CPT I/Y 4.7339 The nominal annual rate is 4.7339% compounded annually.

11.

PV  106, 500; PMT  1710; n  180; c  2  1 12 6 (Set P/Y  12; C/Y  2) 0 FV 106,500 PV 1710  PMT 180 N CPT I/Y 18.616949 The nominal annual rate is 18.6169% compounded semi-annually.

12.

PV  0.90  210, 000   189, 000; PMT  12, 600; n  24; c  4  2 2 (Set P/Y  2; C/Y  4) 0 FV 189, 000 PV 12, 600  PMT 24 N CPT I/Y 8.2355 The nominal annual rate is 8.2355% compounded quarterly.

Exercise 12.6 1.

PMT  500; k  3%  0.03; n  25

 a  Size of the 25th deposit 24  500 1.03  500  2.032794   $1016.40

 b  Total amount deposited 1.0325  1    500

0.03 1.093778  500 0.03  500  36.459264   $18, 229.63

PMT  5000; k  5%  0.05; n  15

 a  Size of the 15th withdrawal 14  5000  0.95  5000  0.487675  $2438.37

 b  Total amount withdrawn  0.95 15  1    5000

0.05  0.536709   5000 0.05  5000 10.734175  $53, 670.88 3.

PMT  1200; k  1.5%  0.015; n  20; i  5% / 2  2.5%  0.025;

(a) Total amount deposited 1.015 20  1   1200  0.015  0.346855  1200 0.015  1200  23.123667 

1.025 20  1.015 20   (b) FV  1200  0.025  0.015 1.638616  1.346855  1200 0.010  1200  29.176143  $35, 011.37

 $27, 748.40 (c) Size of the 12th payment

(d) Interest  35, 011.37  27, 748.40  $7262.97

 1200 1.015 

 1200 1.177949   $1413.54

n n   (a) PV  PMT 1  (1  k ) (1  i)  ik  

1  (1.04) 24 (1.015) 24   1500   0.015  0.04   1  (2.563304165) (0.69954392)   1500   0.025  1  1.793143842   1500   0.025   1500 31.7257537  $47,588.63 (b) Size of the 24th payment  1500 1.04 

 1500  2.464716   $3697.07

(c) Total amount withdrawn 1.04 24  1   1500  0.04 1.563304   1500 0.04  1500  39.082604   $58, 623.91 (d) Interest  58, 623.91  47, 588.63  $11, 035.28

PMT  400; k  0.5%  0.005; n  120; i  6% /12  0.5%  0.005; PV  120  400 1.005 

1

 120  400  0.995025   47, 761.19

Total amount withdrawn 1.005 120  1   400  0.005  0.819397   400 0.005  400 163.879347   65, 551.74 Interest  65, 551.74  47, 761.19  $17, 790.55 6.

PMT  1500; k  1%  0.01; n  60; i  3% / 2  1.5%  0.015; (1.015)60  (1.01)60 0.015  0.010 2.443220  1.816697  1500 0.005  187,956.92

FV  1500

Total amount deposited 1.0160  1   1500  0.01  0.816697   1500 0.01  1500  81.669670   122,504.50 Interest  187,956.92  122,504.50  $65, 452.42 7.

PMT  1250; k  0.5  0.005; n  120; i  4.8% /12  0.40%  0.004; 1  (1  k ) n (1  i)  n  PV  1250   ik   120 1  (0.995) (1.004) 120  PV  1250   0.004  0.005   1  (0.547986)(0.619376)  PV  1250   0.009  PV  1250[73.398931] PV  $91, 748.66

PMT  2000; k  3.0%  0.03; n  20; i  4.0%  0.04;

1  (1  k ) n (1  i)  n  PV  2000   ik   20 20 1  (1.03) (1.04)  PV  2000   0.04  0.03   1  (1.806111)(0.456387)  PV  2000   0.01  PV  2000[17.571441] PV  $35,142.88

Answer: You want to compare the PV of the growing annuity to the PV of receiving $100,000 right now (which is obviously just $100,000). 1  (1.05)16 (1.12) 16  PV = 10,000   0.12  0.05  

PV = 10,000 [9.198941] PV = $91,989.41 It is better to receive $100,000 right now. 10. Answer: PMT = $12,000; k = –2.5% i = 9.2%

n = 15 years

 (1  i ) n  (1  k ) n  FV  PMT   ik    (1.092)15  (0.975)15  FV = 12,000   = 12,000(26.153945)  0.092  0.025 

FV = $313,847.34 11.

1  (1  k ) n (1  i)  n  PV  PMT   ik   1  (0.985) 22 (1.04) 22  PV = 60,000   0.04  0.015   1  (0.717129)(0.421955)  PV = 60,000   0.055 

PV = 60,000 [0.697404 / 0.055] PV = 60,000 (12.680067) PV = $760,804.04

Review Exercise 1.

PMT  375; i  0.25%; n  32; c  12  3 4 3 p  1.0025  1  1.007519  1  0.7519 1.00751932  1 FV  375    375  36.025676   $13,509.62  0.007519 

(Set P/Y  4; C/Y  12) 0 PV 375  PMT 3 I/Y 32 N CPT FV 13,509.62 PMT  145; i  2.5%; n  144; c  2  1 12 6 1 p  1.025 6  1  1.004124  1  0.4124% 1.004124144  1 FV=145    145 196.106335  $28, 435.42  0.004124 

(Set P/Y  12; C/Y  2) 0 PV 145  PMT 5 I/Y 144 N CPT FV 28, 435.42 PMT  4800; i  1.10%; n  22; c  4  2 2

p  (1.011)2  1  1.022121  1  2.2121% 1  1.02212122  PV=4800    4800 17.271188  $82, 901.70  0.022121  (Set P/Y  2; C/Y  4) 0 FV 4800  PMT 4.4 I/Y 22 N CPT PV 82,901.70273 4.

PMT  240; i  3.25%; n  300; c  2  1 12 6 1

p  1.0325 6  1  1.005345  1  0.5345% 1  1.005345300  PV=240    240 149.292996   $35, 830.32  0.005345  (Set P/Y  12; C/Y  2) 0 FV 240  PMT 6.5 I/Y 300 N CPT PV 35,830.32

PMT  500; i  3.0%; n  80; c  2  1 4 2 1 2 p  1.03  1  1.014889  1  1.4889% 1.01488980  1 FV  500    500 151.925181  $75,962.59  0.014889 

(Set P/Y  4; C/Y  2) 0 PV 500  PMT 6 I/Y 80 N CPT FV 75,962.59 PMT  2110; i  2.6%; n  240; c  2  0.16& 12

p = 1.026.16 - 1 = 1.004287 - 1 = 0.4287% é1- 1.004287- 240 ù ú= 2110 (149.708034) = $315, 883.95 PV=2110 ê ê 0.004287 ú ë û (Set P/Y = 12; C/Y = 2) 0 FV 2110 ± PMT 5.2 I/Y 240 N CPT PV 315,883.95 7.

FV  18, 000; i  1.8%; n  8; c  2  2 1 p  1.0182  1  1.036324  1  3.6324% 1.0363248  1  18, 000  PMT    0.036324  18, 000  9.094414PMT PMT  $1979.24 (Set P/Y  1; C/Y  2) 0 PV 18, 000 FV 3.6 I/Y 8 N CPT PMT  1979.24

(a)

PV  15, 750; i  2.125%; n  144; c  2  1 12 6

p  1.02125 6  1  1.003511  1  0.3511% 1  1.003511144  15, 750  PMT    0.003511  15, 750  112.879545 PMT PMT  $139.53 (Set P/Y  12; C/Y  2) 0 FV 15, 750  PV 4.25 I/Y 144 N CPT PMT 139.53

(b)

PV  15, 750; i  2.125%; n  15; c  2  2 1

p  1.021252  1  1.042952  1  4.2952% 1  1.04295215  15, 750  PMT    0.042952  15, 750  10.892393 PMT PMT  $1445.96 (Set P/Y  1; C/Y  2) 0 FV 15, 750 PV 4.25 I/Y 15 N CPT PMT  1445.96

(a)

FVnc  12, 000; PMT  350; i  0.25%; c  12  3 4

p  1.00253  1  1.007519  1  0.7519% 1.007519n  1  12, 000  350    0.007519  1.007519n  1 34.285714  0.007519 n 1.007519  1.257786 n ln1.007519  ln1.257786 0.00749 l n  0.229353 n  30.618640 quarters  7 years, 9 months (Set P/Y  4; C/Y  12) 0 PV 350  PMT 12, 000 FV 3 I/Y CPT N 30.618640

(b)

FVnc  12, 000; PMT  350; i  0.25%; c  12  6 2

p  1.00256  1  1.015094  1  1.5094% 1.015094n  1  12, 000  350    0.015094  1.015094n  1 34.285714  0.015094 n 1.015094  1.517511 n ln1.015094  ln1.517511 0.014981n  0.417071 n  27.839496 semi-annual periods  14 years (Set P/Y  2; C/Y  12) 0 PV 350  PMT 12, 000 FV 3 I/Y CPT N 27.839496

10.

FVnc  276, 000; i  2.9%; n  300; c  2  1 12 6

p  1.029 6  1  1.004776  1  0.4776% 1  1.004776300  276, 000  PMT    0.004776  276, 000  159.244083PMT PMT  $1733.19 (Set P/Y  12; C/Y  2) 0 FV 276, 000 PV 5.8 I/Y 300 N CPT PMT  1733.19

11. (a)

FVnc  14,000; PMT  1500; i  0.75%; c 

12 6 2

p  1.00756  1  1.045852  1  4.5852% 1  1.045852 n  14, 000  1500    0.045852  1  1.045852 n 9.3  0.045852 n 1.045852  0.572046  n ln1.045852  ln 0.572046  0.044832n  0.558536 n  12.458403 half years  6 years, 6 months

Set P/Y  2; C/Y  12  0 FV 14,000  P V 1500 PMT 9 I/Y CPT N 12.458403 (b)

FVnc  14, 000; PMT  1500; i  0.75%; c 

12  12 2

p  1.007512  1  1.093807  1  9.3807% 1  1.093807  n  14, 000  1500    0.093807  1  1.093807  n 9.3  0.093807 n 1.093807  0.124469  n ln1.093807  ln 0.124469  0.089664n  2.083699 n  23.238923 years  24 years

Set P/Y  1;C/Y  12  0 FV 14, 000  P V 1500 PMT 9 I/Y CPT N 23.238923 12.

FVnc  9200; PMT  200; n  32; c 

12 3 4

Set P/Y  4; C/Y  12  0 P V 9200 FV 200  PMT 32 N CPT I/Y 8.8639 The nominal annual rate is 8.8639% compounded monthly.

13.

FVnc  2290; PMT  198; i  1.4125%; c 

12  3. 4

p  1.0141253  1  1.042976  1  4.2976% 1  1.042976 n  2290  198    0.042976  . 1  1.042976 n 11.56  0.042976 n 1.042976  0.502950  n ln1.042976  ln 0.502950 0.042079n  0.687264 n  16.332902 quarters  4 years, 3 months

Set P/Y  4; C/Y  12  0 FV 2290  P V 198 PMT 16.95 I/Y CPT N 16.332902 14.

P V  265, 000; PMT  1672; n  300; c 

2  0.16 12

Set P/Y  12; C/Y  2  0 F V 265, 000 PV 1672  PMT 300 N CPT I/Y 5.850436 The nominal annual rate is 5.8504% compounded semi-annually.

15.

P V  148, 000; PMT  5000; i  0.3583%; c 

12 3 4

p  1.0035833  1  1.010789  1  1.0789% 1  1.010789 n  148, 000  5000    0.010789  1  1.010789 n  29.6     0.010789  0.319342  1  1.010789  n 1.010789 n  0.680658 n ln1.010789  ln 0.680658 0.010731n  0.384695 n  35.849629 quarters  9 years

Set P/Y  4; C/Y  12  148,000 PV 5000  PMT 0 FV 4.3 I/Y CPT N 35.849629

16.

PMT  145; i  0.625%  0.00625; n  15 12  180

1.00625180  1  FV  145    145  331.112276   $48, 011.28  0.00625 

Set P/Y  12; C/Y  12  0 P V 145  PMT 7.5 I/Y 180 N CPT FV 48, 011.28 2 1  4 2 Set P/Y  4; C/Y  2  0 FV 48, 011.28  PV 1200 PMT 48 N CPT I/Y 3.0883 PV  48, 011.28; PMT  1200; n  48; c 

The nominal annual rate is 3.0883% compounded semi-annually.

17. (a)

12 3 4 p  1.0053  1  1.015075  1  1.5075% PMT  345; i  0.5%; n  36; c 

1.01507536  1  FV  345    345  47.342858   $16, 333.29  0.015075 

Set P/Y  4; C/Y  12  0 PV 345  PMT 6 I/Y 36 N CPT FV 16,333.29 (b)

Interest  16, 333.29  345  36  $3913.29

(c)

12 3 4 p  1.004163  1  1.012552  1  1.2552%

P V  16, 333.29; i  0.416%; n  32; c 

1  1.01255232  16,333.29  PMT    0.012552  16,333.29  PMT (26.220410) PMT  $622.92

Set P/Y  4; C/Y  12  16,333.29 PV 0 FV 5 I/Y 32 N CPT PMT  622.92 (d)

Combined interest earned  32  622.92    36  345  19,933.44  12, 420  $7513.44 18.

FV  100, 000; i  2.0%; n  25; c  4

p  1.024  1  1.082432  1  8.2432% 1.08243225  1  100, 000  PMT    0.082432  100, 000  75.754974 PMT PMT  $1320.05

Set P/Y  1; C/Y  4  0 PV 100,000 FV 8 I/Y 25 N CPT PMT 1320.05 19.

FV  9500; PMT  75 i  1.625%; c 

4 1  12 3

p  1.01625 3  1  1.005388  1  0.5388% 1.005388n  1  9500  75    0.005388  1.005388n  1 126.6  0.005388 n 1.005388  1.682428 n ln1.005388  ln1.682428 0.005373n  0.520238 n  96.822182 months  8 years, 1 month

Set P/Y  12; C/Y  4  0 PV 75  PMT 9500 FV 6.5 I/Y CPT N 96.822182 20. P V  337,500; PMT  2435; i  2.46%; c 

2  0.16 12

p  1.02460.16  1  1.004059  1  .4059% 1  1.004059 n  337, 500  2435    0.004059  1  1.004059 n  138.603696     0.004059  0.562536  1  1.004059  n 1.004059 n  0.437464 n ln1.004059  ln 0.437464 0.004050n  0.826762 n  204.119455 months  17 years

Set P/Y  12; C/Y  2  337,500 PV 2435  PMT 0 FV 4.92 I/Y CPT N 204.119455

21.

PMT  1200; k  1.5%  0.015; n  80; i  7% / 4  1.75%  0.0175 1.0175 80  1.015 80   FV  1200  0.0175  0.015  4.006392  3.290663  1200 0.0025  343,549.99 Total amount deposited 1.015 80  1   1200  0.015  2.290663  1200 0.015  1200 152.710852   183, 253.02 Interest  343,549.99  183,253.02  $160, 296.97

22.

PV  250,000; k  0.7%  0.007; n  180; i  0.70% 250, 000  180  PMT 1.007 

1

250, 000  180  PMT  0.993049  250, 000  178.748759  PMT  PMT  1398.61 Total amount received 1.007 180  1  1398.61  0.007  2.509980  1398.61 0.007 1398.61 358.568638  $501, 497.68

Self-Test 1.

PMT  480; i  2.25%; n  96; c 

2 1  12 6

p  1.0225 6  1  1.003715  1  0.3715% 1.00371596  1  FV  480    480 115.096817   $55, 246.47  0.003715 

Set P/Y  12; C/Y  2  0 P V 480  PMT 4.5 I/Y 96 N CPT FV 55,246.47 2.

PMT  1200; i  4.75%; n  20; c 

2 1  4 2

p  1.0475 2  1  1.023474  1  2.3474% 1

1  1.02347420  P V  1200    1200 15.816180   $18,979.42  0.023474  Interest  20 1200   18,979.41  24, 000  18,979.42  $5020.58

Set P/Y  4; C/Y  2  0 F V 1200  PMT 9.5 I/Y 20 N CPT P V 18,979.42 3.

PV  6000; PMT  450; i  1.0%; c 

12 3 4

p  1.013  1  1.030301  1  3.0301%

1  1.030301 n  6000  450    0.030301  1  1.030301 n 13.3  0.030301 0.404013  1  1.030301 n 1.030301 n  0.595987 n ln1.030301  ln 0.595987 0.029851n  0.517537 n  17.337346 quarters  4 years, 6 months

Set P/Y  4; C/Y  12  0 FV 6000  P V 450 PMT 12 I/Y CPT N 17.337346 4.

PV  190, 000; i  4.25%; n  300; c 

2 1  12 6

p  1.0425 6  1  1.006961  1  0.6961% 1

1  1.006961300  190, 000  PMT    0.006961  190, 000  125.728673 PMT PMT  $1511.19

Set P/Y  12; C/Y  2  0 F V 190,000 P V 8.5 I/Y 300 N CPT PMT 1511.190685

2  0.5 4 p  1.0220.5  1  1.010940  1  1.0940%

PV  67, 250; i  2.2%; n  28; c 

1  1.01094028  67, 250  PMT    0.010940  67, 250  PMT  24.005746  PMT  $2801.41

Set P/Y  4; C/Y  2  67,250 PV 0 FV 4.4 I/Y 28 N CPT PMT  2801.41 6.

2 1  12 6 Set P/Y  12; C/Y  2  0 FV 270, 000 PV 2326.32  PMT 240 N CPT I Y 8.5499 PV  270, 000; PMT  2326.32; n  240; c 

The nominal annual interest rate is 8.5499% compounded semi-annually.

F V  20, 000; PMT  491; i  0.5%; c 

12 3 4

p  1.0053  1  1.015075  1  1.5075% 1.015075n  1  20, 000  491    0.015075  1.015075n  1 40.733198  0.015075 n 1.015075  1.614058 n ln 1.015075  ln1.614058 0.014963n  0.478752 n  31.996494 quarters  8 years

Set P/Y  4; C/Y  12  0 PV 491  PMT 20,000 FV 6 I/Y CPT N 31.996494 8.

F V  67, 200; i  3.25%; n  96; c 

2 1  12 6

p  1.0325 6  1  1.005345  1  0.5345% 1.00534596  1  67, 200  PMT    0.005345  67, 200  125.014972 PMT PMT  $537.54

Set P/Y  12; C/Y  2  0 PV 67,200 FV 6.5 I/Y 96 N CPT PMT 537.54

After nine years: 1  0.083 12 p  1.0430.083  1  1.003515  1  0.3515% 1.003515108  1 FV  200    200 131.079875  26, 215.97  0.003515  Annuity payment: 2 P V  26, 215.97; i  2.7%; n  240; c   0.16 12 p  1.0270.16  1  1.004450  1  0.4450% 1  1.004450240  26, 215.97  PMT    PMT 147.298170   0.004450  PMT  $177.98 Set P/Y  12; C/Y  1 0 PV 200  PMT 4.3 I/Y 108 N CPT FV 26, 215.97 PMT  200; i  4.3%; n  108; c 

Set P/Y  12; C/Y  2  26,215.97 PV 0 FV 5.4 I/Y 240 N CPT PMT 177.98 10.

PMT  6000; i  3.25%; n  40; c 

2 1  4 2

p  1.0325 2  1  1.016120  1  1.6120% 1  1.01612040  PV  6000    6000(29.313068)  $175,878.41  0.016120  Set P/Y  4; C/Y  2  0 F V 6000  PMT 6.5 I/Y 40 N CPT P V 175,878.41

11.

PV  250, 000; k  1.75%  0.0175; n  40; i  4% 1  1.0175 40 1.04 40   250, 000   PMT   0.04  0.0175 1   2.001597  0.208289   250, 000   PMT  0.0225 1  0.416911  250, 000   PMT  0.0225 250, 000   PMT  25.915076  PMT  9646.89 Total amount withdrawn 1.0175 40  1   9646.89  0.0175 1.001597    9646.89 0.0175  9646.89  57.234134   $552,131.39 Interest  552,131.39  250,000  $302,131.39

Challenge Problems 1.

i  7.5%; c 

1 12

f  1.07512  1  1.006045  1  0.006045  0.6045% Present value of $300 for the first 12 months 1

1  1.00604512   300    300 11.541501  $3462.45  0.006045  Present value of $350 for the next 24 months 1  1.00604524  12  350   1.006045  0 .006045    350  22.277782  0.930233  $7253.23 Present value of $375 for the next 36 months 1  1.00604536  36  375   1.006045  0.006045    375  32.265020  0.804961  $9739.53 Purchase price  3462.45  7253.23  9739.53  $20, 455.21

PVn  5600; i  9%; c 

1 4

p  1.09 4  1  1.021778  1  0.021778  2.1778% Let the size of the first four payments be $PMT; then the size of the last six payments is $2PMT. The present value of the first four payments 1

1  1.0217784   PMT    3.791355 PMT  0.021778  The present value of the last six payments  deferred for four payment periods  1  1.0217786  4  2 PMT   1.021778  0.021778    2 PMT  5.567971 0.917431  10.216461 PMT 5600  3.791355 PMT  10.216461 PMT 5600  14.007816 PMT PMT  399.78 The first four payments are $399.78 each; the remaining six payments are 2  399.78 = $799.56 each.

Case Study 1.

Dealer financing option: Using the financial calculator―Karim’s monthly payment will be $554.34.

Set P/Y  12; C/Y  12 24, 600 PV 3.9 I/Y 48 N 0 FV CPT PMT  554.34 2.

Bank financing option: (a)

Cash price of car  Total cost  Cash back  24,600 1800  $22,800

(b)

Monthly interest rate for bank financing:

Set P/Y  12; C/Y  12  22,800 PV 554.34  PMT 48 N 0 FV CPT I/Y 7.787202 2nd I Conv  7.787202 Enter   C/Y 12 Enter   CPT Eff  8.07124

The nominal annual rate of interest is 7.7872% compounded monthly.

The effective annual rate of interest is 8.07124% compounded annually. If Karim can obtain a nominal rate of interest from the bank that is less than 7.7872% compounded monthly, he should choose a bank loan and pay cash for the car. 3.

Dealer financing option:

Set P/Y  12; C/Y  12 34, 900 PV 3.9 I/Y 48 N 0 FV CPT PMT  786.45 Karim’s monthly payment will be $786.45. Bank financing option:

Cash price of car  Total cost  Cash back  34,900  2500  $32,400 Monthly interest rate for bank financing:

Set P/Y  12; C/Y  12  32, 400 PV 786.45  PMT 48 N CPT I/Y 7.701713 2nd I Conv  7.701713 Enter   C/Y 12 Enter   CPT Eff  7.9795

The nominal annual rate of interest is 7.7017% compounded monthly. The effective annual rate of interest is 7.9794% compounded annually.

The highest effective annual rate of interest from the bank is 7.9794% (i.e., 7.7017% compounded monthly.)

Chapter 13 Annuities Due, Deferred Annuities, and Perpetuities Exercise 13.1 1.

PMT  300; i  0.5%; n  84 1.00584 –1  FV(due)  300 (1.005)    0.005 

 300 (1.005)(104.073927)  $31,378.29. (Set P/Y 12; C/Y  12)(―BGN‖ Mode) 0 PV 300  PMT 6 I/Y 84 N CPT FV 31,378.29 2.

PMT  360; i  1.75%; n  48 1.017548 –1  FV(due)  360 (1.0175)    0.0175 

 360 (1.0175)(74.262784)  $27,202.46 (Set P/Y 4; C/Y  4)(―BGN‖ Mode) 0 PV 360  PMT 7 I/Y 48 N CPT FV 27,202.46 3.

(a) PMT  530; i  0.98%; n  16 1.009816  1  FV(due)  530 (1.0098)    0.0098 

 530 (1.0098)(17.231536)  $9222.21. (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 PV 530  PMT 3.92 I/Y 16 N CPT FV.9222.21 (b) Interest  9222.21  530(16)  9222.21  8480  $742.21 (c) PMT  530; i  0.98%; n  20 1.009820  1  FV(due)  530 (1.0098)    0.0098 

 530 (1.0098)(21.976192)  $11,761.53. She will have 11,761.53  9222.21  $2539.32 more.

(Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 PV 530  PMT 3.92 I/Y 20 N CPT FV 11,761.53 4.

(a) PMT  1200; i  4.15%; n  18

1.041518  1  FV(due)  1200 (1.0415)    0.0415   1200(1.0415)(26.0014359)  $32,496.59. (Set P/Y  1; C/Y  1)(―BGN‖ Mode) 0 PV 1200  PMT 4.15 I/Y 18 N CPT FV 32,496.59 (b) Interest  32,496.59  1200 (18)  32,496.59  21,600  $10,896.59 (c) PMT  1200; i  4.65%; n  18 1.046518  1  FV(due)  1200 (1.0465)    0.0465 

 1200 (1.0465)(27.231)  $34,196.69. Difference in interest  34,196.69  32,496.59  $1700.10. (Set P/Y  1; C/Y  1)(―BGN‖ Mode) 0 PV 1200  PMT 4.65 I/Y 18 N CPT FV .34,196.69 5.

PMT  1600; i  1.65%; n  4

1  1.0165 –4  PV(due)  1600 (1.0165)    0.0165   1600(1.0165)(3.840292)  $6245.85. (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 FV 1600  PMT 6.6 I/Y 4 N CPT PV 6245.85 6.

(a) PMT  122.50; i  1.75%; n  30

1  1.017530  PV(due)  122.50 (1.0175)    122.50(1.0175)(23.185849)   0.0175  $2889.97 (b) Paid in installments  30(122.50)  $3675. (c) Interest  3675  2889.97  $785.03.

(Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 122.50  PMT 21 I/Y 30 N CPT PV 2889.97 (d) PMT  135; i  1.75%; n  30

1  1.017530  PV(due)  135(1.0175)    0.0175   135(1.0175)(23.185849)  $3184.87. (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 135  PMT 21 I/Y 30 N CPT PV 3184.87 7.

(a) PMT  62.25; i  2%; n  36

1  1.0236  PV(due)  62.25(1.02)   = 62.25(1.02)(25.488843)  $1618.41  0.02  (b) Total paid  36(62.25)  $2241. (c) Interest  2241  1618.41  $622.59. (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 62.25  PMT 24 I/Y 36 N CPT .PV 1618.41 (d) PV(due)  1618.41; i  1.6583%; n  36

1  1.01658336  1618.41  PMT(1.016583)    0.016583  1618.41  PMT(1.016583)(26.94499) 1618.41  27.391828PMT PMT  $59.08. (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 1618.41 PV 19.9 I/Y 36 N CPT PMT 59.08 8.

(a) PMT  3100; i  3.4%; n  18

1.03418  1  FV(due)  3100 (1.034)    0.034   3100 (1.034)(24.277911)  $77,820.41585

FV  77,820.41585(1.034)12 $116,235.83. (b) Total deposits  18(3100)  $55,800. (c) Interest  116,235.83  55,800  $60,435.83 (Set P/Y  2; C/Y  2)(―BGN‖ Mode) 0 PV 3100  PMT 6.8 I/Y 18 N.CPT FV 77,820.41585 77,820.41585  PV 12 N 0 PMT CPT FV 116,235.83 9.

FV(due)  5700; i  0.36%; n  60

1.003660  1  5700  PMT (1.0036)    0.0036  5700  PMT (1.0036)(66.839191) PMT  $84.97. (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 PV 5700 FV 4.32 I/Y 60 N CPT PMT .84.97 10. PV(due)  12,500; i  0.625%; n  48

1  1.0062548  12,500  PMT(1.00625)    0.00625  12,500  PMT(1.00625)(41.358371) PMT  $300.36 (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 12 500 PV 7.5 I/Y 48 N CPT PMT .300.36

11. PV(due)  37,750; i  4.5%; n  10

1  1.04510  37,750  PMT(1.045)    0.045  37,750  PMT(1.045)(7.912718) 37,750  8.268 790 PMT PMT  $4565.36 (Set P/Y  2; C/Y  2)(―BGN‖ Mode) 0 FV 37,750 PV 9 I/Y 10 N CPT PMT .4565.36 12. Value of annuity in 25 years: PMT  900; i  1.013125%; n  80

1  1.01312580  PV(due)  900 (1.013125)    900(1.013125)(49.345804)  0.013125  PV(due)  $44,994.12 Payments made into the annuity: FV(due)  44,994.12; i  1.31255%; n  100

1.013125100  1  44,994.12  PMT(1.013125)    0.013125  44,994.12  PMT(1.013125)(204.484637) 44,994.12  207.168498 PMT PMT  $217.19 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 FV 900  PMT 5.25 I/Y 80 N CPT PV 44,994.12 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 44,994.12 FV 0 PV 5.25 I/Y 100 N CPT PMT. 217.19 13. FV(due)  5000; PMT  240; i  0.41 6 %

1.00416n  1  5000  240 (1.00416)    0.00416  1.00416n  1  20.746888     0.00416  0.086445  1.00416 n  1 1.086445  1.0041 1 year, 8 months ln 1.086445  n ln 1. 00416 0.082911 = 0.004158n n  19.94 months  1 year, 8 months (Set P/Y  12; C/Y  I2)(―BGN‖ Mode) 0 PV 5000 FV 240  PMT 5 I/Y CPT N 19.940123 14. Amount needed for withdrawals: PMT  1000; i  2.5%; n  60

1  1.02560  PV  due   1000 1.025     0.025   1000 1.025  30.908657   $31 , 681.37 Number of deposits: FVn(due)  31,681.37; PMT  450; i  2.5%

1.025n  1  31,681.37  450 (1.025)    0.025 

1.025n  1 68.685897  0.025 1.025n  2.717147 n ln 1.025  ln 2.717147 0.024693n  0.999583 n  40.481038  10 years, 3 months. (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 FV 1000  PMT 10 I/Y 60 N CPT PV .31,681.37

(Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 PV 31,681.37 FV 450  PMT 10 I/Y .CPT .N 40.481038 15. Amount after 10 years: PMT  75; i  0.5%; n  120

1.005120  1  FVn(due)  75 (1.005)    0.005   75 (1.005)(163.879347)  $12,352.41 Withdrawals:

PVn  due   12 , 352.41; PMT  260; i  0.5% 1  1.005 n  12,352.41  260 1.005     0.005  1  1.005 n 47.272905  0.005 1.005n  0.763636 n ln 1.005  ln 0.763636 0.004988n  0.269665 n  54.067666 months  55 monthly withdrawals. Since the first withdrawal is immediate, the last withdrawal is 54 months from the first withdrawal, or 4 years, 6 months.. (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 PV 75  PMT 6 I/Y 120 N CPT FV.12,352.41 (Set P/Y  12; C/Y  12)( ―BGN‖ Mode) 0 FV 12,352.41  PV 260 PMT 6 I/Y.CPT N 54.067666 16. PV(due)  25,000; PMT  1445; i  2% 1  1.02 n  25,000  1445 (1.02)    0.02 

16.961802 

1  1.02 n 0.02

1.02-n  0.0660764 n ln 1.02  ln 0.660764 0.019803 n  0.414359 n  20.924425 quarters  5 years, 3 months. (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 FV 25,000  PV 1445 PMT 8 I/Y CPT N 20.924425 17. FV(due)  14,559; PMT  250; n  40 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 PV 14,559 FV 250  PMT 40 N CPT .I/Y 7.0001 The nominal annual rate is 7.0001% compounded quarterly. 18. FV(due)  250,000; PMT  4220; n  20 (Set P/Y  l; C/Y  l)(―BGN‖ Mode) 0 PV 250,000 FV 4220  PMT 20 N CPT .I/Y 9.4969 The effective annual rate is 9.4969%. . 19. PV(due)  25,000; PMT  2200; n  15 (Set P/Y  2; C/Y  2)(―BGN‖ Mode) 0 FV 25,000  PV 2200 PMT 15 N CPT .I/Y 8.5803 The nominal annual rate is 8.5803% compounded semi-annually. . 20. PV(due)  27,000; PMT  725; n  48 (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 27,000  PV 725 PMT 48 N CPT I/Y 13.6627 The nominal annual rate is 13.6627% compounded monthly. . Exercise 13.2 1.

PMT  50,000; i  1.5%; n  10; c 

4 2 2

p  1.0152  1  1.030225  1  3.0225%

1.03022510  1  FV(due)  50,000(1.030225)    0.030225   50,000(1.030225)(11.475765)  $591,131.02. (Set P/Y  2; C/Y  4)(―BGN‖ Mode) 0 PV 50,000  PMT 6 I/Y 10 N CPT FV 591,131.02 2.

PMT  125; i  3.25%; n  300; c 

2 1  12 6

p  1.0325 6  1  1.005345  1  0.5345%

1.005345300  1  FV(due)  125 (1.005345)    0.005345   125 (1.005345)(738.826477)  $92,846.91 (Set P/Y  12; C/Y  2)(―BGN‖ Mode) 0 PV 125  PMT 6.5 I/Y 300 N CPT FV.92,846.91 3.

PMT  750; i  2%; n  36; c 

4 1  12 3

p  1.02 3  1  1.006623  1  0.6623%

1  1.00662336  PV(due)  750 (1.006623)    0.006623   750 (1.006623)(31.936605)  $24,111.08

(Set P/Y  12; C/Y  4)(―BGN‖ Mode) 0 FV 750  PMT 8 I/Y 36 N CPT PV .24,111.08 4 PMT  12,500; i  2.25%; n  14; c  2 2 p  1.02252  1  1.045506  1  4.5506%

1  1.045506 14  PV(due)  12,500(1.045506)    0.045506   12,500(l.045506)(10.189284)  $133,162.01 (Set P/Y  2; C/Y  4)(―BGN‖ Mode) 0 FV 12,500  PMT 9 I/Y 14 N CPT PV .133,162.01

PMT  121,300; i  0.995%; n  8, c 

4 4 1

p  (1.009954)  1  1.040380  1  4.0380%

1  1.0403808  PV(due)  121,300(1.040380)    0.040380   121,300(1.040380)(6.721695)  $848,279.71 (Set P/Y  1; C/Y  4)(―BGN‖ Mode) 0 FV 121,300  PMT 3.98 I/Y 8 N CPT .PV 848,279.71 6.

FV(due)  10,000; i  1.75%; n  16; c 

4 2 2

p  1.01752  1  1.035306  1  3.5306%

1.03530616  1  10,000  PMT(1.035306)    0.035306  l0,000  PMT(1.035306)(21.022164) PMT  $459.47.

(Set P/Y  2; C/Y  4)(―BGN‖ Mode) 0 PV 10,000 FV 7 I/Y 16 N CPT PMT .459.47 4 1 PV(due)  5400; i  1.375%; n  36; c   12 3 1

p  1.01375 3  1  1.004563  1  0.4563%

1  1.00456336  5400  PMT(1.004563)    0.004563  5400  PMT(1.004563)(33.129447) PMT  $162.26 (Set P/Y  12; C/Y  4)(―BGN‖ Mode) 0 FV 5400 PV 5.5 I/Y 36 N CPT PMT .162.26 8.

FV(due)  14,000; i  3.5%; n  12; c 

2  0.5 4

p  1.0350.5  1  1.017349  1  1.7349%

1.01734912  1  14,000  PMT(1.017349)    0.017349  14,000  PMT(1.017349)(13.21394657)

14,000  13.443202PMT PMT  $1041.42 (Set P/Y  4; C/Y  2)(―BGN‖ Mode) 14,000 FV 0 PV 7 I/Y 12 N CPT PMT .1041.42 9.

PV(due)  640,000; i  1.3925%; n  60; c 

4  0.3 12

p = 1.0139250.3  1  1.004620  1  0.4620% &

1  1.004620 60  640,000  PMT(1.004620)    0.004620  640,000  PMT(1.004620)(52.296778) 640,000  52.538405PMT PMT  $12,181.57. (Set P/Y  12; C/Y  4)(―BGN‖ Mode) 0 FV 640,000 PV 5.57 I/Y 60 N CPT .PMT 12,181.57 1 10. PV(due)  85,000; i  6.125%; PMT  3000; c  4 1

p  1.06125 4  1  1.014973  1  1.4973%

1  1.014973 n  85,000  3000 (1.014973)    0.014973  1  1.014973 n 27.915359  0.014973 1.014973n  1  0.417975 

n ln 1.014973  ln 0.582026 0.014862n  0.541240 n  36.417884  37 quarterly payments Since first payment is now, her last withdrawal will be 36 quarters, or 9 years from now..

(Set P/Y  4; C/Y  1)(―BGN‖ Mode) 0 PV 85,000  PV 3000 PMT 6.125 I/Y CPT N 36.417884 4 11. FV(due)  96,000; PMT  1600; i  2.5%; c   4 1

p  1.0254  1  1.103813  1  10.3813% 1.103813n  1  96, 000  1600 1.103813    0.103813  1.103813n  1 54.357038  0.103813 n 1.103813  6.642962 nln 1.103813  ln 6.642962 0.098770n  1.893558 n  19.171299 years  20 years (Set P/Y  1; C/Y  4)(―BGN‖ Mode) 0 PV 96,000 FV 1600  PMT 10 I/Y CPT N .19.171299 12. PV(due)  242; PMT  25; i  1.75%; c 

12 3 4

p  1.01753  1  1.053424  1  5.3424%

1  1.053424 n  242  25 (1.053424)    0.053424  1  1.053424 n  9.189082     0.053424  0.490917  1  1.053424n 1.053424n  0.509083 n ln 1.053424  ln 0.509083 0.052046n  0.675144 n  12.972144 (quarters)  3 years, 3 months. (Set P/Y  4; C/Y  12)(―BGN‖ Mode) 242 PV 0 FV 25  PMT 21 I/Y CPT .N 12.972144 13. PV(due)  21,600; PMT  680; n  36; c 

1 12

(Set P/Y  12; C/Y  1)(―BGN‖ Mode) 0 FV 21,600  PV 680 PMT 36 N .CPT I/Y 9.1776 The nominal annual rate is 9.1776% compounded annually.. 14. FV(due)  50,000; PMT  1700; n  15; c 

4 4 1

(Set P/Y  1; C/Y  4)(―BGN‖ Mode) 0 PV 50,000 FV 1700  PMT 15 N CPT. I/Y 7.8029 STO 1 2nd I Conv Nom  RCL 1 Enter ↓ ↓ C/Y  4 Enter ↓ ↓ CPT Eff  8.0342 The effective annual rate is 8.0342% compounded annually.. Business Math News Box 1. (a) Government will contribute $400 per year per child if parents contribute a minimum of $2000 contribution. Therefore, total annual contribution is $2400 per child.

16 N 0 PV 2400 PMT 4.85 I/Y Set P/Y  1;C/Y  2  CPT FV  $56,359.34 The child will have $56,359.34 for college tuition. (b) Value of non-RESP plan over a 16-year period:

16 N 0 PV 2000 PMT 4.85 I/Y Set P/Y  1;C/Y  2  CPT FV  $46,966.11 The child will have $46,966.11 for college tuition. (c) Difference in value of RESP plan versus a non-RESP plan is $9393.23($56,359.34  $46,966.11 ). 2. Government will contribute $1 for every $5 contributed. Therefore, $200 is contributed by the government per year. Total contributions are $1200 per year at the beginning of each year. (a)

STEP 1 Calculate the future value of contributions for the first five years:

5 N 0 PV 1200 PMT 4 I/Y Set P/Y  1;C/Y  1;BGN ON  CPT FV  $6759.57 STEP 2 Calculate the future value of invested amount and contributions for the last five years:

5 N 6759.57 PV 1200 PMT 5.2 I/Y Set P/Y  1;C/Y  4;BGN ON  CPT FV  $15,776.64

The child would have $15,766.64.

(b)

Calculate annuity payments from the future value calculated after 10 years:

(―BGN‖ Mode) (Set P/Y = 1; C/Y = 4) 4 N 15,776.64 PV 0 FV 5.2 I/Y CPT PMT

$4254.88 Student will receive payments of $4254.88 per year before tax. After-tax payments are $3148.61 [ $4254.88 × (1  0.26)]. Exercise 13.3 1.

PVn(defer)  14,500; i  3.5%; n  20; d  14 FV  14,500 (1.035)14 FV  14,500 (1.618695) FV  23,471.07058

1  1.03520  23,471.07058  PMT    0.035  23,471.07058  14.212403 PMT PMT  $1651.45 (Set P/Y  2; C/Y  2) 0 PMT 14,500  PV 7 I/Y 14 N CPT FV 23,471.07058 (Set P/Y  2; C/Y  2) 0 FV 23,471.07058  PV 7 I/Y 20 N CPT PMT 1651.45 2.

PMT  220; i  0.675%; n  36; d  15

1  1.0067536  PV  220    220(31.86487)  $7010.272424.  0.00675  PV(defer)  7010.272424(1.0067515)  7010.27(0.904015)  $6337.39 (Set P/Y  12; C/Y  12) 0 FV 220  PMT 8.1 I/Y 36 N CPT PV 7010.272424 (Set P/Y  12; C/Y  12) 7010.272424 FV 0 PMT 8.1 I/Y 15 N CPT PV 6337.39

(a) PMT  8500; i  2.25%; n  32; d  12

1  1.022532  PVn  8500    8500 (22.637674)  192,420.2306  0.0225  PVn(defer)  192,420.2306(l.022512)  192,420.2306(0.765668)  $147,329.91. (Set P/Y  4; C/Y  4) 0 FV 8500 PMT 9 I/Y 32 N CPT PV 192,420.2306 (Set P/Y  4; C/Y  4) 192,420.2306 FV 0 PMT 9 I/Y 12 N CPT PV 147,329.91 (b) Amount repaid  $8500(32)  $272,000. (c) Interest paid  $272,000  $147,329.91  $124,670.09. 4.

PVn(defer)  8000; i  0.35%; n  48; d  66 FV  8000 (1.0035)66 FV  8000 (1.259351) PVn  10,074.80983

1  1.003548  10,074.80983  PMT    0.0035  10,074.80983  44.113764 PMT PMT  $228.38 (Set P/Y  12; C/Y  12) 0 PMT 8000  PV 4.2 I/Y 66 N CPT FV 10,074.80983

(Set P/Y  12; C/Y  12) 0 FV 10,074.80983  PV 4.2 I/Y 48 N CPT PMT 228.38 Amount after 5 years: PV  4000; i  1%; n  20 FV  4000 (l.0l)20  4000 (1.220190)  $4880.76016 Withdrawals: PVn  4880.76016; PMT  500; i  1%

1  1.01 n  4880.76016  500    0.01  9.761520 

1  1.01 n 0.01

1.01n  0.902385 n ln 1.01  ln 0.902385 0.009950n  0.102714 n  10.322697 quarters  2 years 9 months.

(Set P/Y  4; C/Y  4) 0 PMT 4000  PV 4 I/Y 20 N CPT FV 4880.76016 (Set P/Y  4; C/Y  4) 0 FV 4880.76016  PV 500 PMT 4 I/Y CPT N 10.322697 6.

PVn(defer)  6500; PMT  300; i  0.53%; d  48 FV  6500 1.0053

FV  6500 (1.290874) PVn  8390.683035

1  1.0053 n  8390.683035  300    0.0053  0.149168  1  1.0053 n

1.0053 n  0.850532 n ln 1.0053  ln 0.850532 0.005319n  0.161540 n  30.369492 months  2 years, 7 months. (Set P/Y  12; C/Y  12) 0 PMT 6500  PV 6.4 I/Y 48 N CPT FV 8390.683035 Set P/Y  12; C/Y  12) 0 FV 8390.683035  PV 300 PMT 6.4 I/Y CPT N 30.369492 7.

PVn(defer)  10,000; PMT  500; i  0.39%; d  36 FV  10,000,(1.0039)36 FV  10,000(1.150420) PVn  11,504.19812 (Set P/Y  12; C/Y  12) 0 PMT 10,000  PV 4.68 I/Y 36 N CPT FV 11,504.19812 (Set P/Y  12; C/Y  12) 0 FV 11,504.19812  PV 500 PMT 24 N CPT I/Y 4.0842 The nominal rate of interest is 4.0842% compounded monthly.. .

(a) PVn (defer)  1500; PMT  114; i  1.03%; d  12 FV  1500 (1.0103)12

FV  1500 (1.131296) FV  1696.943719 (Set P/Y  12; C/Y  12) 0 PMT 1500  PV 12.4 I/Y 12 N CPT FV 1696.943719 (Set P/Y  12; C/Y  12) 0 FV 1696.943719  PV 114 PMT 18 N CPT I/Y 24.9751 The nominal rate of interest is 24.9751% compounded monthly. (b) 18(114)  1500  2052  1500  $552.

Amount after 18 years: PV  2000; i  1.25%; n  72 FV  2000 (l.012572) FV  2000 (2.445920) FV  4891.840536 Withdrawals: PVn  4891.840536; i  0.5%; n  48

1  1.00548  4891.840536  PMT    0.005  4891.840536  42.580318 PMT PMT  $114.89 (Set P/Y  4; C/Y  4) 0 PMT 2000  PV 5 I/Y 72 N CPT FV 4891.840536 (Set P/Y  12; C/Y  12) 0 FV 4891.840536  PV 6 I/Y 48 N CPT PMT 114.89 10. PMT  800; i  3.75%; n  120; d  84; c 

2 1  12 6

1 6

p  1.0375  1  1.006155  1  0.6155%

1  1.006155120  PVn  800    800 (84.670669)  67,736.53524  0.006155  PVn (defer)  67,736.53524(l.00615584)  67,736.53524(0.597264)  $40,456.61 (Set P/Y  12; C/Y  2) 0 FV 800 PMT 7.5 I/Y 120 N CPT PV 67,736.53524 (Set P/Y  12; C/Y  2) 67,736.53524 FV 0 PMT 7.5 I/Y 84 N CPT PV 40,456.61 11. PMT  5000; i  2%; n  40; d  8

1  1.0240  PVn  5000    5000 (27.355479)  136,777.3962  0.02  PVn(defer)  136,777.3962(l.028)  136,777.3962(0.853490)  $116,738.19

Purchase price  100,000 + 116,738.19  $216,738.19. (Set P/Y  4; C/Y  4) 0 FV 5000 PMT 8 I/Y 40 N CPT PV 136,777.3962 (Set P/Y  4; C/Y  4) 136,777.3962 FV 0 PMT 8 I/Y 8 N CPT PV 116,738.19

12. PVn(defer)  6200; PMT  230; d  36; P/Y  12; C/Y  1; c  1/12; I/Y  7.64; i  7.64% p  1.07641/12  1  1.006154  1  0.006154  0.6154% FV  6200 (1.006154)36  6200 (1.247157)  $7732.372307

(1  1.006154 n ) 0.006154 7732.372307  37,373.84048(1  1.006154n) 7732.372307  230

0.206893  1  1.006154n 1.006154n  0.793107 n 1n 1.006154  1n 0.793107 n 0.006135  0.231797 n  37.781571  38 monthly payments. (Set P/Y  12; C/Y  l) 0 PMT 6200  PV 7.64 I/Y 36 N CPT FV 7732.372307 (Set P/Y  12; C/Y  l) 0 FV 7732.372307  PV 230 PMT 7.64 I/Y CPT N .37.781571 13. PMT  600; i  1.5%; p  0.4975%; n  48; d  24

(1  1.00497548 ) 0.004975  600  42.605060 

PVn  600

 $25,563.03587

PVn  defer   25,563.03597(1.00497524 )  25, 563.03587  0.887711  $22, 692.59 (Set P/Y  12; C/Y  4) 0 FV 600  PMT 6 I/Y 48 N CPT PV 25,563.03587 (Set P/Y  12; C/Y  4) 0 PMT 25,563.03587  FV 24 N 6 I/Y CPT PV 22,692.59 14. PVn (defer)  25,000; PMT  1500; d  12; P/Y  4; C/Y  12; c  12/4  3;

I/Y  2.75; i  0.2292% p  1.0022923  1  1.006891  1  0.006891  0.6891% FV  25,000 (1.006891)12  25,000 (1.085896)  $27,147.40435

(1  1.006891 n ) 0.006891 27,147.40435  217,682.5812 (1  1.006891 n ) 27,147.40435  1500

0.124711  1  1.006891 n 1.006891 n  0.875289 n ln 1.006891  ln 0.875289  n 0.006867  0.133201 n  19.396900 quarters  5 years (Set P/Y  4; C/Y  12) 0 PMT 25,000  PV 2.75 I/Y 12 N CPT FV 27,147.40435 (Set P/Y  4; C/Y  12) 0 FV 27,147.40435  PV 1500 PMT 2.75 I/Y CPT N . 19.396900 15. PVn (defer)  12,000; d  12; P/Y  4; C/Y  2; c  2/4  0.50; I/Y  4.48; i  2.24% p  1.02240.50  1  1.011138  1  0.011138  1.1138% FV  12,000 (1.011138)12  12,000 (1.142155)  $13,705.86 PMT  1000; P/Y  4; C/Y  2; c  2/4  0.50; I/Y  3.82; i  1.91% p  1.01910.50  1  1.009505  1  0.009505  0.9505%

(1  1.009505 n ) 0.009505 13,705.85999  105,209.6769(1  1.009505 n ) 13,705.85999  1000

0.130272  1  1.009505 n 1.009505 n  0.869728  n ln 1.009505  ln 0.869728  n 0.009460  0.139575 n  14.754275  15 quarterly withdrawals (Set P/Y  4; C/Y  2) 0 PMT 12,000  PV 4.48 I/Y 12 N CPT FV 13,705.85999

(Set P/Y  4; C/Y  2) 0 FV 13,705.85999  PV 1000 PMT 3.82 I/Y CPT N 14.754275

16. PVn (defer)  9000; i  1.25%; n  84; d  36; c 

4 1  12 3

p  1.0125 3  1  1.004149  1  0.4149% FV  9000 (1.004149)36 FV  9000 (1.160755) FV  10,446.79066

1  1.004149 84  10,446.79066  PMT    0.004149  10,446.79066  70.800595 PMT PMT  $147.55 (Set P/Y  12; C/Y  4) 0 PMT 9000  PV 5 I/Y 36 N CPT FV 10,446.79066 (Set P/Y  12; C/Y  4) 0 FV 10,446.79066  PV 5 I/Y 84 N CPT PMT 147.55 17. PVn (defer)  10,000; i  4%; n  120; d  48; c 

2 1  12 6

p  1.04 6  1  1.006558  1  0.6558% FV  10,000 (1.006558)48 FV  10,000 (1.368569) FV  13,685.6905

1  1.006558120  13,685.6905  PMT    0.006558  13,685.6905  82.890613 PMT PMT  $165.11 (Set P/Y  12; C/Y  2) 0 PMT 10,000  PV 8 I/Y 48 N CPT FV 13,685.6905 (Set P/Y  12; C/Y  2) 0 FV 13,685.6905  PV 8 I/Y 120 N CPT PMT 165.11

18. PMT  8000; i  2.67%; p  1.3262%; n  16; d  20 PVn  8000

(1  1.01326216 ) 0.013262

 8000 (14.331298)  $114,650.3769 PVn (defer)  114,650.3769(1.013262)20  114,650.3769(0.768359)  $88,092.69 (Set P/Y  4; C/Y  2) 0 FV 8000  PMT 5.34 I/Y 16 N CPT PV 114,650.3769 (Set P/Y  4; C/Y  2) 0 PMT 114,650.3769  FV 20 N 5.34 I/Y CPT PV 88,092.69 19. PV  28,000; i  8.32%; n  24 FV  28,000 (1.0832)2  28,000(1.173322)  $32,853.02272 PV(defer)  32,853.02272; FV0; PMT  679; n  60 The future value of the period of deferment is $32,853.02272. The interest rate is 8.8132% compounded monthly.. (Set P/Y  12; C/Y  1) 28,000  PV 0 PMT 24 N 8.32 I/Y CPT FV 32,853.02272 (Set P/Y  12; C/Y  12) 0 FV 32,853.02272 PV 679  PMT 60 N CPT I/Y 8.813192% 20. Value of project in one year: PV  14,000,000; i  7.25%; n  1 FV  14,000,000 (1.0725)  $15,015,000 Less $3,000,000  $12,015,000 still owing Over next two years: PV(defer)  12,015,000; PMT  1,620,000; n  8 The annual interest rate is 7.03% compounded annually..

(Set P/Y  4; C/Y  1) 0 FV 12,015,000 PV 1,620,000  PMT 8 N CPT I/Y 7.033650693

Exercise 13.4 1.

PVn(defer)  50,000; i  1.75%; n  28; d  12 FV  50,000 (1.0175)12 FV  50,000 (1.231439) FV  61,571.96575

1  1.017528  61,571.96575  PMT(due)(1.0175)    0.0175  61,571.96575  PMT(due)(1.0175)(21.986955) PMT(due)  $2752.22.

(Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 PMT 50,000  PV 7 I/Y 2 N CPT FV 61,571.96575 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 FV 61,571.96575  PV 7 I/Y 28 N CPT PMT 2752.22 PVn(defer)  14,000; i  3.25%; n  6; d  8 FV  14,000 (1.0325)8 FV  14,000 (1.291578) FV  18,082.08549

1  1.03256  18,082.08549  PMT(due)(1.0325)    0.0325 

18,082.08549  PMT(due)(1.0325)(5.372590) PMT(due)  $3259.68. (Set P/Y  2; C/Y  2)(―BGN‖ Mode) 0 PMT 14,000  PV 6.5 I/Y 8 N CPT FV .18,082.08549 (Set P/Y  2; C/Y  2)(―BGN‖ Mode) 0 FV 18,082.08549  PV 6.5 I/Y 6 N CPT .PMT 3259.68 PVn(defer)  12,000; i  0.6 %; n  60; d  12 FV  12,000 ( 1.006 )12 FV  12,000 (1.083) FV  12,995.99408

1  1.00660  12,995.99408  PMT(due)(1.006)    0.006  12,995.99408  PMT(due)(1.006) (49.318433) PMT(due)  $261.77. (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 PMT 12,000  PV 8 I/Y 12 N.CPT FV 12,995.99408

(Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 12,995.99408  PV 8 I/Y 60 N CPT PMT 261.77

PVn(defer)  21,000; PMT(due)  3485; i  2.5%; d  32 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 21,000  PV 10 I/Y 32 N CPT FV 46,278.89569 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 FV 46,278.89569  PV 3485 PMT 10 I/Y CPT . N 15.850821  16 Quarterly payments It will last 3 years, 9 months, since the first payment is at the beginning.

PVn(defer)  32,000; PMT(due)  4000; i  3%; d  12 FV  32,000 (1.03)12 FV  32,000 (1.425761) FV  45,624.34838 1  1.03 n  45, 624.34838  4000 1.03    0.03 

0.332216  1  1.03 n 1.03 n  0.667784  n ln1.03  ln 0.667784  0.029559n  0.403791 n  13.660591 quarters  3 years, 6 months.

(Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 PMT 32,000  PV 12 I/Y 12 N CPT FV .45,624.34838 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 FV 45,624.34838  PV 4000 PMT 12 I/Y.CPT N 13.660591 PVn(defer)  23,000; PMT(due)  1800; i  0.916%; d  9 FV  23, 000(1.00916)9 FV  23,000(1.085591) FV  24,968.58378

1  1.00916 n  24,968.58378  1800(1.00916)    0.00916  0.12600  1  1.00916 n 1.009167 n  0.874000 n ln 1.009167  ln 0.874000  0.009125n  0.134675 n  14.759 months  1 year, 3 months (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 PMT 23,000  PV 11 I/Y 9 N CPT FV. 24,968.58378

(Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 24,968.58378  PV 1800 PMT 11 I/Y CPT N 14.759021 7.

PV  7200; i  1%; n  12 FV  7200 (1.01)12  7200(1.126825)  8113.140217 PV(due)  8113.140217; PMT  450; n  20 The annual interest rate is 4.4797% compounded quarterly. . (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 8113.140217 PV 0 FV 450  PMT 20 N CPT I/Y 4.479666

PV  1384; i  2.4%; n  12 FV  1384(1.024)12  1384(1.329228)  1839.651546 PV(due)  1839.651546; PMT  95; n  24 The annual interest rate is 23.565142% compounded monthly or

23.565142%  1.9638% per month.. 12 (P/Y  12; C/Y  12)(―BGN‖ Mode) 1839.651546 PV 0 FV 95  PMT 24 N CPT I/Y 23.565142 9.

PV  30,000; i  2.9%; n  20 FV  30,000 (1.029)20  30,000(1.771363)  53,140.88098 PV(due)  53,140.88098; PMT  850; n  80 The annual interest rate is 2.6202% compounded quarterly, (P/Y  4; C/Y  4)(―BGN‖ Mode) 53,140.88098 PV 0 FV 850  PMT 80 N CPT I/Y 2.620207

10. (a) PMT(due)  400; i  0.5%; n  48; d  90 1  1.00548  PVn  400(1.005)    400(1.005)(42.580318)  17,117.28775  0.005 

PVn(defer)  17,117.28775(1.00590)  17,117.28775(0.638344)  $10,926.71 . (b) Total withdrawals  48(400)  $19,200.

(c) Interest  19,200  10,926.71  $8273.29. (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 400 PMT 6 I/Y 48 N CPT PV.17,117.28775 (Set P/Y  12; C/Y  12) 17,117.28775 FV 0 PMT 6 I/Y 90 N CPT PV 

11. (a) PMT(due)  1600; i  7.12%/12  0.593 %; n  24; d  18

1  1.0059324  PVn  1600(1.00593)    1600 (1.00593) (22.307984)  35,904.55171  0.00593  PVn(defer)  35,904.55171( 1.00593 18)  35,904.55171(0.898989)  $32,277.80 (b) Total withdrawals  24(1600)  $38,400. (c) Interest  38,400  32,277.80  $6122.20. (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 1600 PMT 7.12 I/Y 24 N CPT PV . 35,904.55171 (Set P/Y  12; C/Y  12) 35,904.55171 FV 0 PMT 7.12 I/Y 18 N CPT PV 32,277.80 12. PMT(due)  2500; i  1%; n  60; d  24 1  1.0160  PVn  2500(1.01)    2500(1.01)(44.955038)  $113,511.472  0.01 

PVn(defer)  113,511.472(l.0124)  113,511.472(0.787566)  $89,397.79. (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 2500 PMT 12 I/Y 60 N CPT PV 113,511.472 (Set P/Y  12; C/Y  12) 113,511.472 FV 0 PMT 12 I/Y 24 N CPT PV 89,397.79 13. PVn (defer)  18,000; i  1.75%; n  16; d  4; c 

4 2 2

p  1.01752  1  1.035306  1  3.5306% FV  18,000(1.035306)4 FV  18,000(1.148881) FV  20,679.87209 1  1.035306 16  20,679.87209  PMT(due)(l.035306)    0.035306 

20 679.87209  PMT  due 1.03530612.066349 PMT  due   $1655.40

(Set P/Y  2; C/Y  4)(―BGN‖ Mode) 0 PMT 18,000  PV 7 I/Y 4 N CPT FV 20,679.87209 (Set P/Y  2; C/Y  4)(―BGN‖ Mode) 0 FV 20,679.87209  PV 7 I/Y 16 N CPT PMT 1655.40

14.

PVn (defer)  250,000; i  0.6916%; n  36; d  8; c 

12 3 4

p  1.0069163  1  1.020894  1  2.0894%

FV  250,000(1.020894)8 FV  250,000(1.179899) FV  294,974.664 1  1.020894 36  294,974.664  PMT(due)(1.020894)    0.020894 

294,974.664  PMT(due)(1.020894)(25.126736) PMT(due)  $11,499.21. (Set P/Y  4; C/Y  12)(―BGN‖ Mode) 0 PMT 250,000  PV 8.3 I/Y 8 N CPT FV 294,974.664 (Set P/Y  4; C/Y  12)(―BGN‖ Mode) 0 FV 294,974.664  PV 8.3 I/Y 36 N CPT PMT 11,499.21 15. PV  75,000; i  7.82%; n  1 FV  75,000(1.0782)  80,865 PV(due)  80,865; i  7.82%; n  10, c 

1  0.5 2

p  (1.0782)0.5  1  1.038364098  1  3.8364% 1  1.038364 10  80,865  PMT(1.038364)    0.038364 

80,865  PMT(1.038364)(8.177354) PMT  $9523.53 (P/Y  2; C/Y  1)(―BGN‖ Mode) 0 FV 80,865 PV 7.82 I/Y 10 N CPT PMT .9523.53 16. PVn (defer)  22,750; PMT(due)  385; i  2.5%; d  12; c  1

p  1.025 6  1  1.004124  1  0.4124% FV  22,750(1.004124)12

2 1  12 6

FV  22,750(1.050626) FV  23,901.71875

1  1.004124 n  23,901.71875  385 1.004124     0.004124  0.254971  1  1.004124  n 1.004124 n  0.745029  nln 1.004124  ln 0.743977 0.004115n  0.294332 n  71.519086 months  6 years (Set P/Y  12; C/Y  2)(―BGN‖ Mode) 0 PMT 22,750  PV 5 I/Y 12 N CPT FV 23,901.71875 (Set P/Y  12; C/Y  2)(―BGN‖ Mode) 0 FV 23,901.71875  PV 385 PMT 5 I/Y . CPT N 71.519086 17. PVn (defer)  3740; PMT(due)  1100; i  0.6 %; d  10; c  12 p  1.00612  1  1.082999  1  8.2999% FV  3740(1.082999)10 FV  3740(2.219640) FV  8301.45

1  1.082999 n  8301.45  1100(1.082999)    0.082999  0.578374  1  1.082999n 1.082999n  0.421626 n ln 1.082999  ln 0.421626 0.079735n  0.863636 n  10.831393 years  11 yearly payments. (Set P/Y  l; C/Y  12)(―BGN‖ Mode) 0 PMT 3740  PV 8 I/Y 10 N CPT FV 8301.45 (Set P/Y  l; C/Y  12)(―BGN‖ Mode) 0 FV 8301.45  PV 1100 PMT 8 I/Y CPT . N 10.831393  11 yearly payments

He will be able to withdraw for 10 years, since the first payment is on his sixty-fifth birthday.

18. PVn (defer)  45,000; PMT  15,000; i  5.5%; d  5; c  2 p  1.0552  1  1.113025  1  11.3025% FV  45,000(1.113025)5 FV  45,000(1.708144) FV  76,866.50063

1  1.113025 n  76,866.50063  15,000 1.113025     0.113025  0.520374  1  1.113025 n 1.113025 n  0.479626 nln 1.113025  ln 0.479626  0.1070815n  0.734748 n  6.861578 years  7 yearly payments (Set P/Y  1; C/Y  2) 0 PMT 45,000  PV 11 I/Y 5 N CPT FV 76,866.50063 (Set P/Y  1; C/Y  2)(―BGN‖ Mode) 0 FV 76,866.50063  PV 15,000 PMT 11 I/Y.. CPT N 6.861578 19. PV  82,000; i  1.8725%; n  6 FV  82,000(1.018725)6  82,000(1.117743)  91,654.88923 PV(due)  91,654.88923; PMT  2200; n  48 The interest rate is 7.4885% compounded quarterly.. (P/Y  12; C/Y  4)(―BGN‖ Mode) 91,654.88923 PV 0 FV 2200  PMT 48 N CPT I/Y 7.4885 20. PV  28,000; i  0.5%; n  24 FV  28,000(1.005)24  28,000(1.12716)  31,560.47373 PV(due)  31,560.47373; PMT  625; n  60 The interest rate is 7.3684% compounded semi-annually.. (P/Y  12; C/Y  2)(―BGN‖ Mode) 31,560.47373 PV 0 FV 625  PMT 60 N .CPT I/Y 7.3684 21. PV  189,000; i  2.645%; n  2

FV 189,000 (1.02645)2  189,000(1.05360)  199,130.3249 PV(due)  199,130.3249; PMT  1169.51; n  300 The interest rate is 5.1435% compounded semi-annually.. (P/Y  12; C/Y  2)(―BGN‖ Mode) 199,130.3249 PV 0 FV 1169.51  PMT 300 N CPT I/Y 5.143478

22. PMT(due)  6000; i  0.75%; n  20; d  8; c 

12 3 4

p  1.00753  1  1.022669  1  2.2669%

 (1  1.02266920 )  PVn  due   6000 1.022669     0.022669   6000 1.022669 15.937958   97,975.55 PV 97,975.55 (1.022669)8  97,975.55(0.835831)  $81,740.59. (Set P/Y  4; C/Y  12)(―BGN‖ Mode) 0 FV 6000  PMT 9 I/Y 20 N CPT PV 97,975.55 (Set P/Y  4; C/Y  12) 97,975.55 FV 0 PMT 9 I/Y 8 N CPT PV  23. PMT(due)  15,000; i  3.275%; n  12; c 

2 2 1

p  (1.03275)2  1  1.066573  1  6.6573%

1  1.06657312  PV(due)  15,000(1.066573)    0.066573   15,000(1.066573)(8.08982124)  $129,425.72 FV  12,9425.72; i  3.275%; n  4; PV(defer)  12,9425.72(1.03275)4  12,9425(0.879061)  $113,773.15 The agreement today is worth 30,000 + 113,773.15  $143,773.15.. (P/Y  1; C/Y  2)(―BGN‖ Mode) 0 FV 15,000  PMT 6.55 I/Y 12 N CPT PV 129,425.72 24. PMT(due)  18,000; i  0.365%; n  12; d  7; c 

12  12 1

p  1.0036512  1  1.044690  1  4.469%

1  1.0446912  PVn  due   18, 000 1.044690     0.04469   18, 000 1.04469  9.134769   171,774.0438 PVn (defer)  171,774.0438 (1.04469)7  171,774.0438 (0.736356)  $126,486.82

(Set P/Y  1; C/Y  12)(―BGN‖ Mode) 0 FV 18,000 PMT 4.38 I/Y 12 N CPT PV . − 171,774.0438 (Set P/Y  l; C/Y  12) 171,774.0438 FV 0 PMT 4.38 I/Y 7 N CPT PV 126,486.8186

Exercise 13.5 1.

PMT  32; i  0.3% PV 

32  $9600. 0.003

PV  $150,000; i  4.5%  0.045 PMT  $150,000(0.045)  $6750.

PMT  4.25; i  4%; c 

2  0.5 4

p  1.040.5  1  1.019804  1  1.9804% PV  4.

4.25  $214.60. 0.019804

PMT  85,000; i  1.75%; c  4 p  1.01754  1  1.071859  1  7.1859% PV 

85,000  $1,182,872 0.071859

PMT  13,000; i  1.1875%; c 

4 2 2

p  1.0118752  1 = 1.023891  1  2.389102% PV  6.

13,000  $544,137.60 0.023891

PMT  2500; i  7.25%; d  4

2500   PV  due   1.07254  2500  0.0725    0.755807  2500  34, 482.76   $27,951.82 7.

 55.65; i  3.1%; c 

2  0.5 4

p  1.0310.5  1  1.015382  1  1.015382% 55.65 

PMT 0.015382

PMT  $0.86.

PMT  1150; i  0.55% PV(due) 

1150 + 1150  209,090.91 + 1150  $210,240.91 0.0055

PV(due)  836,000

3600 i

836,000  3600 + 832,400 

3600 i

i  0.004325 (monthly) The annual yield earned is 5.1898% compounded monthly.. 10. PMT  2225 (due); i  1.375%; c  4 p  1.013754  1  1.056145  1  5.6145% PV(due)  2225 +

2225  2225 + 39,629.67  $41,854.67 0.056145

11. PV(due)  35,000; i  4.25%; c 

2 1   0.16 12 6

p  1.04250.16  1  1.006961  1  0.6961%

PMT 0.006961 35, 000  PMT  143.65546 PMT 144.65546 PMT  35, 000 35, 000  PMT 

PMT  $241.95 12. PMT  480 (due); i  4.3%; PV(due)  480 +

480  480 + 11,162.79  $11,642.79. 0.043

13. PMT  1200; i  0.75% FV36  1200/0.0075  160,000 PV(defer)  160,000(1.007536)  $122,263.83. 14. PV  14,000; i  7%; d  4 PMT  [(14,000)(1.07)4](0.07) PMT  14,000(1.310796)(0.07)  $1284.58. 15. PV 280,000; i  0.5%; d  36 Amount after 3 years: FV  280,000(1.005)36  280,000(1.196681)  $335,070.55

335,070.55  PMT +

PMT 0.005

335,070.55  PMT + 200 PMT 335,070.55  201 PMT PMT  $1667.02 16. Answer: p = (1 + 0.0827/4)4/1 – 1 = 0.08530027 PV = 60,000/0.08530027 = $703,397.56 17. PV = PMT/i i = PMT/PV i = $6500/$1,000,000 i = 0.0065 j = mi j = 12(0.0065) = 0.078 = 7.80% compounded monthly 18. Answer: FV = 238,455 (1.004483)36 = 238,455(1.174731) = $280,120.51 = PV(due) 280,120.51 = PMT + PMT/ 0.004483 280,120.51 = PMT + 223.048327 PMT 280,120.51 = 224.048327 PMT $1250.27 = PMT 19. Answer: p = ( 1 + 0.062/4)4/1 – 1 = 0.063456 PV1 = 5000/0.063456 = $78,794.19 PV2 = 78,794.19(1.063456)–2 = 78,794.19(0.884220) = $69,671.44 20. PV = PMT/(i – k) PV = 25/(0.0475 – 0.01) = $666.67 21. PV = PMT/(i – k) PV = 32/(0.0525 – (– 0.005)) = 32/0.0575 = $556.52 Review Exercise 1.

PMT  50; i  0.385%; n  50

1.0038550  1  FV(due)  50(1.00385)    50(1.00385)(55.020393)  $2761.61.  0.00385  (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 PV 50  PMT 4.62 I/Y 50 N CPT FV 2761.611062 2.

(a) PMT  225; i  0.75%; n  32

1.007532  1  FV(due)  225(1.0075)    225(1.0075)(36.014830)  $8164.11  0.0075  (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 PV 225  PMT 3 I/Y 32 N CPT FV 8164.11 (b) 225(32)  $7200. (c) 8164.11  7200  $964.11. 3.

PMT  25; i  1%; n  360; c 

4  0.3 12

p  1.010.3  1  1.003322 1  0.3322% 1.003322360  1  FV(due)  1.003322(25)    0.003322   1.003322(25)(692.411564)  $17,367.79. (Set P/Y  12; C/Y  4)(―BGN‖ Mode) 0 PV 25  PMT 4 I/Y 360 N CPT FV 17,367.79 4.

PV(due)  18,000; PMT  215; n  108 (P/Y  12; C/Y  4)(―BGN‖ Mode) 18,000 PV 0 FV 215  PMT 108 N CPT I/Y 6.0268 The rate is 6.0268% compounded quarterly..

(a) PMT  82; i  1.375%; n  42

1  1.0137542  PV(due)  82 (1.01375)    82(1.01375)(31.74454)  $2638.84.  0.01375  (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 82  PMT 16.5 I/Y 42 N CPT PV 2638.84 (b) 82(42)  $3444.

FV(due)  10,000; i  0.635%; n  20

1.0063520  1  10,000  PMT(1.00635)    PMT(1.00635)(21.253734)  0.00635  10,000  21.3887PMT PMT  $467.54 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 PV 10,000 FV 2.54 I/Y 20 N CPT PMT 467.54 7.

(a) PMT  375; i  0.3125%; n  32; c 

12 3 4

p  1.0031253  1  1.009404  1  0.9404%

1.00940432  1 FV  375    375(37.134769)  $13,925.54  0.009404  (Set P/Y  4; C/Y  12) (―END‖ Mode) 0 PV 375  PMT 3.75 I/Y 32 N CPT FV 13,925.54

(b) PMT  375; i  0.3125%; n  32; c 

12 3 4

p  1.0031253  1  1.009404  1  0.9404%

1.00940432  1 FV(due)  375(1.009404)    375(1.009404)(37.134781)  0.009404   $14,056.50 (Set P/Y  4; C/Y  12)(―BGN‖ Mode) 0 PV 375  PMT 3.75 I/Y 32 N CPT FV 14,056.50 8.

PMT 2700; i  5.5%; n  40; c 

1  0.25 4

p  1.0550.25  1  1.013475  1  1.3475%

1  1.01347540  PV(due)  2700(1.013475)    0.013475   2700(1.013475)(30.765421)  $84,185.98. (Set P/Y  4; C/Y  1)(―BGN‖ Mode) 0 FV 2700  PMT 5.5 I/Y 40 N CPT PV 84,185.97531 9.

(a) PMT  850; i  0.42%; n  240

1  1.0042 240  PV(due)  850(1.0042)    0.0042   850(1.0042)(151.019208)  $128,905.47. (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 850  PMT 5.04 I/Y 240 N CPT PV 128,905.47 (b) 128,905.47  850(240)  204,000  128,905.47  $75,094.53. 10. (a) FV(due)  32,000; i  3.5%; n  240; c 

2  0.16 12

p  1.0350.16  1  1.005750  1  0.575%

1.00575240  1  32,000  PMT(1.00575)    0.00575  32,000  PMT(1.00575)(514.653861) 32,000  517.613121 PMT PMT  $61.82 (P/Y  12; C/Y  2)(―BGN‖ Mode) 32,000 FV 0 PV 7 I/Y 240 N CPT PMT 61.82 (b) FV  32,000; i  3.5%; n  15; c  2 p  1.0352  1  1.071225  1  7.1225%

1.07122515  1  32,000  PMT    0.071225  32,000  25.367409 PMT PMT  $1261.46 (P/Y  1; C/Y  2)(―BGN‖ Mode) 32,000 FV 0 PV 7 I/Y 15 N CPT PMT 1261.46 11. PV(due)  49,350; i  0.75%; n  12; c 

12 6 2

p  1.00756  1  1.045852  1  4.5852%

1  1.045852 12  49,350  PMT(1.045852)    0.045852  49,350  PMT(1.045852) 9.074288 49,350  9.490364 PMT PMT  $5200.01 (P/Y  2; C/Y  12)(―BGN‖ Mode) 0 FV 49,350 PV 9 I/Y 12 N CPT PMT 5200.01 12. FV(due)  10,000; PMT  300; i  2.25%

1.0225n  1 10,000  300(1.0225)    0.0225  1.0225n  1 10,000  (306.75)    0.0225  10,000  13, 633.3 (1.0225n  1) 0.733496  (1.0225n  1) 1.733496  1.0225n ln 1.733496  n ln 1.0225 0.550140  0.0222510n n  24.7247 (semi-annual payments)  12 years, 6 months. (P/Y  2; C/Y  2)(―BGN‖ Mode) 10,000 FV 0 PV 300  PMT 4.5 I/Y CPT N 24.7247

13. PMT  50; i  0.5%; n  192 1.005192  1  FV(due)  50(1.005)    0.005 

 50(1.005)(321.091337)  $16,134.83967 PV  16,134.83967; PMT  375; i  0.5%; 1  1.005 n  16,134.83967  375    0.005  1  1.005 n  43.026239     0.005  0.215131  1  1.005 – n 1.005– n  0.784869 n ln 1.005  ln 0.784869 0.004988n  0.242238 n  48.568709 months  4 years, 1 month (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 PV 50  PMT 6 I/Y 192 N CPT FV 16,134.83967 (Set P/Y  12; C/Y  12) 0 FV 16,134.83967 PV 375  PMT 6 I/Y CPT N 48.568709 14. PV(due)  698; PMT  34; n  24 The rate is 16.9054% compounded monthly.. (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 698 PV 34  PMT 24 N CPT I/Y 16.9054 15. PV  12,500; i  2.975%; n  20 FV  12,500(1.02975)20  12,500(1.797364)  $22,467.04864 PV(due)  22,467.04864; PMT  500; i  0.43%

1  1.0043 n  22,467.04864  500(1.0043)    0.0043  1  1.0043 n  44.741708     0.0043  0.192389  1  1.0043n 1.0043n  0.807611 n ln 1.0043  ln 0.807611 0.00429 1n  0.213675 n  49.79 months  4 years, 2 months.

(Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 22,467.04864 PV 500  PMT 5.16 I/Y CPT N 49.79 16. FV(due)  8400; PMT  400; n  16 The rate is 6.2093% compounded quarterly.. (Set P/Y  2; C/Y  4)(―BGN‖ Mode) 8400 FV 0 PV 400  PMT 16 N CPT I/Y 6.2093 17. (a) PMT  1630; i  2.025%; n  24

1  1.0202524  PV 1630    0.02025   1630(18.86047)  $30,742.57. (b) PMT  1630, n  4; i  2.025%

1.020254  1  FV  1630    1630(4.1231486)  $6720.73.  0.02025  (Set P/Y  4; C/Y  4) 0 PV 1630  PMT 8.1 I/Y 4 N CPT FV 6720.73 (c) PMT  1630; i  2.025%; n  20

1  1.0202520  PV  1630    1630(16.312053)  $26,588.65  0.02025  Amount needed  6720.73 + 26,588.65  $33,309.38. (Set P/Y  4; C/Y  4) 0 FV 1630  PMT 8.1 I/Y 20 N CPT PV 26,588.65 (d) 6720.73  1630(4)  6720.73  6520  $200.73. 18. (a) PV  20,000; PMT  3500; i  0.75%; c 

12 6 2

p  1.00756  1  1.045852  1  4.5852%

1  1.045852 n  20,000  3500    0.045852  1  1.045852 n  5.714286     0.045852  0.262013  1  1.045852– n

1.045852n  0.737987 n ln 1.045852  ln 0.737987 0.044832n  0.303829 n  6.777038  7 semi-annual payments  3.5 years (Set P/Y  2; C/Y  12) 0 FV 20,000 PV 3500  PMT 9 I/Y CPT N 6.777038 (b) PV(due)  20,000; PMT  3500; i  0.75%; c 

12  12 1

p  1.007512  1  1.093807  1  9.3807% 1  1.093807  n  20, 000  3500 1.093807     0.093807  1  1.093807  n  5.224218     0.093807  0.490068  1  1.093807  n 1.093807  n  0.509932  n ln 1.093807  ln 0.509932  0.089664n  0.673477 n  7.511107  8 years (Set P/Y  1; C/Y  12)(―BGN‖ Mode) 0 FV 20,000 PV 3500  PMT 9 I/Y CPT N 7.511107 (c) PV  20,000; i  0.75%; n  60 FV  20,000(1.0075)60  20,000(1.565681)  $31,313.62054 PV  31,313.62054; i  0.75%; c 

12 3 4

p  1.00753  1  1.022669  1  2.2669%

1  1.022669 n  31,313.62054  3500    0.022669 

1  1.022669 n  8.946749     0.022669  0.202815  1  1.022669 n 1.022669 n  0.797185  n ln 1.022669  ln 0.797185  0.022416n  0.226669 n  10.111908 quarters  2 years, 9 months (Set P/Y  4; C/Y  12) (―END Mode‖) 0 FV 31,313.62054 PV 3500  PMT 9 I/Y CPT N 10.111908

(d) PV  20,000; i  0.75%; n  36 FV  20,000(1.0075)36  20,000(1.308645)  $26,172.90742 PV(due)  26,172.90742; i  0.75%; c 

12 6 2

p  1.00756  1  1.045852  1  4.5852%

1  1.045852 n  26,172.9074  3500(1.045852)    0.045852  1  1.045852 n  7.150124     0.045852  0.327849  1  1.045852 n 1.045852 n  0.672151 n ln 1.045852  ln 0.672151 0.044832n  0.397273 n  8.861343  9 half years  4.5 years (Set P/Y  2; C/Y  12)(―BGN‖ Mode) 0 FV 26,172.9074 PV 3500  PMT .9 I/Y CPT N8.861343 19. PMT  725; i  0.60416; n  36

1  1.006041636  PV(due)  725( 1.0060416 )    0.0060416   725( 1.0060416 )(32.266885)  $23,534.82734 FV  23,534.82734; i  0.60416 ; n  18



PV  23,534.82734 1.0060416



18

 $21,116.58.

(Set P/Y  12; C/Y  12)(―BGN‖ Mode) 0 FV 725  PMT 7.5 I/Y 36 N CPT PV 23,534.82734

20. PMT  900; n  40, i  2.75%; d 10

1.027540  1  FV  900    900(71.268145)  $64,141.33049  0.0275  5 years later: FV  64,141.33049(1.0275)10  64,141,33049(1.311651)  $84,131.04237 PV  84,131.04237; i  0.5%; n  180

1  1.005180  84,131.04237  PMT    0.005  84,131.04237  118.503515 PMT PMT  $709.95 (Set P/Y  2; C/Y  2) 0 PV 5.5 I/Y 40 N 900  PMT CPT FV 64,141.33049 (Set P/Y  12; C/Y  12) 0 FV 84,131.04237 PV 6 I/Y 180 N CPT PMT  709.95 21. PV  40,000; i  1.75% n  84; c 

4  0. 3 ; d  12 12

FV  40,000(1.0175)12  40,000(1.231439)  $49,257.5726 p  1.01750.3  1  1.0058

1  1.005884  49,257.5726  PMT    0.0058  49,257.5726  66.345785PMT PMT  $742.45 (Set P/Y  12; C/Y  4) 0 FV 49,257.5726 PV 7 I/Y 84 N CPT PMT 742.45

22. PV  25,000; i  2.75%; n  60 FV  25,000(1.0275)60  25,000(5.092251)  $127,306.284 PV  127,306.284; PMT  6000; i  1.5%

1  1.015 n  127,306.284  6000    0.015  1  1.015 n  21.217714     1.015  0.318266  1  1.015 n 1.015 n  0.681734  n ln 1.015  ln 0.681734 0.014889n  0.383116 n  25.73 quarters  6.5 years (Set P/Y  4; C/Y  4) 0 FV 127,306.284 PV 6 I/Y 6000  PMT CPT N 25.73 23. PV  33,500; i  0.4116 % ; n 144 FV  33,500( 1.004116 )144  33,500(1.806847)  60,529.36367 PV  60,529.36; PMT  4500; i  0.4116%; c 

12 3 4

p  ( 1.004116 )3  1  1.012401  1  1.2401%

1  1.012401 n  60,529.36  4500    0.012401  1  1.012401 n  13.450969     0.012401  0.166804  1  1.012401 n 1.012401 n  0.833196 n ln 1.012401 ln 0.833196  0.012325n  0.182487 n  14.806643 quarters  3 years, 9 months (Set P/Y  4; C/Y  12) 0 FV 60,529.36 PV 4.94 I/Y 4500  PMT CPT N 14.806643

24. PV  15,000; i  0.403 %; n  45 FV  15, 000(1.00403)45  15,000(1.198577)  $17,978.65386 PV  17,978.65386; PMT  335; n  60 The rate is 4.4784% compounded monthly.. (Set P/Y  12; C/Y  12) 17,978.65386 PV 0 FV 335  PMT 60 N CPT I/Y 4.4784 25. PV  200,000; i  5.29%; n  2 FV  200,000(1.0529)2  200,000(1.108598)  $221,719.682 PV  221,719.682; PMT  35,000; n  10 The rate is 9.296% compounded annually.. (Set P/Y  1; C/Y  1) 221,719.682 PV 0 FV 35,000  PMT 10 N CPT I/Y 9.296 26. (a) PMT  3500; i  2%, n  30; c 

4 2 2

p  (1.02)2  1  1.0404  1  4.04%

1  1.040430  PV  3500    0.0404   3500(17.20836)  $60,229.26. (Set P/Y  2; C/Y  4) 0 FV 8 I/Y 30 N 3500  PMT CPT PV 60,229.26 (b) PMT(due)  3500; i  4.04%; n  20

1  1.040420  PV(due)  3500(1.0404)    0.0404   3500(1.0404)(13.542316)  $49,312.99. (Set P/Y  2; C/Y  4)(―BGN‖ Mode) 0 FV 8 I/Y 20 N 3500  PMT.CPT PV 49,312.99 (c) PMT  3500; i  4.04%; n  16, d  8

1  1.040416  8 PV  3500   (1.0404) 0.0404    3500(11.61798754)(0.728445813)  ( 40,662.960)(0.728445813)  $29,620.76 (Set P/Y  2; C/Y  4) 0 FV 8 I/Y 16 N 3500  PMT CPT PV 40,662.96 (d) PMT(due)  3500; i  4.04%; n  18, d  6

1  1.040418  6 PV  3500(1.0404)   (1.0404) 0.0404    3500(1.0404)5(12.618239)  $36,229.73. (Set P/Y  2; C/Y  4)(―BGN‖ Mode) 0 FV 8 I/Y 18 N 3500  PMT CPT PV 45,948.06 (e) PV 

3500  $86,633.66 0.0404

(f) 3500 +

3500  3500 + 86,633.66  $90,133.66 0.0404

27. (a) PMT(due)  450; i  1.5; n  28

1.01528  1  FV(due)  450(1.015)    0.015   450(1.015)(34.481489)  $15,749.42. (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 PV 6 I/Y 28 N 450  PMT CPT FV 15,749.42 (b) Interest  15,749.42  450(28)  15,749.42  12,600  $3149.42. (c) PV  15,749.42; i  1.25%; n  48

1  1.012548  15,749.42  PMT    0.0125  15,749.42  PMT(35.931481) PMT  $438.32 (Set P/Y  4; C/Y  4) 15,749.42 PV 0 FV 5 I/Y 48 N CPT PMT 438.32

(d) Total interest earned  438.32(48)  450(28)  21,039.36  12,600  $8439.36.

28. PV  37,625; i  1.75%; n  16 FV  37,625(1.0175)16  $37,625(1.319929)  $49,662.34184 PV(due)  49,662.34184; i  1.75%; n  10; c 

4 4 1

p  (1.0175)4  1  1.071859  1  7.1859%

1  1.071849 10  49,662.34184  PMT(1.071859)    0.071859  49,662.34184  PMT(1.071859)(6.963620) 49,662.34184  7.464019PMT PMT  $6653.57 (Set P/Y  1; C/Y  4)(―BGN‖ Mode) 49,662.34184 PV 0 FV 7 I/Y 10 N CPT PMT 6653.57 29. (a) PV  16,750; i  3.25%; n  48; c 

2  0.5 4

p  (1.0325)0.5  1  1.016120  1  1.6120%

1  1.016120 48  16,750  PMT    0.016120  16,750  33.242480 PMT PMT  $503.87 (Set P/Y  4; C/Y  2) 16,750 PV 0 FV 6.5 I/Y 48 N CPT PMT 503.87 (b) PV(due)  16,750; i  3.25%; n  20; c 

2 2 1

p  (1.0325)2  1  1.066056  1  6.6056%

1  1.066056 20  16,750.00  PMT(1.066056)    0.066056  16,750  PMT(1.066056)(10.926658)

16,750  11.648432 PMT PMT  $1437.96 (Set P/Y  1; C/Y  2) (―BGN‖ Mode) 16,750 PV 0 FV 6.5 I/Y 20 N CPT PMT 1437.96 (c) PV  16,750; i  3.25%; n  180; c 

2  0.16 , d  20 12

FV  16,750(1.0325)20  17,750(1.895838)  $31,755.28523 p  (1.0325)0.16  1  1.005345  1  0.5345%

1  1.005345180  31,755.28523  PMT   1  0.005345  31,755.28523  115.424194 PMT PMT  $275.12 (Set P/Y  12; C/Y  2) 31,755.28523 PV 0 FV 6.5 I/Y 180 N CPT PMT 275.12 (d) PV  16,750; i  3.25%; n  48; c 

2  0.5; d  40 4

FV  16,750(1.0325)40  16,750(3.594201)  $60,202.87402 p  (1.0325)0.5  1  1.016120  1  1.6120%

1  1.016120 48  60,202.87402  PMT(1.016120)    0.016120  60,202.87402  PMT(1.016120)(33.24248)  60,202.87402 = 33.778351 PMT  $1782.29 (Set P/Y  4; C/Y  2) (―BGN Mode‖) 60,202.87402 PV 0 FV 6.5 I/Y 48 N CPT PMT 1782.29 (e) PV  16,750; i  3.25%; 

2  0.16 12

p  (1.0325)0.16  1  1.005345  1  0.5345%

PMT  16,750(0.005345)  $89.52 (f) PV(due)  16,750; i  3.25%; c 

2 2 1

p  (1.0325)2  1  1.066056  1  6.6056%

PMT  (16, 750  PMT) 0.066056 PMT  1106.442188  0.066056 PMT 1.066056 PMT  1106.442188 PMT  $1037.88 30. PV  75,000; PMT  6000; i  1.8275%

1  1.018275 n  75,000  6000    0.018275  1  1.018275 n  12.5     0.018275  0.228438  1  1.018275n 1.018275n  1  0.228438  0.771563 n ln 1.018275  ln 0.771563 0.018110n  0.259338 n  14.32 (quarters)  3 years, 9 months. (Set P/Y  4; C/Y  4) 75,000 PV 0 FV 6000  PMT 7.31 I/Y CPT N 14.32 31. PV  20,000; PMT  1223; i  1.95%; d  10

1  1.0195 n  10 20, 000 1.0195  1223    0.0195  20,000 1.213032   62,717.94872(1  1.0195 n ) 24,260.64242  62,717.94872(1  1.0195 n ) 0.386821  1  1.0195 n 1.0195 n  1  0.386821  0.613179  n ln 1.0195  ln 0.613179  0.019312n  0.489099 n  25.33 quarters  6 years, 6 months (Set P/Y  4; C/Y  4) 20,000 PV 0 FV 1223  PMT 7.8 I/Y CPT N 25.33 32. PMT  1.45; i  1.4%; c 

4 2 2

p  (1.014)2  1  1.028196  1  2.8196% 1.45  PV(0.028196) PV 

1.45  $51.43. 0.028196

33. PMT  11,000; i  6.5% PV  11,000 +

11, 000  11,000 + 169,230.77  $180,230.77. 0.065

34. PMT  4000; i  1.205%; d  8, c 

4 4 1

p  (1.01205)4  1  1.049078  1  4.9078% In two years must have

4000  $81,502.52373 0.049078

Today must deposit: PV  81,502.52373(1.01205)8  81,502.52373(0.908624)  $74,055.15. Self-Test 1.

PMT  1800; i  1.31%; n  72

1.013172  1  FV(due)  1800(1.0131)    0.0131   1800(1.0131)(118.511878)  $216,115.89 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 PV 1800  PMT 5.24 I/Y 72 N CPT FV.216,115.89 2.

PMT  960; i  0.50%; n  84

1  1.00584  PV(due)  960(1.005)    0.005   960(1.005)(68.453042)  $66,043.50 (Set P/Y  12; C/Y  124)(―BGN‖ Mode) 0 FV 960  PMT 6 I/Y 84 N CPT PV. 66,043.50 3.

PV(due)  10,104; i  4.4%; n  10

1  1.044 10  10,104  PMT(1.044)    0.044 

10,104  PMT 1.044  7.951768  10,104  8.301645 PMT PMT  $1217.11 (Set P/Y  2; C/Y  2)(―BGN‖ Mode) 10,104 PV 0 FV 8.8 I/Y 10 N CPT PMT .1217.11 4.

PV(due)  5200; PMT  270; n  24 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 FV 5200 PV 270  PMT 24 N CPT I/Y 8.0661 The rate is 8.0661% compounded quarterly.

PMT  145; i  1.06%; n  144; c 

2  0.16 12

p  (1.0106)0.16  1  1.001759  1  0.1759% FV(due)  145(1.001759) C  145(1.001759)(163.71579)  $23,780.54 (Set P/Y  12; C/Y  2)(―BGN‖ Mode) 0 PV 145  PMT 2.12 I/Y 144 N CPT FV 23,780.54 6.

PMT  950; i  3%; n  60; c  p  1.03

0.3

4  0.3 12

 1  1.009902  1  0.9902%

1  1.009902 60  PV(due)  950(1.009902)    0.009902   950(1.009902)(45.075817)  $43,246.03 (Set P/Y  12; C/Y  4) (―BGN‖ Mode) 0 FV 950  PMT 12 I/Y 60 N CPT PV 43,246.03 7.

PMT  1680; i  2.75%; n  16

1.027516  1  FV(due)  1680(1.0275)    0.0275   1680(1.0275)(19.763979)  34,116.5814 PV  34,116.5814; n  240; i  0.05%

34,116.5814  139.580772PMT PMT  $244.42 (Set P/Y  2; C/Y  2) (―BGN‖ Mode) 0 PV 1680  PMT 5.5 I/Y 16 N CPT FV 34,116.5814 (Set P/Y12; C/Y4) 0 FV 6 I/Y 240 N 34,116.5814 PV CPT PMT  244.42

PMT  4000; i  1.25%; n  12, c  4; d  12 p  (1.0125)4  1  1.050945  1  5.0945%

1  1.05094512  PV (defer)  4000    0.050945   4000(8.816185)  $35,264.73969 PV  35,264.73969(1.0125)12  35,264.73969(0.861509)  $30,380.88 . (Set P/Y  1; C/Y  4) 0 FV 4000  PMT 5 I/Y 12 N CPT PV 35,264.73969 9.

FV  39,600(1.00875)16  39,600(1.149574)  $45,523.112 PV  45,523.11268; PMT  6000; n  14 (Set P/Y  2; C/Y  2) 0 FV 45,523.11268 PV 6000  PMT 14 N CPT I/Y 18.927 The rate is 18.927% compounded semi-annually..

10. PMT  6000; i  1.17%; n  80

1  1.0117 80  PV(due)(defer)  6000(1.0117)    0.0117   6000(1.0117)(51.766652)  $314,233.9344 FV(due)  314,233.9344; i  1.17%; n  72

1.0117 72  1  314,233.934  PMT(1.0117)    0.0117  314,233.934  PMT(1.0117)(112.017654) 314,233.934  113.32826PMT PMT  $2772.78 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 0 FV 6000  PMT 4.68 I/Y 860 N CPT PV 314,233.934

(Set P/Y  4; C/Y  4)(―BGN‖ Mode) 314,233.934 FV 0 PV 4.68 I/Y 72 N CPT PMT 2772.78

11. FV  27,350 (1.005)72  27,350(1.432044278)  $39,166.41102 PV(due)  39,166.41102; PMT  1600; i  2.48%

1  1.0248 n  39,166.41102  1600(1.0248)    0.0248  1  1.0248 n  23.88661874     0.0248  0.592388144  1  1.0248n 1.0248n  1  0.592388144  0.407611855 n ln 1.0248  ln 0.407611855 0.0244970  0.897439893 n  36.63 (semi-annual periods)  18.5 years.. (Set P/Y  2; C/Y  2)(―BGN‖ Mode) 39,166.41102 PV 0 FV 1600  PMT 4.96 I/Y CPT. N 36.63 12. PMT  0.75, i  0.346 %; c 

12 3 4

p  (1.00346)3  1  1.010436  1  1.0436% PV(0.010436)  0.75 PV  $71.87 13. FV  50,000(1.04)4  50,000(1.169859)  $58,492.928 PMT  58,492.928(0.04)  $2339.72 14. (a) PMT  240; i  2.375%; n  300; c 

2  0.16 12

p  (1.02375)0.16  1  1.003920  1  0.003920

1  1.003920300  PV  240    0.003920   240(176.225246)  $42, 294.06 (Set P/Y  12; C/Y  2) (―END‖ Mode) 0 FV 240  PMT 4.75 I/Y 300 N CPT PV 42,294.06

(b) PMT  240; i 2.375%; n  180; c =

2  0.16 12

p  (1.02375)0.16  1  1.003920  1  0.003920 1  1.003920 180  PV  240(1.003920)    0.003920   240(1.003920)(128.958522)  $31,071.36. (Set P/Y  12; C/Y  2)(―BGN‖ Mode) 0 FV 240  PMT 4.75 I/Y 180 N CPT PV.31,071.36 (c) PMT  240; i  2.375%; n  240; c 

2  0.16 12

p  (1.02375)0.16  1  1.003920  1  0.003920

1  1.003920240  PV  240    0.003920   240(155.352873)  $37, 284.69 .

PV (defer)  37,284.69(1.02375)20  37,284.69(0.625348)  $23,315.90 . (Set P/Y  12; C/Y  2) (―END‖ Mode) 0 FV 240  PMT 4.75 I/Y 240 N CPT PV 37,284.69 (d) PMT  240; i  2.375%; n  180; c 

2  0.16 ; d  24 12

p  (1.02375)0.16  1  1.003920  1  0.003920

1  1.003920180  PV  240(1.003920)    0.003920   240(1.003920)(128.958522)  $31, 071.36 PV(defer)  31, 071.36(1.02375)24  31, 071.36(0.569306)  $17, 689.12 (Set P/Y  12; C/Y  2)(―BGN‖ Mode) 0 FV 240  PMT 4.75 I/Y 180 N CPT.PV 31,071.36

(e) PMT  240; i  2.375%; c 

2  0.16 12

p  (1.02375)0.16  1  1.003920  1  0.003920 PV 

240  $61,228.84 0.003920

(f) PMT  240; i  2.375%; c 

2  0.16 12

p  (1.02375)0.16  1  1.003920  1  0.003920 PV(due)  240 

240  240  61,228.84  $61,468.84. 0.003920

Challenge Problems 1.

To determine the simple interest rate, use Formula 7.1A, I  Prt PV  100; t  1, 2, 3, 4,............, 20 The amount of interest for Year 1  100 (r) The amount of interest for Year 2  100 (2r) The amount of interest for Year 3  100 (3r) •

•

The amount of interest for Year 20  100 (20r) The total amount of interest for the 20 years is 840  100 (r)  100 (2r)  100 (3r)  . . .  100 (20r) 840  100 (r)(l  2  3  . . .  20)

 20(21)  840  100 (r)   2  840  21,000r r

840  0.04  4% 21, 000

For interest compounded annually PMT(due)  100; i  4%  0.04; n  20

1.0420  1  FVn(due)  100(1.04)    100(l.04)(29.778079)  $3096.92  0.04  The total amount of interest  3096.92  20(100)  $1096.92.

Present value of the payments of $100 per month for the first 2.5 years: PMT  100; i 

9%  0.75%  0.0075; n  30 12

1  1.007530  PVn (due)  100(1.0075)    0.0075   100(1.0075)(26.775080)  $2697.59 Present value 4.5 years from now of the payments of $200 per month for the final year: PMT  200; i 

8.5%  0.7083%  0.007083; n  12 12

1  1.00708312  PVn(due)  200(1.007083)    0.007083   200(1.007083) (11.465291)  $2309.30 Present value of $2309.30 now: FV  2309.30; i 

9%  0.75%  0.0075; n  54 12

PV  2309.30(1.0075)54  2309.30(0.667986)  $1542.58 Herman’s debt  2697.59 + 1542.58  $4240 Case Study 1.

Stephen would receive $1000 at the beginning of each month for four years. Jack would receive $1080(1000  1.08) at the beginning of each month for four years. Danny would receive $1188(1080  1.1) at the beginning of each month for four years.

(a) • Stephen turned 12 years old in April. He would receive $1000 at the beginning of each month for 4 years starting on September 1 of his 18th birthday.

The solution requires a two-step process: Calculate the present value of a 48-month annuity due at 4% per annum compounded monthly. The present value becomes the future value in Step 2. (Set P/Y  12; C/Y  12) (―BGN‖ Mode) 1000  PMT 4 I/Y 48 N 0 FV CPT PV 44,436.46 Determining the initial investment required that the parents are required to invest on the basis of the future value attained in Step 1. (Set P/Y  12; C/Y  12) (―BGN‖ Mode) 44,436.46  FV 4 I/Y 75 N 0 PMT CPT PV . 34,621.54 The parents require a $34,621.54 deposit to satisfy Stephen’s educational requirement. •

Jack turned 9 years old in January. He would receive $1080 (1000  1.08) at the beginning of each month for 4 years starting on September 1 of his 18th birthday.

The solution requires a two-step process: Calculate the present value of a 48-month annuity due at 4% per annum compounded monthly. The present value becomes the future value in Step 2. (Set P/Y  12; C/Y  12) (―BGN‖ Mode) 1080  PMT 4 I/Y 48 N 0 FV CPT PV..47,991.38 Determining the initial investment required that the parents are required to invest on the basis of the future value attained in Step 1. (Set P/Y  12; C/Y  12) (―BGN‖ Mode) 47,991.38  FV 4 I/Y 109 N 0 PMT CPT PV 33,391.20 The parents require a $33,391.20 deposit to satisfy Jack’s educational requirement. •

Danny turned seven years old in March. He would receive $1188(1080 × 1.1) at the beginning of each month for four years.

The solution requires a two-step process: Calculate the present value of a 48-month annuity due at 4% per annum compounded monthly. The present value becomes the future value in Step 2. (Set P/Y 12; C/Y 12) (―BGN‖ Mode) 1180  PMT 4 I/Y 48 N 0 FV CPT PV 52,435.03 Determining the initial investment required that the parents are required to invest on the basis of the future value attained in Step 1. (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 52,435.03  FV 4 I/Y 135 N 0 PMT CPT PV 33,459.08 The parents require a 33,459.08 deposit to satisfy Danny’s educational requirement. The total amount interest on June l is 34,621.54  33,391.20  33,459.08  $101,471.82 Chapter 14 Amortization of Loans, Including Residential Mortgages (b)

Due to rounding, you may find some slight differences between these answers and your calculations.

Exercise 14.1 A. 1.

(a) PVn  12,000; i  2.5%; n  32

1  1.02532  12, 000  PMT    0.025  12, 000  21.849178 PMT PMT  $549.22 (Set P/Y  4; C/Y  4) 12,000 PMT 549.22

I/Y

(b) Interest paid in the first period is $12,000  0.025  $300 Principal repaid in the first period is $549.22  300  $249.22 Balance at the end of the first period is $12,000  249.22  $11,750.78 (c) Interest paid in the second period is $11,750.78  0.025  $293.77 Principal repaid in the second period is $549.22  293.77  $255.45

CPT

Balance at the end of the second period is $11,750.78  255.45  $11,495.33 2.

(a) PVn  8000; i  1%; n  60

1  1.0160  8000  PMT    0.01  8000  44.955038 PMT PMT  $177.96 (Set P/Y  12; C/Y  12) 8000 PMT 177.96

I/Y

CPT

(b) Interest paid in the first period is $8000  0.01  $80 Principal repaid in the first period is $177.96  80  $97.96 Balance at the end of the first period is $8000  97.96  $7902.04 (c) Interest paid in the second period is $7902.04  0.01  $79.02 Principal repaid in the second period is $177.96  79.02  $98.94 Balance at the end of the second period is $7902.04  98.94  $7803.10 3.

(a) PVn  15,000; i  4%; n  20

1  1.0420  15, 000  PMT    0.04  15, 000  13.590326 PMT PMT  $1103.73 (Set P/Y  2; C/Y  2) 15,000 PMT 1103.73

I/Y

CPT

(b) Interest paid in the first period is $15,000  0.04  $600 Principal repaid in the first period is $1103.73  600  $503.73 Balance at the end of the first period is $15,000  503.73  $14,496.27 (c) Interest paid in the second period is $14,496.27  0.04  $579.85 Principal repaid in the second period is $1103.73  579.85  $523.88 Balance at the end of the second period is $14,496.27  523.88  $13,972.39 4.

(a) PVn  9600; i  6%; n  7

1  1.067  9600  PMT    0.06  9600  5.582381 PMT PMT  $1719.70 (Set P/Y  1; C/Y  1) 9600 PMT 1719.70

I/Y

CPT

(b) Interest paid in the first period is $9600  0.06  $576 Principal repaid in the first period is $1719.70  576  $1143.70 Balance at the end of the first period is $9600  1143.70  $8456.30 (c) Interest paid in the second period is $8456.30  0.06  $507.38 Principal repaid in the second period is $1719.70  507.38  $1212.32 Balance at the end of the second period is $8456.30  1212.32  $7243.98 5.

PV = 3600 I/Y = 3.6% I = .036/4 = 0.009 c/y = 4 P/Y = 4 Term = 2 years n = 2 × 4 = 8 8 (a) 3600  PMT 1  (1  0.009) 



0.009



3600 = 7.685485 PMT PMT = $468.42 (Set P/Y  4; C/Y  4) Quit; 3600 ± PV , CPT PMT = 468.42

PV , 3.6

(b) Interest paid in the first period is $3600 × 0.009 = $32.4

I/Y

Principal repaid in the first period is $468.42 − 32.4 = $436.02 Balance at the end of the first period is $3600 − 436.02 = $3163.98 (c) Interest paid in the second period is $3163.98 × 0.009 = $28.48 Principal repaid in the second period is $468.42 − 28.48 = $439.94 Balance at the end of the second period is $3163.98 − 439.94 = $2724.04

B. 1.

(a) PVn  36,000  4000  32,000; i  2%; n  60

1  1.0260  32, 000  PMT    0.02  32, 000  34.760887 PMT PMT  $920.58 (Set P/Y  4; C/Y  4) 32,000 CPT PMT 920.58

I/Y

(b) Interest paid in the first period is $32,000  0.02  $640 Principal repaid in the first period is $920.58  640  $280.58 Balance at the end of the first period is $32,000  280.58  $31,719.42 (c) Interest paid in the second period is $31,719.43  0.02  $634.39 Principal repaid in the second period is $920.58  634.39  $286.19 Balance at the end of the second period is $31,719.42  286.18  $31,433.24 (d) Total paid  4000  60(920.58) 

$59,234.80

(e) Total interest paid  59,234.80  36,000  2.

23,234.80

(a) PVn  3,600,000  1,200,000  2,400,000; i  2%; n  60

1  1.0260  2, 400, 000  PMT    0.02  2, 400, 000  34.76 PMT PMT  $69, 043.12 (Set P/Y  4; C/Y  4) 2,400,000 CPT PMT $69,043.12

I/Y

(b) Interest paid in the first period is $2,400,000  0.02  $48,000 Principal repaid in the first period is $69,043.12  48,000  $21,043.12 Balance at the end of the first period is $2,400,000  21,043.12  $2,378,956.88 (c) Interest paid in the second period is $2,378,956.88  0.02  $47,579.14 Principal repaid in the second period is $69,043.12  47,579.14  $21,463.98

Balance at the end of the second period is $2,378,956.88  21,463.98  $2,357,492.90 (d) Total paid  1,200,000  60(69,043.12)  1,200,000  4,142,587.2  $5,342,587.20 (e) Total interest  5,342,587.20  3,600,000 

$1,742,587.20

(a) PVn  55,000; i  0.06/12 = 0.005 = 0.5%; n  96

1  1.00596  55,000  PMT    0.005  55,000  76.095218 PMT PMT  $722.78 (Set P/Y  12; C/Y  12) 55,000 CPT PMT 722.78

I/Y

(b) PMT  722.78; i  0.005 FV  55,000 (1.005)13  58,684.24 1.00513  1  FV13  722.78    9683.26  0.005  Balance  58,684.24  9683.26  $49,000.98

Alternatively, using the Amortization Worksheet, 96

N 2nd

I/Y

AMORT

55,000

722.78

PMT

P1  13, P2  13  BAL  $49,001

(a) PVn  24,000; i  3.8%; n  16

1  1.03816  24, 000  PMT    0.038  24, 000  11.826111 PMT PMT  $2029.41 (Set P/Y  2; C/Y  2) 24,000 CPT PMT 2029.41

7.6

I/Y

(b) PMT  1494.55; i  0.035 FV  24,000(1.038)7  31,159.66 1.0387  1  FV7  2029.41    15,931.89  0.038  Balance  31,159.66  15,931.89  $15,227.77

Alternatively, using the Amortization Worksheet, (Set P/Y  2; C/Y  2) 16 PMT 0 FV . 2nd

AMORT

7.6

I/Y

24,000

(d) Principal repaid  2029.41  578.66  2nd

AMORT

2029.41

P1  7, P2  7  BAL  15,227.79

(e)

$1450.75

$578.66

P1 9, P2 10  INT 989.83

(a) PMT  2300; i  0.5%; n  36

1  1.00536  PV  2300    0.005   2300(32.87101624)  $75,603.34 (Set P/Y  12; C/Y  12) 0 CPT PV 75,603.34

2300

PMT

I/Y

(b) PV  75,603.34; PMT  2300; i  0.5% FV  75,603.34(1.005)12  75,603.34(1.0616778)  80,266.39 1.00512  1  FV12  2300    28,371.79  0.005 

Balance  80,266.39  28,371.79  36

N 2nd

I/Y

75,603.34

AMORT

$51,894.60

PV .2300

. PMT

FV .

P1  12, P2  12  BAL  51,894.59

(c) Interest paid  2300 (12)  (75,603.34  51,894.59)  27,600  23,708.75  $3891.25 (d) Principal repaid  75,603.34  51,894.59  6.

$23,708.75

(a) PMT  600; i  0.64%; n  24 1  1.006424  PV  600    $13,309.23  0.0064 

(Set P/Y  12; C/Y  12) 0 CPT PV 13,309.23

600

PMT

7.68

I/Y

(b) PV  13,309.23; PMT  600; i  0.64%; FV  13,309.23(1.0064)8  14,006.12 1.00648  1 FV8  600    4908.91  0.0064 

Balance  14, 006.12  4908.91  $9097.21 24

N 2nd

7.68

I/Y

AMORT

13,309.23

.600

PMT

P1  12, P2  12  BAL  9097.21

PVn  10,000; i  7.75%; n  7 1  1.07757  10,000  PMT    0.0775 

10,000  5.251184 PMT PMT 

$1904.33

(Set P/Y  1; C/Y  1) 10,000 PMT 1904.33

7.75

I/Y

CPT

Amortization Schedule Amount paid

Payment

Interest paid

Principal repaid

Outstanding principal

10,000.00

1904.33

775.00

1129.33

8870.67

1904.33

687.48

1216.85

7653.82

1904.33

593.17

1311.16

6342.66

1904.33

491.56

1412.77

4929.88

1904.33

382.07

1522.26

3407.62

1904.33

264.09

1640.24

1767.38

1904.33

136.97

1767.38

Total

$13,330.31

$3330.33

$10,000.00

Note that there is a difference of $2 cents because of rounding. 8.

PVn  8000; i  1.75%; n  8 1  1.01758  8000  PMT    0.0175  8000  7.405053 PMT PMT  $1080.34 . (Set P/Y  4; C/Y  4) 8000 ± 1080.34

I/Y

CPT

PMT

Amortization Schedule Payment number

Amount paid

Interest paid

Principal repaid

Outstanding principal 8000.00

1080.34

140.00

940.34

7059.66

1080.34

123.54

956.80

6102.86

1080.34

106.80

973.54

5129.32

1080.34

89.76

990.58

4138.74

1080.34

72.43

1007.91

3130.83

1080.34

54.79

1025.55

2105.28

1080.34

36.84

1043.50

1061.78

1080.36

18.58

1061.78

—

Total 9.

$8642.74

$642.74

$8000.00

Interest paid

Principal repaid

PVn  9200; PMT  2500; i  11% Payment number

Amount paid

Outstanding principal 9200.00

2500.00

1012.00

1488.00

7712.00

2500.00

848.32

1651.68

6060.32

2500.00

666.64

1833.37

4226.95

2500.00

464.97

2035.04

2191.92

2433.03

241.11

2191.92

Total

$12,433.03

$3233.04

$9200.01

Note that there is a small difference because of rounding. 10. PVn  14,500; PMT  2600; i  3.5% Amortization Schedule Payment number

Amount paid

Interest paid

Principal repaid

Outstanding principal 14,500.00

2600.00

507.50

2092.50

12,407.50

2600.00

434.26

2165.74

10,241.76

2600.00

358.46

2241.54

8000.22

2600.00

280.01

2319.99

5680.23

2600.00

198.81

2401.19

3279.04

2600.00

114.77

2485.23

821.59

27.78

793.81

Total

$16, 421.59

$1921.59

$14,500.00

11. PMT  1904.33; i  7.75% FV  10,000(1.0775)3  12,509.84

793.81 —

1.07753  1 FV3  1904.33    6167.19  0.0775 

Balance  12,509.84  6167.19  $6342.65 Alternatively, using the Amortization Worksheet, (Set P/Y  1; C/Y  1) 7 PMT 0 FV . 2nd

AMORT

7.75

I/Y

10,000

.1904.33

P1  3, P2  3  BAL  6342.66

Interest for payment period 4 is 6342.65(0.0775) 

$491.56

.1080.34

12. PMT  1080.34; i  1.75% FV  8000(1.0175)4  8574.87 1.01754  1 FV4  1080.34    4436.13  0.0175 

Balance  8574.87  4436.13  $4138.74 Alternatively, using the Amortization Worksheet, (Set P/Y  4; C/Y  4) 8 N 7 FV . 2nd

AMORT

I/Y

8000

P1  4, P2  4  BAL  4138.74

Interest for payment period 5 is $4138.75(0.0175)  $72.43 Principal repaid  1080.34  72.43 

$1007.91

PMT

13. Use the retrospective method: PVn  9200; PMT  2500; i  11% Accumulated value of principal after three payments  9200(l.11)3  9200(1.367631)  $12,582.21 Accumulated value of first three payments 1.113  1  2500    2500(3.3421)  $8355.25  0.11 

FV3  12,582.21  8355.25  4226.96 Interest paid in PMT4  4226.96(0.11)  464.97 Principal repaid  2500  464.97  $2035.03 (Set P/Y  1; C/Y  1) 9200 ± CPT FV 12,582.20 0

2500

PMT

I/Y

CPT

8355.25

14. PVn  14,500; PMT  2600; i  3.5% Use the retrospective method: Accumulated value of principal after four payments  14,500 (1.035)4  14,500(1.147523)  $16,639.08 Accumulated value of first four payments 1.0354  1  2600    2600(4.214943)  $10,958.85  0.035 

FV4  16,639.08  10,958.85  $5680.23 Interest paid in PMT5  5680.23(0.035)  $198.81

(Set P/Y  2; C/Y  2) 14,500 CPT FV 16,639.08

PMT

I/Y

2600

PMT

15. (a) PVn  85,000; i  2%; n  40

CPT

I/Y

10,958.85

1  1.0240  85, 000  PMT    0.02  85, 000  27.355479 PMT PMT  $3107.24 (Set P/Y  4; C/Y  4) 85,000 PMT 3107.24

I/Y

PMT

CPT

(b) PMT  3107.24; i  0.02 FV  85,000(1.02)15  114,398.81 1.0215  1 FV15  3107.24    53,734.80  0.02 

Balance  114,398.81  53,734.80  $60,664.01 Alternatively, using the Amortization Worksheet, 32

N 2nd

I/Y

85,000

AMORT

. 3107.24

P1  15, P2  15  BAL  60,664.01

Interest for payment period 16 is $60,664.01(0.02) 

$1213.28

1.0219  1 FV19  3107.24    70,971.10  0.02 

Balance  123,828.95  70,971.10  $52,857.85 Alternatively, using the Amortization Worksheet, 2nd

AMORT

P1  19, P2  19  BAL  52,857.85

Interest for payment period 20 is $52,857.85(0.02)  $1057.16 Principal repaid  $3107.24  $1057.16  $2050.08 (d) FV  85,000(1.02)37  176,858.23 1.0237  1  FV29  3107.24    167,897.40  0.02  Balance  176,858.23  167,897.40  $8960.83 Alternatively, using the Amortization Worksheet, (Set P/Y  4; C/Y  4) 40 N 8 I/Y 85,000 PMT 0 FV . 2nd

AMORT

3107.24

P1  29, P2  29  BAL  $8960.84 Partial Amortization Schedule

Payment number

Amount paid

Interest paid

Principal repaid

Outstanding principal

i  0.02

balance

85,000.00

3107.24

1700.00

1407.24

83,592.76

3107.24

1671.86

1435.38

82,157.38

3107.24

1643.15

1464.09

80,693.29

‫׃‬

8960.91

3107.24

179.22

2928.02

6032.89

3107.24

120.66

2986.58

3046.31

Note that total paid is 6 cents more than the sum of principal and interest, which is due to rounding error. 40 Total

3107.24

60.93

3046.31

$124,289.60

$39,289.54

$85,000

16. (a) PVn  17,500; i  0.06 /12  .005; n  60 1  1.00560  17,500  PMT    0.005  17,500  51.72556 PMT PMT  $338.32

(Set P/Y  12; C/Y  12) 17,500  CPT PMT 338.32 (b) PMT  338.32; i  0.005 FV  17,500(1.005)19  19,239.48

I/Y

PMT

1.00519  1  FV20  338.32    6725.71  0.005 

Balance  19,239.48  6725.71  $12,513.77 60 N 6 I/Y 17,500 PV . 338.32 2nd

AMORT

FV .

P1  19, P2  19  BAL  12,513.77

Interest for payment period 20 is $ 12,513.77 (.005)  $62.57 . (c) PMT  338.32; i  0.005

FV  17,500(1.005)39  21,257.61 1.00539  1  FV20  338.32    14,528.86  0.005 

Balance  21,257.61  14,528.86  $6728.75 60

N 2nd

I/Y

AMORT

17,500

PV . 338.32

PMT

P1  39, P2  39  BAL  6728.75

Interest for payment period 40 is $6728.75 (.005)  $33.64 Principal repaid in period 40 is $338.31  $33.64  $304.68 (d) FV  17,500(1.005)57  23,254.32 1.00557  1  FV45  338.32    22,249.15  0.005 

Balance  23,254.32  22,249.15  $1005.17 60

I/Y

17,500

PV . 338.32

PMT

2nd

AMORT

P1  57, P2  57  BAL  1005.17

Partial Amortization Schedule Payment number

Amount paid

Interest paid

Principal repaid

Outstanding principal

17,500.00

338.32

87.5

250.82

17,249.18

338.32

86.25

252.04

16,977.11

338.32

84.99

253.33

16,723.78

‫׃‬

338.32

5.03

333.29

671.89

338.32

3.36

334.96

336.92

338.32

1.69

336.64



Totals

$20,299.20

$2799.48

$17,500.00

Note that for payment 60, outstanding principal equals , indicating an underpayment. This amount will be added to the last payment to make the total principal repaid to equal 17,500. 17. (a) PVn  24,000; PMT  2500; i  5.5%

1  1.055 n  24, 000  2500    0.055  1  1.055 n 9.60  0.055 0.528  1 –1.055– n 1.055– n  0.472 – n ln 1.055  ln 0.472 –0.05354 1n  –0.750776 n  14.02 half-years (Set P/Y  2; C/Y  2) 24,000 CPT N 14.02

I/Y

15 payments are required to repay the loan (b) Use the retrospective method: PVn  24,000; PMT  2500; i  5.5%; n  5 FV  24,000 (1.055)5  24,000 (1.306960)  $31,367.04

2500

PMT

1.0555  1 FV5  2500    2500 (5.581091)  $13,952.73  0.055  Balance  31,367.04  13,952.73  $17,414.31

Interest paid in PMT6  17,414.31(0.055)  24,000

31,367.04 0 PV FV 13,952.73

PMT

2500

PMT

$957.79

I/Y

N I/Y

CPT

(c) Use the retrospective method: PVn  24,000; PMT  2500; i  5.5%; n  9 FV  24,000 (1.055)9  24,000 (1.619094)  $38,858.26 1.0559  1 FV9  2500    2500 (11.256260)  $28,140.65  0.055 

Balance  38,858.26  28,140.65  $10,717.61 Interest paid in PMT10  10,717.61(0.055)  $589.47 Principal repaid  2500  589.47  $1910.53 24,000

858.26 0 PV 28,140.65

PMT

2500

I/Y

PMT

I/Y

CPT

38 FV

(d) Last three payments are PMT13, PMT14, PMT15. To find the balance outstanding after 12 payments, use the retrospective method: PVn  24,000; PMT  2500; i  5.5%; n  12 FV  24,000(1.055)12  24,000(1.901207)  $45,628.98 1.05512  1 FV12  2500    2500(16.385591)  $40,963.98  0.055 

Balance  45,628.98  40,963.98  $4665 24,000

45,628.98 0 PV FV 40,963.98

PMT

2500

PMT

I/Y

Partial Amortization Schedule

CPT

Payment number

Amount paid

Interest paid i  0.055

Principal repaid

Outstanding principal balance 24,000.00

2500.00

1320.00

1180.00

22,820.00

2500.00

1255.10

1244.90

21,575.10

2500.00

1186.63

1313.37

20,261.73

4665.00

2500.00

256.58

2243.42

2421.58

2500.00

133.19

2366.81

54.77

57.78

3.01

54.77

—

Total

$35,057.78

$11,057.78

$24,000.00

18. (a) PVn  28,000; i  0.5%; PMT  575 1  1.005 n  28,000  575    0.005  1  1.005 n 48.69565  0.005 0.243478  1 – 1.005– n 1.005– n  0.75652 – n ln 1.005  ln 0.75652 –0.004987n  –0.279026 n  55.94 months

(Set P/Y  12; C/Y  12) 28,000 CPT N 55.94

I/Y

575

PMT

It will have to make 56 payments (b) PVn  28,000; PMT  575; i  0.5%; n  17 Retrospective method: FV  28,000(1.005)17  28,000(1.0884865)  $30,477.62 1.00517  1 FV17  575    575(17.697301)  $10,175.95  0.005 

Balance  30,477.62  10,175.95  $20,301.67 Interest paid in PMT18  20,301.67(0.005)  $101.51 28,000 ± .30,477.62

0 PV 575 10,175.95

PMT PMT

6 6

I/Y I/Y

17 17

N N

CPT CPT

(c) Use the retrospective method: PMT  575; PV  28,000; i  0.5%; n  29 FV  28,000(1.005)29  28,000(1.155622)  $32,357.42 1.00529  1 FV29  575    575(31.124395)  $17,896.53  0.005 

Balance  32,357.42  17,896.53  $14,460.89

FV FV

Interest paid in PMT30  14,460.89(0.005)  $72.31 Principal repaid  575  72.31 

$502.70

28,000 ± 32,357.42

0 PV 575 17,896.53

PMT

PMT 6

I/Y

CPT

FV FV

(d) Last three payments are PMT54, PMT55, PMT56. PMT  575; PV  28,000; i  0.5%; n  53 FV  28,000(1.005)53  28,000(1.302571)  $36,471.98 1.00553  1  FV34  575    575(60.51412)  $34,795.62  0.005 

Balance  36,471.98  34,795.62  $1676.36 28,000 ± .36,471.98

0 PV 575 34,795.62

PMT PMT

6 6

I/Y I/Y

53 53

N N

CPT CPT

FV FV

Partial Amortization Schedule Payment number

Amount paid

Interest paid i  0.005

Principal repaid

Outstanding principal balance 28,000.00

575.00

140.00

435.00

27,565.00

575.00

137.83

437.18

27,127.83

575.00

135.64

439.36

26,688.46

1676.36

575.00

8.38

566.62

1109.74

575.00

5.55

569.45

540.29

542.99

2.70

540.29



Total

$32,167.99

$4167.99

$28,000.00

Exercise 14.2 A. 1.

(a) PVn  36,000; i  2%; n  40; c 

4 2 2

p  1.022  1  1.0404  1  4.04%

1  1.040440  36, 000  PMT    0.0404  36, 000  19.675502 PMT PMT  $1829.69 (Set P/Y  2; C/Y  4) 36,000 CPT PMT 1829.69

PV 8

I/Y

(b) Interest paid in the first period is $36,000  0.0404  $1454.40 Principal repaid in the first period is $1829.69  1454.40  $375.29 Balance at the end of the first period is $36,000  375.29  $35,624.71

(c) Interest paid in the second period is $35,624.71  0.0404  $1439.24 Principal repaid in the second period is $1829.69  1439.24  $390.45 Balance at the end of the second period is $35,624.71  390.45  $35,234.26 2.

(a) PVnc  15,000; i  1%; n  40; c 

12 3 4

p  1.013  1  1.030301  1  3.0301%

1  1.03030140  15, 000  PMT    0.030301  15, 000  23.002713 PMT PMT  $652.10 (Set P/Y  4; C/Y  12) 15,000 PMT 652.10

I/Y

CPT

(b) Interest paid in the first period is $15,000  0.030301  $454.52 Principal repaid in the first period is $652.10  454.52  $197.58 Balance at the end of the first period is $15,000  197.58  $14,802.42 (c) Interest paid in the second period is $14,802.42  0.030301  $448.53 Principal repaid in the second period is $652.10  448.53  $203.57 Balance at the end of the second period is $14,802.42  203.57  $14,598.85 3.

(a) PVnc  8500; i  3%; n  60; c 

2 1  12 6

p  1.036  1  1.004939  1  0.4939% 1  1.00493960  8500  PMT    0.004939  8500  51.817308 PMT PMT  $164.04 (Set P/Y  12; C/Y  2) 8500 PMT 164.04

I/Y

(b) Interest paid in the first period is $8500  0.004939  $41.98

0 FV CPT

Principal repaid in the first period is $164.04  41.98  $122.06 Balance at the end of the first period is $8500  122.06  $8377.94 (c) Interest paid in the second period is $8377.94  0.004939  $41.38 Principal repaid in the second period is $164.04  41.38  $122.66 Balance at the end of the second period is $8377.94  122.66  $8255.28

(a) PVnc  9600; i  4.5%; n  7; c 

2  2 1

p  1.0452  1  1.092025  1  0.092025 = 9.2025% 1  1.0920257  9600  PMT    0.092025  9600  4.998937 PMT PMT  $1920.41 (Set P/Y  1; C/Y  2) 9600 PMT 1920.41

I/Y

0 FV

CPT

(b) Interest paid in the first period is $9600  0.092025  $883.44 Principal repaid in the first period is $1920.41  883.44  $1036.97 Balance at the end of the first period is $9600  1036.97  $8563.03 (c) Interest paid in the second period is $8563.03  0.092025  $788.01 Principal repaid in the second period is $1920.41  788.01  $1132.40 Balance at the end of the second period is $8563.03  1132.40  $7430.63 5.

(a) (Set P/Y  4; C/Y  12) 3600 FV PMT 481.06

I/Y

(b) Interest paid in the first period is 2nd

AMORT

P1  1, P2  1  INT  54.27

Principal repaid in the first period is 2nd

AMORT

P1  1, P2  1  PRN  426.79

Balance at the end of the first period is 2nd

AMORT

P1  1, P2  1  BAL  3173.21

AMORT

P1  1, P2  1  INT  47.84

Principal repaid in the second period is 2nd

AMORT

P1  1, P2  1  PRN  433.22

Balance at the end of the second period is 2nd B. 1.

AMORT

P1  1, P2  1  BAL  2739.99

PVnc  16,000; i  2%; n  7; c  4 p  1.024  1  1.08243216  1  8.243216%

CPT

1  1.082432167  16, 000  PMT    0.08243216  16, 000  5.163342 PMT PMT  $3098.77 (Set P/Y  1; C/Y  4) 16,000  PMT 3098.77

I/Y

CPT

Amortization Schedule Payment number

Amount paid

Interest paid p  0.08243216

Principal repaid

Outstanding principal balance

16,000.00

3098.77

1318.92

1779.86

14,220.15

3098.77

1172.20

1926.57

12,293.58

3098.77

1013.39

2085.38

10,208.20

3098.77

841.48

2257.29

7950.91

3098.77

655.41

2443.36

5507.55

3098.77

454.00

2644.77

2862.78

3098.77

235.99

2762.78

—

Total

$21,691.39

$5691.39

$16,000.00

PVnc  22,500; i  4.5%; PMT  2200; c 

2 1  4 2

p  1.045 2  1  1.0222524  1  2.22524% 1

(Set P/Y  4; C/Y  2) N = 11.73

I/Y

2200

PMT

CPT

Amortization Schedule Payment number

Amount paid

Interest paid

Principal repaid p  0.022524

Outstanding principal balance 22,500.00

2200.00

500.68

1699.32

2200.00

462.87

1737.14

19,063.55

2200.00

424.21

1775.79

17,287.76

2200.00

384.69

1815.31

15,472.45

2200.00

344.30

1855.70

13,616.75

2200.00

303.01

1896.99

11,719.76

2200.00

260.79

1939.21

9780.55

2200.33

217.64

1982.36

7798.19

2200.00

173.53

2026.47

5771.72

20,800.68

2200.00

128.44

2071.57

3700.15

2200.00

82.34

2117.66

1582.49

1617.71

35.21

1582.50

Total

$25,817.71

$3317.71

$22,500.00

(a) PVnc  45,000; i  4.5%; n  300; c 

0.00

2  1/ 6 12

p  1.0451/6  1  1.00763  1  0.007363 = 0.7363%

1  1.007363300  45, 000  PMT    0.007363  45, 000  120.7763 PMT PMT  $372.59 (Set P/Y  12; C/Y  2) 45,000 CPT PMT 372.59

I/Y

300

(b) PMT  372.59; p  0.007363 FV  45,000(1.007363)11  48,781.94 1.00736311  1  FV11  372.59    4252.76  0.007363 

Balance  48,781.94  4252.76  $44,529.18 Alternatively, using the Amortization Worksheet, (Set P/Y  12; C/Y  2) 300 N 9 I/Y 45,000 PV 372.59 2nd

AMORT

PMT 0 FV

P1  11, P2  11  BAL  44,529.18

(c) Interest for payment period 12 is $44,529.18 (0.007363)  $327.88 (d) Principal repaid  $372.59  $327.88  $44.71 (e) (Set P/Y  12; C/Y  2) 300 N 9 I/Y 45,000 PV 372.59 2nd 4.

AMORT

PMT 0 FV

P1  49, P2  60    INT  3739.93

(a) PVnc  65,000; i  2.8%; n  180; c 

2 1  12 6

p = 1.028 6  1  1.004613  1  0.4613% 1  1.004613–180  60, 000  PMT    0.004613  65,000  122.102737 PMT PMT  $532.34 (Set P/Y  12; C/Y  2) 65,000 CPT PMT 532.34

5.6 I/Y 180

0 FV

532.34

(b) PMT  532.34; p  0.004613 FV  65,000(1.004613)119  112,402.71 1.004613119  1  FV119  532.34    84,155.54  0.004613  Balance  112,402.71  84,155.54  $28,247.17

Alternatively, using the Amortization Worksheet, (Set P/Y  12; C/Y  2) 180 PMT 0 FV 2nd

AMORT

5.6 I/Y 65,000

P1  119, P2  119  BAL  28,247.17

(c) Interest for payment period 120 is $28,247.17 (0.004613)  $130.31 (d) Principal repaid  $532.34  $130.31  $402.03 (e) (Set P/Y 12; C/Y  2) 180 PMT 0 FV 2nd

AMORT

5.6 I/Y 65,000

532.34

P1  49, P2  60  PRN  3569.21

(a) PVnc  12,000; PMT  960; i  1.19%; c  p  1.01192  1  1.023942  1  2.3942%

4 2 2

1  1.023942 – n  12,000  960    0.023942  1  1.023942 – n 12.5  0.023942 0.299275  1 –1.023942 – n 1.023942 – n  0.700725 – n ln 1.023942  ln 0.700725 –0.02366n  –0.35564 n  15.03 half-years (Set P/Y  2; C/Y  4) 12,000 FV CPT N 15.03

4.76

I/Y

960

PMT

16 payments are required to amortize the loan (b) PMT  960; PV  12,000; p  2.3942%; n  12 FV12  12,000 (1.023942)12  15,939.90 1.02394212  1  FV12  960    13,164.81  0.023942 

Balance  15,939.90  13,164.81  $2766.09 12,000

960

PV 0 PMT 4.76 ±

I/Y

CPT

FV 15,939.83

I/Y

CPT

FV 13,164.78

PMT 4.76

(a) PVnc  22,500; PMT  920; i  3.4%; c  p  (1.034)0.5  1  1.01686  1  1.686%

2 1  4 2

1  1.01686 – n  22,500  920    0.01686  1  1.01686 – n 24.4565  0.01686 0.41229  1 –1.01686 – n 1.01686 – n  0.587714 – nln 1.01686  ln 0.587714 –0.01672n  0.531514 n  31.79 quarters

(Set P/Y  4; C/Y  2) 22,500 CPT N 31.79

6.8

32 payments are needed to repay the loan

I/Y

920 PMT 0 FV

(b) Use retrospective method: PMT  920; PV  22,500; p  1.686%; n  20 FV  22,500(1.01686)20  31,433.15 1.0168620  1  FV20  920    21,667.38  0.01686 

Balance  31,433.15  21,667.38  $9765.77 22,500

0 PMT 6.8 I/Y 20

920

PMT 6.8 I/Y 20

CPT

FV 31,433.15 FV 21,667.38

(c) Total paid  920 (20)  $18,400 Principal repaid  22,500  9765.77  $12,734.23 Interest  18,400  12,734.23  $5665.77 7.

(a) PMT  1850; i  0.005%; n  14; c 

12 3 4

P  (1.005)3  1  1.015075  1  1.5075% 1  1.01507514  PV  1850    $23,192.79  0.015075 

(Set P/Y  4; C/Y  12) 0 FV 6 I/Y 14 N 1850 23,192.79 (b) PMT  1850; i  0.5%; n  14; c 

PMT

12 3 4

P  (1.005)3  1  1.015075  1  1.5075% FV  23,192.79(1.015075)8  26,141.98 1.0150758  1  FV8  1850    15,604.89  0.015075 

Balance  26,141.98  15,604.89  $10.537.10 14 N 7.44 I/Y 23,192.79 2nd

AMORT

1850

PMT 0 FV

P1  8, P2  8  BAL  10,537.10.

CPT

(d) Interest  1850(8)  12,655.69  14,800  12,655.69  $2144.31 8.

(a) PMT  875; i  2.04%; n  60; c  P  1.0204 

0.3

4  0.3 12

 1  1.006754  1  0.6754%

1  1.00675460  PV  875    $43,046.73  0.006754 

(Set P/Y  12; C/Y  4) 0 FV 8.16 I/Y 60 N 875 PV 43,046.73 (b) PMT  875; i  2.04%; n  14; c 

PMT 0 FV

CPT

4  0.3 12

FV  43,046.73(1.006754)14  47,300.85 1.00675414  1  FV14  875    12,802.61  0.006754 

Balance  47,300.85  12,802.61  $34,498.24 14 N 8.16 I/Y 43,046.73 2nd

AMORT

875

PMT 0 FV

P1  14, P2  14  BAL  34,498.24.

(c) Principal repaid  43,046.73  34,498.24  $8548.49 (d) Interest  875(14)  8548.49  12,250  8548.49  $3701.51 9.

(a) PVnc  42,000; PMT  2950; i  0.5%; c 

12 6 2

p  1.0056  1  1.0303775  1  3.0303775%

1  1.0303775 n  42,000  2950    0.0303775  1  1.0303775 n 14.2373  0.0303775 0.432493  1 –1.0303775– n 1.0303775– n  0.56751 – n ln 1.0303775  ln 0.56751 –0.02993n  –0.5665 n  18.93 half-years (Set P/Y  2; C/Y  12) 42,000 FV CPT N 18.93

0 FV 6

I/Y

2950 PMT 0

19 payments are required to retire the debt. (b) Use retrospective method: PMT  2950; PV  42,000; p  3.03775%; n  10 FV  42,000(1.0303775)10  56,651.71 1.030377510  1  FV10  2950    33,877.30  0.0303775 

Balance  56,651.71  33,877.30  $22,774.41 42,000

0 PV 2950

PV ±

0 PMT 6 I/Y 10 N

PMT 6 I/Y 10

CPT CPT

56,651.71

FV 33,877.30

Total paid  2950 (10)  29,500 Principal repaid  42,000  22,774.41  19,225.59 Cost of debt = $10,274.41 (c) PMT  2950; PV  42,000; p  3.03775%; n  9 FV  42,000(1.0303775)9  54,981.51 1.03037759  1  FV9  2950    30,015.50  0.0303775 

Balance  54,981.51  30,015.50  $24,966.01 Interest paid in PMT10  24,966.01(0.0303775)  $758.41 42,000 0

0 PMT 6

PV 2950

PMT 6 I/Y 9

I/Y

N CPT

(d) Last three payments are PMT17, PMT18 and PMT19 PMT  2950; p  3.03775%; n  16 FV  42,000(1.0303775)16  67,793.99 1.030377516  1  FV12  2950    59,640.21  0.0303775 

Balance  67,793.99  59,640.21  $8153.78

FV 54,981.51 FV 30,015.50

42,000

0 PMT 6

2950

PMT 6 I/Y 16

Payment #

Amount paid

I/Y

16 N

FV 67,793.99

CPT FV 59,640.21

Interest paid

Principal repaid

Outstanding principal 42,000.00

2950

... ...

1275.86

1674.15

40,325.86

1725

38,600.85

1172.60

1777.40

36,823.45

... ...

2950

247.69

2702.31

5451.48

2950

165.60

2784.40

2667.08

2748.10

81.02

2868.98

Totals

$55,848.10

$13,848.10

$42,000.00

1225

0.00

10. (a) PV = 680,000 – 70,000 = $610,000 FV = 0 I/Y = 4.24 % C/Y = 2 P/Y = 26 N = 25(26) = 650 (―END‖ Mode) (Set P/Y = 26, C/Y = 2) 0 FV 610,000 

I/Y 650 N CPT PMT = $1516.38 Each bi-weekly payment is $1516.38.

PV 4.24

(b) To calculate total amount of interest paid over the life of this loan, use the amortization function in Texas Instrument BAII PLUS calculator with the following steps: 1-Press 2nd and then Amort buttons. 2-Set P1 equal to 1 and P2 equal to 650. 3-Use the arrow key down to get to INT and then press CPT. 4-INT = 375,644.89

Therefore, the total interest paid over 25 years equals $375,644.89. (c) Use the AMORT function in your Texas Instrument BAII PLUS calculator to complete the amortization table: Payment Number 121* 122 123

Payment Amount 1516.38 1516.38 1516.38

Principle Paid

Interest Paid

644.71 645.75 649.80

871.66 870.62 869.58

Remaining Balance 539,076.25 538,430.49 537,873.69

* For each row, set P1 and P2 in the Amort function to equal the payment number and scroll to next screen of the calculator to get the values for each column for that row. 11. (a) PV = 68,500 FV = 0 I/Y = 5.5% C/Y = 2 P/Y = 12 N = 6(12) = 72 (―END‖ Mode) (Set P/Y = 12, C/Y = 2) 0 FV 68,500 ± PV 5.5 I/Y 72 N CPT PMT = 775.3181 Each monthly payment is $1117.16. (b) (―END‖ Mode) (Set P/Y = 12, C/Y = 2) 0 FV 48,500 ± PV 4.79 I/Y 850 PMT CPT N = 100.51 You need to make 101 monthly payments. Exercise 14.3 A. 1. CPT

(Set P/Y  4; C/Y  4) 9

I/Y

17,500

1100

PMT

FV .

19.914, or 20 payments. Set N  20

Balance after 20 payments, 20

I/Y

17,500

2nd AMORT overpayment,

1100

PMT

FV .

P1  20, P2  20  BAL  93.76, indicating an

Final payment is $1100  $93.76 

$1006.24 .

(―BGN‖ mode) (P/Y  2; C/Y  2) 7 FV .

I/Y

7800

775

PMT

CPT

yields 12.0937, or 13 payments. Set N  13

Balance after 13 payments, 13

I/Y

7800

2nd AMORT overpayment,

775

PMT

P1  13, P2  13  BAL  701.26, indicating an

Final payment is $775  $701.26 

$73.74

(―BGN‖ mode) (P/Y  4; C/Y  4) 7 0

CPT

I/Y

9300

580

PMT

yields 18.598, or 19 payments. Set N  19

Balance after 19 payments, 19

I/Y

9300

2nd AMORT overpayment,

PV .580

(―END‖ mode) (P/Y  2; C/Y  4) 8 0

FV .

CPT

PMT

P1  19, P2  19  BAL  231.74, indicating an

Final payment is $580.00  $231.74  4.

$348.26

I/Y

15,400

1600

PMT

yields 12.433, or 13 payments. Set N  13

Balance after 13 payments, 13

I/Y

15,400

2nd AMORT overpayment,

1600

$700.61

(―END‖ mode) (P/Y  4; C/Y  12) 9

I/Y

FV .

CPT

FV .

P1  13, P2  13  BAL  899.39, indicating an

Final payment is $1600  $899.39 

PMT

29,500

1650

PMT

yields 23.184, or 24 payments. Set N  24

Balance after 24 payments, 24

I/Y

29,500

2nd AMORT overpayment,

1650

PMT

FV .

P1  24, P2  24  BAL  1343.61, indicating an

Final payment is $1650  $1343.61  $306.39 6.

(―BGN‖ mode) (P/Y  12; C/Y  4) 6 FV .

CPT

I/Y

17,300

425

PMT

yields 45.345, or 46 payments. Set N  46

Balance after 46 payments, 46

I/Y

17,300

2nd AMORT overpayment,

425

PMT

FV .

P1  46, P2  46  BAL  278.06, indicating an

Final payment is $425  $278.06 

$146.94

(―END‖ mode) (P/Y  4; C/Y  12) 4.5

I/Y

FV .

CPT

32,500

725

PMT

FV .

 62.84 or 63 payments

Balance after 63 payments, 63

4.5

I/Y

2nd AMORT overpayment,

32,500

725

P1  63, P2  63 BAL  115.15, indicating an

Final payment is $725  $115.15  $609.85

PMT

B. 1. (a) (―END‖ mode) (P/Y  4; C/Y  4) 10 0 FV . V

CPT

I/Y

7800

380

PMT

= 29.15, or 30 payments. Set N  30 payments

(b) Balance after 30 payments, 30

I/Y

2nd AMORT overpayment,

7800

PMT

I/Y

60,000

= 15.90, or 16 payments. Set N 

(b) Balance after 16 payments, 16 PMT 0 FV . 2nd AMORT overpayment,

FV .

$58

(a) (―END‖ mode) (P/Y  2; C/Y  2) 6 PMT 0 FV . CPT

P1  30, P2  30  BAL  322, indicating an

Final payment is $380  $322  2.

.380

I/Y

4800 .

16 payments

60,000

PV .4800

P1  16, P2  16  BAL  470.64, indicating an

Final payment is $4800  $470.64  $4329.36 3.

(a) (―END‖ mode) (P/Y  12; C/Y  12) 12 PMT 0 FV . CPT

2nd AMORT overpayment,

(a) (―END‖ mode) (P/Y  1; C/Y  4) 6 0 FV . N

PV .1100

I/Y

45,000

.1100

P1 53, P2  53  BAL  140.28, indicating an

Final payment is $1100  $140.28 

CPT

45,000

= 52.78, or 53 payments. Set N  53 payments

(b) Balance after 53 payments, 53 PMT 0 FV .

I/Y

$959.72 I/Y

3600

.600

PMT

= 7.71, or 8 payments. Set N  8 payments

(b) Balance after 8 payments, 8 0N . 2nd AMORT overpayment,

I/Y

3600

.600

P1 8, P2  8 BAL  170.46, indicating an

Final payment is $600  $170.46 

$429.54

(a) (―BGN‖ mode) (P/Y  4; C/Y  2) 8 0 FV . CPT

I/Y

12,500

950

PMT

.950

= 15.05, or 16 payments. Set N  16 payments

(b) Balance after 16 payments, 16 PMT 0 FV . 2nd AMORT overpayment,

I/Y

12,500

P1  16, P2  16  BAL  905.48, indicating an

Final payment is $950  $905.48  $44.52 6.

(―END‖ mode) (P/Y  4; C/Y  12) 6 0 FV . CPT

I/Y

28,000

1450

PMT

= 22.99, or 23 payments. Set N  23

Balance after 23 payments, 23 PMT 0 FV . 2nd AMORT overpayment,

I/Y

28,000

PV .1450

P1  23, P2  23  BAL  8.78, indicating an

Final payment is $1450  $8.78 

$1441.22

(―END‖ mode) (P/Y  12; C/Y  4) 8 0

FV .

CPT

40,000

500

PMT

= 114.32, or 115 payments. Set N  115

Balance after 115 payments, 115 PMT 0 FV . 2nd AMORT overpayment,

I/Y

40,000

500

P1  115, P2  115  BAL  338.07, indicating an

Final payment is $500  $338.07  $161.92 8.

(―BGN‖ mode) (P/Y  4; C/Y  4) 6 FV . CPT

I/Y

20,000

1200

PMT

= 18.99, or 19 payments. Set N  19.

Balance after 19 payments, 19 PMT 0 FV . 2nd AMORT overpayment,

I/Y

20,000

1200

P1  19, P2  19  BAL  9.25, indicating an

Final payment is $1200  $9.25  $1190.75 9.

(―BGN‖ mode) (P/Y  12; C/Y  12) 6 0

FV .

CPT

10. (―BGN‖ mode) (P/Y  2; C/Y  4) 8 FV . N

1180

PMT

I/Y

125,000

.1180

P1  151, P2  151  BAL  1041.67, indication an

Final withdrawal is $1180  $1041.67 

CPT

125,000

= 150.12, or 151 payments. Set N  151

Balance after 151 payments, 151 PMT 0 FV . 2nd AMORT overpayment,

I/Y

$138.33 64,000

6200

PMT

6200

= 12.93, or 13 payments. Set N  13

Balance after 13 payments, 13 PMT 0 FV . 2nd AMORT overpayment,

I/Y

64,000

P1  13, P2  13  BAL  406.36, indicating an

Final payment is $6200  $406.36  $5793.64

11.

(―BGN‖ mode) (P/Y  4; C/Y  1) 8 . CPT

I/Y

26,800

1600

PMT

1600

PMT

= 19.98, or 20 payments. Set N  20

Balance after 20 payments, 20 FV . 2nd AMORT overpayment,

I/Y

26,800

P1  20, P2  20  BAL  26.98, indicating an

Final payment is $1600  $26.98  $1573.02 12. (a) Determining the growth of the $18,600 during the deferment, (―END‖ mode) (P/Y  4; C/Y  4) 40 . 0

PMT

CPT

FV ±

54 payments. Set N 

54 payments

2nd AMORT overpayment,

I/Y

18,600

CPT

41,069.54

8 I/Y 41,069.54 = 53.29, or

(b) Balance after 54 payments, 54 1260 PMT 0 FV .

1260

PMT

I/Y

PV .

P1  54, P2  54  BAL  893.55, indicating an

Final payment is $1260  $893.55 

$366.45

$67,146.45

$48,546.45

13. (a) Determining the growth of the $36,800 during the deferment, (―BGN‖ mode) (P/Y  2; C/Y  2) 8 0

PMT

CPT

I/Y

36,800

PMT

FV .

CPT

I/Y

54,370.36

 54,370.36

10 I/Y 54,370.36 14.12, or

5200

15 payments. Set N 

15 payments

(b) Balance after 15 payments, 15 PMT 0 FV .

. 10

. 5200

2nd AMORT overpayment,

P1  15, P2  15  BAL  4558.94, indicating an

Final payment is $5200  $4558.94  $641.06 (c) Total paid  $5200(14)  $641.06  $73,441.06 (d) Interest  $73,441.06  $36,800  $36,641.06 14. (a) PV = 3,800,000 – 400,000 = $3,400,000 FV = 0 I/Y = 4.25% C/Y = 2 P/Y = 12 N = 25(12) = 300 (―END‖ mode) (Set P/Y  12; C/Y  2) 0 4.25 I/Y 300 N CPT

3,400,000

PMT = 18,348.45

Each monthly payment is $ 18,348.45. (b) To calculate total amount of interest paid over the life of this loan, use the amortization function in Texas Instrument BAII PLUS calculator with the following steps: 1234-

Press 2nd and then Amort buttons. Set P1 equal to 1 and P2 equal to 300. Use the arrow key down to get to INT and then press CPT. INT = 2,104,533.67

Therefore, the total interest paid over 25 years equals $2,104,533.67. (c) Use the AMORT function in your Texas Instrument BAII PLUS calculator to complete the amortization table: Payment Numbers

Payment Amount

Principle Paid

Interest Paid

Remaining Balance

101

18,348.45

9103.26

9245.19

2,624,320.46

102

18,348.45

9135.22

9213.23

2,615,185.24

103

18,348.45

9167.29

9181.16

2,606,017.95

* For each row, set P1 and P2 in the Amort function to equal the payment number and scroll to next screen of the calculator to get the values for each column for that row. (d) At the end of the ten-year mortgage term, 120 payments have been made. Use the Amort function and set P1 = 1, P2 = 120 and get the balance after 120 payments. It equals:

BAL = $2,445,156.49 (e) (―END‖ Mode) (Set C/Y = 2 and P/Y = 12) 0 FV 2,445,156.49 

PV 5.25

I/Y 180 N CPT PMT = $19,583.48. Each monthly payment would be $19,583.48. Business Math News Box 1.

PV  250,000; n  25(12)  300; P/Y  12; C/Y  2; I/Y  5.5 c  2/12  1/6; i  5.5%/2  2.75%  0.0275 p  1.02751/6  1  1.004532  1  0.004532  0.4532% 250,000  PMT [(1  (1.004532)300)/(0.004532)] 250,000  PMT (163.823181) PMT  $1526.04

FV of mortgage at 5 years: FV  250,000 (1.004532)60  $327,918.99 FV of payments at 5 years: FV  1526.04[(1.00453260  1) / (0.004532)]  $104,949.25 Balance at the end of the 5-year term: 327,918.99  104,949.25  $222,969.74 The outstanding balance after 5 years is $222,969.74

FV of mortgage at 5 years: FV  250,000 (1.004532)60  $327,918.99 New monthly payment:



 1526.04  (0.40)(500)



 1526.04  200



 $1726.04

FV of payments at 5 years: FV  1726.04[(1.00453260  1) / (0.004532)]  $118,703.70 Balance at the end of the 5-year term: 327,918.99  118,703.70  $209,215.29 The outstanding balance after 5 years is $209,215.29. $222,969.74  209,215.29  $13,754.45 The homeowner will have reduced the balance of her mortgage by $13,754.45 by contributing an additional $12,000 over 5 years. In addition, she will have

contributed $30,000 to her RRSP. 4.

i = 0.055/2 = 0.0275 1 6

P  (1  i )  1  (1  0.0275)  0.004531682 c

1  (1  p)  n  PV  PMT   p   1  (1  0.004531682)  n  250, 000  1726.04   0.004531682   Simplify: 0.34363 = 1.004531682–n Ln 0.34363 = –n Ln 1.004531682 n = 236.25 Months This means that if the homeowner pays an extra $200 per month on the mortgage, she can reduce the number of payments from 300 to 237 and can be mortgage free about 5 years sooner. Exercise 14.4 1.

(a) PV  180,000; n  12(25)  300; i  2.445%; c  0.1667 P  1.024450.1667  1  1.004034  1  0.004034 = 0.4034% 1  1.004034300  180,000  PMT    173.804552PMT  0.004034  PMT  $1035.65

(b) PMT  1035.65; p  0.4034% FV  180,000(1.004034)36  208,072.95 1.00403436  1 FV36  1035.65    40,039.86  0.004034 

Balance is 208,072.95  40,039.86  $168,033.09 (―END‖ mode)(P/Y  12; C/Y  2) 300 1035.65 ± PMT 0 FV . 2nd

AMORT

(c ) 264 N 5.24 I/Y 1068.19 or $1068.19 2.

4.89

I/Y

180,000

P1  36, P2  36  BAL  168,033.73 168,033.73

PV .0

(a) PV  150,000; n  240; i  3%; c  0.16

CPT

PMT



P  1.03

0.16

 1  1.004939  1  0.4939%

1  1.004939240  150, 000  PMT    140.407133PMT  0.004939  PMT  1068.32

After 18 months: FV  150,000 (1.004939)18  163,909.05 1.00493918  1 FV18  1068.32    20,058.72  0.004939 

Balance  163,909.05  20,058.72 

$143,850.33

(―END‖ mode) (P/Y  12; C/Y  2) 240

1068.32 ± PMT BAL  143,850.33

2nd

AMORT P1  18, P2  18 

(Set P/Y  12; C/Y  2) 150,000 PMT  1068.28

I/Y

240

150,000

CPT

(b) PV  143,851.15; n  222; i  3.3%; c  0.16& P  (1.033)0.16  1  1.005426  1  0.5426% 1  1.005426222  143,851.15  PMT    128.862936PMT  0.005426  PMT  $1116.31

143,851.15

. 6.6

I/Y

222

CPT

 1116.31

PMT

(c) Balance at the end of 2 years: PMT  1116.31; i  0.5426%; n  6 FV  143,851.15(1.005426)6  148,598.24 é1.005426 6 - 1ù ú 6789.37 FV6  1116.31 ê ê 0.005426 ú ë û

Balance at end of 2 years  148,598.24  6789.37  $141,808.87 After 4-year term: PV  141,808.87; PMT  1116.31; i  2.825%; n  48; c  0.16 P  (1.02825)0.16  1  1.00465385  1  0.4654% FV  141,808.87(1.004654)48  177,212.01 é1.004654 48- 1ù ú  59,883.99 FV48  1116.31 ê ê 0.004654 ú ë û

Balance  177,212.01  59,883.99  6

N 2nd

48 .

2nd 3.

6.6

I/Y

143,851.15

$117,328.02 PV

.1116.31

. ±

PMT

AMORT

P1  6, P2  6  BAL  141,808.87

5.65

141,808.87

I/Y

AMORT

.1116.31

P1  48, P2  48  BAL  117,328.02

(a) PVnc  40,000; n  12(10)  120; i 

7.15% 1  3.575%; c  6 2

p  1.03575 6  1  1.005871  1  0.5871%

PMT

FV .

1  1.005871120  40, 000  PMT    0.005871  40, 000  PMT  85.951596  PMT  465.38

The rounded monthly payment is $500 PVnc  40,000; PMT  500; p  0.5871% (Set P/Y  12; C/Y  2) 40,000 CPT PMT 465.38

7.15

I/Y

120

1  1.005871 n  40, 000  500    0.005871  0.469719  1 –1.005871– n 1.005871– n  0.530281 – n ln 1.005871  ln 0.530281 –0.005855n  –0.634347 n  108.36 months or 109 payments

40,000

500

PMT

7.15

I/Y

CPT

108.36

(b) PMT  500; p  0.005871 Balance after 108 payments: FV  40,000(1.005871)108  75,274.78 1.005871108  1 FV108  500    75,097.69 0.005871  

Balance  75,274.78  75,097.69  177.09 The last payment is $177.09(1.005871)  $178.13 Alternatively, using the Amortization Worksheet, 109 N 7.15

I/Y 40,000

2nd AMORT overpayment,

500

PMT

0 FV

P1  109, P2  109  BAL  321.87, indicating an

The final payment is 500  321.87  $178.13 (c) Total amount paid with unrounded payments  120(465.38)  55,845.60

Total amount paid with rounded payments  108(500)  178.13  54,178.13 Amount of interest saved  55,845.60  54,178.13  $1667.47 4.

4.6% 1  2.3%; c  6 2 1 6 p = 1.023  1  1.0037971  1  0.0037971  0.37971%

(a) PVnc  23,960; PMT  250; i 

1  1.0037971 n  23, 960  250    0.0037971  0.363914  1  1.0037971 n 1.0037971 n  0.636086 n ln 1.0037971  ln 0.636086 0.0037899n  0.452422 n  119.38 months or 120 payments (Set P/Y  12; C/Y  2) 23,960 FV CPT N 119.38

250

PMT

4.6

I/Y

(b) PMT  440; p  0.0037971 Balance after 119 payments: FV  23,960(1.0037971)119  37,614.33 1.0037971119  1  FV65  250    37,520.74  0.0037971 

Balance  37,614.33  37,520.74  93.59 The last payment is $93.59(1.0037971)  $93.95 Alternatively, using the Amortization Worksheet, 120 N AMORT

4.6 I/Y P1  120,

23,960

250

PMT

2nd

0 FV

P2  120  BAL  156.05, indicating an overpayment, The final payment is $250  156.05  $93.95 (c) PVnc  23,960; n  12(10)  120; p  0.0037971 1  1.0037971120  23,960  PMT    0.0037971 

23,960  96.236132 PMT PMT  248.97 23,960 248.97

120

4.6

I/Y

CPT

PMT

The total amount required to pay with contractual payments  248.97(120)  29876.52 The total amount paid with requested payments  119(250)  93.95  29,843.95 Difference in payments  29,876.52  29,843.95  $32.57 5.

PV  180,000; PMT  611.31; n  24(20)  480 (Set P/Y  24; C/Y  2) 180,000 FV CPT I/Y 5.42

611.31

PMT

480

300

The nominal annual rate is 5.42% compounded semi-annually. 6.

PV  162,000; PMT  1017.31; n  300 (Set P/Y  12; C/Y  2) 162,000 ± FV CPT I/Y 5.80

1017.31

PMT

The nominal annual rate is 5.80% compounded semi-annually. 7.

(a) PVnc  105,000; n  52(20)  1040; i 

2 4.39%  2.195%; c  52 2

p = 1.02195 52  1  1.000835  1  0.000835  0.0835% 1  1.0008351040  105,000  PMT    0.000835 

105,000  PMT(694.742233) PMT  151.14 (Set P/Y  52; C/Y  2) 105,000 FV CPT PMT 151.14

1040

4.39

I/Y

PMT

Balance at end of 3 years. PMT  151.14; p  0.000835 Balance after 3(52)  156 payments: FV  105,000(1.000835)156  119,609.92 1.000835156  1 FV156  151.14    25,172.06  0.000835 

Balance  119,609.92  25,172.06  94,437.86 New balance is 94,437.86  7000  $87,437.86 Alternatively, using the Amortization Worksheet, 1040 2nd

4.39 AMORT

I/Y

105,000

151.14

P1  156, P2  156  BAL  94,437.86

Balance after 4 years, or 52 more payments, FV  87,437.86(1.000835)52  91,319.55 1.00083552  1 FV52  151.14    8029.07  0.000835 

Balance  91,319.55  8029.07  $83,290.48 (b) PVnc  83,290.48; PMT  151.14; p  0.000835

1  1.000835 n  83, 290.48  151.14    0.000835  0.460400  1 –1.000835– n 1.000835– n  0.539600 – n ln 1.000835  ln 0.539600 –0.000835n  –0.616927 n  738.75 weeks  739 payments

83,290.48 738.75

151.14

PMT

4.39

I/Y

CPT

(c) Balance after another 738 payments, FV  83,290.48(1.000835)738  154,259.63 1.000835738  1 FV738  151.14    154,146.80  0.000835 

Balance  154,259.63  154,146.80  $112.83 Final payment is 112.83(1.000835)  112.93 Total contractual amount  1040(151.14)  157,185.60 Total actually paid  (156  52  738)(151.14)  7000  112.93  $150,091.37 Difference in cost  157,185.60  150,091.37  $7094.23 8.

(a) PVnc  36,000; n  12(10)  120; i 

8.75% 1  4.375%; c  6 2

p = 1.04375 6  1  1.007162  1  0.007162  0.7162% 1  1.007162120  36, 000  PMT    0.007162  36, 000  PMT  80.326174  PMT  448.17

The rounded monthly payment is $500. (Set P/Y  12; C/Y  2) 36,000 ± FV CPT PMT 448.17

120

8.75

I/Y

To determine the balance after 30 months, use the Retrospective Method: The accumulated value of $36,000 after 30 months  36,000 (1.007162)30  36,000(1.238747)  $44,594.88 The accumulated value of the 30 monthly payments made

1.00716230  1  FVnc  500    0.007162   500  33.334278   16, 667.14

The mortgage balance after 30 months  44,594.88  16,667.14  $27,927.74 36,000 ± 44,594.88

0 PV 500 16,667.14

PMT PMT

8.75 8.75

I/Y I/Y

30 30

N N

CPT CPT

FV FV

The increased payment  110% of $500  $550 The accumulated value of $27,927.74 at the end of the 5-year term  27,927.74(1.007162)30  27,927.74(1.2387468)  $34,595.40 The accumulated value of the increased payments made during the remaining 30 months of the 5-year term 1.00716230  1 FVnc  550    0.007162   550  33.334278  18,333.85

The mortgage balance at the end of the 5-year term  34,595.40  18,333.85  $16,261.55 27,927.74 34,595.40 0

± 550

PV ±

0 PMT

PMT

8.75

I/Y

8.75

I/Y

30 N CPT

CPT FV

18,333.85

(b) PVnc  16,261.55; PMT  550; p  0.7162%

1  1.007162 n  16, 261.55  550    0.007162  0.211761  1 –1.007162 – n 1.007162 – n  0.788239 – n ln 1.007162  ln 0.788239 –0.007137n  –0.237954 n  33.342388 months  34 payments 16,261.55 33.342388

550

PMT

8.75

I/Y

CPT

(c) Determine the number of payments of $500 required after the initial 30 payments: PVnc  27,927.74; PMT  500; p  0.7162%

1  1.007162 n  27,927.74  500    0.007162  0.400048  1 –1.007162 – n 1.007162 – n  0.599952 – n ln 1.007162  ln 0.599952 –0.007137n  –0.510906 n  71.588771 months 27,927.74 71.588771

500

PMT

8.75

I/Y

CPT

The total amount paid after the first 30 months with the initial rounded payments  500 (71.588771)  $35,794.39 The total amount paid with the increased rounded payments  550 (30  33.342388)  550 (63.342388)  $34,838.29 Amount saved  35,794.39  34,838.29  $956.10 9.

PVnc  40,000; n  12(12)  144; i 

5.5% 1  2.75%; c  2 6

p = (1.0275) 6  1  1.004532  1  0.004532  0.4532%

1  1.004532144  40, 000  PMT    0.004532  40, 000  PMT 105.59465  PMT  378.81

The rounded monthly payment is $380.00 (Set P/Y  12; C/Y  2) 40,000 ± FV CPT PMT 378.81

0 FV 144

5.5

I/Y

Amortization Schedule for the First 6 Months Payment date

Amount paid

Interest paid P  0.004532

Principal repaid

June 1

Balance 40,000.00

July 1

380.00

181.27

198.73

39,801.27

Aug 1

380.00

180.37

199.63

39,601.64

Sept 1

380.00

179.46

200.54

39,401.10

Oct 1

380.00

178.55

201.45

39,199.65

Nov 1

380.00

177.64

202.36

38,997.29

Dec 1

380.00

176.72

203.28

38,794.01

10. Mortgage statement: PMT  380; Since P/Y is different from C/Y, you need to calculate periodic rate of return using formula 12.1. Then the annual rate of interest  12(0.45317%)  5.4380% Payment date

Number of days

Amount paid

Interest paid

Principal

June 1

Balance repaid 40,000.00

July 1

380.00

178.78

201.22

39,798.78

Aug 1

380.00

183.81

196.19

39,602.59

Sept 1

380.00

182.91

197.09

39,405.50

Oct 1

380.00

176.13

203.87

39,201.63

Nov 1

380.00

181.06

198.94

39,002.69

Dec 1

380.00

174.33

205.67

38,797.02

The mortgage statement balance on December 1 of $38,797.02 differs from the amortization schedule balance of $38,794.01 by $3.01. This difference is due to the calculation of interest on a daily basis. 11. Mortgage statement: PMT  190; Since P/Y is different from C/Y, you need to calculate a periodic rate of return using formula 12.1. Then the annual rate of interest  12(0.45317%)  5.4380% Copyright © 2025 Pearson Canada Inc.

Payment date

Number of days

Amount paid

Interest paid

Principal

June 1

Balance repaid 40,000.00

190.00

89.39

100.61

39,899.39

July 1

190.00

89.17

100.83

39,798.56

190.00

88.94

101.06

39,697.50

Aug 1

190.00

94.63

95.37

39,602.13

190.00

88.50

101.50

39,500.63

Sept 1

190.00

94.16

95.84

39,404.79

190.00

88.06

101.94

39,302.85

Oct 1

190.00

87.83

102.17

39,200.68

190.00

87.61

102.39

39,098.29

Nov 1

190.00

93.20

96.80

39,001.49

190.00

87.16

102.84

38,898.65

Dec 1

190.00

86.93

103.07

38,795.58

The mortgage statement balance on December 1 of $38,795.58 differs from the amortization schedule balance of $38,794.01 by $1.57. The difference is reduced from $3.01 in the answer to Question 10 by $1.44 due to making semi-monthly payments. 12. Mortgage statement: PMT  190; annual rate of interest  12(0.4532%)  5.4380% Payment date

Number of days

Amount paid

Interest paid

Principal

June 1

Balance repaid 40,000.00

190.00

89.39

100.61

39,899.39

190.00

83.22

106.78

39,792.61

July 14

190.00

83.00

107.00

39,685.61

190.00

82.78

107.22

39,578.39

Aug 11

190.00

82.55

107.45

39,470.94

190.00

82.33

107.67

39,363.27

Sept 8

190.00

82.10

107.90

39,255.37

190.00

81.88

108.12

39,147.25

Oct 6

190.00

81.65

108.35

39,038.90

190.00

81.43

108.57

38,930.33

Nov 3

190.00

81.20

108.80

38,821.53

190.00

80.97

109.03

38,712.50

Dec 1

190.00

80.75

109.25

38,603.25

The mortgage statement balance on December 1 of $38,603.25 differs from the amortization schedule balance of $38,794.01 by $190.76. The large difference from the balances in Questions 9, 10, and 11 is due to the additional payment made. Review Exercise 1.

(a) PVn  45,000  10,000  35,000; i  1.5%; n  32

1  1.01532  35, 000  PMT    0.015  35, 000  25.267138 PMT PMT  $1385.20 (Set P/Y  4; C/Y  4) 35,000 CPT PMT 1385.20

I/Y

0 FV

PMT

(b) Total paid  1385.20(32)  $44,326.35 Amount borrowed  35,000 Cost of financing  $9326.35 (c) FV  35,000(1.015)20  47,139.93 1.01520  1  FV20  1385.20    32,030.90  0.015 

Balance  47,139.93  32,030.90  $15,109.03 Alternatively, using the Amortization Worksheet, 32

I/Y

35,000

1385.20

(d)

2nd

AMORT

P1  20, P2  20  BAL  15,109.02

2nd

AMORT

P1  20, P2  20  BAL  15,109.02  PRN 

1141.44  INT   $243.76 (e)

(f)

P1  24, P2  24  BAL  10,369.45  PRN  

2nd AMORT $1211.49 2nd

AMORT

P1  9, P2  9  BAL  26,777.06

2nd

AMORT

P1  29, P2  29  BAL  4033.92 Partial Amortization Schedule

Payment number

Periodic payment

Interest paid

Principal repaid

Outstanding balance 35,000.00

1385.20

525.00

860.20

34,139.80

1385.20

512.10

873.10

33,266.70

1385.20

499

886.20

32,380.50

1385.20

26,777.06

1385.20

401.66

983.54

25,793.52

1385.20

386.90

998.30

24,795.22

1385.20

371.93

1013.27

23,781.95

1385.20

4033.92

1385.20

60.51

1324.69

2709.23

1385.20

40.64

1344.56

1364.66

1385.13

20.47

1364.66

—

Total

$44,326.33

$9326.33

$35,000.00

1  1.004560  8000  PMT    0.0045  8000  52.47955 PMT PMT  $152.44 (Set P/Y  12; C/Y  12) 8000 CPT PMT 152.44

5.4

I/Y

0 FV

(b) Total paid  152.44(60)  $9146.42 Amount borrowed  $8000 Cost of financing  $1146.41 (c) FV  8000(1.0045)18  8673.39 1.004518  1  FV18  152.44    2851.44  0.0045 

Balance  8673.39  2851.44  $5821.95 Alternatively, using the Amortization Worksheet, 60

5.4

I/Y 8000

152.44

PMT

2nd

AMORT

P1  18, P2  18  BAL  5821.95

(d)

2nd

AMORT

P1  36, P2  36    INT  $16.19

(e)

2nd

AMORT

P1  48, P2  48   PRN  $143.80

(f) Balance after 23 payments 2nd

AMORT

P1  23, P2  23  BAL  5185.04

2nd

AMORT

P1  57, P2  57  BAL  453.26 Partial Amortization Schedule

Payment number

Periodic payment

Interest paid

Principal repaid

Outstanding balance 8000.00

152.44

36.00

116.44

7883.56

152.44

35.48

116.96

7766.60

152.44

34.95

117.49

7649.11

5185.04

152.44

23.33

129.11

5055.93

152.44

22.75

129.69

4926.25

152.44

22.17

130.27

4795.97

453.26

152.44

2.04

150.40

302.86

152.44

1.36

151.08

151.78

152.46

0.68

151.78

—

Total

$9146.42

$1146.42

$8000.00

(a) PVn  50,000; PMT  1600; i  1.2%

1  1.012 n  50, 000  1600    0.012  1  1.012 n 31.25  0.012 –n 1.0125  0.625 – n ln 1.012  ln 0.375 –0.011929n  –0.47 n  39.40 quarters (Set P/Y  4; C/Y  4) 50,000 FV CPT N 39.40

1600

PMT

4.8

I/Y

A total of 40 payments are needed. (b) Use the retrospective method: FV  50,000(1.012)8  55,006.51 1.0128  1  FV8  1600    13,350.70  0.012 

Balance  55,006.51  13,350.70  $41,655.81 50,000 ± 55,006.51

PMT

0 PV 16,000

PMT

4.8

I/Y

CPT

FV 13,350.70

CPT

FV 57,010.60

Balance  57,010.60  18,694.94  $38,315.66 Interest in 12th payment  38,315.66(0.012)  $459.79 50,000

0 PV 1600 18,694.94

PV ±

PMT PMT

11N 4.8 11

I/Y 4.8

(d) Use the retrospective method: FV  50,000(1.012)19  62,719.09 1.01219  1  FV19  1600    33,917.57  0.012 

I/Y

CPT

Balance  62,719.09  33,917.57  $28,801.52 Interest in 20th payment  28,801.52(0.012)  $345.62 Principal repaid  1600  345.62  $1254.38 50,000 ± 62,719.09

0 PV 1600 33,917.57

PMT

4.8 I/Y

4.8

I/Y

CPT

(e) The last three payments are PMT38, PMT39, PMT40. Balance after 37th payment: n  37 FV  50,000(1.012)37  77,740.79 1.01237  1  FV22  1600    73,975.45  0.012 

Balance  77,740.79  73,975.45  $3765.34 50,000 ± 77,740.79

0 PV 1600 73,975.45

Payment number

PMT

4.8

I/Y

Partial Amortization Schedule Periodic Interest Principal payment paid repaid

0 1 2 3 : : 37 38 39 40

1600.00 1600.00 1600.00 : : : 1600.00 1600.00 644.71

600 588 575.86 : : : 45.18 26.53 7.65

1000 1012 1024.14 : : : 1554.82 1573.47 637.06

Total

$63,044.71

$13,044.71

$50,000.00

(a) PVnc  198,000; i  2.325%; n  240; c 

2 1  12 6

CPT

Outstanding balance 50,000.00 49,000.00 47,988.00 4696.86 : : 3765.34 2210.53 637.06 

p  1.02325 6  1  1.003838  1  0.3838% 1

1  1.003838240  198, 000  PMT    0.003838  198, 000  156.650725 PMT  $1263.96 (Set P/Y  12; C/Y  2) 198,000 ± FV CPT PMT 1263.96

240

4.65

I/Y

PMT

(b) Balance after one year PMT  1263.96; p  0.003838 FV  198,000(1.003838)12  207,314.03 1.00383812  1 FV12  1263.96    15,491.82  0.003838 

Balance  207,314.03  15,491.82  $191,822.21 Alternatively, using the Amortization Worksheet, 240 2nd

4.65

I/Y

198,000

1263.96

P1  12, P2  12  BAL  191 822.21

AMORT

Total paid  1263.96(12)  $15,167.52 Principal repaid  198,000  191,822.21  $6177.79 Interest paid  15,167.52  6177.79  $8989.73 (c)

2nd AMORT $33,934.97

(d) 180 N 5.36 I/Y 1323.30 or $1323.30 (e) Payment #

P1  1, P2  60  BAL  164,065.14  PRN  

164,065.14

Amount paid

Interest paid

Principal repaid

CPT

Outstanding principal 198,000.00

1263.96

759.92

504.04

197,495.96

1263.96

757.99

505.97

196,989.99

PMT

1263.96

756.05

507.91

196,482.08

164,065.14

1323.30

724.77

598.52

163,466.62

1323.30

722.13

601.17

162,865.45

1323.30

719.47

603.82

162,261.63

(a) PVnc  27,500; i  7%; n  60; c 

1 4

p = 1.07 4  1  1.017059  1  1.7059% 1  1.017059 60  27,500  PMT    0.017059 

27,500  37.374532 PMT PMT  $735.80 (Set P/Y  4; C/Y  1) 27,500 ± PV 60 N 7 I/Y CPT PMT 735.80 (b) Balance after three payments PMT  735.80, p  0.017059 FV  27,500(1.0170585)3  28,931.47 1.01705853  1 FV3  735.8    2245.27  0.0170585  Balance  28,931.47  2245.27  $26,686.20 Interest in fourth payment  26,686.20(0.017059)  $455.23 Principal repaid  735.80  455.23  $280.57

0 FV

Alternatively, using the Amortization Worksheet, 60

N 2nd

I/Y

AMORT

27,500

735.80

PMT

P1  4, P2  4   PRN  280.57

P1  12, P2  12  BAL  $23,981.71 2 1 (d) PVnc  23,981.71; i  3.75%; n  32; c   4 2 (c)

2nd

AMORT

p  1.0375 2  1  1.018578  1  1.8578% 1

1  1.01857832  23,981.71  PMT    0.018578  23,981.71  23.960847 PMT PMT  $1000.87 (Set P/Y  4; C/Y  2) 23,981.71 ± PV 32 CPT PMT 1000.87 (e) Alternatively, using the Amortization Worksheet,

7.5

I/Y

0 FV

N 2nd

7.5

I/Y

AMORT

23,981.71

1000.87

PMT

P1  13, P2  13  BAL  15,899.97 Partial Amortization Schedule

Payment interval

Periodic payment

Interest paid

Principal repaid

Outstanding balance 27,500.00

735.80

469.11

266.69

27,233.31

735.80

464.56

271.24

26,962.07

735.80

459.93

275.87

26,686.20

23,981.71

1000.87

445.52

555.35

23,426.36

1000.87

435.20

565.67

22,860.69

1000.87

424.69

576.18

22,284.51

15,899.97

1000.87

295.38

705.49

15,194.48

1000.87

282.27

718.60

14,475.88

1000.87

268.92

731.95

13,743.94

Total

$24,843.52

$11,087.46

$13,756.06

(a) PVnc  17,500; PMT  2650; i  4%; c  2 p  1.042  1  1.0816  1 816 = 8.16% 1  1.0816 n  17,500  2650    0.0816  1  1.0816 n 6.60377  0.0816

1.0816 – n  0.461132 – n ln 1.0816  ln 0.461132 –0.078441n  –0.774071 n  9.87 years or 10 payments (Set P/Y  1; C/Y  2) 17,500 CPT N 9.87

2650

PMT

I/Y

(b) Use the retrospective method: FV  17,500 (1.0816)3  22,143.08 1.08163  1 FV3  2650    8616.37 0.0816  

Balance  22,143.08  8616.376  $13,526.71 Total paid  2650(3)  $7950 Principal repaid  17,500  13,526.71  $3973.29 Interest paid  $3976.71 17,500 ± 22,143.08

0 PV 8616.37

2650

PMT PMT

3 3

I/Y

CPT

I/Y

CPT

I/Y

CPT

Balance  23,949.96  11,969.46  $11,980.50 Interest in PMT5  11,980.50(0.0816)  $977.61 Principal repaid  2650  977.61  $1672.39 17,500

0 PV 2650 11,969.46

PV ±

PMT PMT

4 4

N N

8 8

I/Y

(d) Use the retrospective method:

CPT

0 FV

FV  17,500 (1.0816)7  30,304.34 1.08167  1  FV6  2650    23,761.55  0.0816 

Balance  30,304.34  23,761.55  $6542.79 17,500 ± 28,018.06

0 PV 2850 20,991.93

Payment number 0 1 2 3 : : 7 8 9 10 Total 7.

PMT PMT

6 6

N N

8 8

I/Y

CPT

I/Y

CPT

Partial Amortization Schedule Periodic Interest Principal payment paid repaid 2650 2650 2650 : : : 2650 2650 2312.35 $26,162.35

1428.00 1328.29 1220.43 : : : 533.89 361.22 174.45 $8662.35

1222.00 1321.72 1429.57 : : : 2116.11 2288.78 2137.90 $17,500.00

Outstanding balance 17,500.00 16,278.00 14,956.29 13,526.72 : : 6542.79 4426.68 2137.90

(a) PVn  25,000; PMT  3500; i  3.2% 1  1.032 n  25, 000  3500    0.032  1  1.032 n 7.142857  0.032 –n 1.032  0.771429 – n ln 1.032  ln 0.771429 –0.03149871n  –0.2595112 n  8.24 half-years 9 payments

(Set P/Y  2; C/Y  2) 25,000 FV CPT N 8.24

3500

(b) Balance after 9 payments

PMT

6.4

I/Y

PMT  3500; i  0.032 FV  25,000(1.032)8  32,164.56 1.0328  1  FV9  3500    31,344.94  0.032 

Balance  32,164.56  31,344.94  $819.62 Final payment  819.62(1.032)  $845.85 9

6.4

I/Y

2nd AMORT overpayment

25,000

3500

PMT

P1  9, P2  9  BAL  2654.16, indicating an

Final payment is $3500  $2654.16  $845.85

(a) PVn(due)  72,500; PMT  900; i  0.525% 1  1.00525 n  72,500  900 1.00525    0.00525  1  1.00525 n 80.134848  0.00525 –n 1.00525  0.579292 – n ln 1.00525  ln 0.579292 –0.005236n  –0.545949 n  104.262931 months

(―BGN‖ Mode) (Set P/Y  12; C/Y  12) 72,500 6.3 I/Y

CPT

900

PMT

104.262931

Jane will receive 105 payments. (b) PMT  900, i  0.00525 FV  72,500(1.00525)104  124,980.57 1.00525104  1 FV104  900(1.00525)    124,743.48  0.00525 

Balance  124,980.57  124,743.48  $237.09 105

6.3

I/Y

2nd AMORT overpayment,

72,500

900

PMT

P1  105, P2  105  BAL  662.91, indicating an

Final payment is $900  $662.91  $237.09 9.

(a) PVn(defer)  33,000; PMT  4300(due); i  2.5%; d  12  1  11 1  1.025 n  33,000  4300(1.02511)    0.025  1  1.025 n  33,000  4300(0.762145)    0.025 

1  1.025 n 0.025 1.025 n  0.748262 n ln 1.025  ln 0.748262 0.024693n  0.290002 10.06502 

n  11.744464 quarters 12 quarters

(―BGN‖ mode) (Set P/Y = 4; C/Y = 4) 33,000 CPT FV

I/Y

4300

PMT

44,381.33

(―BGN‖ mode) (Set P/Y  4; C/Y  4) 44,381.33 10 I/Y

CPT

11.744464

PMT

(b) Deferred for three years, FV  33,000(1.02512)  44,381.33 PMT  4300, i  0.025 FV  44,381.33(1.025)11  58,232.15 1.02511  1  FV12  4300(1.025)    55,020.88  0.025 

Balance  58,232.15  55,020.88  $3211.27 12

I/Y

2nd AMORT overpayment,

44,381.33

4300

PMT

P1  12, P2  12  BAL  1088.73, indicating an

Final payment is $4300  $1088.73  $3211.27 10. (a) PVnc(defer)  52,000;

PMT  20,000; i  1.2%; d  10; c  4

p  1.0124  1  1.048871  1  4.8871%

1  1.048871 n  52, 000  20, 000 1.048871     0.048871  1  1.048871 n  52, 000  20, 000  0.620554     0.048871  10

1  1.048871 n 4.18981  0.048871 –n 1.048871  0.79524 – n ln 1.048871  ln 0.79524 –0.047714n  0.229111 n  4.802 years  5 payments (Set P/Y  1; C/Y  4) 10 N CPT FV  83,796.11

4.8

(Set P/Y  1; C/Y  4) 83,796.11 0 FV CPT N 4.802

I/Y

52,000

(b) FV  83,796.11(1.048871)4  101,417.3033 1.0488714  1  FV19  20, 000    86,057.92  0.048871 

20,000

PMT

4.8

I/Y

Balance  101,417.3033  86,057.92 

$15,359.39

Size of the last benefit payment 15,359.39 1.048871  $16,110.02 Alternatively, using the Amortization Worksheet, (―END‖ mode) (P/Y  1; C/Y  4) 4.802 PV

20,000

PMT

4.8

I/Y

83,796.11

2nd AMORT P1  5, P2  5  BAL  −3889.98 indicating an overpayment. Size of the last payment 20,000 − 3889.98 = $16,110.02 11. (a) PVnc  235,000; n  12(25)  300; i 

4.6% 1  2.3%; c  2 6

1 6

p  1.023  1  1.003797  1  0.3797% 1  1.003797 300  235,000  PMT    0.003797 

235,000  PMT(178.876237) PMT 

$1313.76

(Set P/Y  12; C/Y  2) 235,000 CPT PMT 1313.76

300

$206,661.46

4.6

I/Y

0 FV

PMT

(b) PMT  1313.76; p  0.003797 FV  235,000(1.003797)60  295,001.48 1.00379760  1 FV60  1313.76    88,340.02  0.003797 

Balance  295,001.48  88,340.02 

Alternatively, using the Amortization Worksheet, 60

4.6

2nd

I/Y

AMORT

235,000

1313.76

P1  60, P2  60  BAL  206,661.46

6.6% 1  3.3%; c  6 2

p = 1.033 6  1  1.005426  1  0.5426% 1

1  1.005426240  206, 661.46  PMT    0.005426  206, 661.46  PMT 134.0082  PMT  $1542.16

206,661.46 ± PMT 1542.16

240

12. (a) PVnc  180,000; n  12(25)  300; i 

6.6

I/Y

CPT

5.62% 1  2.81%; c  6 2

p  1.02816  1  1.004629  1  0.4629% 1  1.004629300  180, 000  PMT    0.004629  180, 000  PMT 161.970 907  PMT  $1111.31

(Set P/Y  12; C/Y  2) 180,000 ± FV CPT PMT 1111.31

300

5.62

I/Y

PMT

(b) PMT  1111.31; p  0.004629 FV  180,000 (1.004629)48  224,675.33 1.00462948  1 FV48  1111.31    59,580.43  0.004629 

Balance  224,675.33  59,580.43  $165,094.90 Alternatively, using the Amortization Worksheet, 48

N 2nd

5.62

I/Y

AMORT

180,000

1111.31

P1  48, P2  48  BAL  165,094.90

1 5.3%  2.65%; c  6 2

p = 1.0265 6  1  1.004369  1  0.4369% 1

1  1.004369252  165,094.90  PMT    0.004369 

165,094.90  PMT(152.592959) PMT  $1081.93

165,094.90 ± PMT 1081.93

252

5.3

13. (a) PVnc  80,000; n  12(10)  120; i 

I/Y

CPT

1 4.78%  2.39%; c  6 2

p = 1.0239 6  1  1.003944  1  0.003944 = 0.3944% 1

1  1.003944120  80, 000  PMT    0.003944  80, 000  PMT  95.451132  PMT  $838.13

The rounded monthly payment is $850. PVnc  80,000; PMT  850; p  0.3944

1  1.003944 n  80, 000  850    0.003944  0.371 222  1  1.003944 n 1.003 944 n  0.628778 n ln 1.003944  ln 0.628778  0.003936n  0.463977 n  117.87 or 118 payments (Set P/Y  12; C/Y  2) 80,000 ± FV CPT PMT 838.13 80,000 117.87

4.78

I/Y

PV 850

120

PMT

N 0

4.78 FV

I/Y CPT

0 N

(b) PMT  850; p  0.3944% FV  80,000(1.003944)117  126,797.91 1.00394117  1 FV117  850    126,064.41  0.00394 

Balance  126,797.91  126,064.41  $733.50 Final payment  733.50(1.003944)  $736.39 Alternatively, using the Amortization Worksheet, 117 2nd

4.78

I/Y

AMORT

80,000

850

PMT

P1  118, P2  118  BAL  113.61 Overpayment

Size of final payment = 850  113.61 = 736.39 (c) The total amount paid with unrounded payments  120 (838.13)  100,575.60 The total amount paid with rounded payments  117(850)  736.39  100,186.39 Amount of interest saved  100,575.60  100,186.39  $389.21 14. (a) PVnc  160,000; n  12 (20)  240; i 

5.44% 1  2.72%; c  2 6

p  1.0272 6  1  1.004483  1  0.4483% 1

1  1.004483240  160, 000  PMT    0.004483  160, 000  PMT 146.823275  PMT  1089.75

The rounded monthly payment is $1100 PVnc  160,000; PMT  1100; p  0.004483

1  1.004483 n  160, 000  1100    0.004483  0.652073  1 –1.004483– n 1.004483– n  0.347927 – n ln 1.004483  ln 0.347927 –0.004473n  –1.055763 n  236.03 months or 237 payments (Set P/Y  12; C/Y  2) 160,000 FV CPT PMT 1089.75

240

5.44

I/Y

(b) PMT  1100; p  0.004483 FV  160,000(1.004483)236  459,780.30 1.004483236  1 FV236  1100    459,755.68  0.004483 

Balance  459,780.30  459,755.68  $24.62 Final payment  24.62(1.004483)  $24.73 Alternatively, using the Amortization Worksheet, 236.03 2nd

5.44

AMORT

I/Y

160,000

1100

PMT

P1  237, P2  237  BAL  1075.25 Overpayment

Size of final payment = 1100  1075.25 = 24.74 (c) The total amount paid with unrounded payments  240(1089.75)  261,540 The total amount paid with rounded payments  236(1100)  24.73  259,624.73 Amount of interest saved  261,540  259,624.73  $1915.27 15. (a) PVn  6500; i  0.75%; n  48

1  1.007548  6500  PMT    0.0075  6500  40.184782 PMT PMT  $161.75 (Set P/Y  12; C/Y  12) 6500 ± CPT PMT 161.7527754

(b) Total paid  161.7527754(48)  $7764.13 Original principal  6500 Total interest  $1264.13 (c) PMT  161.75; i  0.0075 FV  6500(1.0075)12  7109.74

I/Y

0 FV

1.007512  1 FV12  161.75    2023.10  0.0075 

Balance  7109.74  2023.10  $5086.64 Alternatively, using the Amortization Worksheet, 12

N 2nd

I/Y

AMORT

6500

161.75

PMT

P1  12, P2  12  BAL  5086.64

(d) FV  6500(1.0075)29  8072.72 1.007529  1 FV29  161.75    5218.21  0.0075 

Balance  8072.72  5218.21  $2854.51 Alternatively, using the Amortization Worksheet, 29

2nd

I/Y

6500

161.75

PMT

P1  29, P2  29  BAL  2854.51

AMORT

Interest for payment period 30  2854.51(0.0075)  $21.41 (e) Last three payments are PMT46, PMT47 and PMT48. FV  6500(1.0075)45  9097.89 1.007545  1 FV45  161.75    8619.68  0.0075 

Balance  9097.89  8619.68  $478.21 Alternatively, using the Amortization Worksheet, 45

I/Y

6500

2nd AMORT to rounding)

161.75

PMT

P1  45, P2  45  BAL  478.22 (one cent error due Partial Amortization Schedule

Payment interval

Periodic payment

Interest paid

Principal repaid

Outstanding balance 6500.00

161.75

48.75

113.00

6387.00

161.75

47.90

113.85

6273.15

161.75

47.05

114.70

6158.45

478.21

161.75

3.59

158.16

320.05

161.75

2.40

159.35

160.70

161.91

1.21

160.70

—

Total

$7764.16

$1264.16

$6500.00

Note that for payment 48, BAL  0.16, indicating an underpayment, so the actual payment is 161.75  0.16  161.91.

16. (a) PVn  46,000; PMT  4500; i  4.5% 1  1.045 n  46,000  4500    0.045 

10.22222 

1  1.045 n 0.045

1.045n  0.54 n ln 1.045  ln 0.54 0.04402 n   0.616186 n  13.9988 half-years or 14 payments (Set P/Y  2; C/Y  2) 46,000 CPT N 13.9988

4500

PMT

I/Y

0 FV

(b) Use the retrospective method: FV  46,000(1.045)4  54,855.86 1.0454  1  FV4  4500    19,251.86  0.045 

Balance  54,855.86  19,251.86  $35,604 Interest in PMT5  35,604(0.045)  $1602.18 Principal repaid  4500  1602.18  46,000 ± 54,855.86

0 PV 4500 19,251.86

PMT PMT

$2897.82

I/Y

CPT

I/Y

CPT

4 4

(c) The last three payments are PMT12, PMT13, and PMT14. FV  46,000(1.045)11  74,651.24 1.04511  1  FV7  4500    62,285.31  0.045 

Balance  74,651.24  62,285.31  $12,365.94 46,000 ± 74,651.24

0 PV 4500 62.285.31

Payment

PMT PMT

11 11

N N

9 9

I/Y I/Y

CPT CPT

Partial Amortization Schedule Periodic Interest Principal

FV FV

Outstanding

interval 0 1 : : 11 12 13 14

payment

paid

4500.00 : : : 4500.00 4500.00 4494.97 $62,994.97

Total

repaid

2070.00 : : : 556.47 379.01 193.56 $16,994.97

2430.00 : : : 3943.53 4120.99 4301.41 $46,000.00

1 6

17. PVnc  95,000; PMT  753.25; n  12(25)  300; c  (Set P/Y  12; C/Y  2) 0 FV 95,000 N CPT I/Y 8.462

balance 46,000.00 43,570.00 : : 12,365.94 8422.40 4301.41 —

753.25

PMT

300

PMT

300

4.8

I/Y

The nominal annual rate is 8.46% compounded semi-annually. 18. PVnc  135,000; PMT  1023.12; n  12(25)  300 (Set P/Y  12; C/Y  2) 0 N CPT I/Y

FV 135,000 7.9161

1023.12

The nominal annual rate is 7.92% compounded semi-annually. 19. (a) PVnc  28,000; i  2.4%; n  80; c 

2 1  4 2

p  1.024 2  1  1.0119289  1  1.19289%

1  1.011928980  28, 000  PMT    0.0119289  28, 000  51.3663 PMT PMT  $545.11. (Set P/Y  4; C/Y  2) 0 FV CPT PMT 545.11

28,000

(b) PMT  545.11; p  0.0119289 FV  28,000(1.0119289)4  29,360.13 1.01192894  1  FV4  545.11    2219.77  0.0119289 

Balance  29,360.13  2219.77  $27,140.36 Alternatively, using the Amortization Worksheet,

4.8

2nd

I/Y

AMORT

28,000

PV 545.11

PMT

FV .

P1  4, P2  4  BAL  27,140.36

Total paid  545.11(4)  $2180.44 Principal repaid  28,000  27,140.36  859.64 Interest paid  $1320.80 (c) FV  28,000(1.0119289)12  32,281.80 1.011928912  1  FV12  545.11    6988.02  0.0119289 

Balance  32,281.80  6988.02  $25,293.78 Alternatively, using the Amortization Worksheet, 12

4.8

2nd

I/Y

AMORT

28,000

545.11

PMT

P1  12, P2  12  BAL  25,293.78

(d) PVnc  25,293.78; i  5.4%; n  68; c 

1 4

p  1.054 4  1  1.0132349  1  1.32349%

1  1.013234968  25,293.78  PMT    0.0132349  25,293.78  44.65554 PMT PMT  $566.42 (Set P/Y  4; C/Y  1) 25,293.78 CPT PMT 566.42

20. (a) PVnc  16,000; PMT  1000; i  0.625%; c 

12 3 4

p  1.006253  1  1.018867  1  1.8867%

5.4

I/Y

0 FV

1  1.018867  n  16, 000  1000    0.018867  1  1.018867  n 16  0.018867 –n 1.018867  0.698122 – n ln 1.018867  ln 0.698122 –0.018692n  –0.359362 n  19.225843 quarters or 20 payments (Set P/Y  4; C/Y  12) 16,000 ± FV CPT N 19.225843

1000

PMT

7.5

PMT

I/Y

(b) PMT  1000, p  0.018867 FV  16,000(1.018867)19  22,822.12 1.01886719  1 FV19  1000    22,598.85  0.018867 

Balance  22,822.12  22,598.85  $223.27 Final payment  223.27(1.018867)  $227.48 Alternatively, using the Amortization Worksheet, 19

7.5

I/Y

2nd AMORT overpayment

16,000

1000

P1  20, P2  20  BAL  772.52, which is an

1000  $227.48

Self-Test 1.

PV  9000; n  42; i 

7.48%  0.6233% 12

1  1.00623342  9000  PMT    0.006233  9000  36.85203 PMT PMT  $244.22 FV  9000 1.006233  10,190.93 20

1.00623320  1  FV22  244.22    5184.73  0.006233  Balance outstanding  10,190.93  5184.73  $5006.21 (―END‖ mode)(Set P/Y  12; C/Y  12) 42 PV CPT PMT 244.22 2nd 2.

AMORT

7.48

I/Y

P1  20, P2  20  BAL  5006.21

PVn  24,000; PMT  800; i  1.55% 1  1.0155 n  24,000  800    0.0155 

1  1.0155 n 0.0155 0.465  1  1.0155 n 30 

1.0155 n  0.535 n ln1.0155  ln0.535 0.01538 n  0.62549 n  40.67 quarters (41 payments) Balance after 18th: PMT  800; i  1.55% FV  24,000(1.0155)18  31,655.56 1.015518  1  FV  800    16,463.56  0.0155 

Balance  31,655.56  16,463.56  $15,192 Interest  15,192(0.0155)  235.48

9000

Principal repaid  800  235.48  $564.52 (Set P/Y  4; C/Y  4) 24,000 CPT FV 31,655.56 0 PV 800 16,463.56

PMT

6.2

I/Y

CPT

6.2

I/Y

PVnc  50,000; i  2.63%; n  240; c 

2 1  12 6

p  1.02636  1  1.004336  1  0.4336% 1  1.004336240  50,000  PMT    0.004336 

50,000  148.9788 PMT PMT  335.62 (Set P/Y  12; C/Y  2) 0 FV CPT PMT 335.62

50,000

240

5.26

I/Y

5.4

I/Y

0 FV

PMT

FV  50,000(1.004336)24  55,471.17 1.00433624  1 FV36  335.62    8469.60  0.004336 

Balance  55,471.17  8469.60  $47,001.57 Alternatively, using the Amortization Worksheet, (―END‖ mode) (P/Y  12; C/Y  2) 24 PV . 335.62 2nd

AMORT

PMT

5.26

I/Y

50,000

P1  24, P2  24  BAL  47,001.57

Total paid in 2 years  335.62  24

 $8054.88

Principal repaid after 2 years  50,000  47,001.57  2998.43 Interest paid  8054.88  2998.43 4.

 $5056.45

1 5.4%  2.27%; c  6 2 1 p  1.027 6  1  1.0044502  1  0.44502%

(a) PVnc  190,000; n  12(25)  300; i 

1  1.0044502300  190, 000  PMT    0.0044502  190, 000  PMT 165.40325 PMT  $1148.71 (Set P/Y  12; C/Y  2) 190,000 CPT PMT 1148.71

300

(b) PMT  1148.71; p  0.0044502 FV  190,000(1.0044502)36  222,933.98 1.004450236  1  FV36  1148.71    44,742.68  0.0044502 

Balance  222,933.98  44,742.68  $178,191.30 Alternatively, using the Amortization Worksheet, 36

5.4

I/Y

190,000

1148.71

2nd

AMORT

P1  36, P2  36  BAL  178,191.30

6.25% 1  3.125%; c  6 2 1 6 p  1.03125  1  1.005142  1  0.5142%

1  1.005142264  178,191.30  PMT    0.005142  178,191.30  PMT 144.2656  PMT  $1235.16

0 FV 178,191.30 PMT 1235.16 5.

264

6.26

I/Y

CPT

1 4.35%  2.175%; c  6 2 1 p  1.02175 6 1  1.003593  1  0.3593%

(a) PVnc  140,000; n  12(15)  180; i 

1  1.003593180  140, 000  PMT    0.003593  140, 000  PMT 132.384728 PMT  1057.52 The rounded monthly payment is $1100 PVnc  140,000; PMT  1100; p  0.3593

1  1.003593 n  140, 000  1100    0.003593  0.457237  1 –1.003593– n 1.003593– n  0.542763 – nln 1.003593  ln 0.542763 –0.003586n  –0.611083 n  170.40 or 171 payments (Set P/Y  12; C/Y  2) 140,000 CPT PMT 1057.52 140,000 170.40

1100

180

PMT

4.35

I/Y

0 FV

(b) PMT  1100; p  0.003593 FV  140,000(1.003593)170  257,568.64 1.003593170  1 FV170  1100    257,128.33  0.003593 

4.35 CPT

I/Y

0 FV N

179 2nd

4.35

I/Y

AMORT

140,000

PV .1100

PMT

P1  171, P2  171  BAL  658.10 Overpayment

Size of final payment = 1100  658.10 = 441.90

(c) Total amount paid with unrounded payments  180(1057.52)  190,353.60 Total amount paid with rounded payments  170 (1100)  441.89  187,000  441.89  187,441.89 Amount of interest saved  190,353.60  187,441.89  $2911.71 6.

PVn  24,000; PMT  1100; i  1.5% 1  1.015 n  24,000  1100    0.015  1  1.015 n 21.818182  0.015 0.327273  1  1.015n

1.015n  0.672727 n ln 1.015  ln 0.672727 0.014889n  0.396415 sn  26.625401 quarters (Set P/Y  4; C/Y  4) 24,000 N 26.625401

1100

PMT

I/Y

CPT

PMT  1100; i  1.5% FV  24,000(1.015)26  35,345.03 1.01526  1 FV26  1100    34,665.37  0.015  Balance  35,345.03  34,665.37  $679.66

Final payment  679.66(1.015)  $689.85 Alternatively, using the Amortization Worksheet, 26

N 2nd

I/Y

24,000

AMORT

PV .1100

PMT

P1  27, P2  27  BAL  410.14, overpayment

Final payment = 1100    $689.86 7.

PVnc  145,000; PMT  1297; n  12(25)  300; c  (Set P/Y  12; C/Y  2) 145,000 CPT I/Y

1 6

1297

PMT

9.999933646 The nominal annual rate is 10.0% compounded semi-annually.

300

0 FV

PVn  12,000; i  0.625%; n  120

1  1.00625120  12, 000  PMT    0.00625  12, 000  84.244743PMT PMT  $142.44 (Set P/Y  12; C/Y  12) 12,000 PMT 142.44

120

7.5

I/Y

CPT

PMT  142.44; i  0.00625 Balance after 39th payment: FV  12,000 (1.00625)39  15,300.69 1.0062539  1 FV39  142.44    6268.68  0.00625 

Balance  15,300.69  6268.68  $9032.01 Alternatively, using the Amortization Worksheet, 39

7.5

I/Y

12,000

PV . 142.44

PMT

P1  39, P2  39  BAL  9032.02 (one cent error due to

2nd AMORT rounding)

Interest in the 40th period = 9032.01(0.00625) = 56.45 Principal repaid in the 40th period = 142.44  56.45 = 85.99 Balance after 117th payment: FV  12,000(1.00625)117  24,875.44 1.00625117  1 FVI17  142.44    24,453.04  0.00625 

Balance  24,875.44  24,453.04  $422.40 Alternatively, using the Amortization Worksheet, 117 2nd

7.5

I/Y

AMORT

Payment

12,000

142.44

PMT

P1  117, P2  117  BAL  $422.40 Partial Amortization Schedule Monthly Interest Principal

Outstanding

interval 0 1 2 3 : : 39 40 : : 117 118 119 120 Total

payment

paid

repaid

142.44 142.44 142.44 : :

75.00 74.58 74.15 : :

67.44 67.86 68.29 : :

142.44 : :

56.45 : :

85.99 : :

142.44 142.44 142.82 $17, 093.18

2.64 1.77 0.89 $5093.18

139.80 140.67 141.93

balance 12 000.00 11 932.56 11 864.70 11 796.41 : : 9032.01 8946.02 : : 422.40 282.60 141.93 —

$12,000.00

FVn(due)  165,000; i  0.4833%; n  72 é1.00483372  1ù ú 165,000  PMT (1.004833) ê ê 0.004833 ú ë û

165,000  PMT (1.004833)(85.8715) PMT  $1912.23 Balance after 15th interval: PMT  1912.23; i  0.4833%; n  15 é1.00483315  1ù ú FVn(due)  1912.23(1.004833) ê ê 0.004833 ú ë û

 1912.23(1.004833)(15.5183)  $29,817.96 Interest earned in 16th payment interval  (29,817.96)(0.004833)  $144.12 (Set P/Y  12; C/Y  12)(―BGN‖ Mode) 165,000 PV CPT PMT . 1912.23

5.8

I/Y

(Set P/Y  12; C/Y  12)(―BGN‖ Mode) 1912.23 ± PMT 15 0 PV CPT FV . 29,818.01 (there is 5 cents of difference due to rounding error)

5.8

I/Y

10. PV  10,000; i  4%; n  40 FV  10,000 (1.04)40  10,000(4.801021)  $48,010.21 After 20 years: FVn  48,010.21; PMT  10,000; i  4% é1  1.04  n ù ú 48,010.21  10,000 ê ê 0.04 ú ë û

4.801021 

1  1.04 n 0.04

0.192041  1  1.04n 1.04n  0.807959 n ln 1.04  ln 0.807959  0.03922n  0.213244 n  5.44  6 payments Balance after 5th payment: PMT  10,000; i  4%; n  0.44 1  1.040.44  FVn  10,000    $4277.27  0.04 

Final payment  4277.27(1.04)  $4448.36 (Set P/Y  2; C/Y  2) 10,000 CPT FV 48,010.21 (Set P/Y  2; C/Y  2) 48,010.21 FV CPT N .5.44

10,000

I/Y

PMT

I/Y

Challenge Problems 1.

Let the loan balance just after the previous payment be $x. Then the interest to be paid by the next payment 8% x  $0.006667x 12

Previous balance  Interest due  Payment  Current balance x  0.006667x  250  3225.68

1.006667x  3475.68 x

3475.68 1.006667

x  3452.66 The loan balance just after the previous payment was $3452.66 2.

Trust Company A option:

1 6.75%  3.375%; c  6 2 1 6 p  1.03375  1  1.005547  1  0.5547% PV  130,000; n  12(25)  300; i 

1  1.005547 –300  130, 000  PMT    0.005547  130, 000  PMT 145.974824  PMT  890.56 The monthly payment is $890.56. (Set P/Y  12; C/Y  2) 130,000 ± CPT PMT 890.56 Outstanding balance after 4 years:

300

6.75

I/Y

PMT  890.57; p  0.5548% FV  130,000(1.005547)48  169,538.19 1.005547 48  1 FV48  890.56    48,824.74  0.005547 

Balance  169,538.19  48,824.74  $120,713.45 Two months’ interest  2(120,713.45)(0.005547)  $1339.31 Payout amount after 4 years  120,713.45  1339.31  $122,052.76 Trust Company B option: PV  130,000; n  12(25)  300; i 

7% 1  3.5%; c  2 6

p  1.035 6 1  1.0057500  1  0.575% 1  0.00575–300  130, 000  PMT    0.00575  130, 000  PMT 142.772345  PMT  910.54

The monthly payment is $910.54. 130,000

300

I/Y

0 FV

CPT

PMT

910.54

Outstanding balance after 4r years: PMT  910.54; p  0.575% FV  130,000(1.00575)48  171,185.17  1.0057548  1 FV48  910.54    50,167.88  0.00575 

Balance  171,185.17  50,167.88  $121,017.29 Payout amount after 4 years is $121,017.29. Difference in the monthly payments  910.54  890.56  $19.98 Investment value of difference: PMT  19.98; n  4(12)  48; i  3%; c 

1 12

p  1.0312 1  1.002466  1  0.2466%  1.00246648  1 FV  19.98    19.98(50.890138)  1016.78  0.002466 

(Set P/Y  12; C/Y  1) 19.98 FV 1016.78

PMT

I/Y

0 FV

The investment value of the difference in the monthly payments is $1016.78. The adjusted payout amount for Trust Company A  $122,052.76  1016.78  $121,035.98 The net difference in the two options  121,035.98  121,017.29  $18.69 (in favour of Trust Company B).

CPT

Case Study 1.

Monthly payments at original rate: (Set P/Y  12; C/Y  2) 300 N CPT PMT 1316.80

7.5

I/Y

180,000

1316.80

PMT

The monthly payment is $1316.80. Outstanding balance after 3 years: 36

N 2nd

7.5

I/Y

180,000

P1  36; P2  36; BAL  171,606.70.

AMORT

The balance after 3 years is $171 606.70. Monthly payments at new interest rate: 264 N 5.5 1115.91

I/Y

171,606.70

CPT

PMT

The new monthly payment is $1115.91. Monthly savings would be 1316.80  1115.91  $200.89. 2.

(a) Three months of interest at the original rate of interest: (Set P/Y  12; C/Y  2)37 PMT 0 FV . 2nd

AMORT

7.5

I/Y

180,000

1316.80

Pl  37; P2  37; INT  1056.16

Three months of interest is $3(1056.16)  $3168.48 (b) Interest from Years 3–5 based on the original rate: 300 N 1316.80 2nd

7.5

I/Y

AMORT

180,000

PV 0

CPT

PMT

Pl  37; P2  60; INT  24,884.41

Interest paid during Years 3–5 based on the original rate of interest is $24,884.41. Interest on the balance of $171,606.70 with new interest rate from Years 3–5: 264 N 1115.91 2nd

5.5

I/Y

AMORT

171,606.70

CPT

PMT

Pl  1; P2  24; INT  18,226.55

Interest paid during Years 3–5 based on the new rate of interest is $18,226.55. Interest differential is the difference between $24,884.41  $18,226.55  $6657.86

If the penalty is paid in full at the beginning of the new 5-year term, the monthly payment is $1115.91 as calculated in Question 1.

New principal  $171,606.70  $6657.86  $178,264.56. 264 N 5.5 1159.20

I/Y

178,264.56

CPT

PMT

If the penalty is added to the existing principal, then the monthly payments will be $1159.20

Chapter 15 Bond Valuation and Sinking Funds Exercise 15.2 1.

FV  500; PMT  500 (0.03)  $15; i  3.75% The purchase date is 5.5 years before maturity: n  11 é1  1.037511 ù ú Purchase price  500 (l.037511)  15 ê ê 0.0375 ú ë û

 500 (0.667008)  15 (8.879795)  333.50  133.20 

$466.70

(Set P/Y  2; C/Y  2) 500 PV  466.70 2.

. FV

PMT

7.5

I/Y

CPT

FV  15,000; PMT  15,000 (0.0125)  187.50; i  1.5%; n  13 é1  1.01513 ù ú Purchase price  15,000 (1.01513)  187.50 ê ê 0.015 ú ë û

 15,000 (0.824027)  187.50(11.731532)  12,360.41  2199.66 

$14,560.07

(Set P/Y  2; C/Y  2) 15,000 CPT PV  14,560.07 3.

187.50

PMT

I/Y

FV  5000; PMT  5000 (0.03)  150; i  3.25%; n  26 26

Purchase price  5000 (1.0325

é1  1.032526 ù ú )  150 ê ê 0.0325 ú ë û

 5000 (0.435370)  150 (17.373233)  2176.85  2605.99 

$4782.84

(Set P/Y  2; C/Y  2) 5000 PV  4782.84 4.

. FV

150

PMT

6.5

FV  2000; PMT  2000 (0.0275)  55; n  12; i  3.75%

I/Y

CPT

1  1.037512  Purchase price  2000 (l.037512)  55    0.0375 

 2000 (0.642899)  55 (9.522694)  1285.80  523.75 

$1809.55

(Set P/Y  2; C/Y  2) 2000 PV  1809.55

. FV

PMT

7.5

I/Y

CPT

FV  10,000; PMT  10,000 (0.015)  150; n  19; i  1.0% é1  1.0119 ù ú Purchase price  10,000 (1.0119)  150 ê ê 0.01 ú ë û

 10,000 (0.82774)  150 (17.226008)  8277.40  2583.90 

$10,861.30

(Set P/Y  2; C/Y  2) 10,000 PV 10,861.30 6.

150

PMT

I/Y

CPT

FV  25,000; PMT  25,000 (0.035)  $875; i  2.5%; n  24 é1  1.02524 ù ú Purchase price  25,000 (1.02524)  875 ê ê 0.025 ú ë û

 25,000 (0.552875)  875(17.884986)  13,821.88  15,649.36 

$29,471.24

(Set P/Y  2; C/Y  2) 25,000 PV 29,471.24 7.

875

PMT

I/Y

FV  1000; PMT  1000 (0.025)  25; i  2.0%; n  17 é1  1.02 17 ù ú Purchase price  1000 (l .0217)  25 ê ê 0.02 ú ë û

 1000 (0.714163)  25 (14.291872)  714.16  357.30 

$1071.46

(Set P/Y  2; C/Y  2) 1000 PV 1071.46 8.

. FV

PMT

I/Y

FV  5,000,000; PMT  5,000,000 (0.0425)  212,500 i  2.25%; n  25; c  2; p  1.02252  1  1.045506  1  4.5506% Purchase price  5,000,000 (1.045506

25

é1  1.045506 25 ù ú )  212,500 ê ê 0.045506 ú ë û

CPT

 5,000,000 (0.328728)  212,500(14.751285)  1,643,640  3,134,648.06 

$4,778,288.06

(Set P/Y  1; C/Y  2) 5,000,000 CPT PV . 4,778,270.32

. FV

212,500

PMT

4.5

I/Y

FV  100,000; PMT  100,000 (0.03375)  3375; i  7.35%; n  16; c 

1 2

p  1.0735 2  1  1.036098  1  3.6098% 1

é1  1.03609816 ù ú Purchase price  100,000 (l.03609816)  3375 ê ê 0.036098 ú ë û  100,000 (0.567005)  3375 (11.995)  56,700.07  40,482.97  $97,183.04 .

(Set P/Y  2; C/Y  l) 100,000 CPT PV . 97,183.04

3375

PMT

7.35

I/Y

10. FV  40,000; PMT  40,000 (0.02)  800; n  30; i  3.4%; c 

2 1  4 2

p  1.034 2  1  1.016858  1  1.6858% 1

é1  1.01685830 ù ú Purchase price  40,000 (l.016858 )  800 ê ê 0.016858 ú ë û  40,000 (0.605608)  800 (23.39505)  24,224.34  18,716.04  $42,940.38 . 30

(Set P/Y  4; C/Y  2) 40,000 CPT PV 42,940.38

800

PMT

6.8

I/Y

FV  100,000; PMT  100,000 (0.0375)  3750; i  4%; The interest date immediately preceding the purchase date is 15 years before maturity. n  30

11. (a)

é1  1.0430 ù ú Market price  100,000 (l.0430)  3750 ê ê 0.04 ú ë û

 100,000 (0.308319)  3750 (17.292033)  30,831.87  64,845.12 

$95,676.99

(Set P/Y  2; C/Y  2) 100,000 CPT PV 95,676.99

3750

PMT

I/Y

Number of months in the interest payment interval is 6 months; number of months between 14 years 10 months to 15 years is 2 months. Cash price on purchase date

 95,676.99(1.04)(2/6)  $96,936.04

(b)

æ2 ö Accrued interest  100,000 (0.0375) çç ÷  ÷ çè 6 ÷ ø

$1250

(c)

Quoted price on purchase date: Quoted price  Cash price  Accured interest  96,936.04  1250

 $95, 686.04 12. (a)

FV  25,000; PMT  25,000 (0.05)  1250; i  3.8% The interest date immediately preceding the purchase date is June 1, 2019. Time period June 1, 2019, to December 1, 2030, is 11.5 years: n  23. Market price on June 1, 2019 is é1  1.03823 ù ú  25,000 (l.03823)  1250 ê ê 0.038 ú ë û

 25,000 (0.424093)  1250 (15.155453)  10,602.32  18,944.32  CPT

$29,546.64

(Set P/Y  2; C/Y  2) 25,000 PV  29,546.64

1250

PMT

7.6

I/Y

Number of days in the interest payment interval June 1, 2019, to December 1, 2019, is 183; number of days from June 1, 2019, to September 25, 2019, is 116. Cash price on September 25, 2019 is  29,546.64(1.038)(116/183)  $30, 253.48

(b)

æ116 ö÷ Accrued interest  25,000 (0.05) çç ÷ çè183 ø÷

(c)

Quoted price on September 25, 2017: 30,253.48  792.35  $29,461.13

(d)

SDT = 09.2519 CPN = 10 RDT = 12.0130 RV = 100 ACT 2/Y YLD = 7.6 To compute the market price (quoted price): PRI = press CPT = 117.8445. Multiply this number by 250 to obtain a clean price of $29,461.12. The actual sale price of the bond, which is known as the DIRTY Price or cash price, includes both the Clean Price plus the accrued interest. Accrued interest equals: AI = 3.1694 * 250 = $792.35 Cash price = 29,461.12 + 792.35 = $30,253.48

13. (a)

$792.35

FV  6(1000)  6000; PMT  6000 (0.012)  72; n  16; i  0.1%;

12 6 2 p  1.0016  1  1.006015  1  0.6015% c

é1  1.00601516 ù ú Market price (preceding)  6000 (l.00601516)  72 ê ê 0.006015 ú ë û

 6000 (0.908508)  72 (15.210659)  5451.05  1095.17 

$6546.21

(Set P/Y  2; C/Y  12) 6000 CPT PV  6546.21

(difference due to rounding)

PMT

1.2

I/Y

Number of months in the interest payment interval is 6 months; number of months between date of purchase and last coupon maturity is 3 months. Purchase price on purchase date

 6546.21(1.006015)(3/6)  $6565.87 (b)

æ3 ö Accrued interest  6000 (0.012) çç ÷  ÷ çè 6 ÷ ø

(c)

Market price on purchase date: $6565.87  36  $6529.87

$36

FV  100,000; PMT  100,000 (0.0295)  2950; i  4.5% The interest date preceding the purchase date is October 1, 2018; the time period

14. (a)

October 1, 2018, to October 1, 2040, is 22 years: n  44. Market price on October 1, 2018 is é1  1.04544 ù ú  100,000 (l.04544)  2950 ê ê 0.045 ú ë û

 100,000 (0.144173)  2950 (19.018383)  14,417.28  56,104.23 

$70,521.51

(Set P/Y  2; C/Y  2) 100,000 CPT PV 70,521.51

2950

PMT

I/Y

Number of days in the interest payment interval October 1, 2018, to April 1, 2019  182 days The number of days from October 1, 2018, to January 15, 2019, is 106. Purchase price on January 15, 2019  70,521.51(1.045)(106/182)  $72,352.79

(b)

æ106 ö÷ Accrued interest  100,000 (0.0295) çç ÷  $1718.13 . çè182 ø÷

(c)

Market price on January 15, 2017  72,352.79  1718.13  $70,634.66

(d)

SDT = 01.1519 CPN = 5.9 RDT = 10.0140 RV = 100 ACT 2/Y YLD = 9 To compute the market price (quoted price): PRI = press CPT = 70.63465161. Multiply this number by 1000 to obtain a clean price of $70,634.65. The actual sale price of the bond, which is known as the DIRTY Price or cash price, includes both the Clean Price plus the accrued interest. Accrued interest equals: AI = 1.718131868 * 1000 = $1718.13 Cash price = 70,634.65 + 1718.13 = $72,352.78

15. Face value  100,000; b  2% Principal  100,000; i  1.625% Since b > i, the bond is expected to sell at a premium. (a)

n  60 Premium

é1  1.01625 60 ù ú  [100,000 (0.02)  100,000 (0.01625)] ê ê 0.01625 ú ë û

 100,000 (0.00375)(38.143997) 

$14,304

Purchase price

 100,000  14,304 

(Set P/Y  4; C/Y  4) 100,000 CPT PV 114,304 (b)

$114,304

PMT

6.5

2000

I/Y

n  20 Premium

é1  1.01625 20 ù ú  [100,000 (0.02)  100,000 (0.01625)] ê ê 0.01625 ú ë û

 100,000 (0.00375)(16.958934) 

$6359.60

Purchase price

 100,000  6359.60 

(Set P/Y  4; C/Y  4) 100,000 CPT PV 106,359.60

2000

$106,359.60

PMT

I/Y

6.5

16. Face value  5000; b  3.75% Principal  5000; i  3% (a)

n  20 Since b > i, the bond is expected to sell at a premium. Premium

é1  1.0320 ù ú  [5000 (0.0375)  5000 (0.03)] ê ê 0.03 ú ë û

 (187.50  150)(14.877475)  (37.50)(14.877475)  $5557.91 . Purchase price

 5000  557.91 

CPT

(Set P/Y  2; C/Y  2) 5000 PV 5557.91

(b)

n  12 Premium

$5557.91

187.50

PMT

é1  1.0312 ù ú  [5000 (0.0375)  5000 (0.03)] ê ê 0.03 ú ë û

 37.50(9.954004) 

$373.28

Purchase price

 5000  373.28 

(Set P/Y  2; C/Y  2) 5000 CPT PV 5373.28

$5373.28

187.50

17. Face value  25,000; b  4% Principal  25,000; n  6 (a)

i  2%

PMT

. 6

Since b > i, the bond will sell at a premium. Premium

é1  1.02 6 ù ú  [25,000 (0.04)  25,000 (0.02)] ê ê 0.02 ú ë û

 25,000 (0.02)(5.601431) 

$2800.72

Purchase price

 25,000  2800.72 

(Set P/Y  l; C/Y  l) 25,000 CPT PV 27,800.72

$27,800.72

PMT

I/Y

1000

(b)

i  6% Since b < i, the bond will sell at a discount. Discount

é1  1.06 6 ù ú  [25,000 (0.04)  25,000 (0.06)] ê ê 0.06 ú ë û

 25,000 (0.02)(4.917324) 

−$2458.66

Purchase price

 25,000  2458.66 

(Set P/Y  l; C/Y  l) 25,000 CPT PV 22,541.34

1000

$22,541.34

PMT

I/Y

18. Face value  1000; b  8% Principal  1000; n  7 (a)

i  6.5%; since b > i, premium is expected. é1  1.0657 ù ú  [1000 (0.08)  1000(0.065)] ê ê 0.065 ú ë û  (80  65)(5.484520)

Premium

 (15)(5.484520) 

$82.27

Purchase price

 1000  82.27 

(Set P/Y  l; C/Y  l) 1000 CPT PV 1082.27 (b)

$1082.27

PMT

6.5

I/Y

i  7.5%; b > i, premium is expected. é1  1.0757 ù ú  [1000 (0.08)  1000 (0.075)] ê ê 0.075 ú ë û  (80  75)(5.296601)  (5)(5.296601)

Premium



$26.48

Purchase price

 1000  26.48 

(Set P/Y  l; C/Y  l) 1000 CPT PV 1026.48 (c)

$1026.48

PMT

7.5

I/Y

i  8.5; b < i, discount is expected. Discount

é1  1.0857 ù ú  [1000 (0.08)  1000 (0.085)] ê ê 0.085 ú ë û

 (80  85)(5.118514)  (5)(5.118514)

$25.59 .



Purchase price

 1000  25.59 

(Set P/Y  1; C/Y  1) 1000 CPT PV 974.41

$974.4l 80

PMT

8.5

I/Y

19. Face value  5,000,000; b  3.625% Principal  5,000,000; i  0.7%; c 

12  6; n  20 2

p  1.0076  1  1.042742  1  4.2742% Since b < p, discount is expected. é1  1.042742 20 ù ú Discount  [5,000,000 (0.03625)  5,000,000 (0.042742)] ê ê 0.042742 ú ë û

 5,000,000 ( 0.006492)(l3.266241)  $430,615.29 Purchase price  $5,000,000  430,615.29  (Set P/Y  2; C/Y  12) 5,000,000 CPT PV .  4,569,384.71

$4,569,384.71

181,250

PMT

. 8.4

I/Y

20. Face value  3000; b  0.75% Principal  3000; i  1%; n  18 Since b < i, the bond is expected to sell at a discount. é1  1.0118 ù ú Discount  [3000 (0.0075)  3000 (0.01)] ê ê 0.01 ú ë û

 (22.50  30)(16.398269)  (7.50)(16.398269)  $122.99 Purchase price  3000  122.99 

$2877.01

(Set P/Y  2; C/Y  2) 3000 PV 2877.01

22.50

PMT

21. Face value  5000 (20)  100,000; b  2.1% Principal  100,000; i  8.0%; c 

1 ; n  32 4

1 4

p  1.08  1  1.01942655  1  1.942655%

I/Y

CPT

Since b > p, premium is expected. é1  1.0194265532 ù ú Premium  [100,000 (0.021)  100,000 (0.01942655)] ê ê 0.01942655 ú ë û

 (2100  1942.655)(23.665093)  (157.345)(23.665093)  $3723.58 Purchase price  100,000  3723.58 

$103,723.58

(Set P/Y  4; C/Y  l) 100,000 CPT PV 103,723.59

2100

PMT

I/Y

22. Face value  60,000; b  2% Principal  60,000; i  2.75%; n  14 Since b < i, the bond is expected to sell at a discount. é1  1.027514 ù ú Discount  [60,000 (0.02)  60,000 (0.0275)] ê ê 0.0275 ú ë û  (1200  1650)(11.491008)  ( 450)(11.491008)  5170.95 Purchase price  60,000  5170.95 

$54,829.05

(Set P/Y  2; C/Y  2) 60,000 CPT PV 54,829.05

1200

PMT

. 5.5

23. Because the bond is purchased on an interest payment date, the purchase price equals the present value of the principal plus the present value of the periodic interest payments. That is: Purchase price = present value of the principal + present value of the periodic payments Periodic interest = 10,000 

0.0888  444 2

Purchase price =  1  1.0227 18  10, 000(1.0227) 18  444    6676.232  6501.1153  13,177.3473  .0227 

The purchase price of the bond on April 17, 2025 was $13,177.35. 24. Since the maturity date is June 11, 2031, the semi-annual interest payment dates are June 11 and December 11. The interest date immediately preceding the date of purchase is June 11, 2018. We first calculate the purchase price of the bond on June

11, 2018, and then calculate the accrued interest for the period of June 11, 2018 to August 10, 2018. Purchase price on June 11, 2018 = present value of the principal + present value of the periodic payments Periodic interest = 10,000 

0.065  325 2

 1  1.022226  Purchase price = 10, 000(1.0222) 26  325    12, 018.13  0.0222  The purchase price of the bond on June 11, 2016 was $12,018.13.

(Set P/Y  2, C/Y  2) 10,000 CPT PV 12,018.13.

325

PMT

4.44

I/Y

The number of days from June 11, 2018 to August 10, 2018 is 60, and the number of days in the interest payment interval from June 11, 2018 to December 11, 2018 is 183. Therefore, PV = 12,018.13 i = 0.0222 n = 60/183 60

Cash price = 12, 018.13(1.0222) 188  12,104.96 The cash price of the bond on August 10, 2018 was $12,104.96.

Business Math News Box 1.

For the Series 2019-1:

b. Semi-annual interest payment = 290,500,000 (0.0353) / 2= $5,127,325 2.

3. 4.

For the Series 2016-1:

b. a. b. a.

Semi-annual interest payment = 204,000,000 (0.0408) / 2= $4,161,600 For the Series 2019-1: Total interest paid = 40 (2) (5,127,325) = $410,186,000 For the Series 2016-1: Total interest paid = 40 (2) (4,161,600) = $332,928,000 There are 71 six-month periods from June 30, 2024 to December 31, 2059. (Set P/Y = 2; C/Y = 2) 71 N 3.2 I/Y 5,127,325 PMT 0 PV CPT FV −668,605,580

b. Hydro Ottawa needs $290,500,000 on December 31, 2059 to pay back the bond principal and the company will have more than that amount in the sinking fund. Exercise 15.3 A. 1.

Face value  5000; b  3% Principal  5000; i  3.25%; n  7 Since b < i, discount is expected. Discount

é1  1.03257 ù ú  [5000 (0.03)  5000 (0.0325)] ê ê 0.0325 ú ë û  5000 (0.0025)(6.1720)



$77.15 .

Purchase price

 5000  77.15 

(Set P/Y  2; C/Y  2) 5000 PV  4922.85 Payment interval 0 1 2 3 4 5 6 7 Total

$4922.85 . 150

PMT

6.5

I/Y

Schedule of Accumulation of Discount Coupon Interest on book Discount Book accumulated value b  3% i  3.25% 4922.85 150.00 159.99 9.99 4932.84 150.00 160.32 10.32 4943.16 150.00 160.65 10.65 4953.81 150.00 161.00 11.00 4964.81 150.00 161.36 11.36 4976.17 150.00 161.73 11.73 4987.90 150.00 162.10 12.10 5000.00 $1050.00 $1127.15 $77.15

CPT

Discount balance 77.15 67.16 56.84 46.19 35.19 23.83 12.10 —

Face value  25,000; b  0.6% Principal  25,000; i  0.5%; n  8 Since b > i, premium is expected. é1  1.0058 ù ú  [25,000 (0.006)  25,000 (0.005)] ê ê 0.005 ú ë û

Premium

 25,000 (0.001)(7.822959) 

$195.57 .

Purchase price

 25,000  195.57 

(Set P/Y  4; C/Y  4) 25,000 PV 25,195.57

PMT

I/Y

Face value  1000; b  2.5% Principal  1000; i  2% March 1, 2020, to September 1, 2023, is 3.5 years: n  7. Since b > i, premium is expected. Premium

Schedule for Amortization of Premium Interest on Coupon book value Premium Book amortized value b  0.6% i  0.5% 25,195.57 150.00 125.98 24.02 25,171.55 150.00 125.86 24.14 25,147.41 150.00 125.74 24.26 25,123.14 150.00 125.62 24.38 25,098.76 150.00 125.49 24.51 25,074.25 150.00 125.37 24.63 25,049.62 150.00 125.25 24.75 25,024.87 150.00 125.12 24.88 25,000.00 $1200.00 $1004.43 $195.57

Payment interval 0 1 2 3 4 5 6 7 8 Total 3.

FV 150

$25,195.57

é1  1.02 7 ù ú  [1000 (0.025)  1000 (0.02)] ê ê 0.02 ú ë û

 (25  20)(6.471991)  $32.36 . Purchase price

 1000  32.36  $1032.36 .

CPT

Premium balance 195.57 171.55 147.41 123.14 98.76 74.25 49.62 24.87 —

(Set P/Y  2; C/Y  2) 1000 PV 1032.36

Payment interval 0 1 2 3 4 5 6 7 Total 4.

25 

PMT

I/Y

Schedule for Amortization of Premium Interest on Coupon book value Premium Book b  2.5% i  2% amortized value 1032.36 25.00 20.65 4.35 1028.01 25.00 20.56 4.44 1023.57 25.00 20.47 4.53 1019.04 25.00 20.38 4.62 1014.42 25.00 20.29 4.71 1009.71 25.00 20.19 4.81 1004.90 25.00 20.10 4.90 1000.00 $175.00 $142.64 $32.36

CPT

Premium balance 32.36 28.01 23.57 19.04 14.42 9.71 4.90 —

Face value  10,000; b  7.75% Principal  10,000; i  7.25%; n  7 Since b > i, premium is expected. é1  1.07257 ù ú  [10,000 (0.0775)  10,000 (0.0725)] ê ê 0.0725 ú ë û

Premium

 (775  725)(5.342633) 

$267.13

Purchase price

 10,000  267.13 

(Set P/Y  l; C/Y  l) 10,000 PV 10,267.13

Payment interval 0 1 2 3 4 5 6

775

$10,267.13 PMT

7.25

I/Y

Schedule of Accumulation of Premium Interest on Coupon book value Premium Book b  7.75% i  7.25% accumulated value 10,267.13 775.00 744.37 30.63 10,236.50 775.00 742.15 32.85 10,203.65 775.00 739.76 35.24 10,168.41 775.00 737.21 37.79 10,130.62 775.00 734.47 40.53 10,090.09 775.00 731.53 43.47 10,046.62

CPT

Premium balance 267.13 236.50 203.65 168.41 130.62 90.09 46.62

7 Total

775.00 $5425.00

728.38 $5157.87

46.62 $267.13

10,000.00

—

B. 1.

Proceeds  25,000 (0.9925)  $24,812.50 Face value  25,000; b  3.25% Principal  25,000; i  3.5%; n  8 b < i  discount é1  1.0358 ù ú Discount  [25,000 (0.0325)  25,000 (0.035)] ê ê 0.035 ú ë û

 25,000 ( 0.0025)(6.873956)  429.62 Book value  25,000  429.62  $24,570.38 Gain on sale  24,812.50  24,570.38  $242.12 . (Set P/Y  2; C/Y  2) 25,000 CPT PV 24,570.30 2.

812.50

PMT

I/Y

Face value  5000 (4)  20,000; b  4.25% Principal  20,000; i  3.75%; n  (20  3)(2)  34 b > i  premium é1  1.037534 ù ú Premium  [20,000 (0.0425)  20,000 (0.0375)] ê ê 0.0375 ú ë û

 20,000 (0.005)(19.039326)  $1903.93 Book value  20,000  1903.93  $21,903.93 Proceeds  20,000 (1.03625)  $20,725

Loss on sale = 21,903.93  20,725 =

$1178.93

(Set P/Y  2; C/Y  2) 20,000 CPT PV 21,903.93

850

PMT

. 7.5

I/Y

Face value  5000; b  4% Principal  5000; i  4.5% The interest date immediately preceding the selling date is June 1, 2021.

The time from June 1, 2021, to June 1, 2031, is 10 years: n  20. b < i  discount Discount on June 1, 2021 é1  1.04520 ù ú  [5000 (0.04)  5000 (0.045)] ê ê 0.045 ú ë û

  $325.20 Book value on June 1, 2021  5000  325.20  $4674.80 Accumulated value on September 22, 2021: The period June 1, 2021, to September 22, 2021, is 113 days. The period June 1, 2021, to December 1, 2021, is 183 days. Accumulated book value on September 22, 2021  4674.80 (1  0.045 )

113/183

=$4803.60

Accrued interest on September 22, 2021 æ113 ö  5000 (0.04) çç ÷  $123.50 ÷ çè183 ÷ ø

Proceeds  5000 (1.01375)  123.5 = $5192.25 Gain on sale  5192.25  4803.60  (Set P/Y  2; C/Y  2) 5000 PV 4674.80 4.

$388.65 200

PMT

I/Y

Face value  3(10,000)  30,000; b  2.625% Principal  30,000; i  3% Interest payment dates are August 1, November 1, February 1, and May 1. The interest date preceding the date of sale is November 1, 2018. The period November 1, 2018, to August 1, 2025, is 6.75 years: n  27. b < i  discount Discount on November 1, 2018

CPT

é1  1.0327 ù ú  [30,000 (0.02625)  30,000 (0.03)] ê ê 0.03 ú ë û

 (787.50  900)(18.327031)  $2061.79 Book value on November 1, 2018  30,000  2061.79  $27,938.21 Interest interval November 1, 2018, to February 1, 2019, is 92 days. Interest period November 1, 2018, to January 16, 2019, is 76 days. Accumulated book value on January 16, 2019 æ 76 ö  27,938.21 çç1  0.03  ÷ ÷ ÷  27,938.21(1.024783)  $28,630.59 çè 92 ø

Accrued interest to January 16, 2019 é76 ù  30,000 (0.02625) ê ú  $650.54 êë92 úû

Proceeds  30,000 (0.935)  650.54  28,050  650.54  $28,700.54 Gain on sale  28,700.54  28,630.59  (Set P/Y  4; C/Y  4) 30,000 CPT PV 27,938.21

$69.95

787.50

PMT

I/Y

Exercise 15.4 1.

Quoted price (initial book value)  10,000 (1.01375)  $10,137.50 Principal  $10,000 Average book value 

1 (10,137.50  10,000)  $10,068.75 2

The semi-annual interest  10,000 (0.03)  $300 The number of interest payments to maturity  30 The total interest payments  30(300)  $9000 The bond premium  10,137.50  10,000  $137.50 Average income per interest payment interval 

1 1 (9,000  137.50)  (8862.50)  $295.42 30 30

Approximate value of i 

295.42  0.029340 = 2.9340% 10, 068.75

The approximate yield rate  2(0.029340)  0.05868 = 2.

5.868%

Quoted price  5000 (0.9475)  $4737.50 Principal  $5000

1 (4737.50  5000)  $4868.75 2 Semi-annual coupon  5000 (0.0525)  $262.50 Average book value 

The number of coupons to maturity  14 The total value of the coupons  14(262.50)  $3675 The bond discount  5000  4737.50  $262.50 Average income per interest payment interval

1 1 (3675  262.50)  (3937.50)  $281.25 14 14 281.25 Approximate value of i   0.057766 = 5.7766% 4868.75 

Approximate yield rate  2(0.057766)  0.115533 = 3.

11.5533%

Initial book value  25,000 (0.97125)  $24,281.25 Principal  25,000 Average book value 

1 (24,281.25  25,000)  $24,640.63 2

Semi-annual coupon  25,000 (0.0375)  $937.50 Number of coupons to maturity  20 Total value of coupons  20(937.50)  $18,750 The bond discount  25,000  24,281.25  $718.75 Average income per interest payment interval 

1 1 (18,750  718.75)  (19,468.75)  $973.44 20 20

Approximate value of i 

973.44  0.039505 = 3.9505% 24,640.63

Approximate yield rate  2(0.039505)  0.07901 = 4.

7.901%

Initial book value  1000 (1.01)  $1010 Principal  1000 Average book value 

1 (1010  1000)  $1005 2

Semi-annual interest payment  1000 (0.0425)  $42.50 Number of payments to maturity  16 Total value of coupons  16(42.50)  $680 Bond premium  1000  1010  $10 Average income per interest payment interval 

1 (680  10)  $41.88 16

Approximate value of i 

41.88  0.041672 = 4.1672% 1005

Approximate yield rate  2(0.041672)  0.083343 =

8.3343%

Initial book value  50,000 (0.98875)  $49,437.50 Principal  $50,000 Average book value 

1 (50,000  49,437.50)  $49,718.75 2

Nearest interest payment date is 5.5 years before maturity. Number of interest payments  11 Semi-annual interest  50,000 (0.045)  $2250 Total interest  11(2250)  $24,750 Bond discount  50,000  49,437.50  $562.50 Average income per interest payment interval 

1 1 (24,750  562.50)  (25,312.50)  $2301.14 11 11

Approximate value of i 

2301.14  0.046283 = 4.6283% 49, 718.75

Approximate yield rate  2(0.046283)  0.092566 = 6.

9.2566%

Initial book value  20,000 (1.0925)  $21,850 Principal  20,000 Average book value 

1 (21,850  20,000)  $20,925 2

Nearest interest payment date is 9.5 years before maturity. Number of interest payments  19 Semi-annual interest  20,000 (0.035)  $700 Total interest  19(700)  $13,300 Bond premium  21,850  20,000  $1850 Average income per interest payment interval 

1 1 (13,300  1850)  (11,450)  $602.63 19 19

Approximate value of i 

602.63  0.028800 = 2.8800% 20,925

Approximate yield rate  2(0.028800)  0.057599 =

5.7599%

Quoted price = 7500(1.091724) = $8187.93 Average book value of the bond = (7500 + 8187.93)/2 = $7843.965 Semi-annual interest payment = 7500(0.0735/2) = $275.625 Number of interest payments = 15(2) = 30 Total interest payments = 30(275.625) = $8268.75 Premium = 8187.93 – 7500 = $687.93

8268.75  687.93  252.694 30

Average income per interest payment interval = Approximate value of i =

252.694  0.0322 7843.965

Approximate yield rate = 2(0.0322) = 0.0644 or 6.44% Exercise 15.5 1.

(a) FVn  75,000; i  1%; n  24 é1.0124  1ù ú 75,000  PMT ê ê 0.01 ú ë û

75,000  26.973465 PMT PMT 

$2780.51

(Set P/Y  4; C/Y  4) 75,000 PMT 2780.51 (b) Total paid  2780.51(24) 

$66.732.25

I/Y

CPT

$8267.50

(a) FVn(due)  100,000; i  2.75%; n  20 é1.027520  1ù ú 100,000  PMT (1.0275) ê ê 0.0275 ú ë û

100,000  PMT(1.0275)(26.197398) PMT  $3715.01 . (Set P/Y  2; C/Y  2)(―BGN‖ Mode) 100,000 FV CPT PMT . 3715.01 (b) Total deposits  3715.01(20)  $74,300.21 . (c) Interest  100,000  74,300.20  $25,699.79 . 3.

FVn  20,000; i  5.5%; n  7 é1.0557  1ù ú 20,000  PMT ê ê 0.055 ú ë û

5.5

I/Y

20,000  8.266894 PMT PMT 

$2419.29

(Set P/Y 1; C/Y  1) 20,000 PMT 2419.29

Payment number 0 1 2 3 4 5 6 7 Total

Periodic payment

5.5

I/Y

Sinking Fund Schedule Interest Increase earned in fund — 133.06 273.44 421.54 577.79 742.63 916.53 $3064.99

2419.29 2419.29 2419.29 2419.29 2419.29 2419.29 2419.29 $16,935.03

2419.29 2552.35 2692.73 2840.83 2997.08 3161.92 3335.82 $20,000.02

CPT

Balance after payment — 2419.29 4971.64 7664.37 10,505.20 13,502.28 16,664.20 20,000.02

Total amount in the fund is $20,000.02 at the end of seven years, which is 2 cents more than the $20,000 needed in the fund. This small difference is due to rounding. 4.

FVn(due)  15,000; i  6.25%; n  8 é1.06258  1ù ú 15,000  PMT (1.0625) ê ê 0.0625 ú ë û

15,000  PMT (1.0625)(9.986722) PMT 

$1413.64

(Set P/Y  2; C/Y  2)(―BGN‖ Mode) 15,000 CPT PMT . 1413.64

12.5

I/Y

Sinking Fund Schedule Payment number 0 1 2 3 4 5 6 7 8 Total

Periodic payment

Interest earned

Increase in fund

1413.64 1413.64 1413.64 1413.64 1413.64 1413.64 1413.64 1413.64 $11,309.12

88.35 182.23 281.97 387.94 500.54 620.18 747.29 882.35 $3690.85

1501.99 1595.87 1695.61 1801.58 1914.18 2033.82 2160.93 2295.99 $14,999.97

Balance at end — 1501.99 3097.86 4793.47 6595.05 8509.23 10,543.05 12,703.98 14,999.97

PMT  2419.29; i  5.5%; n  3 é1.0553  1ù ú FVn  2419.29 ê ê 0.055 ú ë û

 2419.29(3.168025)  $7664.37 Interest in Year 4  7664.37(0.055)  $421.54 Increase in Fund  2419.29  421.54 

$2840.83

(Set P/Y  l; C/Y  l) (―END‖ Mode) 20,000 PV CPT PMT 2419.29 (Set P/Y  l; C/Y  l) 2419.29  CPT FV 7664.37

PMT

5.5

I/Y

Balance at the end of fourth quarter = (Set P/Y  l; C/Y  1) 2419.29  4 N 5.5 I/Y 0 PV CPT FV 10,505.20 6.

PMT

PMT  1413.64; i  6.25%; n  4 é1.06254  1ù ú FVn(due)  1413.64 (1.0625) ê ê 0.0625 ú ë û

 1413.64(1.0625)(4.390869)  $6595.05 Interest in Period 5  (6595.05  1413.64)(0.0625) 

$500.54

(Set P/Y  2; C/Y  2)(―BGN‖ Mode) 15,000 PV CPT PMT . 1413.64

(Set P/Y  2; C/Y  2)(―BGN‖ Mode) 1413.64  0 PV CPT FV 6595.05 .

(a) FVn  45,000; i  2%; n  48 é1.0248  1ù ú 45,000  PMT ê ê 0.02 ú ë û

45,000  79.353519 PMT PMT 

$567.08

PMT

. 12.5

I/Y

12.5

I/Y

é1.0216  1ù ú (b) FV16  567.08 ê ê 0.02 ú ë û

 567.08(18.639285) 

$10,569.97

(Set P/Y  4; C/Y  4) (―END‖ Mode) 45,000 PV CPT PMT 567.08

I/Y

(Set P/Y  4; C/Y  4) 567.08  CPT FV 10,569.97

I/Y

PMT

(a) FVn (due)  72,000; i  3.5%; n  30 é1.03530  1ù ú 72,000  PMT(1.035) ê ê 0.035 ú ë û

72,000  PMT(1.035)(51.622677) PMT 

$1347.57

é1.03520  1ù ú (b) FV20  1347.57(1.035) ê ê 0.035 ú ë û

 1347.57(1.035)(28.279682) 

$39,442.66

(Set P/Y  2; C/Y  2)(―BGN‖ Mode) 72,000 PV CPT PMT . 1347.57 (Set P/Y  2; C/Y  2)(―BGN‖Mode) 1347.57  0 PV CPT FV . 39,442.66 9.

PMT

(a) P  95,000; i  4.5% Semi-annual interest  95,000 (0.045) 

$4275

(b) FVn  95,000; i  3.5%; n  40 é1.03540  1ù ú 95,000  PMT ê ê 0.035 ú ë û

95,000  84.55027775 PMT

I/Y

PMT  $1123.59 rounded to

$1124

. to answer parts c and d

(Set P/Y  2; C/Y  2) 95,000 PMT 1123.59

$5399

I/Y

CPT

é1.03530  1ù ú (d) FV30  1124 ê ê 0.035 ú ë û

 1124 (51.622677)  $58,023.89 rounded to $58,024 Book value after 15 years  95,000  58,024  $36,976 . (Set P/Y  2; C/Y  2) 1124  CPT FV 58,023.89

PMT

10. (a) p  80,000; i  0.5% Monthly interest  80,000 (0.005) 

$400

(b) FVn  80,000; i  0.625%; n  144 é1.00625144  1ù ú 80,000  PMT ê ê 0.00625 ú ë û

80,000  232.43581 PMT PMT  $344.18 rounded to

$344

(Set P/Y  12; C/Y  12) 80,000 CPT PMT 344.18 (c) Monthly cost  400  344 

. FV

$744

144

7.5

I/Y

é1.0062596  1ù ú (d) FV96  344 ê ê 0.00625 ú ë û

 344 (130.995147)  $45,062.33 rounded to $45,062 Book value after 8 years  80,000  45,062  (Set P/Y  12; C/Y  12) 344  PMT CPT FV 45,062.33 11. (a) FVn  100,000; i  0.625%; n  180 é1.00625180  1ù ú 100,000  PMT ê ê 0.00625 ú ë û 100,000  331.11228 PMT PMT  $302.01 (Set P/Y  12; C/Y 12) 100,000 CPT PMT 302.01 (b) Balance after 5 years: é1.0062560  1ù ú FV60  302.01 ê ê 0.00625 ú ë û

$34,938

7.5

I/Y

180

7.5

I/Y

 302.01(72.527105)  $21,903.91 (Set P/Y  12; C/Y  12) 302.01  CPT FV 21,903.91

PMT

 302.01(136.48548)

7.5

I/Y

 $41,219.98 Interest in 100th interval  41,219.98(0.00625)  $257.63 (Set P/Y  12; C/Y  12) 302.01  CPT FV 41,219.98

PMT

7.5

I/Y

(d) Balance after 149 intervals: é1.00625149  1ù ú FV149  302.01 ê ê 0.00625 ú ë û  302.01(244.85368)  $73,948.26 Interest in interval 150  73,948.26(0.00625)  $462.18 Increase in fund  302.01  462.18  $764.19 (Set P/Y  12; C/Y  12) 302.01  PMT 149 N 7.5 I/Y 0 PV CPT FV 73,948.26 (e) Last three payments are PMT178, PMT179, PMT180. Balance after 177 payments: é1.00625177  1ù ú FV177  302.01 ê ê 0.00625 ú ë û

 302.01(322.01784)  $97,252.61 (Set P/Y  l2; C/Y  12) 302.01  PMT 177 N 7.5 I/Y 0 PV CPT FV 97,252.61 Partial Sinking Fund Schedule Payment interval

Periodic payment

Interest earned

Increase in fund

Balance at end — 302.01 605.91 911.71 : : 97,252.61 98,162.45 99,077.98 99,999.23

0 1 2 3 : : 177 178 179 180

302.01 302.01 302.01 : : : 302.01 302.01 302.01

— 1.89 3.79 : : : 607.83 613.52 619.24

302.01 303.90 305.80 : : : 909.84 915.53 921.25

Total

$54,361.80

$45,637.43

$99,999.23

12. (a) FVn(due)  120,000; i  1.5%; n  40

é1.01540  1ù ú 120,000  PMT(1.015) ê ê 0.015 ú ë û 120,000  PMT(1.015)(54.267894)

PMT  $2178.57 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 120,000 FV 40 N 6 I/Y 0 PV CPT PMT 2178.57 (b) Balance after 6 years: é1.01524  1ù ú FV24(due)  2178.57(1.015) ê ê 0.015 ú ë û  2178.57(1.015)(28.633521)

 $63,315.83 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 2178.57  PMT 24 N 6 I/Y 0 PV CPT FV 63,315.83 (c) Balance after 27th payment interval: é1.01527  1ù ú FV27(due)  2178.57(1.015) ê ê 0.015 ú ë û

 2178.57(1.015)(32.986679)  $72,941.74 Interest in 28th interval  (72,941.74  2178.57)(0.015)  $1126.80 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 2178.57  PMT 27 N 6 I/Y 0 PV CPT FV 72,941.74 (d) Balance at end of the 32nd interval: é1.01532  1ù ú FV32(due)  2178.57(1.015) ê ê 0.015 ú ë û

 2178.57(1.015)(40.688288)  $89,971.92 Interest in the 33rd interval  (89,971.92  2178.57)(0.015)  $1382.26 Increase in fund  1382.26  2178.57  $3560.83 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 2178.57  PMT 32 N 6 I/Y 0 PV CPT FV 89,971.92 Copyright © 2025 Pearson Canada Inc.

(e) The last three payments are PMT38, PMT39, PMT40. Balance at end of the 37th interval: é1.01537  1ù ú FV37(due)  2178.57(1.015) ê ê 0.015 ú ë û

 2178.57(1.015)(48.985109)  $108,318.25 (Set P/Y  4; C/Y  4)(―BGN‖ Mode) 2178.57  PMT 37 N 6 I/Y 0 PV CPT FV 108,318.25

Payment internal 0 1 2 3 : : 37 38 39 40 Total

Partial Sinking Fund Schedule Periodic Interest Increase payment earned in fund 2178.57 2178.57 2178.57 : : : 2178.57 2178.57 2178.57 $87,142.80

32.68 65.85 99.51 : : : 1657.45 1714.99 1773.40 $32,857.00

2211.25 2244.42 2278.08 : : : 3836.02 3893.56 3951.97 $119,999.80

Balance at end — 2211.25 4455.67 6733.75 : : 108,318.25 112,154.27 116,047.83 119,999.80

13. (a) p  300,000; i  8.25% Annual interest  300,000 (0.0825)  $24,750 (b) FVn  300,000; i  5.5%; n  20 é1.05520  1ù ú 300,000  PMT ê ê 0.055 ú ë û

300,000  34.868318 PMT PMT  $8603.80 rounded up to $8604 (Set P/Y  l; C/Y  l) 300,000 FV 20 N 5.5 I/Y 0 PV CPT PMT 8603.80 (c) Annual cost  24,750  8604  $33,354 (d) Balance after 9 years:

é1.0559  1ù ú FV9  8604 ê ê 0.055 ú ë û

 8604 (11.256260)  $96,848.86 rounded up to $96,849 Interest in year 10  96,849 (0.055)  $5327 Increase in fund  5327  8604  $13,931 (Set P/Y  l; C/Y  l) 8604  PMT 9 N 5.5 I/Y 0 PV CPT FV 96,848.86

(e) Balance after 15 years: 1.05515  1  FV15  8604    0.055 

 8604(22.408664)  $192,804.14 rounded down to $192,804 Book value of debt  300,000  192,804  $107,196 (Set P/Y  1; C/Y  l) 8604 PMT 15 N 5.5 I/Y 0 PV CPT FV –192,804.14 (f ) Last three payments are PMT18, PMT19, PMT20. Balance after 17 years: é1.05517  1ù ú FV17  8604 ê ê 0.055 ú ë û

 8604 (26.996403)  $232,277.05 rounded down to $232,277 (Set P/Y  l; C/Y  l) 8604  PMT 17 N 5.5 I/Y 0 PV CPT FV 232,277.05 Partial Sinking Fund Schedule Payment Periodic Interest interval payment earned 0 1 8604.00 — 2 8604.00 473.00 3 8604.00 973.00 : : : : : : 17 : : 18 8604.00 12,775.00 19 8604.00 13,951.00 20 8597.00 15,192.00 Total $172,073.00 $127,927.00

Increase in fund

Balance in fund

8604.00 9077.00 9577.00 : : : 21,379.00 22,555.00 23,789.00 $300,000.00

8604.00 17,681.00 27,258.00 : : 232,277.00 253,656.00 276,211.00 300,000.00

14. (a) p  225,000; i  6.5% Semi-annual interest  225,000 (0.065)  $14,625 (b) FVn  225,000; i  5.5%; n  30 é1.05530  1ù ú 225,000  PMT ê ê 0.055 ú ë û

225,000  72.435478 PMT

Book value of debt 300,000.00 291,396.00 282,319.00 272,742.00 : : 67,723.00 46,344.00 23,789.00 —

PMT  $3106.21 rounded down to $3106 (Set P/Y  2; C/Y  2) 225,000 FV 30 N 11 I/Y 0 PV CPT PMT 3106.21 (c) Annual cost  2(14,625  3106)  $35,462 (d) Balance at the end of the 19th interval: é1.05519  1ù ú FV19  3106 ê ê 0.055 ú ë û

 3106 (32.102671)  99,710.90 rounded up to $99,711 Interest earned in 20th interval  99,711 (0.055)  $5484 (Set P/Y2;C/Y2) 3106  PMT 19 N 11 I/Y 0 PV CPT FV 99,710.90 (e) Balance after 12 years: é1.05524  1ù ú FV24  3106 ê ê 0.055 ú ë û

 3106 (47.537998)  $147,653.02 rounded down to $147,653 Book value of debt  225,000  147,653  $77,347 (Set P/Y  2; C/Y  2) 3106  PMT 24 N 11 I/Y 0 PV CPT FV 147,653.02 (f ) The last three payments are PMT28, PMT29, PMT30. Balance after 27 intervals: é1.05527  1ù ú FV27  3106 ê ê 0.055 ú ë û

 3106 (58.989109)  $183,220.17 rounded down to $183,220 (Set P/Y  2; C/Y  2) 3106  PMT 27 N 11 I/Y 0 PV CPT FV 183,220.17

Payment interval 0 1 2 3 : : 27 28 29 30 Total

Partial Sinking Fund Schedule Interest Increase earned in fund

Periodic payment

— 171.00 351.00 : : : 10,077.00 10,802.00 11,567.00 $131,804.00

3106.00 3106.00 3106.00 : : : 3106.00 3106.00 3122.00 $93,196.00

3106.00 3277.00 3457.00 : : : 13,183.00 13,908.00 14,689.00 $225,000.00

Balance in fund — 3106.00 6383.00 9840.00 : : 183,220.00 196,403.00 210,311.00 225,000.00

Book value of debt 225,000.00 221,894.00 218,617.00 215,160.00 : : 41,780.00 28,597.00 14,689.00 —

Review Exercise 1.

FV  5000; PMT  5000 (0.0225) 112.50; n  24 (a)

i  1. 5% é1  1.01524 ù ú PV  5000 (1.01524)  112.50 ê ê 0.015 ú ë û  5000 (0.699544)  (112.50)(20.030405)

 3497.72  2253.42 

$5751.14

(Set P/Y  2; C/Y  2) 5000 CPT PV 5751.14 (b)

112.50

i  2.5% é1  1.02524 ù ú PV  5000 (1.02524)  112.50 ê ê 0.025 ú ë û

 5000 (0.552875)  112.50(17.884986)

PMT

1/Y

 2764.38  2012.06 

$4776.44

(Set P/Y  2; C/Y  2) 5000 CPT PV 4776.44

112.50

PMT

I/Y

FV  10,000; PMT  10,000 (0.03)  300; i  3.75% n  18

(a)

é1  1.037518 ù ú PV  10,000 1.037518   300 ê ê 0.0375 ú ë û

 10,000 (0.515483)  300 (12.920461)  5154.83  3876.14 

$9030.97

(Set P/Y  2; C/Y  2) 10,000 CPT PV 9030.97

300

PMT

7.5

I/Y

7.5

1/Y

n  30

(b)

é1  1.037530 ù ú PV  10,000 1.037530   300 ê ê 0.0375 ú ë û

 10,000 (0.331403)  300 (17.829245)  3314.03  5348.77 

$8662.80

(difference due to rounding)

(Set P/Y  2; C/Y  2) 10,000 CPT PV 8662.80 3.

300

PMT

FV  25,000; PMT  25,000 (0.0225)  562.50; i  8.25%; c 

1 ; n  24 4

1 4

p  1.0825  1  1.020016  1  0.020016 = 2.0016% é1  1.020016 24 ù ú PV  25,000 1.020016 24   562.50 ê ê 0.020016 ú ë û

 25,000 (0.621488)  562.50(18.910502)  15,537.19  10,637.16 

$26,174.35

(Set P/Y  4; C/Y  l) 25,000 CPT PV 26,174.35

562.50

PMT

8.25

I/Y

FV  1000; PMT  1000 (0.0475)  47.50; i  3.5% The interest dates are March 1, September 1. The interest date immediately preceding the purchase date is September 1, 2017. The period September l, 2017, to March l, 2026, is 8.5 years: n  17. Purchase price on September 1, 2017 é1  1.03517 ù ú  1000(1.035)17  47.50 ê ê 0.035 ú ë û

 1000 (0.557204)  47.50(12.651321)  557.20  600.94  $1158.14 (Set P/Y  2; C/Y  2) 1000 PV 1158.14

47.50

PMT

I/Y

CPT

The period September 1, 2017, to March l, 2018, is 181 days. The interest period September 1, 2017, to September 19, 2017, is 18 days. Purchase price on September 19, 2017 = 1158.14(1.035)(18/181) = $1162.11 5.

Face value  4(5000)  20,000; b  3.5% Principal  20,000; i  3%; n  14 b > i → premium Premium

é1  1.0314 ù ú  [20,000 (0.035)  20,000 (0.03)] ê ê 0.03 ú ë û

 (20,000)(0.005)(11.296073) 

$1129.61

Purchase price

 20,000  1129.61 

(Set P/Y  2; C/Y  2) 20,000 PV 21,129.61 6.

700

$21,129.61 PMT

Face value  9(1000)  9000; b  4% Principal  9000; n  20 (a)

i  3%; b > i → premium Premium

é1  1.0320 ù ú  [9000 (0.04  0.03)] ê ê 0.03 ú ë û

 9000 (0.01)(14.877475) 

$1338.97

. N

I/Y

CPT

Purchase price

 9000  1338.97 

(Set P/Y  2; C/Y  2) 9000 CPT PV 10,338.97

360

$10,338.97 PMT

. N

I/Y

(b)

i  4%; b  i → sold at par é1  1.0420 ù ú Premium  [9000 (0.04  0.04)] ê ê 0.04 ú ë û

Purchase price 

$9000

(Set P/Y  2; C/Y  2) 9000 CPT PV 9000 (c)

360

PMT

I/Y

i  5%; b < i → discount é1  1.0520 ù ú  [9000 (0.04  0.05)] ê ê 0.05 ú ë û

Discount

 (9000)( 0.01)(12.462210) 

$1121.60

Purchase price

 9000  1121.60 

(Set P/Y  2; C/Y  2) 9000 CPT PV 7878.40 7.

(a)

360

$7878.40

PMT

20 N

Face value  100,000; b  2.5% Principal  100,000; i  3.5% The interest dates are July 15 and January 15. The interest date preceding the purchase date is January 15, 2020. The period January 15, 2020, to July 15, 2031, is 11.5 years: n  23. b < i → discount Discount

on January 15, 2020

é1  1.03523 ù ú  [100,000 (0.025  0.035)] ê ê 0.035 ú ë û

 100,000 (0.01)(15.620410)   $15,620.41 . To calculate the discount on the purchase date, calculate the purchase price in part b and subtract it from the face value of the bond. Discount = 100,000 – 85,892.23 = $14,107.77 (b)

$84,379.59

(Set P/Y  2; C/Y  2) 100,000 CPT PV 84,379.59

2500

PMT

I/Y

The interest payment interval January 15, 2020, to July 15, 2020, is 182 days. The interest period January 15, 2020, to April 18, 2020, is 94 days.

Purchase price on April 18, 2020 equals the purchase price on January 15, 2020 plus accrued interest for 94 days to April 18, 2020.  84,379.59(1.035)(94/182)  $85,892.23 8.

(a)

Face value  4(10,000)  40,000; b  1.5% Principal  40,000; i  1.25% The interest dates are September 1, December 1, March 1, and June 1. The interest date preceding the date of purchase is December 1, 2019. The period December 1, 2019, to September l, 3032, is 12.75 years: n  51. b > i → premium expected

Premium on December 1, 2019  [40,000 (0.015)  40,00 1  1.012551  (0.0125)]    0.0125   (600  500)(37.543581)  (b)

$3754.36

Purchase price on December 1, 2019  40,000  3754.36 

CPT

$43,754.36

(Set P/Y  4; C/Y  4) 40,000 PV 43,754.36

. 600

PMT

I/Y

The interest payment interval from December 1, 2019, to March 1, 2020, is 91 days. The interest period December 1, 2019, to January 23, 2020, is 53 days. Purchase price on January 23, 2020  43,754.36(1.0125)(53/91)  $44,072.08 Premium on January 23, 2020 = 44,072.08  40,000 = $4072.08 9.

FV  5000; PMT  5000 (0.04)  200; i  5%; n  20

é1  1.0520 ù ú Purchase price  5000(l.0520)  200 ê ê 0.05 ú ë û

 5000(0.376889)  200 (12.462210)  1884.45  2492.44 

$4376.89

(Set P/Y  2; C/Y  2) 5000 PV 4376.89

(difference due to rounding) FV

200

PMT

I/Y

CPT

10. FV  1000; PMT  1000 (0.04)  40; i  3.5% Interest dates are February 1 and August 1. The interest date preceding the purchase date is August 1, 2021. The period August l, 2021, to February l, 2028, is 6.5 years: n  13. Purchase price on August 1, 2021 é1  1.03513 ù ú  1000 (1.03513)  40 ê ê 0.035 ú ë û

 1000 (0.639404)  40 (10.302738)  639.40  412.11  $1051.51 (Set P/Y  2; C/Y  2) 1000 PV 1051.51

PMT

I/Y

CPT

The interest payment interval August 1, 2021, to February 1, 2022, is 184 days. The interest period August 1, 2021, to October 12, 2021, is 72 days. Purchase price on October 12, 2021  1051.51(1.035)(72/184)  $1065.76 11. Quoted price  50,000 (0.92375)  $46,187.50 Principal  $50,000 Average book value 

1 (50,000  46,187.50)  $48,093.75 2

Payment dates are April 15 and October 15. The nearest preceding interest payment date is April 15, 2019. The time interval from April 15, 2019, to April 15, 2026, is 7 years: n  14. Semi-annual interest  50,000 (0.055)  $2750 Total interest  14(2750)  $38,500 Bond discount  50,000  46,187.50  $3812.50

Average income per interest payment interval 

1 1 (38,500  3812.50)  (42,312.50)  $3022.32 14 14 3022.32  0.062842 = 6.2842% 48, 093.75

Approximate value of i 

The approximate yield rate is 2(0.062842)  0.125685 =

12.5685%

12. Face value  1000; b  8.5% Principal  1000; i  10.5%; n  6 b < i → discount Discount

é1  1.1056 ù ú  [1000 (0.085  0.105)] ê ê 0.105 ú ë û

 1000 (0.020)(4.292179) 

$85.84

Purchase price

 l000  85.84 

(Set P/Y  l; C/Y  l) 1000 PV 914.16

Payment interval 0 1 2 3 4 5 6 Total

$914.16

PMT

10.5

I/Y

Schedule of Accumulation of Discount Interest on Coupon book Discount Book b  8.5% i  10.5% accumulated value 914.16 85.00 95.99 10.99 925.15 85.00 97.14 12.14 937.29 85.00 98.42 13.42 950.71 85.00 99.82 14.82 965.53 85.00 101.38 16.38 981.91 85.00 103.09 18.09 1000.00 $510.00 $595.84 $85.84

13. Face value  5000; b  4% Principal  5000; i  4.75%; n  7; b < i → discount Discount

é1  1.04757 ù ú  [5000 (0.04  0.0475)] ê ê 0.0475 ú ë û

 (5000)(0.0075)(5.839166)

CPT

Discount balance 85.84 74.85 62.71 49.29 34.47 18.09 —



$218.97

Purchase price PV

 5000  218.97 

(Set P/Y  l; C/Y  l) 5000 4781.03

200

$4781.03 PMT

. 7

4.75

I/Y

CPT

Payment interval 0 1 2 3 4 5 6 7 Total

Schedule for Amortization of Premium Interest on Coupon book value Discount Book accumulated value b  4% i  4.75% 4781.03 200.00 227.10 27.10 4808.13 200.00 228.39 28.39 4836.52 200.00 229.73 29.73 4866.25 200.00 231.15 31.15 4897.40 200.00 232.63 32.63 4930.02 200.00 234.18 34.18 4964.20 200.00 235.80 35.80 5000.00 $1400.00 $1618.97 $218.97

Premium balance 218.97 191.87 163.48 133.75 102.60 69.98 35.80 0.00

14. FV  3(25,000)  75,000; b  5.5% Principal  75,000; i  6%; n  2(8  5)  6 b < i → discount é1  1.06 6 ù ú Discount  [75,000 (0.055  0.060)] ê ê 0.06 ú ë û

 75,000 (0.005)(4.917324)  $1844 Book value at time of sale  75,000  1844  $73,156 Proceeds from sale  75,000 (0.89375)  $67,031.25 Loss

on sale  73,156  67,031.25 

(Set P/Y  2; C/Y  2)75,000 PV 73,156

4125

$6124.75

PMT

I/Y

CPT

15. Face value  10,000; b  1.25% Principal  10,000; i  2.25% Interest dates are November 15, February 15, May 15, and August 15. The interest date preceding the date of sale is August 15, 2024. The time period from August 15, 2024, to November 15, 2034, is 10.25 years: n  41. b < i → discount Discount on August 15, 2024

é1  1.022541 ù ú  [10,000 (0.0125)  10,000 (0.0225)] ê ê 0.0225 ú ë û

 (125  225)(26.595132)   $2659.51 Book value on August 15, 2024  10,000  2659.51  $7340.49 The interest payment interval from August 15, 2024, to November 15, 2024, is 92 days. The interest period from August 15, 2024, to September 10, 2024, is 26 days. Interest on September 10, 2024 æ 26 ö  10,000 çç0.0125  ÷ ÷ çè ø 92 ÷

 $35.33 Cash price on September 10, 2020  7340.49  35.33  7375.82 Proceeds  10,000 (0.9275)  $9275 Gain

on sale  9275  7375.82 

(Set P/Y  4; C/Y  4) 10,000 PV 7340.49

$1899.18 125

PMT

16. Quoted price  25,000 (0.7825)  $19,562.50 Principal  $25,000 Average book value 

1 (25,000  19,562.50)  $22,281.25 2

Semi-annual interest  25,000 (0.0475)  $1187.50 Number of interest payments to maturity  32 Total interest  32(1187.50)  $38,000 Discount on bond  25,000  19,562.50  $5437.50 Average income per interest payment interval 

1 1 (5437.50  38,000)  (43,437.50)  $1357.42 32 32

I/Y

CPT

Approximate value of i 

1357.42  0.060922 = 6.0922% 22, 281.25

The approximate yield rate  2(0.060922)  0.121844 =

12.1844%

17. Quoted price  10,000 (0.9875)  $9875 Principal  10,000 Average book value 

1 (10,000  9875)  $9937.50 2

Interest payment dates are October 15, January 15, April 15, and July 15. The nearest preceding interest payment date is April 15, 2020. The time period April 15, 2020, to October 15, 2032, is 12.5 years. The number of interest payments to maturity is 50. Quarterly interest payment  10,000 (0.01875)  $187.50 Total interest  50(187.50)  $9375 Bond discount  10,000  9875  $125 Average income per interest period 

1 1 (9375  125)  (9500)  $190 50 50

Approximate value of i 

190  0.019119 = 1.9119% 9937.50

The approximate yield rate  4(0.019119)  0.076478 =

7.6478%

18. Quoted price  10,000 (0.9875)  $9875 Selling price  10,000 (0.92)  $9200 Average book value 

1 (9200  10,000)  $9600 2

Interest dates are October 15, January 15, April 15, and July 15. The closest interval dates are April 15, 2020, and July 15, 2025. The time interval April 15, 2020, to July 15, 2025, is 5.25 years. The number of interest payments is 21. Quarterly interest  10,000 (0.01875)  $187.50 Interest paid  21(187.50)  $3937.50

Bond discount  10,000  9875  $125 Average income per payment period 

1 1 (3937.50  125)  (4062.50)  $193.45 21 21

Approximate value of i 

193.45  0.020151 = 2.0151% 9600

The approximate yield rate realized is 4(0.020151)  0.080604 = 19. (a)

8.0604%

Face value  50,000; b  1.625% Principal  50,000; i  2%; n  48 Since b < i → discount é1  1.0248 ù ú Discount  [50,000 (0.01625  0.02)] ê ê 0.02 ú ë û

 50,000 (0.00375)(30.673120)  $5751.21 Purchase price  50,000  5751.21  (Set P/Y  4; C/Y  4) 50,000 CPT PV .

$44,248.79 812.50

PMT

I/Y

PMT

44,248.79 (b)

After 9 years, n  (12  9)(4)  12 é1  1.0212 ù ú Discount  50,000 (0.00375) ê ê 0.02 ú ë û

 50,000 (  0.00375)(10.575341)   $1982.88 Book value  50,000  1982.88  (Set P/Y  4; C/Y  4) 44,248.79  CPT FV 48,017.12 (c)

$48,017.12 PV

812.50

Proceeds from sale of bond  50,000 (0.99625)  $49,812.50 Gain

on sale  49,812.50  48,017.12 

20. Quoted price  5000 (0.955)  $4775 Principal  $5000

$1795.38

I/Y

Average book value 

1 (5000  4775)  $4887.50 2

Interest dates are August 1, February 1. The interest date closest to the date of purchase is February 1, 2019. The time interval February 1, 2019, to August 1, 2030, is 11.5 years. The number of interest payments is 23. Semi-annual interest  5000 (0.0725)  $362.50 Total interest  23(362.50)  $8337.50 Bond discount  5000  4775  $225 Average income per interest interval 

1 1 (8337.50  225)  (8562.50)  $372.28 23 23

Approximate value of i 

372.28  0.076170 = 7.6170% 4887.50

The approximate yield rate  2(0.076170) 

15.234%

21. (a)

FV = $150,000,000, PV = 0, P/Y = 1, i = 0.0349/2 = 0.01745 p  (1  i)  1 and c  c

c y p y

N = 20(1) = 20,

I/Y = 3.49%, C/Y = 2,

 12  2

p  (1  0.01745) 2  1  0.0352045

 (1  p ) n  1  FV  PMT   p    (1  0.0352045) 20  1  150, 000, 000  PMT   0.0352045  

150, 000, 000  28.33918097PMT PMT  $5, 293, 025.235

Programmed Solution (―END‖ Mode) (Set P/Y = 1; C/Y =2) 0 PV , 150,000,000 FV , 3.49 I/Y , 20 N , CPT PMT = –5,293,025.235 The Company needs to make annual payments of $5,293,025.24. (b) Payment Numbers

Payment Amount

Interest Earned on Fund

5,293,025.235

2,188,075.23

7,481,100.46

69,634,378.42

5,293,025.235

2,451,443.65

7,744,468.88

77,378,847.31

5,293,025.235

2,724,083.82

8,017,109.06

85,395,956.37

22. (a)

Total Increase in Fund

Balance in Fund

FV = $155,000,000, PV = 0, N = 25(1) = 25, I/Y = 2.85%, C/Y = 2, P/Y = 1, i = 0.0285/2 = 0.01425 p  (1  i)  1 and c  c

c y p y

 12  2

p  (1  0.01425)2  1  0.028703063

 (1  p ) n  1  FV  PMT   p  

 (1  0.028703063)25  1 155, 000, 000  PMT   0.028703063  

155,000,000  35.84475032PMT PMT  $4,324, 203.65

Programmed Solution (―END‖ Mode) (Set P/Y = 1; C/Y =2) 0 PV , 155,000,000 FV , 2.85 I/Y , 25 N , CPT PMT = −4,324,203.645 The Company needs to make annual payments of $4,324,203.65. (b) Payment Numbers

Payment Amount

Interest Earned on Fund

4,324,203.645

2,476,387.452

6,800,591.098

93,076,661.89

4,324,203.645

2,671,585.244

6,995,788.889

100,072,450.8

4,324,203.645

2,872,385.809

7,196,589.455

107,269,040.2

Total Increase in Fund

Balance in Fund

23. (a) The selling price of a bond depends on the relationship between the coupon or bond rate (the interest the bond pays) and the bond’s YTM. In this case, the coupon rate is 5.05 percent and the YTM is 2.74 percent. Since the coupon rate is higher than the YTM, the bond sells at a premium. (b) Because the bond was purchased on an interest payment date, the purchase price equals the present value of the principal plus the present value of the periodic interest payments. That is: Purchase price on an interest payment date (clean price) = present value of the principal + present value of the periodic payments Periodic interest = 15,000 

0.0505  378.75 2

 1  1.0137 12  Purchase price = 15, 000(1.0137) 12  378.75    16,905.1189  .0137 

The purchase price or the clean price of the bond on July 23,2018 was $16,905.12. (c) Cash price (dirty price) = market price + accrued interest

To calculate the purchase price of the bond on September 28, 2018, we first calculate the purchase price on the interest payment date immediately preceding the purchase date, and then calculate the accumulated value from that day to the settlement date. The purchase price for this bond on July 23, 2018 was calculated in part (ii) and equals to $16,905.12.

Purchase price on July 23, 2018 = $16,905.1189 The number of days from July 23, 2018 to September 28, 2018 is 67 and the number of days in the interest payment interval from July 23, 2018 to January 23, 2019 is 184. Therefore, PV = 16,905.1189 i = .0137 n = 67/184 67

Cash price = 16,905.1189  16,905.1189(.0137)184  16989.4516 The cash price of the bond on September 28, 2018 was $16,989.09. 24. This bond pays interest twice per year, so the interest payment dates would be December 22 and June 22 of each year until maturity. Since we want to find the purchase price of the bond on June 22, 2026 – the calculation is on an interest payment date. Using the Texas Instrument BAII PLUS calculator ―Bond‖ worksheet: SDT = 06.2226 (mm.ddyy) CPN = 6.25 RDT = 12.2232 (mm.ddyy) RV = 100 ACT 2/Y YLD = 5.42 To compute the market price: PRI = press CPT = 104.4964913 Note that this calculation is in terms of $100 of face value of the bond. Therefore, multiply this amount by 250 to compute the selling price of a bond with a $25,000 face value. The bond should sell for 250  104.4964913 = $26,124.12 in the market. Since this bond was purchased on a coupon payment date, the remaining entries on the worksheet would be: AI or Accrued Interest = 0 25. Since this settlement date is between interest payment dates, the next coupon payment must be prorated between the old and new owners of the bond, based on the number of days the bond is held by each of them in the coupon payment period (which in this case is six months). The first step is to compute the PV of all payments not yet received. This is known as the CLEAN Price and it is the price quoted in bond markets. To compute the clean price, we can use the Bond worksheet. The data input will look like the following: SDT = 07.2222 CPN = 7.25

RDT = 12.1534 RV = 100 ACT 2/Y YLD = 4.55 To compute the market price: PRI = press CPT = 125.3644986. Multiply this number by 200 to obtain a clean price of $25,072.90. However, the actual sale price of the bond, which is known as the DIRTY price, includes both the clean price plus the accrued interest. By the arrow down key button, you can find the accrued interest amount of 0.7329 (which must be multiplied by 200) to obtain $146.59. The dirty price or the cash price of the bond is then $25,072.90 + $146.59 = $25,219.49. 26.

(a) FVn  110,000; i  3.75%; n  10 é1.037510  1ù ú 110,000  PMT ê ê 0.0375 ú ë û

110,000  11.867838 PMT PMT 

$9268.75

(Set P/Y  2; C/Y  2) 110,000 CPT PMT 9268.75

7.5

I/Y

PMT

7.5

I/Y

Interest in R6  49,952.34(0.0375) 

$1873.21

(Set P/Y  2; C/Y  2) 9268.75  CPT FV 49,952.34

PMT

7.5

I/Y

é1.03753  1ù ú (b) FV3  9268.75 ê ê 0.0375 ú ë û

 9268.75(3.113906) 

$28,862.02

(Set P/Y  2; C/Y  2) 9268.75  CPT FV 28,862.02 é1.03755  1ù ú (c) FV5  9268.75 ê ê 0.0375 ú ë û

 9268.75(5.389328)  $49,952.34

(d) Payment interval 0 1 2 3 4 5 6 7 8 9 10 Total

Periodic payment

Sinking Fund Schedule Interest Increase earned in fund

9268.75 9268.75 9268.75 9268.75 9268.75 9268.75 9268.75 9268.75 9268.75 9268.75 $92,687.50

— 347.58 708.19 1082.33 1470.49 1873.21 2291.04 2724.53 3174.28 3640.89 $17,312.54

Balance at end — 9268.75 18,885.08 28,862.02 39,213.10 49,952.34 61,094.30 72,654.09 84,647.37 97,090.40 110,000.04

9268.75 9616.33 9976.94 10,351.08 10,739.24 11,141.96 11,559.79 11,993.28 12,443.03 12,909.64 $110,000.04

27. (a) FVn(due)  65,000; i  3%; n  32 é1.0332  1ù ú 65,000  PMT (1.03) ê ê 0.03 ú ë û

65,000  PMT (1.03)(52.502759) PMT 

$1201.97

(Set P/Y  4; C/Y  4)(―BGN‖ Mode) 65,000 PV CPT PMT .

I/Y

PMT

1201.97 (b) Balance after 3 years: n  12 é1.0312  1ù ú FV12(due)  1201.97(1.03) ê ê 0.03 ú ë û

 1201.97(1.03)(14.192030) 

$17,570.15

(Set P/Y  4; C/Y  4) (―BGN‖ Mode) 1201.97  I/Y 0 PV CPT . FV 17,570.15 (c) Balance at end of 23rd payment interval: é1.0323  1ù ú FV23(due)  1201.97(1.03) ê ê 0.03 ú ë û

 1201.97(1.03)(32.452884)

 $40,177.61

Interest earned in 24th interval  (40,177.61  1201.97)(0.03)  $1241.39 Increase in fund  1201.97  1241.39 

$2443.36

(―BGN‖ Mode) (Set P/Y  4; C/Y  4) 1201.97  PMT I/Y 0 PV CPT . FV 40,177.61

(d) Last three payments are PMT30, PMT31, PMT32. Balance at end of 29th interval: é1.0329  1ù ú FV29(due)  1201.97(1.03) ê ê 0.03 ú ë û

 1201.97(1.03)(45.21885)  $55,982.25 (―BGN‖ Mode) (Set P/Y  4; C/Y  4) 1201.97  PMT I/Y 0 PV CPT . FV 55,982.25 Payment interval 0 1 2 3 : : 29 30 31 32 Total

Partial Sinking Fund Schedule Periodic Interest Increase payment earned in fund 1201.97 1201.97 1201.97 : : : 1201.97 1201.97 1201.97 $38,463.04

36.06 73.20 111.46 : : : 1715.53 1803.05 1893.20 $26,536.90

28. (a) PV  100,000; i  13.75% Annual interest  100,000 (0.1375)  $13,750 (b) FVn  100,000; i  11.5%; n  8 é1.1158  1ù ú 100,000  PMT ê ê 0.115 ú ë û

1238.03 1275.17 1313.43 : : : 2917.50 3005.02 3095.17 $64,999.94

Balance at end 1238.03 2513.20 3826.63 : : 55,982.25 58,899.75 61,904.77 64,999.94

(Set P/Y  l; C/Y  l) 100,000 CPT PMT 8279.90

11.5

I/Y

11.5

I/Y

 8279.90(3.358225)  $27,805.77 Book value  100,000  27,805.77  $72,194.23 (Set P/Y  l; C/Y  l) 8279.90  CPT FV 27,805.77

PMT

(e) Balance after 5 years: é1.1155  1ù ú FV5  8279.90 ê ê 0.115 ú ë û

 8279.90(6.290029)  $52,080.81 Interest earned  52,080.81(0.115)  $5989.29 (Set P/Y ; C/Y  l) 8279.90  CPT FV 52,080.81

PMT

11.5

I/Y

(f ) Payment Periodic interval payment 0 1 8279.90 2 8279.90 3 8279.90 4 8279.90 5 8279.90 6 8279.90 7 8279.90 8 8279.93 Total $66,239.23

Sinking Fund Schedule Interest Increase earned in fund — 952.19 2013.88 3197.66 4517.58 5989.29 7630.25 9459.92 $33,760.77

8279.90 9232.09 10,293.78 11,477.56 12,797.48 14,269.19 15,910.15 17,739.85 $100,000.00

Balance in fund 8279.90 17,511.99 27,805.77 39,283.33 52,080.81 66,350.00 82,260.15 100,000.00

Book value of debt 100,000.00 91,720.10 82,488.01 72,194.23 60,716.67 47,919.19 33,650.00 17,739.85 —

29. (a) PV  300,000; i  4.625% Semi-annual interest  300,000 (0.04625)  $13,875 (b) FVn  300,000; i  4%; n  30 é1.0430  1ù ú 300,000  PMT ê ê 0.04 ú ë û

300,000  56.084938 PMT PMT  $5349.03 rounded down to $5349 (Set P/Y  2; C/Y  2) 300,000 PMT 5349.03

I/Y

CPT

I/Y

CPT

I/Y

 5349 (12.006107)  $64,220.67 rounded up to $64,221 Book value  300,000  64,221  $235,779 (Set P/ 2; C/Y2) 5349  FV 64,220.67

PMT

(e) Balance after 19th payment: é1.0419  1ù ú FV19  5349 ê ê 0.04 ú ë û

 5349 (27.671229)  $148,013.41 rounded down to $148,013 Interest in 20th payment  148,013 (0.04)  $5921 Increase in fund  5349  5921  $1l,270 (Set P/Y  2; C/Y  2) 5349 CPT FV 148,013.41

PMT

(f ) Last three payments are PMT28, PMT29, PMT30. Balance after 27th payment:

é1.0427  1ù ú FV27  5349 ê ê 0.04 ú ë û

 5349 (47.084214)  $251,853.46 rounded down to $251,853 (Set P/Y  2; C/Y2) 5349  CPT FV 251,853.46

Payment Periodic interval payment 0 1 5349.00 2 5349.00 3 5349.00 : : : : 27 5349.00 28 5349.00 29 5349.00 30 5351.00 Total $160,472.00

PMT

Partial Sinking Fund Schedule Interest Increase earned in fund — 214.00 436.00 : : : 10,074.00 10,691.00 11,333.00 $139,528.00

5349.00 5563.00 5785.00 : : : 15,423.00 16,040.00 16,684.00 $300,000.00

I/Y

Balance in fund — 5349.00 10,912.00 16,697.00 : : 251,853.00 267,276.00 283,316.00 300,000.00

Book value of debt 300,000.00 294,651.00 289,088.00 283,303.00 : : 48,147.00 32,724.00 16,684.00 —

30. (a) FVn  600,000; i  2%; n  20 é1.0220  1ù ú 600,000  PMT ê ê 0.02 ú ë û

600,000  24.297370 PMT PMT  $24,694.03 (Set P/Y  4; C/Y  4) 600,000 PMT 24,694.03

I/Y

CPT

(b) Total payments  24,694.03(20)  $493,880.60 Interest  600,000  493,880.60  $106,119.40 (c) Balance after 2 years: n  8 é1.028  1ù ú FV8  24,694.03 ê ê 0.02 ú ë û

 24,694.03(8.582969)  $211,948.09 (Set P/Y  4; C/Y  4) 24,694.03  CPT FV 211,948.09

PMT

(d) Balance after 14th payment: é1.0214  1ù ú FV14  24,694.03 ê ê 0.02 ú ë û

 2469.40(15.973938)  $394,460.09 Interest in 15th interval  $39,446.04(0.02)  $7889.20 (Set P/Y  4; C/Y  4) 24,694.03  CPT FV 394,460.09

PMT

7.5

I/Y

31. (a) FVn (due)  10,000; i  7.5%; n  7 é1.0757  1ù ú 10,000  PMT(1.075) ê ê 0.075 ú ë û 10,000  PMT(1.075)(8.787322) PMT  $1058.61 (Set P/Y  l; C/Y  l)(―BGN‖ Mode) 10,000 CPT PMT . 1058.61

I/Y

(b) Payment interval 0 1 2 3 4 5 6 7 Total 32.

Periodic payment 1058.61 1058.61 1058.61 1058.61 1058.61 1058.61 1058.61 $7410.27

Sinking Fund Schedule Interest Increase earned in fund 79.40 164.75 256.50 355.13 461.16 575.15 697.68 $2589.77

1138.01 1223.36 1315.11 1413.74 1519.77 1633.76 1756.29 $10,000.04

PVn (defer)  150,000; PMT  35,000 (due); i  3%; d  50  1  49 é1  1.03 n ù ú 150,000  35,000 (1.0349) ê ê 0.03 ú ë û  é1  1.03 n ù ú 150,000  35,000(0.234950) ê ê 0.03 ú ë û

Balance at end — 1138.01 2361.37 3676.48 5090.22 6609.99 8243.75 10,000.04

0.029559n  0.292367 n  26.81 or 27 payments Set P/Y  2; C/Y  2)(―BGN‖ Mode) 150,000  CPT FV . 657,585.90

(Set P/Y  2; C/Y  2)(―BGN‖ Mode) 657,585.90  I/Y 0 FV CPT N 26.81

N 35,000

I/Y

PMT

33. (a) PV  96,000; i  14.5% Annual interest  96,000 (0.145)  $13,920 (b) FVn  96,000; i  13%; n  10 é1.1310  1ù ú 96,000  PMT ê ê 0.13 ú ë û

96,000  18.419749 PMT PMT  $5211.80 (Set P/Y  l; C/Y  l) 96,000 FV PMT 5211.80 (c) Annual cost  13,920  5211.80 

I/Y

CPT

$19,131.80

(d) Balance after 4 years: é1.134  1ù ú FV4  5211.80 ê ê 0.13 ú ë û

 5211.80(4.849797)  $25,276.17 Book value  96,000  25,276.17  $70,723.83 (Set P/Y  l; C/Y  l) 5211.80  CPT FV 25,276.17

PMT

N 13

I/Y 0

(e) Last three payments are PMT8, PMT9, PMT10. Balance after 7 years: é1.137  1ù ú FV7  5211.80 ê ê 0.13 ú ë û

 5211.80(10.404658)  $54,226.99 (Set P/Y  l; C/Y  1) 5211.80  CPT FV 54,226.99 Payment interval 0 1 :

Periodic payment 5211.80 :

PMT

I/Y

Partial Sinking Fund Schedule Interest Increase Balance earned in fund in fund — — 5211.80 5211.80 : : :

Book value of debt 96,000.00 90,788.20 :

: 7 8 9 10 Total

: : 5211.80 5211.80 5211.75 $52,117.95

: : 7049.51 8643.48 10,444.67 $43,882.05

: : 12,261.31 13,855.28 15,656.42 $96,000.00

: 54,226.99 66,488.30 80,343.58 96,000.00

: 41,773.01 29,511.70 15,656.42 —

Self-Test 1.

FV  10,000; PMT  10,000 (0.025)  250 i  2.75%; n  60 é1  1.027560 ù ú Purchase price  10,000 (1.0275)60  250 ê ê 0.0275 ú ë û  10,000 (0.196377)  250 (29.222662)  1963.77  7305.67  $9269.43 (difference due to rounding) (Set P/Y  4; C/Y  4) 10,000 FV 250 PMT 60 N CPT PV 9269.43 FV  1000; PMT  1000 (0.0375)  37.50 i  3%; n  20 é1  1.0320 ù 20 ú Purchase price  1000 (l.03)  37.50 ê ê 0.03 ú ë û  1000 (0.553676)  37.50(14.877475)  553.68  557.90  $1111.58 . (Set P/Y  2; C/Y  2) 1000 FV 37.50 PMT 20 N PV 1111.58 FV  5000; PMT  5000 (0.04)  200 i  3.25%; n  12 Premium (or discount) é1  1.032512 ù ú  (5000  0.04  5000  0.0325) ê ê 0.0325 ú ë û  (200  162.50)(9.807076)



$367.77 (premium)

I/Y

CPT

6.5

I/Y

CPT

(Set P/Y  2; C/Y  2) 5000 PV 5367.77

200

PMT

5367.77  5000  $367.77 premium 4.

FV  20,000; PMT  20,000 (0.05)  1000; i  4.5% Interest dates: March 1, September 1 The interest date preceding the date of purchase is September 1, 2018. The time interval September 1, 2018, to March 1, 2025, is 6.5 years: n  13. Purchase price on September 1, 2018 é1  1.04513 ù ú  20,000 (1.045)13  1000 ê ê 0.045 ú ë û

 20,000 (0.564272)  1000 (9.682852)

 11,285.43  9682.85  $20,968.29 (difference due to rounding) (Set P/Y  2; C/Y  2) 20,000 CPT PV  20,968.29

1000

PMT

I/Y

The interest payment interval September 1, 2018, to March 1, 2019, contains 181 days. The interest period September 1, 2018, to November 15, 2018, is 75 days. Purchase price on November 15, 2018  20,968.29(1.045)(75/181)  $21,354.24 5.

Face value  5000; b  3.5% Principal  5000; i  4.25% Interest dates: December 15, June 15 The interest date immediately preceding the date of purchase is June 15, 2019. The time interval June 15, 2019, to December 15, 2030, is 11.5 years: n  23. Premium or discount on June 15, 2019 é1  1.042523 ù ú  [5000 (0.035)  5000 (0.0425)] ê ê 0.0425 ú ë û

 (175  212.50)(14.495796)  (37.50)(14.495796)  $543.59 (discount) Market price on June 15, 2019  5000  543.59  $4456.41 (Set P/Y  2; C/Y  2) 5000 PV  4456.41

175

PMT

8.5

I/Y

The payment interval June 15, 2019, to December 15, 2019, is 183 days. The interest period June 15, 2019, to November 9, 2019, is 147 days. Cash price on November 9, 2019  4456.41(1.0425)(147/183)  $4607.92 6.

Face value  5000; b  2.75% Principal  5000; i  3.5%; n  8 b < i → discount expected

CPT

é1  1.0358 ù ú Discount  [5000  0.0275  5000  0.035] ê ê 0.035 ú ë û é1  1.0358 ù ú  (137.50  175) ê ê 0.035 ú ë û

 (37.50) (6.873956)   $257.77 Purchase price  5000  257.77  $4742.23 (Set P/Y  2; C/Y  2) 5000 PV  4742.23

Payment interval 0 1 2 3 4 5 6 7 8 Total

137.50

PMT

Schedule of Accumulation of Discount Interest on Coupon book value Discount Book b  2.75% i  3.5% accumulated value 4742.23 137.50 165.98 28.48 4770.71 137.50 166.97 29.47 4800.18 137.50 168.01 30.51 4830.69 137.50 169.07 31.57 4862.26 137.50 170.18 32.68 4894.94 137.50 171.32 33.82 4928.76 137.50 172.51 35.01 4963.77 137.50 173.73 36.23 5000.00 $1100.00 $1357.77 $257.77

Quoted price  100,000 (1.02625)  $102,625 Principal  100,000; b  6.5% Average book value 

1 (100,000  102,625)  $101,312.50 2

Payment dates are July 15, January 15. The interest payment date preceding purchase is July 15, 2019. The time interval July 15, 2019, to July 15, 2026, is 7 years: n  14. Semi-annual interest  100,000 (0.065)  $6500 Total interest  14  6500  $91,000 Premium paid  102,625  100,000  $2625 Copyright © 2025 Pearson Canada Inc.

I/Y

CPT

Discount balance 257.77 229.29 199.82 169.31 137.74 105.06 71.24 36.23 —

Average income per interest payment interval

91,000  2625 88,375   $6312.50 14 14 Approximate value of i 

6312.50  0.062307 = 6.2307% 101,312.50

The approximate yield rate  2(0.062307)  0.124614 = 12.4614%

Face value  25,000; b  3% Principal  25,000; i  4% n  2(20  7)  26; b < i é1  1.0426 ù ú Discount  [25,000 (0.03)  25,000 (0.04)] ê ê 0.04 ú ë û é1- 1.04- 26 ù ú  (750  1000) ê ê 0.04 ú ë û

 (250)(15.982769)  $3995.69 Book value  25,000  3995.69  $21,004.31 Proceeds  25,000 (0.9825)  $24,562.50 Gain

on sale  24,562.50  21,004.31 

(Set P/Y  2; C/Y  2) 25,000 PV 21,004.31 9.

750

$3558.19

PMT

I/Y

Quoted price  10,000 (0.93875)  $9387.50 Principal  $10,000 Average book value 

1 (10,000  9387.50)  $9693.75 2

Interest payment dates are December 1, June 1. The interest payment date closest to purchase is June 1, 2017. The time interval June 1, 2017, to December 1, 2028, is 11.5 years: n  23. Semi-annual coupon  10,000 (0.06)  $600 Total interest  23(600)  $13,800 Bond discount  9387.50  10,000  $612.50 Average income per interest payment interval 

1 1 (13,800  612.50)  (14,412.50)  $626.63 23 23

Approximate value of i 

626.63  0.064642 = 6.4642% 9693.75

CPT

The approximate yield rate  2(0.064642)  0.129285 =

12.9285%

10. FV = 25,000, N = 6, i = 0.095 (―BGN mode‖) (Set P/Y = 1, C/Y = 1) Quit. 25,000 FV , 6 N , 9.5 I/Y , 0 PV , CPT PMT –2996.65 Annual payment to the fund equals $2996.65 Payment Interval

Interest for payment interval

number

Periodic payment

2996.65

284.68

3281.33

3,281.33

2996.65

596.41

3593.06

6,874.39

2996.65

937.75

3934.40

10,808.79

2996.65

1,311.52

4308.17

15,116.96

2996.65

1,720.79

4717.44

19,834.40

2996.65

2,168.95

5165.60

$25,000.00

i = 0.095

Increase in Fund

Balance in Fund at the End of Payment Interval 0

11. FV = 25,000, N = 6, i = 0.045 (―END mode‖) (Set P/Y = 1, C/Y = 1) Quit. 25,000 FV , 6 N , 4.5 I/Y , 0 PV , CPT PMT –3721.96 Annual payment to the fund equals $3721.96 Interest for payment interval

Balance in Fund at the End of Payment Interval

Payment Interval number

Periodic payment

3721.96

3,721.96

3721.96

167.49

3889.45

7,611.41

3721.96

342.51

4064.47

11,675.88

3721.96

524.42

4247.38

15,923.26

3721.96

716.55

4438.51

20,361.76

3721.96

916.28

4638.24

$25,000.00

i = 0.045

Increase in Fund

Challenge Problems 1.

Value of each annual coupon: Year 1

400

Year 2

75% of $400



300

Year 3

75% of $300



225

Year 4

75% of $225



$168.75

Year 5

75% of $168.75



$126.56

FV  2000; PMT  specific year coupon; n  5; i  10% Purchase price  2000 (1.105)  400 (1.101)  300 (l.l02)  225 (1.103)  168.75(1.104)  126.56(1.105)  2000 (0.620921)  400 (0.90)  300 (0.826446)  225 (0.751315)  168.75(0.683014)  126.56(0.620921)  1241.84  363.64  247.93  169.05  115.26  78.58  2216.30 The purchase price of the bond is $2216.30 2.

Consider a $1000 bond. Principal at 15% premium is $1150; b  4.5%; PMT  45; yield  1000i; the number of conversion periods  2n é1  (1  i ) 2 n ù ú 1150  1000  (45  1000i) ê ê ú i ë û é1  (1  i ) 2 n ù ú 150  (45  1000i) ê ê ú i ë û

1  (1  i)2 n 150  ← Equation 1 i 45  1000i Principal at 25% premium is $1250; b  5%; PMT  50; yield  1000i; the number of conversion periods  2n é1  (1  i ) 2 n ù ú 1250  1000  (50  1000i) ê ê ú i ë û

é1  (1  i ) 2 n ù ú 250  (50  1000i) ê ê ú i ë û

1  (1  i)2 n 250  ← Equation 2 i 50  1000i From Equation 1 and Equation 2

150 250  45  1000i 50  1000i 150(50  1000i)  250(45  1000i) 7500  150,000i  11,250  250,000i 100,000i  3750 i  0.0375 Since i is a semi-annual rate, j  2(0.0375)  0.075  7.5% The two offers are equivalent at 7.5% compounded semi-annually

Case Study 1.

Approximate market value of bonds is $1155.35 Assume each unit of bonds  $1000 Number of periods in bond life  20  2  40 Annual Coupon rate = (1.0275)2 1 = 0.05575625 Bond interest payments  (1000  0.05575625)/2  $27.88 (Set P/Y  2; C/Y  2) 40 N 4.4 FV CPT PV 1155.35

(a)

I/Y

27.88 

PMT

1000

The size of the semi-annual payment is $110,822.36

(Set P/Y  2; C/Y  2) 40 CPT PMT .

1.6

I/Y

5,200,000

110,822.36 (b)

Total deposited into the sinking fund  40(110,822.36)  $4,432,894.40

(a)

Balance of sinking fund after 10 years is $2,393,264.94

(Set P/Y  2; C/Y  2) 40 CPT PMT 110,822.36

1.6

I/Y

2nd PV P1  1 P2  20 BAL  2,393,264.94

5,200,000 

(b)

New sinking fund payments when fund earns 2.5% after January 21, 2030 is $94,479.78

(Set P/Y  2; C/Y  2) 20 5,200,000 ± (c)

(d)

CPT

N PMT

2.5

I/Y

2,393,264.57

94,479,78

Total amount deposited is $4,106,042.80 Year 110 payments (110,822.36  20)

2,216,447.20

Year 1120 payments (94,479.78  20)

1,889,595.60

Total amount deposited

4,106,042.80

The amount of interest in the sinking fund will be $1,093,957.20 (5,200,000  4,106,042.80  $1,093,957.20)

Chapter 16 Investment Decision Applications Exercise 16.1 A. 1.

Alternative 1: PV of $20,000 in three years  20,000(l.l23)  20,000(0.711780)  $14,236 PV of $60,000 in six years  60,000(1.126)  60,000(0.506631)  .30,398 PV of Alternative 1 

$44,634

Alternative 2: PV of $13,000 at the end of each of the next six years 1  1.126   13,000    13,000(4.111407)  $53,448  0.12 

Since PV of Alternative 2 > PV of Alternative 1, Alternative 2 is preferred at 12%. PROGRAMMED SOLUTION (Set P/Y  l; C/Y  l) Alt.1 0 PMT 14,236

20,000

I/Y

CPT

Alt.1 0 PMT 30,398

60,000

I/Y

CPT

PMT

I/Y

CPT

Alt.2 0 FV 53,448 2.

13,000

Alternative 1: PV of $50,000  50,000(1.154)  50,000(0.571753)  $28,588 PV of $40,000  40,000(l.l57)  40,000(0.375937)  15,037 PV of $30,000  30,000(1.1510)  30,000(0.247185)  7416 PV of Alternative 1 

51,041

Alternative 2: PMT  750; i  15%; n  120

c

1 , p  1.15  1  1.011715  1  1.1172% 12 1 12

1  1.011715120  PV of Alternative 2  750    750(64.26100)  $48,196  0.011715 

Since PV of Alternative 1 > PV of Alternative 2, Alternative 1 is preferred. Alt.1 (Set P/Y  l; C/Y  1) 0 CPT PV 28,588

PMT

50,000

I/Y

Alt.1 0 PMT 15,037

40,000

I/Y

CPT

Alt.1 0

30,000

I/Y

CPT

Alt.2 (Set P/Y  l2; C/Y  l) 0 CPT PV 48,196

750

PMT

I/Y

120

7416 N

Alternative 1: PV of $15,000 now  $15,000 PV of $20,000 in five years  20,000(l.02520)  20,000(0.610271)  12,205 PV of Alternative 1 

$27,205

Alternative 2: PV of $1500 at the end of every three months for five years 1  1.02520   1500    1500(15.589162)  $23,384  0.025 

In this case, ―the smaller the better‖ principle applies. Since PV of Alternative 2 < PV of Alternative 1, Alternative 2 is preferred.

Alt.1 (Set P/Y  l; C/Y  4) 0 CPT PV 12,205

PMT

20,000

Alt.2 (Set P/Y  4; C/Y  4) 0 CPT PV 23,384

1500

I/Y

PMT

I/Y

Alternative 1: PMT  2500; i  7%; n  14; c 

1 2

p  1.07 2  1  1.034408  1  3.4408% 1

1  1.03440814  PVn  2500    2500(10.964017)  $27,410  0.034408 

PV of $10,000 now  . 10,000 PV of Alternative 1  $10,000 + $27,410 = $37,410 Alternative 2: PMT  600; i  7%; n  84; c 

1 12

p  1.07 12  1  1.005654  1  0.5654% 1  1.005654 84  PVn(due)  600(1.005654)    0.005654 

 600(1.005654)(66.721382)  $40,259 For cost, since PV of Alternative 1 < PV of Alternative 2, Alternative 1 is preferred. Alt.1 (Set P/Y  2; C/Y  l) 0 FV CPT PV 27,410.03

2500

AIt.2 (Set P/Y  12; C/Y  1) (―BGN‖ Mode) 0 N CPT PV 40,259

PMT

I/Y

600

PMT

I/Y

Alternative 1: Pay cash now  $2000

Alternative 2: PV of $108 at the end of every month for 24 months 1  1.006524   85    85(22.1552)  $1883  0.0065 

In this case, ―the smaller the better‖ principle applies. Since PV of Alternative 2 < PV of Alternative 1, Alternative 2 is preferred. Alt.2 (Set P/Y  12; C/Y  12) 0 CPT PV 1883 6.

PMT

7.8

I/Y

Alternative 1: Pay cash now  $613

Alternative 2: PV of $26.99 at the end of every month for 30 months 1  1.013530   26.99    26.99 (24.534343)   0.0135 

$662

In this case, the ―smaller the better‖ principle applies Since PV of Alternative 1 < PV of Alternative 2, Alternative 1 is preferred. Alt.2 (Set P/Y  12; C/Y  12) 0 CPT PV 662 B. 1.

26.99

PMT

16.2

I/Y

$73,601

Alternative 1: PV of $25,000 now  $25,000 PV of $50,000  50,000(l.065)  50,000(0.747258)  . 37,363 PV of Alternative 1  $62,363 Alternative 2: 1  1.0610  PV of Alternative 2  10,000    10,000(7.360087)   0.06 

At 6%, Alternative 2 is preferred. Alt.1 (Set P/Y  1; C/Y  1) 0 CPT PV 37,363

PMT

Alt.2 0 FV 73,601

10,000

PMT

50,000

I/Y

CPT

Alternative 1: 1  1.0736  PV of Alternative 1  400,000    400,000(4.722714)   0.073  .

$1,889,085

Alternative 2:  $100,000

PV of $100,000 now

PV of $200,000  200,000(1.0732)  200,000(0.868562)  173,712 PV of $200,000  200,000(l.0733)  200,000(0.809470)  161,894 PV of $800,000  800,000(l.0734)  800,000(0.754399)  603,519 PV of $800,000  800,000(l.0735)  800,000(0.703075)  562,460 PV of $800,000  800,000(1.0736)  800,000(0.655242)  524,194 Total PV of Alternative 2

$2,125,779

At 7.3%, Alternative 2 is preferred.

Alt.1 (Set P/Y  1; C/Y  1) 0 FV CPT PV 1,889,085

400,000

PMT

7.3

I/Y

Alt.2 0 PMT 173,712

200,000

7.3

I/Y

CPT

Alt.2 0 PMT 161,894

200,000

7.3

I/Y

CPT

Alt.2 0 PMT 603,519

800,000

7.3

I/Y

CPT

Alt.2 0 PMT 562,460

800,000

7.3

I/Y

CPT

Alt.2 0 PMT 524,194

800,000

7.3

I/Y

CPT

Offer A: PV of $15,000 now  $15,000 PV of $20,000  20,000(1.083)  20,000(0.793832)  15,877 PV of Offer A

15,000 + 15,877 

$30,877

Offer B: PV of $3000 now

 $3000

PMT  3000; i  8%; n  12; c 

1 2

p  1.08 2  1  1.039231  1  0.039231 = 3.9231% 1  1.03923112  PVn  3000    3000(9.4271)  .$28,281  0.039231 



PV of Offer B = 28,281 + 3000

$31,281

At 8%, Offer B is preferred because they will receive a higher value for their business. .

A (Set P/Y  1; C/Y  1) 0 PV 15,876.65

PMT

B (Set P/Y  2; C/Y  1) 0 PV 28,281.35

20,000

3000

PMT

I/Y

CPT

Alternative 1:  $250,000

PV of $250,000 now

PV of $750,000  750,000(l.0641)  750,000(0.939849)  704,887 PV of $500,000  500,000(l.0642)  500,000(0.883317)  441,659 Total PV

$1,396,546

Alternative 2:  $600,000

PV of $600,000 now

PV of $300,000  300,000(l.0641)  300,000(0.939850)  281,955 PV of $600,000  600,000(l.0642)  600,000(0.883317)  529,990 Total PV

$1,411,945

At 6.4%, Alternative 2 is preferred.

Alt.1 (END Mode)-(Set P/Y  1; C/Y  1) 0 CPT PV 704,887.22 0 PMT 441,658.66

500,000

6.4

PMT

I/Y

PMT

600,000

6.4

I/Y

CPT

300,000

Alt.2 (END Mode)-(Set P/Y  1; C/Y  1) 0 CPT PV 281,954.89 PMT

750,000

CPT

6.4

I/Y

6.4

I/Y

529,990.39

Buying:  $90,000

PV of cash payment

Less PV of 30,000  30,000(l.0820)  30,000(0.214548)  . 6437 

PV of cost of buying

$83,563

Leasing: 1  1.0820  PVn(due)  10,000(1.08)    10,000(1.08)(9.818147)  $106,036  0.08 

Since PV of the cost of leasing > PV of cost of buying, the warehouse should be purchased. A (Set P/Y  1; C/Y  1) 0 PMT CPT PV 6436.45

30,000

I/Y

B (―BGN‖ Mode) 0 FV PV 106,035.99

PMT

I/Y

CPT

10,000

Buying: PV of buying  $16,500 Leasing: PV of lease payments PMT  360; i  3%; n  36; c 

2 12

1 6

p  1.03  1  1.004939  1  0.4939% 1  1.004939 36  PVn(due)  360(1.004939)    0.004939 

 360(1.004939)( 32.9069) 

$11,905

PV of buy back  6750 (l.036)  6750 (0.837484)  5653 PV of cost of leasing



$17,558

Since the PV of cost of buying < PV of cost of leasing, the car should be purchased. (Set P/Y  12; C/Y  2) (―BGN‖ Mode) 360 N CPT

PMT

6750

PMT

6750

I/Y

. 17,558.08

(Set P/Y  2; C/Y  2) 6 PV 5653.02

I/Y

Exercise 16.2 A. 1.

PV of inflows: PMT  3500; i  3%; n  28 1  1.0328  PVn  3500    3500(18.764108)  $65,674  0.03 

PV of outflows: Immediate outlay

 $50,000

PV of 30,000  30,000(l.0312)  30,000(0.701380)  . 21,041

CPT

 $71,041 Net present value  65,674  71,041

 $5367

Since NPV < 0, the investment should be rejected.

(Set P/Y  4; C/Y  4) 0 PV 65,674

3500

PMT

I/Y

PMT

30,000

. 12

I/Y

CPT

$84,770

CPT 21,041

PV of savings: PMT  3000; i  10%; n  48; c 

1 4

p  1.10 4  1  1.024114  1  2.4114% 1

1  1.024114 48  PVn  3000    3000(28.256524)   0.024114 

PV of costs: Immediate outlay

 $50,000

PV of 30,000  30,000(1.10)6  30,000 (0.564474)  . 16,934  $66,934

PV of outlays Net present value  84,770  66,934  $17,836 Since NPV > 0, the replacement should be made.

(Set P/Y  4; C/Y  1) 0 PV 84,770

PMT

I/Y

0 PMT

(Set P/Y  1; C/Y  1)30,000 PV 16,934 3.

3000

Alternative 1: PVIN  7000(l.08514)  7000(0.319142)  $2234 PVOUT

 . 2000

NPV in Alternative 1

 $234

Alternative 2:

CPT

1  1.08514  PVIN  250    250(8.010097)  $2003  0.085 

PVOUT

. 1800

NPV of Alternative 2

 $203

Since the net present value of Alternative 1 is greater than the net present value of Alternative 2, Alternative 1 is preferred. (Set P/Y  2; C/Y  2) 0 PV 2234

PMT

250

PMT

I/Y

7000

I/Y

CPT

2003

CPT

Alternative 1:

2 4

PMT  500; n  40; c  ; i  6% p  1.06 2  1  1.029563  1  2.9563% 1

1  1.02956340  PVIN  500    500(23.278933)  $11,639  0.029563 

PVOUT

 . 9000

NPV

 $2639

Alternative 2: PVIN  30,000(l.0620)  30,000(0.311805)  $9354 PVOUT  4000  5000 (l .064)  4000  5000(0.792094)  4000  3960

 . 7960  $1394

NPV

Since the net present value of Alternative 1 is greater than the net present value of Alternative 2, Alternative 1 is preferred.

(Set P/Y  4; C/Y  2) 0 PV 11,639

(Set P/Y  2; C/Y  2) 0 PV 3960

PMT

5000

I/Y

PMT

30,000

500

. PMT

I/Y

CPT

I/Y

CPT

9354

CPT

Alternative 1:

1  1.017512  PVIN  900    900(10.739550)  $9666  0.0175  PVOUT

 $8000

NPV

 $1666

Alternative 2:

1  1.017512  PVIN  300    300(10.739550)  $3222  0.0175  PVOUT  2000  1000(l.01758)  2000  1000(0.870412)  2000  870

 2870 

NPV

$352

Since the net present value of Alternative 1 is greater than the net present value of Alternative 2, Alternative 1 is preferred.

(Set P/Y  4; C/Y  4) 0 PV 9666

PMT

1000

I/Y

PMT

300

PMT

I/Y

900

PMT

I/Y

870

CPT N

CPT

3222

Alternative 1 (Buy the new car): PVOUT  20,000  4000  $16,000

1  1.007560  PVOUT  240(1  0.0075)    240(48.534674)  11,648  0.0075  

NPV

$27,648

Alternative 2 (Keep the old car):

1  1.007560  NPV  PVOUT  600(1  0.0075)    600(48.534674)   0.0075 

$29,121

Since the net present value of Alternative 1 is less than the net present value of Alternative 2, Alternative 1 is preferred. You should buy a new car.

(Set P/Y  12; C/Y  12; ―BGN‖ Mode) 0 CPT PV 11,648

240

PMT

I/Y

(Set P/Y  12; C/Y  12; ―BGN‖ Mode) 0 CPT PV 29,121

600

PMT

I/Y

1. Project A: PVIN 4000(1.124)  4000(0.635518)  $2542 9000(1.129)  9000(0.360610)  . 3245 Total PVIN

 $5787

 . 4000

PVOUT

 $1787

NPV Project B:

1  1.129  PVIN  1500    1500(5.328250)  $7992  0.12  PVOUT  4000  2000(1.123)  4000  2000(0.711780)  4000  1,424  . 5424  $2568

NPV

Project B is preferred at 12%.

(Set P/Y  1; C/Y  1)

PMT

4000

I/Y

CPT

2542

PMT

9000

I/Y

CPT

3245

PMT

2000

I/Y

CPT

1424

PMT

I/Y

CPT

7992

1500

Project 1: PVIN  8000(1.144)  8000(0.592080)  $4737 8000(1.1410)  8000(0.269744)

 . 2158

PVIN

 $6895

PVOUT

 . 5000

NPV

 $1895

Project 2

1 2

PMT  950; n  20; c  ; i  14% p  1.140.5  1  1.067708  1  6.7708%

1  1.06770820  PVIN  950    950(10.785388)  $10,246  0.067708  PVOUT  5000  5000(1.143)  5000  5000(0.674972)  5000  3,375  . 8375  $1871

NPV Project 1 is preferred at 14%.

(Set P/Y  1; C/Y  1) 0

PMT

8000

I/Y

CPT

4737

PMT

8000

I/Y

CPT

2158

PMT

5000

I/Y

CPT

3375

(Set P/Y  2; C/Y  1) 0 PV 10,246

950

PMT

I/Y

CPT

End of year, numbers on the time graph are in thousands of dollars.

PMT  33,000; n  12; i  14%; d  3

1  1.1412  PVIN  33,000(l.l4 )    0.14  3

 33,000(0.674972)(5.660292) PVOUT

 $126,078

 $60,000

60,000 now

50,000(1.141)  50,000(0.877193) 

43,860

40,000(1.142)  40,000(0.769468) . 30,779

 134,639

 $8561

NPV

Since NPV < 0, the project will not return 14% on the investment and therefore should not be undertaken. (Set P/Y  1; C/Y  1)

PMT

50,000

I/Y

CPT

43,860

PMT

40,000

I/Y

CPT

30,779

PMT

I/Y

CPT

186,790

PMT

I/Y

CPT

126,078

33,000 186,790

End of year, numbers on the time graph are in thousands of dollars.

PVIN : PMT  15,000; n  5; i  12%

1  1.125  15,000    15,000(3.604776)  0.12 

 $54,072

PMT  10,000; n  10; d  5; i  12%

1  1.1210  10,000(1.125)    10,000(0.567427)(5.650223)  . 32,061  0.12   $86,133

PVIN PVOUT: 60,000 now  $60,000 60,000(1.125)  60,000(0.567427)

 . 34,046

PVOUT

 $94,046

NPV

 $7913

Since NPV < 0, the project will not provide the required return on investment of 12% and therefore should not be undertaken. . (Set P/Y  1; C/Y  1)

PMT

I/Y

CPT

34,046

15,000

PMT

I/Y

CPT

54,072

10,000

PMT

I/Y

CPT

56,502

I/Y

CPT

32,061

60,000

56,502

End of year, numbers on the time graph are in thousands of dollars.

PVIN: PMT  15,000; n  4; i  20%

1  1.24  15,000    15,000(2.588735)  0.2 

 $38,831

PMT  10,000; n  3; d  4; i  20%

1  1.23  10,000(1.24)    10,000(0.482253)(2.106482)  10,159  0.2   $48,990 PVOUT: 36,000 now

 $36,000

10,000(l.23)  10,000(0.578704)



5787

10,000(l.25)  10,000(0.401878)



4019



9000 l.2–7

  9000  0.279082

 <2512>

PVOUT

 $43,294

NPV

 $ 5696

Since NPV > 0, the new product provides the required return on investment of 20% and therefore the new product should be distributed. (Set P/Y  1; C/Y  1)

PV 38,831

15,000

PMT

I/Y

CPT

10,000

PMT

I/Y

CPT

21,065

PMT

21,065

I/Y

CPT

10,159

PMT

10,000

I/Y

CPT

5787

PMT

10,000

I/Y

CPT

4019

PMT

9,000



I/Y

CPT

2512

End of year, numbers on the time graph are in thousands of dollars.

PVIN: 60,000(1.164)  60,000(0.552291)

 $33,138

20,000(1.165)  20,000(0.476113)



9522

 $42,660

1  1.164  PVOUT: 15,000(1.16)    15,000(1.16)(2.798181)  $48,688  0.16 

l0,000  l.l6–5   10, 000  0.476113

 <4761>

PVOUT

 $43,927

NPV

 $1267

Since NPV < 0, the project will not return 16% on the investment and therefore should not be undertaken.

(Set P/Y  1; C/Y  1) (―BGN‖ Mode)

FV 15,000

CPT

PV 48,688

CPT

PMT

I/Y

PMT

10,000 

I/Y

PMT

60,000

I/Y

CPT

33,137

PMT

20,000

I/Y

CPT

9522

PV 4761

End of year, numbers on the time graph are in thousands of dollars.

1  1.1210  PVIN: 30,000    30,000(5.650223)  0.12 

 $169,507

PVOUT: 140,000 now

$140,000

20,000(1.124)  20,000(0.635518)

12,710

40,000(l.127)  40,000(0.452349)

18,094

20,000  l.12–10   20,000  0.321973 

<6440>

164,364 

NPV

$5143

Since NPV > 0, the investment will return more than 12% on the investment and should therefore be made. (Set P/Y  l; C/Y  l) (―END‖ Mode)

PMT

I/Y

CPT

 169,507

20,000

I/Y

CPT

 12,710

PMT

40,000

I/Y

CPT

 18,094

PMT 6439

20,000



CPT

PMT

0 0

30,000

I/Y

End of year, numbers on the time graph are in thousands of dollars.

1  1.186  PVIN: 80,000    80,000(3.497603)  0.18 

 $279,808

PVOUT: 250,000 now

 $250,000

1  1.185  20,000    20,000(3.127171)  0.18 



40,000 1.18–6   40,000  0.370432 

 <14,817>

62,543 

297,726

 $17,918

NPV

Since NPV < 0, the component would not produce a return of 18% on the investment



and therefore should not be marketed. (Set P/Y  l; C/Y  l) (―END‖ Mode) 0

80,000

PMT

I/Y

CPT

279,808

20,000

PMT

I/Y

CPT

62,543

PMT 40,000 14,817

CPT



I/Y

Option A Now

<20K>

13K

<2K> 13K

18K

14K <3.5K> 14K

NPVA  20  13 1.15  2 1.15   13 1.15   14 1.15   3.5 1.15  2

3

4

5

14 1.15  18 1.15   18 1.15   5 1.15  7

8

9

10

6

 18.89495

Therefore, the NPV for option A is 18.89495*1000 = $18,895 Option B Now

<19K> <19K> <19K>

16K

29K

 1  1.152   1  1.156  10 3 NPVA  19  19   16   1.15  29 1.15   2.90637  .15   .15  Therefore, the NPV for option B is 2.90637*1000 = $2906 .

Because option A has a positive NPV, it is the preferred option. Business Math News Box 1.

P/Y = 1, C/Y =2, PMT = $100,000, I/Y = 2%, N = 41

c /y 2 p /y p  (1  i)c  1  (1  0.01) 2  1  0.0201 1  (1  p)  n  1  (1  0.0201) 41  pv  pmt    100, 000    $2,774,979.44 p 0.0201     c

The present value of $100,000 per year for life would be $2,774,979.44. 2.

He has to pay tax on his annuity after he has received $2,000,000, which is after 20 years of winning the lottery. Therefore, he would be 61, when he is required to pay tax on his winning annuity.

(Set P/Y = 1 and C/Y = 1) 0 CPT

1,500,000 PV I/Y 6.072

100,000 

PMT

At an interest rate of 6.072% compounded annually the two alternatives are equivalent. 4.

Under normal conditions, people at younger age have higher probability of living longer than older people. Mr. Schwager should have accepted the $100,000 per year for life if he was 20 years old at the time of winning the lottery. He should have accepted the cash payout if he was 65.

Exercise 16.3 1.

End of year, numbers on the time graph are in thousands of dollars.

For i  14%

For i  20%

For i  18%

23,084

20,833

21,546

26,999

23,148

24,345

35,525

28,935

30,947

60,000(1  i)

25,968

20,094

21,855

50,000(1  i)5

9112

6698

7409

20,000(1  i)6

120,688

99,708

106,102

100,000

$20,688

$292

$6102

PVIN 30,000(1  i)2 40,000(1  i)3 4

PVIN PVOUT NPV

d 6102  2 6102  292 d

12,204  1.91% 6394

R.O.I.  18%  1.9%  19.9% CF

CFo  100,000 

C01  0 F01  1 C02  30,000 F02  1 C03  40,000 F03  1 C04  60,000 F04  1 C05  50,000 F05  1 C06  20,000 F06  1 2.

IRR

CPT

 19,905

End of year, numbers on the time graph are in thousands of dollars.

Try i  20%

For i  16%

For i  14%

PVIN

1  (1  i) 10  12,000   i  

50,310

57,999

62,593

PVOUT

60,000

NPV

$9690

$2001

$2593

d 2593  2 2593  2001 d

5186  1.13% 4594

R.O.I.  14%  1.1%  15.1% CF

CFo  60,000 

C01  12,000 F01  10

IRR

CPT

 15,098

End of year, numbers on the time graph are in thousands of dollars.

For i  16% PVIN

For i  20%

For i  18%

4492

4213

4349

1  (1  i) 5  5000 (1  i)   i  

10,489

8653

9516

1  (1  i) 4  3000 (1  i) 8   i  

2561

1806

2147

17,542 15,000

14,672 15,000

16,012 15,000

2000(1  i) 12

<337>

<224>

<274>

PVOUT NPV

14,663 $2879

14,776 $104

14,726 $1286

1  (1  i) 3  2000   i   3

PVOUT

d

PVIN 15,000 now

2572 2  1286   1.85% 1286  104 1390

R.O.I.  18%  1.85%  19.85% CF

CFo  15,000 

C01  2000 F01  3 C02  5000 F02  5 C03  3000 F03  3 C04  5000 F04  1

IRR

CPT

 19.84

End of year, numbers on the time graph are in thousands of dollars.

PVIN

PVOUT

d

Try i  16%

For i  14%

<8621>

<8,772>

103,975

112,100

64,527

74,039

30,000(1  i ) 9

159,881 150,000 22,092 <7889>

177,367 150,000 23,683 <9225>

PVOUT NPV

164,203 $4322

164,458 $12,909

10,000  (1  i ) 1

1  (1  i) 4  50,000 (1  i) 2   i   3 1  (1  i)  70,000 (1  i) 6   i   PVIN 150,000 now 40,000 (1  i) 4

2  12,909 25,818   1.50% 12,909  4322 17, 231

R.O.I.  14%  1.5%  15.5% CF

CFo  150,000 

C01  10,000  F01  1 C02  0 F02  1 C03  50,000 F03  1 C04  10,000 F04  1 C05  50,000 F05  2 C06  70,000 F06  2 C07  100,000 F07  1

IRR

CPT

 15.47

End of year, numbers on the time graph are in thousands of dollars.

PVIN

PVIN PVOUT

Try i  16%

For i  26%

For i  22%

For i  24%

2000(1  i ) 1

<1724>

<1587>

<1639>

<1613>

2000 (1  i) 2

1486

1260

1344

1301

1  (1  i) 13  6000 (1  i) 2   i  

23,821

13,815

16,942

15,267

1  (1  i) 10  3000 (1  i ) 15   i  

1565

325

16,000 now

25,148 16,000

13,813 16,000

8000(1  i)10

1813

PVOUT NPV

d

17,813 16,793 $7335 $2980

296 2  148   0.18 148  1538 1686

R.O.I.  22%  0.2%  22.2% CF

793

CFo  16,000 

C01  2000  F01  1 C02  2000 F02  1 C03  6000 F03  7 C04  2000  F04  1 C05  6000 F05  5 C06  3000 F06  10

IRR

CPT

 22.16

596 17,243 16,000

438

15,393 16,000

1095 17,095 16,931 $148 $1538

931

End of year, numbers on the time graph are in thousands of dollars.

Try i  16%

For i  18%

1  (1  i) 5  40,000   i  

130,972

125,087

1  (1  i) 10  70,000 (1  i) 5   i  

161,081

137,508

292,053

262,595

220,000 now

220,000

1  (1  i) 3  30,000 (1  i )   i  

78,157

76,969

80,000(1  i ) 15

<8634>

<6681>

PVOUT

289,523

290,288

NPV

$2530

$27,693

PVIN

PVIN PVOUT

d

5060 2  2530   0.17% 2530  27,693 30,223

R.O.I.  16%  0.17%  16.17% CF

CFo  250,000 

C01  10,000 F01  2 C02  40,000 F02  3 C03  70,000 F03  9 C04  150,000 F04  1

IRR

CPT

 16.15

Option A, numbers on the time graph are in thousands of dollars. Now

<25K>

<2K> 3K

<3.5K> 4K

As shown below, enter the periodic cash flow in cells B2 to B12 and in cell B13 enter the IRR function (i.e., IRR(B2:B12) and press enter. A

IRR Function in EXCEL

Initial cash outflow

Cash inflow period 1

Cash inflow period 2

Cash inflow period 3

Cash inflow period 4

Cash inflow period 5

Cash inflow period 6

Cash inflow period 7

Cash inflow period 8

Cash inflow period 9

Cash inflow period 10

Internal Rate of Return

–25,000 0 3000 –2000 3000 4000 –3500 4000 8000 8000 5000 2.21%

Option B, numbers on the time graph are in thousands of dollars. Now

<7K> <7K> <7K>

1 2 3 4 5 6 7 8 9 10 11 12 13

IRR Function in EXCEL

–7000 –7000 –7000 0 5000 5000 5000 5000 5000 5000 9000 10.65%

Initial cash outflow Cash inflow period 1 Cash inflow period 2 Cash inflow period 3 Cash inflow period 4 Cash inflow period 5 Cash inflow period 6 Cash inflow period 7 Cash inflow period 8 Cash inflow period 9 Cash inflow period 10 Internal Rate of Return

Both options have an internal rate of return that is smaller than the company’s 13 percent expected rate. Therefore, none of these two options is acceptable. Review Exercise

Alternative A: 20,000(1.143)  20,000(0.674972)  $13,499 60,000(l.l46)  60,000(0.455587) 

27,335

40,000(l.l410)  40,000(0.269744) 

10,790

PV of Alternative A:

$51,624

Alternative B: 1  1.1410  PV of Alternative B  10,000    10,000(5.216116)  $52,161  0.14 

Since the PV of Alternative B > PV of Alternative A, .

Alternative B is preferred.

Alt.1 (Set P/Y  1; C/Y  1) 0 CPT PV 13,499

PMT

20,000

Alt.1 0 PMT 27,335

60,000

I/Y

CPT

Alt.1 0 PMT 10,790

40,000

I/Y

CPT

Alt.2 0 FV 52,161

10,000 

I/Y

PMT

I/Y

CPT

Alternative 1: $25,000 now  $25,000 37,500(l.0756)  37,500(0.647962)  24,299 50,000(l.07510)  50,000(0.485194)  24,260 PV of Alternative 1:

$73,559

Alternative 2: PMT  5150; i  7.5%; n  20; c 

1 2

p  1.075 2  1  1.036822  1  0.036822 = 3.6822% 1  1.036822 20  PVn(due)  5150(1.036822)    0.036822 

 5150(1.036822)(13.98092)  $74,653

Since in the case of costs, smaller values are preferred, .

Alternative 1 is preferred.

Alt.1 (Set P/Y  2; C/Y  2) 0 CPT PV 24,299

PMT

37,500

Alt.1 0 PMT 24,260

I/Y

Alt.2 (Set P/Y  4; C/Y  2) (―BGN‖ Mode) 0 20 N CPT PV 74,653

5150 

50,000

CPT

Alternative 1: 1  1.0336  PVIN  500    500(21.832253)  0.03 

 $10,916

PVOUT



NPV of Alternative 1

 $3916

7000

Alternative 2 PVIN  26,000 (1.0332 )  26,000(0.388337)

 $10,097

PVOUT



6500

NPV of Alternative 2



$3597

I/Y

PMT

I/Y

Since the NPV of Alternative 1 is greater than the NPV of Alternative 2, Alternative 1 is preferred. . Alt.1 (Set P/Y  4; C/Y  4) 0 CPT PV 10,916

500

PMT

Alt.2 0 PMT 10,097

I/Y

26,000

12 N

I/Y CPT

N PV

PV of savings: PMT  8000; i  14%; n  20; c 

1 2

p  1.14 2  1  1.067708  1  6.7708% 1  1.06770820  PVn  8000    8000(10.785405)  $86,283  0.067708 

PVOUT Immediate cost

$65,000

40,000(1.145)  40,000(0.519369)  20,775

85,775

NPV = 86,283  85,775

$

508

Since at 14% the NPV of the replacement is positive, be replaced. .

the old equipment should

(Set P/Y  2; C/Y  1) 0 PV 86,283

I/Y

(Set P/Y  1; C/Y  1) 0 PV 20,775

PMT

8000

PMT

40,000

I/Y

CPT

End of year, numbers on the time graph are in thousands of dollars.

PVIN PMT  250,000; i  18%; n  15; d  8 1  1.1815  PVn(defer)  250,000(1.18 )    0.18  8

 250,000(0.266038)(5.091578)

 $338,638

PVOUT PMT  75,000; i  18%; n  8 1  1.188  PVn(due)  75,000(1.18)    0.18 

 75,000(1.18)(4.077566)

 360,865

 $22,227

NPV At 18%, the net present value is $22,227. (Set P/Y  1; C/Y  1) (―BGN‖ Mode) 0 CPT PV 360,865

75,000

PMT

(Set P/Y = 1; C/Y =1) (―END‖ Mode) 0 N CPT PV 1,272,894

250,000

PMT

1,272,894

I/Y

PMT

I/Y

338,638

CPT

N 15

OR: CF CFo  75,000  C01  75,000  F01  7 C02  0 F02  1 C03  250,000 F03  15 6.

NPV

I  18

CPT

NPV  $22,226

End of year, numbers on the time graph are in thousands of dollars.

PVIN 60,000(1.144)  60,000(0.592080)

 $35,525

40,000(l.l45)  40,000(0.519369)

 20,775

1  1.145  20,000(1.145)    20,000(0.519369)(3.433081)  0.14 

 . 35,661

1  1.144  PVOUT30,000(1.14)    30,000(1.14)(2.913712)  0.14 

 $99,649

30,000 1.1410   30, 000(0.269744)



<8092>

NPV



$91,557 $ 404

CPT

At 14%, the net present value is $404. (Set P/Y  1; C/Y  1) 0 PV 35,525 0

PMT

40,000 20,000 68,662

PMT

60,000

I/Y

CPT

20,775

PMT

I/Y

CPT

68,662

I/Y

CPT

35,661

(―BGN‖ Mode) 0 99,649 0

PMT CF

30,000 

30,000

PMT

I/Y

NPV

I  14

I/Y

CPT

CFo  30,000 

OR: C01  30,000  F01  3 C02  60,000 F02  1 C03  40,000 F03  1 C04  20,000 F04  4 C05  50,000 F05  1

CPT

NPV  404

PV  8092

PVIN PVOUT NPV

d

1  (1  i) 8  14,000   i  

45,000

Try i  20%

For i  26%

For i  28%

53,720

45,370

43,061

45,000 $8720

45,000 $370

45,000 $1939

740 370  2   0.32% 2309 370  1939

R.O.I.  26%  0.3%  CF

26.3%

IRR

CPT

CFo  45,000 

C01  14,000 F01  8

 26.31

PVIN PVOUT

1  (1  i) 12  2000   i   10,000 4000(1  i ) 12

PVOUT NPV

d

Try i  20%

For i  18%

8878

9586

10,000 <449>

10,000 <549>

9551 $673

9451 $135

270 2  135   0.33% 808 135  673

R.O.I.  18%  0.3%  CF

18.3%

CFo  10,000 

C01  2000 F01  11 C02  6000 F02  1

IRR

CPT

 18.32

PVIN

PVOUT

For i  16%

338,639

425,164

360,865

377,892

$22,226

$47,272

1  (1  i) 15  250,000(1  i)8   i   1  (1  i) 8  75,000(1  i)   i  

NPV

d

For i  18%

47, 272  2 94,544   1.36% 47, 272  22, 226 69, 498

R.O.I.  16%  1.4%  CF

17.4%

CFo  75,000 

C01  75,000  FO1  7 C02  0 F02  1 C03  250,000 F03  15

IRR

CPT

 17.29

10.

PVIN

60,000(1  i)4

For i  14% 35,525

For i  16% 33,137

40,000(1  i)5

20,775

19,045

35,661

31,179

91,961

83,361

1  (1  i) 4  30,000(1  i)   i  

99,649

97,377

30, 000(1  i ) 10

<8092>

<6801>

91,557 $404

90,576 $7215

20,000(1  1  (1  i) 5   i)5  i  

PVOUT

NPV

d

808 404  2   0.11 404  7215 7619

R.O.I.  14%  0.1  CF

14.1%

CFo  30,000 

. C01  30,000  F01  3

OR:

C02  60,000 F02  1 C03  40,000 F03  1 C04  20,000 F04  4 C05  50,000 F05  1

IRR

CPT

 14,098

11. End of year, numbers on the time graph are in thousands of dollars.

PVIN 8000(1  i)1 1 1  (1  i )

12,000(1  i)

 

5

  

1  (1  i) 4  6000(1  i)6   i  

PVOUT 36,000 now 9000(1  i) 10

For i  20% 6667 29,906 5202

For i  26% 6349 25,096 3479

For i  24% 6452 26,568 3968

41,775 36,000 <1454>

34,924 36,000 <892>

36,988 36,000 <1047>

34,546 $7229

35,108 $184

34,953 $2035

NPV d

4070 2035  2   1.83 2219 2035  184

R.O.I.  24%  1.83%  CF

25.83%

CFo  36,000 

C01  8000

F01  1

C02  12,000

F02  5

C03  6000

F03  3

C04  15,000

F04  1

IRR

CPT

 25.83

12. End of year, numbers on the time graph are in thousands of dollars.

1  (1  i) 4  PVIN 8000   i   1  (1  i) 6  4   10,000(1  i) i  

PVOUT

For i  12% 24,299

For i  14% 23,310

16,037

26,129

23,024

36,747 50,428 46,334 60,000 60,000 60,000 <6460> <12,879> <10,790> 53,540 47,121 49,210 $3307 $16,793 $2876

60,000 now 40, 000(1  i ) 10

NPV

d

For i  20% 20,710

6614 3307  2   1.07 6183 3307  2876

R.O.I.  12%  1.07%  CF

13.07%

CFo  60,000 

C01  8000 F01  4 C02  10,000 F02  5 C03  50,000 F03  1

IRR

CPT

 13.03

13. Project A: 1  1.208  5800    5800(3.837160)  $22,256  0.20 

Project B: 13,600(1.201)  13,600(0.83)

 $11,333

17,000(l.205)  17,000(0.401878)



20,400(l.208)  20,400(0.232568)

 . 4744

6832

 $22,909 Since at 20%, the PV of Project B is greater than the PV of Project A, Outway Ventures should choose Project B. A. (Set P/Y  1; C/Y  1) (―END‖ Mode) 0 N CPT PV 22,256

. FV

5800

PMT

I/Y

B. 0

PMT

13,600

I/Y

CPT

11,333

B. 0

PMT

17,000

I/Y

CPT

6832

B. 0

PMT

20,400

I/Y

CPT

4744

14. Project A: PVIN 4000(1.122)  4000(0.797194)

 $ 3189

12,000(1.124)  12,000(0.635518)



7626

8000(1.126)  8000(0.506631)



4053 $14,868

PVOUT 8000 now  8000 6000 (1.123)  6000(0.71178)



4271 $12,271

 $ 2597

NPV of Project A Project B:

1  (1.12 –7 )  PVIN 3400    3400(4.56375654)  0.12 

 $15,517

PVOUT 4000 now

$4000

6000(1.122)  6000(0.797194)

4783

4000(1.124)  4000(0.635518) = 2542

11,325  $ 4192

NPV of Project B

Since at 12%, the NPV of Project B is greater than the NPV of Project A, Project B should be preferred. A. (Set P/Y  1; C/Y  1) (―END‖ Mode) 4000 N CPT PV 3188.78 12,000 8000

FV FV

0 0

PMT PMT

(―BGN‖ Mode) 6000 PV 4270.68 B. 0

3400

PMT

12 12

I/Y I/Y

4 6

PMT

I/Y

PMT

CPT

CPT I/Y

PV N

CPT

I/Y

7626.22 4053.05 CPT

15,516.77

(―BGN‖ Mode) 6000 PV 4783.16

PMT

I/Y

CPT

(―BGN‖ Mode) 4000 PV 2542.07

PMT

I/Y

CPT

CFo  8000 

C01  0 F01  1 C02  4000 F02  1 C03  6000  F03  1 C04  12,000 F04  1 C05  0 F05  1 C06  8000 F06  1 B.

NPV

I  12

CPT

NPV  2597.36

I  12

CPT

NPV  4191.54

CFo  4000 

C01  3400 F01  1 C02  2600  F02  1 C03  3400 F03  1 C04  600  F04  1 C05  3400 F05  3

NPV

15. End of year, numbers on the time graph are in thousands of dollars.

1  1.20 –12  PVIN 13,000    13,000(4.439217)  0.20 

 $57,710

PVOUT 50,000 now

$50,000

30,000(l.206)  30,000(0.334898)

10,047

10,000 1.2012   (10,000(0.112157))

<1122>

58,925

NPV = $57,710  $58,925

 $1215

(Set P/Y  1; C/Y  1) (―END‖ Mode) 0 N CPT PV 57,710

13,000

PMT

(―BGN‖ Mode) 30,000 10,047

I/Y

CPT

PMT

I/Y

12 PV

(―BGN‖ Mode) 10,000  PV  1122 CF

PMT

I/Y

CPT

CFo  50,000 

C01  13,000 F0l  5 C02  17,000  F02  1 C03  13,000 F03  5 C04  23,000 F04  1 16.

NPV

I  20

CPT

NPV  1216

1  (1  i) 10  PVIN 12,500   i  

PVOUT 45,000 NPV

2046 1023  2   2713 1023  1690 0.75% d

R.O.I.  24%  0.8%  CF

24.8%

For i  16% 60,415

For i  24% 46,023

For i  26% 43,310

45,000 $15,415

45,000 $1023

45,000 $1690

CFo  45,000 

C01  12,500 F01  10

IRR

CPT

 24.73

17. End of year, numbers on the time graph are in thousands of dollars.

PVIN 64,000(1.162)  64,000(0.743163)

$ 47,562

256,000(1.163)  256,000(0.640658)

164,008

128,000(1.164)  128,000(0.552291)

70,693

32,000(1.165)  32,000(0.476113)

15,236 $297,499

1  1.16 –5  PVOUT 32,000(1.16)    32,000(1.16)(3.274294) $121,542  0.16 

64,000(1.161)  64,000(0.862069)

55,172

96,000(l.l62)  96,000(0.743163)

71,344

32,000(l.l63)  32,000(0.640658)

20,501 $268,559

NPV = $297,499  $268,559

$ 28,940

Since at 16%, the NPV is positive, the return on investment will be greater than 16% and

the product should be marketed.

(Set P/Y  1; C/Y  1) 64,000 PV 47,562

. 0

PMT

I/Y

CPT

256,000

PMT

I/Y

CPT

164,008

128,000

PMT

I/Y

CPT

70,693

32,000

PMT

I/Y

CPT

15,236

PMT

CPT

(―BGN‖ Mode) 0 FV PV 121,542

32,000

(―BGN‖ Mode) 64,000 55,172

0 PMT

(―BGN‖ Mode) 96,000 PV 71,344

(―BGN‖ Mode) 32,000 FV 0 20,501 CF

I/Y

PMT

I/Y

PMT

I  16

CPT

CFo  32,000 

C01  96,000  F0l  1 C02  64,000  F02  1 C03  192,000 F03  1 C04  96,000 F04  1 C05  32,000 F05  1

NPV

NPV  28,941

CPT N

CPT CPT

18. End of year, numbers on the time graph are in thousands of dollars.

1  (1  i) 12  PVIN 100,000(1  i)3   i  

For i  16%

For i  14%

1  (1  i) 3  PVOUT 80,000(1  i)   i  

332,957

382,054

208,419

211,733

240,000(1  i)3

153,758

161,993

200,000 1  i 

<21,585> 340,592 $7635

<28,019> 345,707 $36,347

15

NPV d

72, 694 36,347  2   1.65% 36,347  7635 43,982

R.O.I.  14%  1.65%  15.65% CF

CFo  80,000 

C01  80,000  F0l  2 C02  240,000  F02  1 C03  100,000 F03  11 C04  300,000 F04  1

IRR

CPT

 15.62%

19. Investing in new machinery option Now

<20K> <3K> 5K

10K

Step 1: First option, purchase machinery: Enter the initial cash flow, then the net cash flow for each year. Note that all of the cash flows are in thousands of dollars. CFo = -20

Cash flow at beginning

C01 = -3

F01 = 1

Net cash flow at end of Year 1

C02 = 5

F02 = 3

Net cash flow at end of Year 2-4

C03 = 3

F03 = 1

Net cash flow at end of Year 5

C04 = 8

F04 = 1

Net cash flow at end of Year 6

C05 = 0

F05 = 1

Net cash flow at end of Year 7

C06 = 10

F06 = 2

Net cash flow at end of Year 8-9

C07 = 2

F07 = 1

Net cash flow at end of Year 10

Step 2: Press IRR and CPT, which results in IRR = 13.8785% Outsourcing option

<7K> <2K> <2K> <2K> <2K> <2K> 5K

12K

Now

Step 1: Second option, the outsourcing option: Enter the initial cash flow, then the net cash flow for each year. Note that all of the cash flows are in thousands of dollars. CFo = -7

Cash flow at beginning

C01 = -2

F01 = 5

Net cash flow at end of Year 1-5

C02 = 5

F02 = 4

Net cash flow at end of Year 6-9

C03 = 12

F03 = 1

Net cash flow at end of Year 10

Step 2: Press IRR and CPT, which results in IRR = 9.87245% Comparing the IRR for each option, we see that investing in new machinery provides a higher return on investment. Therefore, it is the preferred option for the firm. Self-Test 1.

Present value of Alternative A: 1  1.15–12  2500    2500(5.420619)  $13,552  0.15 

Present value of Alternative B: 10,000(1.15)4  10,000(0.571753) 

$5718

10,000(1.15)8  10,000(0.326902) 

3269

10,000(1.15)12  10,000(0.186907) 

1869 $10,856

Programmed Solution A. (Set P/Y  1; C/Y  1)0 FV CPT PV 13,552

2500

PMT

I/Y

B. 10,000

PMT

I/Y

CPT

5718

10,000

PMT

I/Y

CPT

3269

10,000

PMT

I/Y

CPT

1869

I  15

CPT

At 15%, PV(A) > PV(B). Alternative A is preferred.

Cash Flow analysis A.

CFo  0

C01  2500 F01  12 B.

NPV

CFo  0

C01  0 F01  3 C02  10,000 F02  1 C03  0 F03  3

NPV  13,552

C04  10,000 F04  1 C05  0 F05  3 C06  10,000 F06  1

NPV

11  15

CPT

NPV  10,856

End of year, numbers on the time graph are in thousands of dollars.

1  1.14 –14  2 PVIN  60,000   (1.14) 0.14  

 60,000(6.002072)(0.769468)  360,124.32(0.769468)

 $277,104.14

PVOUT 100,000 now

 $100,000

1  1.14 –4  50,000    50,000(2.9137123)  145,685.62  0.14 

$245,685.62 NPV 

$31,418.52

(Set P/Y  1; C/Y  1) 0 FV 60,000 CPT PV 360,124.29

PMT

I/Y

328,052 FV 0 

PMT

I/Y

CPT

PMT

I/Y

CPT

NPV

I  14

FV CF

50,000



CFo  100,000 

C01  50,000  F01  2 C02  10,000 F02  2 C03  60,000F03  12 3.

CPT

NPV  31,418.33

End of year, numbers on the time graph are in thousands of dollars.

1  (1  i) 10  PVIN 20,000   i  

PVOUT 100,000 now

For i  12%

For i  18%

For i  16%

113,004

89,882

96,665

100,000 <9659>

100,000 <5732>

100,000 <6801>

30,000 1  i  NPV

10

90,341

94,268

93,199

$22,663

$4386

$3466

d

6932 3466  0.88% 2  7852 3466  4386

R.O.I.  16%  0.88%  CF

16.88%

CFo  100,000 

C01  20,000 F01  9 C02  50,000 F02  1 4.

 16.85

IRR

Present value of Purchase: Purchase price

$6600

Less disposable value: 1200(1.01)60

661

PV of purchase:

$5939

Lease Immediate payment

$1500

1  1.01–60  Monthly payments  100(1.01)     0.01 

4540

PV of lease:

$6040

Purchase: (Set P/Y  12; C/Y  12) 1200 CPT PV 661

PMT

I/Y

Lease: (Set P/Y  12; C/Y  12; ―BGN‖ Mode) 0 N CPT PV 4540

Since PV of purchase is less than PV of lease, .

the system should be purchased.

100

PMT

CFo  6600 

Purchase: C01 0 F01  59 C02  1200 F02  1 CF

NPV

I  1.0

CPT

NPV  5939

CFo  1600 

C01   100 F01  59

NPV

I  1.0

CPT

NPV  6040

I/Y

Proposal A: 1  1.210  PVIN 20,000    0.2 

$83,849

PVOUT Immediate outlay

$60,000

After 3 years: 40,000(l.2)3

23,148

83,148

NPV(A)= $83,849  $83,148

701

Proposal B: 1  1.27  3 PVIN 40,000   (1.20) 0.2  

$83,440

1  1.24  PVOUT 29,000(1.2)    0.2 

$90,088

Less 50,000(1.2)10

<8075>

82,013

NPV(B)= $83,440  $82,013

$1427

A. (Set P/Y  l; C/Y  l) 0 FV CPT PV  83,849 (―BGN‖ Mode) 40,000  23,148 B. 0

144,184

20,000

PMT 20

40,000

PMT

I/Y

CPT

 144,184

PMT

I/Y

CPT

 83,440

PMT

I/Y

29,000

50,000  8075

PMT

I  20

CPT

PV 

Proposal B is preferred at 20%.

CPT

C02  20,000  F02  1 NPV

I/Y

CFo  60,000 

C03  20,000 F03  7

CPT

C01  20,000 F0l  2

PMT

(―BGN‖ Mode) 0 90,088

I/Y

Since NPV(B) is greater than NPV(A), A.

CPT

CFo  29,000 

NPV  701

C0l   29,000 F0l  3 C02  40,000 F02  6 C03  90,000 F03  1

NPV

I  20

CPT

NPV  1427

End of year For i  16%

For i  22%

For i  26%

For i  28%

1  1  i  4  75,000   i  

209,864

187,023

174,014

168,073

1  1  i  3  4 50,000   (1  i) i  

62,019

46,093

38,156

34,802

PVIN

271,883

233,116

212,170

202,875

PVOUT Now 180,000

180,000

50,000(1  i)3

32,033

27,535

24,995

23,842

50,000(1  i)5

23,806

18,500

15,744

14,552

<45,000(1  i)7>

<15,922> <11,186>

<8925>

<7994>

PVIN

PVOUT

219,917

214,849

211,814

210,400

NPV

$51,966

$18,267

$356

$7525

d

712 356 2  0.09% 7881 356  7525

R.O.I.  26%  0.1%  CF

26.1%

CFo  180,000 

C01  75,000 F01  2 C02  25,000 F02  1 C03  75,000 F03  1 C04  0 F04  1 C05  50,000 F05  1 C06  95,000 F06  1

IRR

CPT

 26.09%

Challenge Problems 1.

Purchase the conveyor only: 1  1.1410  NPV at 14%  17,000    85,000  88,674  85,000  3674  0.14 

Purchase the conveyor with loader: 1  1.1410  NPV at 14%  22,000    114,000  114,755  114,000  755  0.14 

Although both systems have a positive net present value at 14% and yield a return of more than the required 14%, purchase of the system without a loader has higher NPV and is preferable. Note that the loader itself has a negative net present value (see calculation below). The owners should investigate if they can obtain a better yield on the additional $29,000. Loader only (initial cost of $29,000 that will generate additional revenue of $5000 per year): 1  1.1410  NPV at 14%  5000    29,000  26,081  29,000  2919  0.14 

Present value of cash flows at 12%: PV of Project A  150,000(l.l21)  120,000(l.l22)  120,000(l.l23)  133,929  95,663  85,414  $315,006 PV of Project B  40,000(l.l22)  200,000(l.l23)  200,000(l.l24)  200,000(l.l25)  31,888  142,356  127,104  113,485  $414,833 PV of Project C  10,000(l.l21)  10,000(l.l22)  100,000(l.l23)  120,000(1.124)  120,000(l.125)  8929  7972  71,178  76,262  68,091  $232,432 1  1.124  PV of Project D  30,000(1.121)  40,000(1.121)    0.12 

 26,786  108,477  $135,263 NPV of Project A  315,006  300,000  $15,006 NPV of Project B  414,833  360,000  $54,833 NPV of Project C  232,432  210,000  $22,432

NPV of Project D  135,263  125,000  $10,263 The possible investment combinations, if $700,000 is available, are: (a) Project A & B Total cost  300,000  360,000  $660,000 Total PV of cash flows  315,006  414,833  $729,839 Total NPV  729,839  660,000  $69,839 Profitability index 

729,839 100   110.6 660, 000

(b) Project A & C & D Total cost  300,000  210,000  125,000  $635,000 Total PV of cash flows  315,006  232,432  135,263  $682,701 Total NPV  682,701  635,000  $47,701 Profitability index 

682, 701 100   107.5 635, 000

(c) Project B & C & D Total cost  360,000  210,000  125,000  $695,000 Total PV of cash flows  414,833  232,432  135,263  $782,528 Total NPV  782,528  695,000  $87,528 Profitability index 

782,528 100   112.6 695, 000

The combination of Projects B, C, and D has the highest investment amount, the highest net present value and the highest profitability index at 12%. This combination will yield the highest rate of return. The owners should select the combination of Projects B, C, and D. Case Study 1.

Cost of buying the car including taxes  37,500  1.13  $42,375

(a) Leasing cost at 7.8%: Three-step process: STEP 1

Calculate the present value of lease payments. Assume that tax is paid on monthly lease payments. Value of payments (594  1.13)  $671.22 (Set P/Y12; C/Y12) 48 N CPT PV 27,600.47

7.8

I/Y

671.22 

PMT

0 FV

STEP 2 Calculate the present value of buyout option. Assume that tax is paid on the buyout option. Value of payment (16,155  1.13)  $18,255.15 48 N 7.8 13,375.93

I/Y

PMT

 18,255.15

CPT

STEP 3 Include down payment of $1330. Total  $27,600.47  $13,375.93  $1330  $42,306.40

At the 7.8% financing rate, leasing the car is cheaper than financing it from Honda.

(b) Leasing cost at 8%: STEP 1 Calculate present value of lease payments. Assume that tax is paid on monthly lease payments. Value of payments (594 1.13)  $671.22 48 N 8 27,494.46

I/Y

671.22 

PMT

CPT

STEP 2 Calculate present value of buyout option. Assume that tax is paid on the buyout option. Value of payment (16,155 1.13)  $18,255.15 48 N 8 13,270.04

I/Y

PMT

18,255.15 

CPT

STEP 3 Include down payment of $1330. Total  $27,494.46  $13,270.04  $1330  $42,094.50 At the 8% financing rate, leasing the car is cheaper than buying it.

(a) Purchase price of car, net of down payment,  ($37,500  1.13)  4000  $38,375 (b) Leasing cost at 3.8% (assume lease option still requires $1330 down payment): STEP 1 Calculate present value of lease payments. Assume that tax is paid on monthly lease payments. Value of payments (594  1.13)  $671.22 48 N 3.8 29,845.71

I/Y

671.22 

PMT

CPT

STEP 2 Calculate present value of buyout option. Assume that tax is paid on the buyout option. Value of payment (16,155  1.13)  $18,255.15 48 N 3.8 15,684.73

I/Y

18,255.15 

PMT

CPT

STEP 3 Include down payment of $1330. Total  $29,845.71  $15,684.73  1330  $46,860.44 At the 3.8% financing rate, with a down payment of $1330, purchasing the car for $42,375 is cheaper than leasing it. Solution to Comprehensive Case (Lux Resources Inc.) 1. CHAPTERS 9 and 10

a. I/Y  6; P/Y  12;C/Y  12; i 

6%  0.5%  0.005 12

E1  35,000(1.005)-12  35,000(0.941905)  $32,966.69

(Set P/Y  12, C/Y = 12) 35,000 FV 6 I/Y 12 N CPT PV  32,966.69 E2  50,000(1.005)24  50,000(0.887186)  $44,359.28

(Set P/Y  12, C/Y = 12) 50,000 FV 6 I/Y 24 N CPT PV  44,359.28 E3  30,000(1.005)36 30,000

(0.835645)  $25,069.35

(Set P/Y  12, C/Y = 12) 30,000 FV 6 I/Y 36 N CPT PV  25,069.35 E1  E2  E3  32,966.69  44,359.28  25, 069.35  $102,395.32

b. 102,395.32  x(1.005) 6  x(1.005) 18

102,395.32  0.970518 x  0.914136 x 102,395.32  1.884654 x x  54,331.09

(Set P/Y  12, C/Y = 12)1 FV 6 I/Y 6 N CPT PV  0.970518

(Set P/Y  12, C/Y = 12)1 FV 6 I/Y 18 N CPT PV  0.914136 Each of the equal payments is $54,331.09.

2. CHAPTERS 11 and 12

a. PV  230,000; i  7.5%; n  24; c 

1  0.25 4

p  1.0750.25  1  1.018245  1  1.8245% 1  1.01824524  230,000  PMT    0.018245  230,000  PMT(19.295488) PMT  $11,919.89

Set P/Y  4;C/Y  1 230,000  PV 0 FV 7.5 I/Y 24 N CPT PMT 11,919.89 b. PV  230,000; i  7.5%; c 

1  0.25; n  5 4

p  1.0750.25  1  1.018245  1  1.8245% FV  230,000(1.018245)5  $251,760.98

Set P/Y  4;C/Y  1 230,000  PV 0 PMT 7.5 I/Y 5 N CPT FV 251,760.98 c. PMT  11,919.89; i  7.5%; c 

1  0.25; n  5 4

1.0182455  1  FV  11,919.89    11,919.89[5.185805]  $61,814.23  0.018245 

Set P/Y  4;C/Y  1 0 PV 11,919.89 PMT 7.5 I/Y 5 N CPT FV 61,814.23 d. $251, 760.98  61,814.23  $189,946.75

Set P/Y  4;C/Y  1 230, 000  PV 11,919.89 PMT 7.5 I/Y 5 N CPT FV 189,946.75 e. PV  189,946.75; i 

5.5% 12  0.4583%  0.004583; n  72  15  57; c   1 12 12

p  1.0045831  1  0.4583% 1  1.00458357  189,946.75  PMT    0.004583  189,946.75  50.062220 PMT PMT  $3794.21

Set P/Y  12;C/Y  12  0 FV 189,946.75  PV 5.5 I/Y 57 N CPT PMT 3794.21 57 payments of $3794.21 are required in the renegotiated term to repay the loan

3. CHAPTER 13

a. PV  450,000; i  7.0%; c 

1  8.3333%  0.08333; n  36; d  8 12

p  1.07 12  1  0.5654%  0.005654 1

FV(defer)  450,000 (1.0056548)  450,000 (1.046138)  $470,762.32 (Set P/Y  12; C/Y  1) 450,000 PV 0 PMT 7 I/Y 8 N CPT PV  470,762.32 b. PV  470,762.32; PMT  8000; p  0.005654

é1- 1.005654- n ù ú 470,762.32  8000 ê ê 0.005654 ú ë û 470,762.32  1,414,891.102(1  1,005654n) 0.332720  (1  1.005654n) 1.005654n  0.667280

n ln 1.005654  ln 0.667280 n 0.005638  0.404546 n

0.404546 = 71.75 monthly payments 0.005638

.(Set P/Y  12; C/Y  1) 0 FV 470,762.32  PV 8000 PMT 7 I/Y CPT N 71.750520 72 month-end payments will have to be made c. 10,000 =

PMT = $500 per year 0.05

4. CHAPTER 14

a. PV  360,000; i  2.0%; n  240; c =

2 1 = 12 6

1 6

p = (1.02) - 1 = 0.3306% = 0.003306 é1- 1.003306- 240 ù ú 360,000 = PMT ê ê 0.003306 ú ë û 360,000 = 165.495383 PMT PMT = $2175.29

(Set P/Y  12; C/Y  2) 360,000  PMT 2175.29 b.

I/Y

240

CPT

FV = 360, 000(1.003306)12 = 374,544 æ1.00330612 - 1÷ ö ÷ FV12 = 2175.29 çç = 26,583.37 ÷ çè 0.003306 ÷ ø Balance  374,544  26,583.37  $347,960.63 (Set P/Y  12; C/Y  2) 360,000  CPT FV 347,960.63

I/Y

2175.29 PMT

c. (2175.29  12)  (360,000  347,960.63)  26,103.48  12,039.77  $14,063.71 2nd Amort P11; P2  12 BAL  347,960.63; PRN  12,039.37; INT  14,064.11* *(Rounding difference) d.

FV = 360, 000(1.003306)36 = 405, 418.47 æ1.00330636 - 1ö ÷ ÷ FV36 = 2175.29 çç ÷= 83, 015.41 çè 0.003306 ø ÷ Balance  405,418.47  83,015.41  $322,403.06 (Set P/Y  12; C/Y  2) 360,000  CPT FV 322,403.06

e. PV  322,403.06; i  2.4%; n  204; c =

I/Y

2175.29 PMT

2 1 = 12 6

p = (1.024) 6 - 1 = 0.3961% = 0.003961 é1- 1.003961- 204 ù ú 322, 403.06 = PMT ê ê 0.003961 ú ë û 322, 403.06 = 139.757558 PMT PMT = $2306.87 (Set P/Y  12; C/Y  2) 322,403.06  PMT 2306.87

4.8

5. CHAPTER 15

a. $500, 000´ .06 = $30, 000

æ1  1.05410 ö ÷ ÷ b. 500, 000(1.054- 10 )  30, 000 çç ÷ çè 0.054 ø ÷ 500, 000(0.591009)  30, 000(7.573913) 295,504.36  227, 217.38  $522, 721.74

I/Y

204

CPT

(Set P/Y  1; C/Y  1) 500,000 CPT PV 522,721.74 Price =

5.4

I/Y

30,000

PMT

522,721.74 ´ 100 = 104.5443 500,000

c. The bonds are being sold at a premium of $522,721.74  500,000  $22,721.74. End of Interest Payment Interval 0 1 2 3 4 5 6 7 8 9 10 TOTALS

Bond Interest b  6%

Interest on Book Value at Yield Rate I  5.4%

Amount of Premium Amortized

$30,000.00 30,000.00 30,000.00 30,000.00 30,000.00 30,000.00 30,000.00 30,000.00 30,000.00 30,000.00 $300,000.00

$28,226.97 28,131.23 28,030.32 27,923.95 27,811.85 27,693.69 27,569.15 27,437.88 27,299.53 27,153.70 $277,278.26

$1773.03 1868.77 1969.68 2076.05 2188.15 2306.31 2430.85 2562.12 2700.47 2846.30 $21.74

Book Value of Bond $522,721.74 520,948.71 519,079.94 517,110.26 515,034.21 512,846.06 510,539.75 508,108.90 505,546.78 502,846.30 500,000.00

Remaining Premium $22,721.74 20,948.71 19,079.94 17,110.26 15,034.21 17,222.36 10,539.75 8108.90 5546.78 2846.30 0

6. CHAPTER 16

Note: All dollar results rounded to the nearest dollar. i  10% Loader A a. Initial cash outflow: $150,000

Cash inflows: Years 13: $30,000; Year 4: $40,000; Years 58: $30,000; Year 9: $14,000; Year 10: $14,000  20,000  $34,000 Initial cash outflow $ 150,000 PV of $30,000 at the end of each of the next 3 years

é1- 1.10- 3 ù ú  30,000(2.486852)  $74,606  30,000 ê ê 0.10 ú ë û PV of $40,000 in 4 years  40,000(l.l04)  40,000(0.683013)  $27,321 PV of $30,000 at the end of each of the next 4 years Copyright © 2025 Pearson Canada Inc.

é1- 1.10- 4 ù ú  30,000(3.169865)  $95,096  30,000 ê ê 0.10 ú ë û PV of $95,096 in 4 years  95,096(l.l04)  95,096(0.683013)

 $64,952

PV of $14,000 in 9 years  14,000(l.l09)  14,000(0.424098)

 $ 5937

PV of $34,000 in 10 years  34,000(l.l010)  34,000(0.385543)  $13,108 

NPV of Loader A

$35,924

Calculator: Using TVM: Initial cash outflow

$150,000

(P/Y, C/Y  1) 30,000  PV

PMT

I/Y

CPT 74,606

(P/Y, C/Y  1) 0 PV

40,000 

I/Y

CPT 27,321

(P/Y, C/Y  1) 30,000  PV

PMT

I/Y

CPT 95,096

(P/Y, C/Y  1) 0 PV

95,096 

I/Y

CPT 64,952

(P/Y, C/Y  1) 0 PMT 14,000  FV 10 I/Y 9 N CPT PV

5937

(P/Y, C/Y  1) 0 PMT 34,000  FV 10 I/Y 10 N CPT PV

13,108

PMT

$ 35,924 b. Calculator: Using Cash Flow, NPV, and IRR:

CF CFo  150,000  C01  30,000 F01  3 C02  40,000 F02  1 C03  30,000 F03  4 C04  14,000 F04  1 C05  34,000 F05  1 NPV I  10; NPV  $35,924 IRR CPT  15.487% IRR for Loader A is 15.48% c. Profitability Index =

185,924´ 100 = 123% 150,000

Profitability index for Loader A is 123%

Loader B a. Initial cash outflow: $ 180,000

Cash inflows: Years 12: $20,000; Years 35: $40,000; Years 69: $30,000; Year 10: $30,000  11,000  $41,000 PV of $20,000 at the end of each of the next 2 years

é1- 1.10- 2 ù ú  20,000(1.735537)  $34,711  20,000 ê ê 0.10 ú ë û PV of $40,000 at the end of each of the next 3 years

é1- 1.10- 3 ù ú  40,000(2.486852)  $99,474  40,000 ê ê 0.10 ú ë û PV of $99,474 in 2 years  99,474(l.l02)  99,474(0.826446)  $82,210 PV of $30,000 at the end of each of the next 5 years

é1- 1.10- 5 ù ú  30,000(03.790787)  $113,724  30,000 ê ê 0.10 ú ë û PV of $113,724 in 5 years  113,724(l.l05)  113,724(0.620921)  $70,614 PV of $11,000 in 10 years  11,000(l.l010)  11,000(0.385543)

 $ 4241

NPV of Loader B  $11,776 .11 Calculator: Using TVM: Initial cash outflow

$ 180,000

(P/Y, C/Y  1) 20,000  PMT 0 FV 10 I/Y 2 N CPT PV

34,711

(P/Y, C/Y  1) 40,000  PMT 0 FV 10 I/Y 3 N CPT PV

99,474

(P/Y, C/Y  1) 0 PMT 99,474  FV 10 I/Y 2 N CPT PV

82,210

(P/Y, C/Y  1) 30,000  PMT 0 FV 10 I/Y 5 N CPT PV

113,724

(P/Y, C/Y  1) 0 PMT 113,724  FV 10 I/Y 5 N CPT PV

70,614

(P/Y, C/Y  1) 0 PMT 11,000  FV 10 I/Y 10 N CPT PV

4241 $ 11,776

b. Calculator: Using Cash Flow, NPV, and IRR:

CF CFo  180,000  C01  20,000 F01  2 C02  40,000 F02  3 C03  30,000 F03  4 C04  41,000 F04  1 NPV I  10; NPV  $11,776 IRR CPT  11.401% IRR for Loader B is 11.401% c. Profitability Index =

191,776´ 100 = 106.5% 180,000

Profitability index of Loader B is 106.5% d. Since NPV of Loader A > NPV of Loader B, Loader A is preferred at 10%.

Since IRR of Loader A > 10%, Loader A is acceptable. Since IRR of Loader B > 10%, Loader B is acceptable, but IRR of Loader A is higher; therefore Loader A is preferred at 10%.