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C H A P T E R
6 Financial arithmetic How do we determine the new price when discounts or increases are applied? How do we determine the percentage discount or increase applied, given the old and new prices? How do we determine the old price, given the new price and the percentage discount or increase? What do we mean by simple interest, and how is it calculated? What do we mean by compound interest, and how is it calculated? How do we calculate appreciation and depreciation of assets? What is hire-purchase, and how is the interest rate determined?
6.1
Percentage change In many financial situations you will need to calculate percentages. Suppose that the price of an item is discounted, or marked down, by 10%. This means that you will pay a reduced price. The amount of the reduction or discount is:
and
Discount = 10% of original price = 0.10 × original price New price = 100% of old price − 10% of old price = 90% of old price = 0.90 × old price
In general, if r% discount is applied: r Discount = × original price 100 New price = original price − discount =
209
(100 − r ) × original price 100
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Example 1
Calculating the discount and the new price
a How much is saved if a 10% discount is offered on an item marked $50.00? b What is the new discounted price of this item? Solution a Evaluate the discount. b Evaluate the new price by either: r subtracting the discount from the original price, or r directly calculating 90% of the original price.
Discount =
10 × $50.00 = $5.00 100
New price = original price − discount = $50.00 − $5.00 = $45.00 90 × $50.00 = $45.00 New price = $ 100
Sometimes, prices are increased, or marked up. If a price is increased by 10%:
and
Increase = 10% of original price = 0.10 × original price New price = 100% of old price + 10% of old price = 110% of old price = 1.10 × old price
In general, if r % increase is applied: r × original price Increase = 100 New price = original price + increase =
Example 2
(100 + r ) × original price 100
Calculating the increase and the new price
a How much is added if a 10% increase is applied to an item marked $50.00? b What is the new increased price of this item? Solution a Evaluate the increase. b Evaluate the new price by either: r adding the increase to the original price, or r directly calculating 110% of the original price.
Increase =
10 × 50.00 = $5.00 100
New price = original price + increase = 50.00 + 5.00 = $55.00 110 × 50.00 = $55.00 New price = 100
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Calculating the percentage change Given the original price and the new price of an item, we can work out the percentage change. To do this, the amount of the decrease or increase is determined and then converted to a percentage of the original price. Percentage discount =
discount 100 × % original price 1
Percentage increase =
increase 100 × % original price 1
Example 3
Calculating the percentage discount or increase
a If the price of an item is reduced from $50 to $45, what percentage discount has been applied? b If the price of an item is increased from $50 to $55, what percentage increase has been applied? Solution a 1 Determine the amount of the discount. 2 Express this amount as a percentage of the original price. b 1 Determine the amount of the increase. 2 Express this amount as a percentage of the original price.
Discount = original price − new price = 50.00 − 45.00 = $5.00 Percentage discount =
100 5.00 × = 10% 50.00 1
Increase = new price − original price = 55.00 − 50.00 = $5.00 Percentage increase =
100 5.00 × = 10% 50.00 1
Calculating the original price Sometimes we are given the new price and the percentage increase or decrease (r%), and asked to determine the original price. Since we know that: (100 − r ) × original price 100 (100 + r ) for an increase: New price = × original price 100 we can rearrange these formulas to give rules for determining the original price, as follows: for a discount:
New price =
When r % discount has been applied: When r % increase has been applied:
100 (100 − r ) 100 Original price = new price × (100 + r )
Original price = new price ×
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Example 4
Calculating the original price
Suppose that Cate has a $50 gift voucher from her favourite shop. a If the store has a ‘10% off’ sale, what is the original value of the goods she can now purchase? b If the store raises its prices by 10%, what is the original value of the goods she can now purchase? Solution a Substitute new price = 50 and r = 10 into the formula for a r% discount. b Substitute new price = 50 and r = 10 into the formula for a r% increase.
100 = $55.56 90 100 Original price = 50 × = $45.45 110 Original price = 50 ×
Exercise 6A 1 Calculate the amount of the discount in each of the following, to the nearest whole cent. a c e g
24% discount on $360 6% discount on $9.60 12% discount on $86.90 2.6% discount on $900
b d f h
72% discount on $250 9% discount on $812 19% discount on $38 14.2% discount on $1650
2 Calculate the amount of the increase in each of the following, to the nearest whole cent. a c e g
1.08% increase on $26 000 3 14 % increase on $1520 9 12 % increase on $18 650 0.2% increase on $10 000
b 15.9% increase on $4760 d 12 12 % increase on $9460 f 2.8% increase on $1 000 000
3 Calculate the following as percentages, correct to 2 decimal places. a $19.56 of $400 d $24 of $2600 g 38c/ of $10.50
b 60c/ of $2 e 30c/ of 90c/ h 25c/ of $186
c $6.82 of $24 f $1.50 of $13.50
4 Calculate the new increased price for each of the following. a c e g
$260 marked up by 12% $42.50 marked up by 60% $42.80 marked up by 35% $258 marked up by 14.2%
b d f h
$580 marked up by 8% $5400 marked up by 17% $9850 marked up by 8% $1960 marked up by 17.4%
5 Calculate the new discounted price for each of the following. a $2050 discounted by 9% c $154 discounted by 82% e $980 discounted by 13.5%
b $11.60 discounted by 4% d $10 600 discounted by 3% f $2860 discounted by 8%
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g $15 700 discounted by 22.7% i $674 discounted by 12%
213
h $147 discounted by 2.8%
6 Find the original prices of the items that have been marked down as follows. a b c d
Marked down by 10%, now priced $54.00 Marked down by 25%, now priced $37.50 Marked down by 30%, now priced $50.00 Marked down by 12.5%, now priced $77.00
7 Find the original prices of the items that have been marked up as follows. a b c d
Marked up by 20%, now priced $15.96 Marked up by 12.5%, now priced $70.00 Marked up by 5%, now priced $109.73 Marked up by 2.5%, now priced $5118.75
8 Mikki has a card that entitles her to a 7.5% discount at the store where she works. How much will she pay for boots marked at $230? 9 The price per litre of petrol is $1.15 on Friday. When Rafik goes to fill up his car, he finds that the price has increased by 2.3%. If his car holds 50 L of petrol, how much will he pay to fill the tank?
6.2
Simple interest When you borrow money, you have to pay for the use of that money. When you invest money, someone else will pay you for the use of your money. The amount you pay when you borrow, or the amount you are paid when you invest, is called interest. There are many different ways of calculating interest. The simplest of all is called, rather obviously, simple interest. Simple interest is a fixed percentage of the amount invested or borrowed and is calculated on the original amount. Suppose we invest $1000 in a bank account that pays simple interest at the rate of 5% per annum. This means that, for each year we leave the money in the account, interest of 5% of the original amount will be paid to us. That is, Interest, I = $1000 Ă—
5 = $50 100
If the money is left in the account for several years, the interest will be paid each year. To calculate simple interest we need to know: the amount of the investment, the interest rate and the length of time for which the money is invested.
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Example 5
Calculating simple interest
How much interest will be earned if we invest $1000 at 5% interest per annum for 3 years? Solution 1 Calculate the interest for the first year. 2 Calculate the interest for the second year. 3 Calculate the interest for the third year. 4 Calculate the total interest.
5 = $50 100 5 Interest = 1000 × = $50 100 Interest = 1000 ×
5 = $50 100 Interest for 3 years = 50 + 50 + 50 = $150 Interest = 1000 ×
The same principles apply when the simple interest is applied to a loan rather than an investment.
The simple interest formula Since the amount of simple interest earned is the same every year, we can apply a general rule. amount invested × interest rate (per annum) × length of time (in years) 100 P ×r ×t Prt or I = = 100 100 where the amount invested or borrowed ($P) is known as the principal, r% is the interest rate per annum and t is the time in years. Interest =
How does this relationship look graphically? Suppose we were to borrow $1000 at 5% per annum simple interest for a period of years. A plot of interest against time is shown.
500 Interest ($)
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400 300 200 100 0 1 2 3 4 5 6 7 8 9 10 Year
From this graph we can see that the relationship is linear. The amount of interest paid is directly proportional to the time for which the money is borrowed or invested. The slope, or gradient, of a line that could be drawn through these points is numerically equal to the interest rate. To determine the amount of the investment, the interest is added to the amount invested. Amount of the investment Prt 100 where $P is the amount invested or borrowed, r% is the interest rate per annum and t is the time in years. A=P+I =P+
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If the money is invested for more or less than 1 year, the amount of interest payable is proportional to the length of time for which it is invested. Example 6
Calculating simple interest for periods other than one year
Calculate the amount of simple interest that will be paid on an investment of $5000 at 10% simple interest per annum for 3 years and 6 months. Solution Apply the formula with P = $5000, r = 10% and t = 3.5 (since 3 years and 6 months is equal to 3.5 years). Example 7
I=
10 Prt = 5000 × × 3.5 100 100 = $1750
Calculating the total amount borrowed or invested
Find the total amount owed on a loan of $16 000 at 8% per annum simple interest at the end of 2 years. Solution 1 Apply the formula with P = $16 000, r = 8% and t = 2 to find the interest. 2 Find the total owed by adding the interest to the principal.
8 Prt = 16 000 × ×2 100 100 = $2560 A = P + I = 16 000 + 2 560 = $18 560
I=
The graphics calculator enables us to investigate simple interest problems using both the tables and graphing facilities of the calculator. How to solve simple interest problems using a graphics calculator How much interest is earned if $10 000 is invested at 8.25% simple interest for 10 years? Steps 1 Substitute P = $10 000 and r = 8.25% in the formula for interest rate. 2 Press o to display the Y= editor.
I=
Prt 10 000 × 8.25 × t = 100 100 = 825t
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3 Because the calculator expects formulas to be defined in terms of X and Y, we will enter this rule as shown. Y1 represents I and X represents t.
4 Select 7 to view a table of values. By scrolling down the table we can see the total interest earned after each year.
After 10 years, I = $8250 5 Note that a graph of the simple interest function can also be constructed using the graphics calculator. Select p and set both Xmin and Ymin to 0. Then press q, followed by 0:ZoomFit, to construct the graph shown here. Select r to find values from the graph. The interest rate is equal to the slope of this line.
Interest on bank accounts One very useful application of simple interest is in the calculation of the interest earned on a bank account. When we keep money in the bank, interest is paid. The amount of interest paid depends on: the rate of interest the bank is paying and the amount on which the interest is calculated. Generally, banks will pay interest on the minimum monthly balance, which is the lowest amount the account contains in each calendar month. When this principle is used, we will assume that all months are of equal length, as illustrated in the following example. Example 8
Calculating interest on a bank account
The following table shows the entries in Tom’s bank book. Date 30 June 3 July 15 July 1 August
Transaction Pay Cash Cash
Debit
Credit 400.00
50.00 100.00
Total 400.00 350.00 450.00 450.00
If the bank pays interest at a rate of 3% per annum on the minimum monthly balance, find the interest payable for the month of July.
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Solution The minimum balance in the account for July was $350.00 Prt 3 1 I= = 350 × × 100 100 12 = $0.88 or 88 cents
1 Determine the minimum monthly balance for July. 2 Determine the interest payable for July.
Exercise 6B 1 Calculate the amount of simple interest for each of the following: Principal
Interest rate
Time
a
$400
5%
4 years
b
$750
8%
5 years
c
$1000
8 years
d
$1250
e
$2400
7 12 % 10 14 % 12 34 %
15 years
f
$865
15%
2 12 years
g
$599
10%
6 months
h
$85.50
22.5%
9 months
i
$15000
33 13 %
1 14 years
3 years
2 Calculate the amount to be repaid for each of the following: Principal
Interest rate
Time
b
$500 $780
5% 6 12 %
4 years 3 years
c
$1200
7 14 %
6 months
d
$2250
10 34 %
8 months
e
$2400
12%
250 days
a
3 A sum of $10 000 was invested in a fixed term account for 3 years. Calculate: a the simple interest earned if the rate of interest is 6.5% per annum. b the total value of the investment at the end of 3 years. 4 A personal loan of $20 000 is taken out for a period of 5 years, at a simple interest rate of 12% per annum. Find the amount still owing after 1 year, if payments of $100 are made every month of that year.
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5 A loan of $1200 is taken out at a simple interest rate of 14.5% per annum. How much is owing after 3 months? 6 A company invests $1 000 000 in the short-term money market at 11% per annum simple interest. How much interest is earned by this investment in 30 days? Give your answer to the nearest cent. 7 A building society offers the following interest rates for its cash management accounts. Interest rate (per annum) on term (months) Balance 1–<3 3–<6 6–<12 12–<24 24–<36 $20 000–$49 999 2.85% 3.35% 3.85% 4.35% 4.85% $50 000–$99 999 3.00% 3.50% 4.00% 4.50% 5.00% $100 000–$199 999 3.40% 3.90% 4.40% 4.90% 5.40% $200 000 and over 4.00% 4.50% 5.00% 5.50% 6.00% Using this table, find the interest earned by each of the following investments. Give your answers to the nearest cent. a $25 000 for 2 months c $37 750 for 18 months e $74 386 for 8 months
b $125 000 for 6 months d $200 000 for 2 years f $145 000 for 23 months
8 An account at a bank is paid interest of 4% per annum on the minimum monthly balance, credited to the account at the beginning of the next month. In October, the following transactions took place: 7 October 12 October
$1000 withdrawn $500 deposited
If the opening balance for October was $5000: a what was the balance of the account at the end of October? b how much interest was paid for the month? 9 The minimum monthly balances for four consecutive months are: $240.00
$350.50
$478.95
How much interest is earned over the 3-month period if it is calculated on the minimum monthly balance at a rate of 3.5% per annum? 10 The bank statement below shows transactions for a savings account that earns simple interest at a rate of 4.5% per annum on the minimum monthly balance. Date 1 March 15 March 31 March 1 April
Transaction Cash Cash
How much interest was earned in March?
Debit
Credit 250.00 250.00
Balance 500.00 750.00 1000.00 1000.00
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11 The bank statement below shows transactions over a 3-month period for a savings account that earns simple interest at a rate of 3.75% per annum on the minimum monthly balance. Date 1 March 8 April 21 May 1 June
Transaction
Debit
Cash Cash
Credit 250.00 250.00
Balance 650.72 900.72 1150.72 1150.72
a What were the minimum monthly balances in March, April and May? b How much was earned over this 3-month period?
6.3
Rearranging the simple interest formula The formula for simple interest can be rearranged to find any one of the variables when the values of the other three variables are known. Then either the formula for simple interest, or the 7 facility of a graphics calculator, may be used to determine the unknown value.
Calculating the interest rate To find the interest rate per annum, r%, given the values of P, I and t: r=
100I Pt
where $P is the principal, $I is the amount of interest and t is the time in years. Example 9
Calculating the interest rate
Find the rate of simple interest if a principal of $8000 increases to $11 040 in 4 years. Solution 1 Find the interest earned on the investment.
Interest, I = 11 040 − 8000 = $3040
2 Apply the formula with P = $8000, I = $3040 and t = 4 to find the value of r.
r =
3 Since the unit of time was years, the interest rate can be written as the interest per annum.
Interest rate = 9.5% per annum
100I 100 × 3040 = Pt 8000 × 4 = 9.5%
Calculating the time period To find the number of years or term of an investment, t years, given the values of P, I and r: 100I t= Pr where $P is the principal, $I is the amount of interest and r% is the interest rate per annum.
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Example 10
Calculating the time period of a loan or investment
Find the length of time it would take for $5000 invested at an interest rate of 12% per annum to earn $1800 interest. Solution Apply the appropriate formula with P = $5000, I = $1800 and r = 12 to find the value of t.
t=
100 × 1 800 100I = Pr 5 000 × 12 = 3 years
Calculating the principal To find the value of the principal, $P, given the values of I, r and t: P=
100I rt
where $I is the amount of the interest, r% is the interest rate per annum and t is the time in years. To find the value of the principal, $P, given the values of A, r and t: A P= rt 1+ 100 where $A is the amount of the investment, r% is the interest rate per annum and t is the time in years.
Example 11
Calculating the principal of a loan or investment
a Find the amount that should be invested in order to earn $1500 interest over 3 years at an interest rate of 5% per annum. b Find the amount that should be invested at an interest rate of 5% per annum if you require $15 600 in 4 years time. Solution a Since we are given the value of the interest, I, we will use the first formula with I = $1500, r = 5 and t = 3 to find the principal, P. b Here we are not given the value of the interest, I, but the value of the total investment, A. We will use the second formula with A = $15 600, r = 5 and t = 4 to find the principal, P.
P=
100 × 1500 100I = rt 5×3 = $10 000
15 600 A = P= rt 5×4 1+ 1+ 100 100 15 600 = $13 000 = 1.2
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Exercise 6C 1 Calculate the time taken for $2000 to earn $975 at 7.5% simple interest. 2 Calculate the principal that earns $514.25 in 10 years at 4.25% simple interest. 3 Calculate the rate of simple interest if a principal of $5000 amounts to $6500 in 2 12 years. 4 Calculate the principal that earns $780 in 100 days at 6.25% per annum simple interest. 5 Calculate the annual rate of simple interest if a principal of $500 amounts to $550 in 8 months. 6 Calculate the time in days for $760 to earn $35 at 4 34 % simple interest. 7 Calculate the answers to complete the following table. Principal $600 $880 $1290 g $3600 $980 m
Rate 6% 6 12 % e 10% i 1 72% 7 14 %
Time 5 years c 6 months 4 months 200 days k 6 months
Simple interest a $171.60 $45.15 $150 $98.63 l $52.50
Total investment b d f h j $1200.50 n
8 If Geoff invests $30 000 at 10% per annum simple interest until he has $42 000, for how many years will he need to invest the money? 9 Josh decides to put $5000 into an investment account that is paying 5.0% per annum simple interest. If he leaves the money there until it doubles, how long will this take, to the nearest month? 10 A personal loan of $15 000 over a period of 3 years costs $500 per month to repay. a How much money will be repaid in total? b What equivalent rate of simple interest is being charged over the 3 years? 11 To buy her first car, Cassie took out a loan for $4000. She paid it back over a period of 2 years, and this cost her $1450 in interest. What simple interest rate was she charged? 12 Over a period of 4 years, an investment earnt $913.50, at a simple interest rate of 5.25% per annum. What was the original amount deposited? 13 How long does it take for $2400 invested at 12% per annum simple interest to earn $360 interest?
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6.4
Compound interest
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We have seen that simple interest is calculated on the original amount borrowed or invested. A more common form of interest, called compound interest, calculates the interest each time period on a sum of money to which the previous amount of interest has been added. The interest is said to compound. Consider, for example, $250 invested at 10% per annum, where the interest is added to the account each year. In the first year: Interest = $250 × 10% × 1 = $25.00 so at the end of the first year the amount of money in the account is $250 + $25 = $275 In the second year: Interest = $275 × 10% × 1 = $27.50 so at the end of the second year the amount of money in the account is $275 + $27.50 = $302.50 In the third year: Interest = $302.50 × 10% × 1 = $30.25 so at the end of the third year the amount of money in the account is $302.50 + $30.25 = $332.75 And so on. Continuing in this way, we can build a table of values. For comparison, the amount of the money if invested at 10% per annum simple interest is also shown.
Year (n) 0 1 2 3 4 5 6 7 8 9 10
Amount of investment ($) 10% simple interest 10% compound interest 250 250 275 275 300 302.50 325 332.75 350 366.03 375 402.63 400 442.89 425 487.18 450 535.90 475 589.49 500 648.44
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Amount of investment ($)
From the table, we can see that after the first month, the compound interest is higher, and this advantage to the investor becomes more obvious as time increases. The difference between the two investment strategies is even clearer when viewed graphically. Compound interest at 10% per year Simple interest at 10% per year
650 600 550 500 450 400 350 300 250 1 2 3 4 5 6 7 8 9 10 Year
The growth in the investment over time when interest is compound is clearly not linear, but rather curved, as can been seen from the graph. The points have been joined to show this.
The compound interest formula Calculation of compound interest is very tedious if carried out for many years. We can, however, soon see a pattern to the calculations. In the example above, the principal was increased by 10% of each year. at the end 10 = 1.1. The multiplying factor to increase a quantity by 10% is 1 + 100 Thus, the amount accrued at the end of: 1st year = $250 × 1.1 = $275 2nd year = $250 × 1.1 × 1.1 = $250 × (1.1)2 = $302.50 3rd year = $250 × 1.1 × 1.1 × 1.1 = $250 × (1.1)3 = $332.75 nth year = $250 × (1.1)n It can be seen that a formula for calculating the amount of an investment earning compound interest can be written as follows. In general, the amount of the investment is given by r t A = P 1+ 100 where: $A = final amount $P = initial amount (principal) r % = interest rate per annum t = number of years. That is, if an amount $P is invested at r% per annum compound interest for t years, it will grow to $A.
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To find the amount of interest earned, we need to subtract the initial investment from the final amount. The interest ($I ) that would result from investing $P at r% per annum, compounded annually for t years, is given by: r t I = P − A = P − P × 1+ 100 where $A is the amount of the investment after t years. Example 12
Calculating the amount of the investment and interest
a Determine, to the nearest dollar, the amount of money accumulated after 3 years if $2000 is invested at an interest rate of 8% per annum, compounded annually. b Determine the amount of interest earned. Solution a Substitute P = $2000, t = 3, r = 8 into the formula giving the amount of the investment.
r A = P × 1+ 100
b Subtract the principal from this amount to determine the interest earned.
t
8 3 = 2000 × 1 + 100 = $2519 to the nearest dollar
I = A − P = 2519 − 2000
= $519
Another way of determining compound interest is to enter the appropriate formula into a graphics calculator, and examine the interest earned using the 7 facility of the calculator. How to investigate compound interest problems using a graphics calculator a Determine, to the nearest dollar, the amount of money accumulated after 3 years if $2000 is invested at an interest rate of 8% per annum, compounded annually. b Determine the amount of interest earned. Steps 1 Substitute P = $2000 and r = 8 into the formula for compound interest. 2 Press o to display the Y1= editor.
3 Because the calculator expects formulas to be defined in terms of X and Y, we will enter this rule as shown. Y1 represents A and X represents t.
8 t A = 2000 × 1 + 100
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4 Select 7 to view a table of values. By scrolling down the table we can see the amount of the investment after each year.
After 3 years: a A = $2519 b I = $2519 − $2000 = $519 5 Note that a graph of the compound interest function can also be constructed using the graphics calculator. Select p and set Xmin to 0 and Ymin to 2000. Then press q, followed by 0:ZoomFit, to construct the graph shown here. Select r to find values from the graph. How would our answer to Example 12 change if, instead of the interest being added to the account at the end of each year (called compounded annually), the interest is added every 3 months (called compounded quarterly)? Compounding 8% quarterly means that the interest rate for the 3-month period is reduced to 2%. However, the number of times the interest is added increases from three to 12, since the interest is added 4 times each year. This situation is shown in the following table.
Year 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3
Amount of investment ($) Compounding annually Compounding quarterly 2000 2040 2000 2080.80 2000 2122.42 2160 2164.86 2160 2208.16 2160 2252.32 2160 2297.37 2332.80 2343.32 2332.80 2390.19 2332.80 2437.99 2332.80 2486.75 2519.42 2536.48
The final entry in the table may be determined by evaluating: $2000 × (1 + 0.02)12 = $2536.48 Thus, we need to modify the formula previously stated to take into account situations where interest is compounded, or adjusted, other than annually.
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In general:
r/n nt Amount of the investment, A = P × 1 + 100
where : $A = amount of the investment after t years $P = initial amount (principal) r % = interest rate per annum n = number of times per year interest is compounded t = number of years. Example 13
Interest compounding monthly
a Determine the amount reached if $5000 is invested at an interest rate of 12% per annum for a period of 2 years and interest is compounded monthly. b Determine the amount of interest earned. Solution a Substitute P = $5000, r = 12, n = 12 and t = 2 into the formula giving the amount of the investment. b Subtract the principal from this final amount to determine the interest earned.
12/12 12×2 r /n nt = 5000 × 1 + A = P × 1+ 100 100 = 5000 × (1.01)24 = $6349 to the nearest dollar I = A − P = $6349 − $5000
= $1349
Calculating the principal Sometimes we are interested in calculating the amount of money that should be invested, the principal P, to reach a particular amount in the investment A. The formula for compound interest can be rearranged as follows. In general, the principal is given by A P= r/n nt 1+ 100 where : $A = amount of the investment after t years $P = initial amount (principal) r % = interest rate per annum n = number of times per year interest is compounded t = number of years. Example 14
Calculating the principal
How much money should be invested at 8% per annum compound interest, compounding monthly, if $5000 is needed in 2 years time?
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Solution Substitute A = $5000, r = 8, n = 12 and t = 2 into the formula giving the principle.
$5000 A P = n t = r /n 8/12 12Ă&#x2014;2 1+ 1+ 100 100 = $4263 to the nearest dollar
Exercise 6D 1 Calculate the final amount if $3500 is invested at 5% compound interest per annum for 5 years. 2 Calculate the amount of compound interest earned from investing $7000 at 8% per annum for 4 years. 3 Find the total amount needed to repay a loan of $1250 at 7 12 % compound interest per annum over 3 years. 4 Calculate the difference between the simple interest and the compound interest on a loan of $2000 at 7% per annum over 5 years. 5 Calculate the amount of money that would need to be invested at 6 34 % per annum compound interest to achieve a sum of $36 000 in 4 years. 6 Calculate the interest paid on a loan of $2750 at 11% per annum, compounded quarterly for 4 years. 7 Calculate the final amount to be paid on a loan of $10 000 at 12% per annum for 5 years, for each of the following conditions. a Simple interest c Compound interest calculated quarterly e Compound interest calculated daily
b Compound interest calculated annually d Compound interest calculated monthly
8 A bank offers various investment possibilities to a customer wishing to invest $24 000 for 12 years. a Calculate the final amount for each of the following. i Simple interest at 13% per annum ii Compound interest at 12% per annum, calculated annually iii Compound interest at 11% per annum, calculated quarterly iv Compound interest at 10% per annum, calculated monthly v Compound interest at 9% per annum, calculated daily b Which investment would you recommend? 9 Suppose that $20 000 is invested at 4.2% per annum, compounding monthly. How much interest is paid in the fourth year of the investment?
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10 Joe buys a skateboard costing $530 on his credit card, knowing that he will not be able to pay it off for 6 months. If he is charged 18% per annum interest for 6 months, compounded monthly, how much will the skateboard end up costing him? (Assume that no payments are made within this time, and give your answer to the nearest cent.) 11 How much money must you invest at 12.5% per annum, compounded monthly, if you know that you will need $10 000 in 3 years time? Give your answer to the nearest $10. 12 Sueâ&#x20AC;&#x2122;s parents open an investment account when she is born, in which they deposit $5000. They intend to withdraw the money on her 18th birthday. If the account pays 8.0% per annum, compounded quarterly, for the first 10 years, and then 6.0% per annum, compounded monthly, for the rest of the term of the investment, how much will Sue receive? Give your answer to the nearest $10.
6.5
Flat rate depreciation Depreciation and book value In the previous sections we looked at the ways in which investments grow or appreciate. It is also possible, of course, for the value of investments to decrease. For example, many of the items purchased by businesses, such as motor vehicles and office equipment, decrease in value, or depreciate, and this is allowed for by accountants when estimating the value of a business. Depreciation is also important for tax purposes, where it is allowed as a deductible expense for a business. The government has prescribed the maximum rates of depreciation allowed on various types of equipment. The longer an item can be expected to last, the lower the rate of depreciation. Depreciation is an estimate of the annual reduction in the value of items, caused by such things as age and wear and tear. The book value of an item is its value at any given time. Book value = purchase price â&#x2C6;&#x2019; depreciation Once the item has reached the stage where it can no longer be used profitably by the company, it is sold off. The price that the item is expected to fetch at this point is called the scrap value of the item. The amount of time for which the item is in use is called the effective life of that item. Two common methods for calculating depreciation will be considered here: flat rate depreciation reducing balance depreciation.
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Flat rate depreciation Using this method the equipment or assets are depreciated in value by a fixed amount each year, normally a percentage of the original value. This is an equivalent, but opposite, situation to simple interest. We have already established a formula for simple interest. This formula can also be applied to the calculation of flat rate depreciation. purchase price × depreciation rate (per annum) × length of time (in years) 100 Prt D= 100
Depreciation = or
where $D is the flat rate depreciation of the item after t years, $P is the purchase price of the item, r% is the flat rate of depreciation per annum and t is the time in years. To determine the book value of the item, the amount of depreciation is subtracted from the purchase price. The book value is given by Prt 100 where $V is the book value of the item after t years, $P is the purchase price of the item, r% is the flat rate of depreciation per annum and t is the time in years. V =P−D=P−
If the item is kept for longer than the original estimated life, the flat rate method will sooner or later give the item a book value of zero.The item is then said to be written off. As with simple interest, the formulas for flat rate depreciation and book value can be rearranged to find the value of any one of the variables, when the values of other variables are known. Example 15
Determining the flat rate depreciation and book value
Michael purchases a new car for $24 000. If it decreases in value by 10% of the purchase price each year: a what is the amount of the annual depreciation? b what is the amount of the depreciation after 4 years? c what is its book value after 4 years? Solution a The annual depreciation is 10% of the purchase price. b Substitute P = $24 000, r = 10 and t = 4 into the formula for flat rate depreciation. c The book value at the end of 4 years is the cost of the item less the depreciation.
24 000 × 10 100 = $2 400 24 000 × 10 × 4 Prt = D = 100 100 = $9600 After 4 years V = 24 000 − 9600 = $14400
Annual depreciation =
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Because the depreciation occurs at a constant rate, a negative straight line relationship exists between the book value and the time over which the depreciation occurs. For this reason, this form of depreciation is also sometimes called straight line depreciation. The graphics calculator can be set up to generate a table for flat rate depreciation over time. How to determine flat rate depreciation and book value using a graphics calculator Michael purchases a new car for $24 000. If it decreases in value by 10% of the purchase price each year: a what is the amount of the annual depreciation? b what is the amount of the depreciation after 4 years? c what is its book value after 4 years? Steps 1 Substitute P = $24 000 and r = 10 into the formulas for depreciation and book value under flat rate depreciation.
24 000 × 10 × t 100 24 000 × 10 × t V = 24 000 − 100
D =
2 Press o to display the Y= editor.
3 Enter the rule for the depreciation (D) in Y1 , and the rule for the book value (V) in Y2 , as shown. Since we wish to generate a table that shows yearly values, time is the variable in each of the formulas, and thus represented by X. 4 Use 2 7 to access the table of values. By scrolling down the table we see the total depreciation and book value after each year.
a After 1 year, D = $ 2400 b After 4 years, D = $ 9600 c After 1 year, V = $ 14400 Example 16
Flat rate depreciation and book value
A factory manager assesses that a particular machine, purchased for $30 000, has a useful life of 10 years. Its scrap value is estimated as $6000. a What is the annual amount of depreciation? b What is the flat rate of depreciation? c Draw a graph of the book value of the machine over time.
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Solution a 1 Work out the total depreciation over the 10-year period. 2 Determine how much this is per year.
Total depreciation = 30 000 − 6000 = $24 000 Depreciation per year =
24 000 = $2400 10
100 2 400 × = 8% 30 000 1
b To find the flat rate of depreciation, the amount of the annual depreciation is expressed as a percentage of the original cost.
r =
c We can draw a graph of the book value over time using a graphics calculator. 1 Substitute P = $30 000 and r = 8 into the formula for flat rate depreciation. 2 Enter this rule into the Y= editor as shown.
v = 30 000 −
30 000 × 8 × t 100 = 30 000 − 2400 t
3 Select p and set Xmin and Ymin to 0. Then press q, followed by 0:ZoomFit, to construct the graph shown here. Select r to find values from the graph. From Example 16, we can see that the amount by which an item is depreciated each year using flat rate depreciation is constant. Thus we may write: Flat rate book value after t years = purchase price − depreciation per year × t When calculations need to be repeated several times, it is useful to generate these as a table on the graphics calculator. This can be achieved by entering the equation for the flat rate book value as above, and then requesting a table.
Exercise 6E 1 A farmer who purchased a tractor for $17 250 estimates that it will have an effective life of 5 years, when its value will be $5500. a What is the amount of the annual depreciation? b Find its book value after 3 years.
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2 Geoffreyâ&#x20AC;&#x2122;s new motorbike, which he purchased for $8500, decreases in value by 12% of the purchase price each year. a What is the amount of the annual depreciation? b What is its book value after 2 years? 3 A machine purchased for $55 400 decreases in value by 13.5% of the purchase price each year. What is its book value after 5 years? 4 Some office furniture is purchased for $1950, and it is estimated that it will be used for 6 years. If at the end of 6 years the furniture is worth $546, use the straight line method to: a calculate the amount of annual depreciation b calculate the depreciation rate as a percentage of the cost price. 5 The cooking and catering equipment purchased for a newly opened restaurant cost $150 000. It is estimated that it will have a book value of $15 000 after 8 years. To allow the restaurant owner to prepare a budget: a calculate the annual depreciation rate as a percentage of the cost price b calculate the book value of the equipment after 6 years c draw a graph to show book value against time. 6 A publisher buys a new phone system. The cost is $12 500, and it is depreciated by 9% per year by the flat rate method. a b c d
Calculate the annual depreciation of the equipment. Calculate the book value of the equipment after 5 years. Draw a graph to show book value against time. Use the graph to estimate the number of years it would take for the equipment to be written off; that is, for the book value to be zero.
7 A tractor purchased for $85 000 has a useful life of 8 years. Its scrap value is estimated as $15 000. a What is the annual amount of depreciation? b What is the flat rate of depreciation? c Draw a graph of the book value of the tractor over time. 8 A developer installs a new air-conditioning system into an office block she is building. The cost is $122 870, and its value depreciates by 12.5% per year by the flat rate method. a b c d
Calculate the annual depreciation of the system. Calculate the book value of the system after 4 years. Draw a graph to show book value against time. Estimate the number of years it would take for the book value of the system to be zero.
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Reducing balance depreciation In reducing balance depreciation, the annual depreciation is not a fixed amount but is calculated as a percentage of the book value at the beginning of each year. The amount of depreciation is largest in the first year, becoming smaller in each subsequent year. Reducing balance depreciation is the equivalent, but opposite, situation to compound interest, where an investment increases by a constant percentage each year. We can apply some of our knowledge of compound interest to this situation. Earlier we established the following formula: The amount of money ($A) that would result from investing $P at r% per annum, compounded annually for a period of t years, is: r t A = P × 1+ 100 Extending this to reducing balance depreciation we can say Book value is given by r t V = P × 1− 100 where : $V = book value of the item after t years $P = purchase price of the item r % = reducing balance depreciation rate per annum, compounded annually t = number of years. That is, the book value of an item that has a purchase price of $P and depreciates at r% per annum, compounded annually, will reduce to $V after t years. To find the amount of depreciation, we need to subtract the book value from the purchase price. The amount of depreciation resulting from depreciating an item with a puchase price of $P at r% per annum, compounded annually for t years, is given by: Depreciation is given by r t D = P −V = P − P × 1− 100 where $V is the book value of the item after t years. Example 17
Determining book value
The factory manager in Example 16 decided that it was better to depreciate the machine, purchased for $30 000, using the reducing balance method. If he depreciates the machine at 15% per annum, what is its book value after 4 years? Solution Substitute P = $30 000, r = 10 and t = 4 into the formula for book value with reducing balance depreciation.
r V = P × 1− 100
t
15 4 = $30 000 × 1 − 100 = $15 660 to the nearest dollar
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The graphics calculator can be set up to generate a table for reducing balance depreciation over time.
How to determine reducing balance depreciation and book value using a graphics calculator The factory manager in Example 16 decided that it was better to depreciate the machine, purchased for $30 000, using the reducing balance method. If he depreciates the machine at 15% per annum, what is its book value after 4 years? Steps 1 Substitute P = $30 000 and r = 15 into the formulas for book value and depreciation under reducing balance depreciation.
15 t V = 30 000 × 1 − 100 15 t D = 30 000 − 30 000 × 1 − 100
2 Press o to display the Y= editor.
3 Enter the rule for the book value (V ) in Y1 , and the rule for the depreciation (D) in Y2 , as shown. Since we wish to generate a table that shows yearly values, time is the variable in each of the formulas, and thus represented by X. 4 Use 2 7 to access the table of values. By scrolling down the table we see the book value and total depreciation after each year.
After 4 years: a V = $ 15660 to the nearest dollar b D = $ 14340 to the nearest dollar 5 We can also use the graphics calculator to construct a graph of the book value over time. Select p and set Xmin and Ymin to 0. Then press q, followed by 0:ZoomFit, to construct the graph shown here. Select r to find values from the graph.
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How do the two methods of depreciation compare? The graph below shows book value plotted against time for both the flat rate method and the reducing balance method, to allow comparison, using the values from Examples 16 and 17. It can be clearly seen that reducing balance depreciation is not linear. This is not unexpected, since reducing balance depreciation is the equivalent but opposite situation to compound interest. 30000
Book value ($)
25000
Straight line depreciation at 8% per year
20000 15000 10000 5000
Reducing balance depreciation at 15% per year 0
2
4
6 Years
8
10
12
From the graph we can also see that, although the rate of depreciation is higher for reducing balance depreciation (15% compared to 8% for flat rate depreciation), the amount of depreciation reduces from year to year until eventually the two lines cross. The points have been joined to show this.
Exercise 6F 1 A new car was purchased for $37 500. It is estimated that a new car depreciates by 30% during its first year after registration and then by 20% in each subsequent year, calculated on the previous yearâ&#x20AC;&#x2122;s value. What would you expect its book value to be after 3 years? 2 A small business bought a new computer system for $3500. If the tax office allows a reducing balance annual depreciation rate of 40%, calculate the book value of the system at the end of 5 years. 3 A boat is purchased for $35 750, and depreciated on a reducing balance basis at an annual depreciation rate of 23%. Calculate, to the nearest dollar, the book value of the boat at the end of 6 years. 4 A stereo system purchased for $1500 incurs 12% per annum reducing balance depreciation. a Find, to the nearest dollar, the book value after 7 years. b What is the total depreciation after 7 years? c Draw a graph to show book value against time. 5 A washing machine purchased for $768 incurs 30% per annum reducing balance depreciation. a Calculate the book value of the washing machine after 4 years. b Draw a graph to show book value against time.
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6 A hire-car company can claim a 40% reducing balance depreciation on minibuses designed to carry nine or more people. If the hire-car company purchases a new 12-seater minibus for $45 000, find, to the nearest dollar: a the book value of the minibus after 7 years b the total depreciation after 7 years. 7 A new computer system, purchased by an engineering company, has an initial value of $75 000. a Calculate the value of the system after 3 years if the annual depreciation rate is 30% using the reducing balance method. b Calculate the value of the system after 3 years if the annual depreciation rate is 15% using the flat rate method. 8 For tax purposes, a taxi owner-driver can claim 40% depreciation for his vehicle using the reducing balance method or 17% depreciation using the flat rate method. a For each method, prepare a table showing the book value for the first 5 years of the life of a taxi originally purchased for $41 000. b Use your tables to decide which method gives the greatest cumulative depreciation after 5 years.
6.7
Hire-purchase One way of purchasing goods when you have insufficient cash available is to enter into a hire-purchase agreement. This means the purchaser agrees to hire the item from the seller and to make periodic payments of an agreed amount. At the end of the period of the agreement, the item is owned by the purchaser. If the purchaser stops making payments at any stage of the agreement, the item is returned to the vendor and no money is refunded to the purchaser. We are interested in being able to calculate the interest rate being charged in these contracts, as it is not always stated explicitly.
Flat interest rate If we calculate the total interest paid as a proportion of the original debt, and express this as an annual rate, this is called the flat rate of interest. The interest rate is exactly the same as the simple interest rate, but is generally called by this name in the hire-purchase context. Earlier in this chapter we established that, for simple interest: 100I Pt where r% is the interest rate per annum, $P is the principal, $I is the amount of interest and t is the time in years. In the case of hire-purchase, we will define the formula as follows: r=
100I Pt where $I is the total interest paid, $P is the principal owing after the deposit has been deducted and t is the number of years. Flat rate of interest per annum, r f =
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Example 18
237
Calculating the interest and flat rate when t = 1
A student buys a notebook computer priced at $1850. She pays a deposit of $370 and repays the loan in 12 monthly instalments of $141.50 each. a How much interest is paid on the computer? b What is the flat rate of interest, correct to 2 decimal places? Solution a 1 First we need to determine how much the student has paid in total. 2 Determine the interest paid. This is the difference between the total paid and the purchase price. b 1 Apply the appropriate formula with P = $1480 (since only $1480 is owing after the deposit is paid), I = $218 and t = 1 (since the computer is paid off in 1 year) to find the value of rf . 2 Since the unit of time is years, the interest rate can be written as the interest rate per annum.
Total paid = deposit + repayments = 370 + 12 × 141.50 = $2068 Interest paid = 2068 − 1850 = $218
rf =
100 × 218 100I = Pt 1480 × 1 = 14.73
Interest rate = 14.73% per annum
The formula for rf given above gives the flat rate of interest over the period of the contract. Since the contract in Example 18 is for 1 year, the flat rate of interest calculated there is also the flat rate of interest per annum. The distinction should be clear in the next example, where the contract is not for 1 year. Example 19
Calculating the interest and flat rate when t = 1
A hire-purchase contract for a sound system, priced at $1400, requires Josh to pay a deposit of $400 and then make six monthly payments of $185. a How much interest does he pay? b What is the flat rate of interest per annum that this represents? Solution a 1 Determine how much Josh pays in total.
Total paid = deposit + repayments = 400 + 6 × 185 = $1510
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2 Determine the interest paid. This is the difference between the total paid and the purchase price. b 1 Apply the appropriate formula with P = $1000 (since only $1000 is owing after the deposit is paid), I = $110 and t = 0.5 (since the sound system is paid off in 6 months) to find the value of rf . 2 Since the unit of time is years, the interest rate can be written as the interest rate per annum.
Interest paid = 1510 â&#x2C6;&#x2019; 1400 = $110
rf =
100I 100 Ă&#x2014; 110 = Pt 1000 Ă&#x2014; 0.5 = 22%
Interest rate = 22% per annum
Exercise 6G 1 The cash price of a tennis racquet is $330. To buy it on hire-purchase requires a deposit of $30 and 12 equal monthly instalments of $28. Calculate: a the total cost of buying the racquet by hire-purchase b the extra cost of buying by hire-purchase.
2 A bicycle has a marked price of $300. It can be bought through hire-purchase with a deposit of $60 and 10% interest on the outstanding balance, to be repaid in 10 monthly instalments. Calculate: a the amount of each monthly instalment b the total cost of buying the bicycle by hire-purchase. 3 A hire-purchase agreement offers hi-fi equipment, with a marked price of $897, for $87 deposit and $46.80 a month payable over 2 years. Calculate: a the total hire-purchase price
b the amount of interest charged.
4 A second-hand car is advertised for $5575 cash or $600 deposit and 24 monthly intalments of $268.75. Calculate the flat rate of interest per annum. 5 Exercise gym equipment, which normally costs $750, can be bought through hire-purchase with a $200 deposit and $26.40 a month for 30 months. Calculate: a the amount of interest being charged
b the flat rate of interest per annum.
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6 A microwave oven is advertised with a marked priced of $576 and the opportunity to buy it on hire-purchase, with no deposit and an interest rate of 10% repayable over a year with four equal instalments. Calculate: a the amount of interest c the amount of each instalment.
b the total amount to be repaid
7 A student buys a tent for bushwalking, priced at $200. He pays a deposit of $50 and agrees to repay the balance of $150 plus interest at 14% over the period of 1 year, in two half-yearly instalments. Calculate: a the amount of each half-yearly instalment
b the total price of the tent.
8 A customer bought a new car for $36 010 on hire-purchase. A deposit of $4000 was paid and the loan plus interest is to be repaid over 18 months in six quarterly repayments of $6295.30. Calculate: a the total amount to be repaid
b the flat interest rate per annum.
9 A student bought a new computer costing $2225 on hire-purchase. She traded in an old computer for $200 and paid a deposit of $150. The balance was paid by monthly instalments of $112.50 over 2 years. Calculate: a the total interest paid
b the flat rate of annual interest.
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Key ideas and chapter summary Percentage increase or decrease
Percentage incresase or decrease is the amount of the increase or decrease expressed as a percentage of the original value. amount of change 100 × Percentage change = original value 1
Simple interest
Simple interest is paid on an investment or loan on the basis of the original amount invested or borrowed, called the principal (P). The amount of simple interest is constant from year to year, and thus is linearly related to the term of the investment. The amount of interest earned ($I ) when a principal ($P) is invested at r % per annum for t years is given by: Prt I = 100 The total amount owed or invested ($A) after simple interest has been added to a principal ($P) for t years at r % per annum is given by: Prt A=P+I =P+ 100 where $I is the amount of interest earned.
Amount of the investment or loan (simple interest)
Minimum monthly balance
The lowest amount an account contains in each calendar month is its minimum monthly balance.
Compound interest
Under compound interest, the interest paid on a loan or investment is credited or debited to the account at the end of each period, and the interest for the next period is based on the sum of the principal and previous interest. The amount of the compound interest increases each year, and thus there is a non-linear relationship between compound interest and the term of the investment. The amount of interest earned ($I ) when a principal ($P) is invested at r % per annum for t years, compounded n times a year, is given by: r/n nt −P I = A− P = P × 1+ 100 where $A is the amount of the investment after t years.
Amount of the The total amount owed or invested ($A) after compound interest has been added to a principal ($P) for t years at r % per annum, investment or loan (compound interest) compounded n times a year, given by: r/n nt A = P × 1+ 100 Depreciation
Depreciation is the amount by which the value of an item decreases over time.
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The book value of an item is its depreciated value.
Scrap value
The scrap value of an item is the value at which it is no longer of value to the business, so it is replaced.
Flat rate depreciation
Flat rate depreciation is when the value of an item is reduced by the same percentage of the purchase price, or amount, for each year the item is in use. It is equivalent, but opposite, to simple interest. The depreciation ($D) of an item that has a purchase price of $P and is depreciating at a flat rate of r % per annum for t years is given by: Prt D= 100 The book value ($V ) of an item that has a purchase price of $P and is depreciating at a flat rate of r % per annum for t years is given by: Prt V =P−D=P− 100 where $D is the amount of depreciation.
Book value (flat rate depreciation)
Reducing balance depreciation
Reducing balance depreciation is when the value of an item is reduced by a constant percentage for each year it is in use. It is the equivalent, but opposite, situation to compound interest. The depreciation ($D) of an item that has a purchase price of $P and is depreciating at a rate of r % per annum, compounded annually for t years, is given by: r t D = P −V = P − P × 1− 100 where $V is the book value of an item after t years.
Book value (reducing balance depreciation)
The book value ($V ) of an item that has a purchase price of $P and is depreciating at a rate of r % per annum, compounded annually for t years, is given by: r t V = P × 1− 100
Hire-purchase
Under a hire-purchase agreement, the purchaser hires an item from the vendor and makes periodic payments at an agreed rate of interest. At the end of the period of the agreement, the item is owned by the purchaser.
Flat rate of interest
Flat rate of interest is when the total interest paid is given as a percentage of the original amount owed, and annualised. 100I Flat rate of interest per annum, rf = Pt where $I is the total interest paid, $P is the principal owing after the deposit has been deducted and t is the number of years.
Review
Book value
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Skills check Having completed this chapter you should be able to: calculate the amount of the discount and the new price when r % discount is applied calculate the amount of the increase and the new price when r % increase is applied calculate the percentage discount or increase that has been applied, given the old and new prices calculate the original price, given the new price and the percentage discount or increase that has been applied use the formula for simple interest to find the value of any one of the variables I, P, r or t when the values of the other three are known determine the interest payable on a bank account, paid on the minimum monthly balance, over a period when up to three transactions have been made calculate the amount of an investment after simple interest has been added plot the value of simple interest (I) against time (t) to show a linear relationship use the formula for compound interest to find the amount of an investment after compound interest has been added calculate the amount of compound interest payable on an investment or loan plot the value of compound interest (I) against time (t) to show a non-linear relationship calculate flat rate depreciation, book value and the length of time for which an asset will be in use under flat rate depreciation calculate the amount of depreciation of an asset under reducing balance rate depreciation determine the flat rate of interest per annum for a hire-purchase agreement.
Multiple-choice questions 1 The amount saved if a 10% discount is offered on an item marked $120 is: A $20 B $12 C $1.20 D $10.90 E $10 2 If a 20% discount is offered on an item marked $30, the new discounted price of the item is: A $10 B $24 C $6 D $25 E $28 3 If a 15% increase is applied to an item marked $60, the new price of the item is: A $69 B $9 C $75 D $67 E $70 4 How much interest is earned if $2000 is invested for 1 year at a simple interest rate of 4% per annum? A $2080 B $160 C $8 D $800 E $80
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6 What is the interest rate, per annum, if a deposit of $1500 earns interest of $50 over a period of 6 months? A 0.56% B 5.45% C 5.55% D 6.45% E 6.67% 7 $2400 is invested at a rate of 4.25% compound interest, paid annually. The value of this investment after 6 years is: A $3080.83 B $3074.41 C $3012 D $680.33 E $674.41 8 How much interest is earned if $5000 is invested for 3 years at an interest rate of 4%, if the interest is compounded quarterly? A $600 B $624.32 C $634.13 D $643.32 E $654.13 9 Machinery costing $145 000 depreciates by 10% in its first year, then by 7.5% in each subsequent year. The book value of the machinery after 3 years is approximately: A $73 400 B $86 500 C $96 700 D $108 800 E $111 700 10 The value of $8500 compounded annually for 5 years at 6% per annum is closest to: A $2550 B $10 731 C $11 050 D $11 375 E $11 700 11 An investment account is opened with a deposit of $5000. Compound interest is paid on the investment at a rate of 12% per annum, credited monthly. The amount in the account after 18 months, if no withdrawals have been made, is closest to: A $5981 B $38 450 C $5926 D $5900 E $6000 12 A new computer costs $3600. If depreciation is calculated at 15% per annum (reducing balance), the computer’s value at the end of 4 years will be closest to (in dollars): B 3600(0.854 ) C 3600/(1.154 ) A 3600(0.154 ) 4 D 3600 × 0.6 E (3600 × 0.85) 13 Office equipment is purchased for $12 000. It is anticipated that it will last for 10 years and have a scrap value of $1250. The amount of depreciation that would be allowed per year, assuming a flat rate of depreciation, would be: A $1200 B $1075 C $1325 D $1250 E $1500
The following information relates to Questions 14 and 15 To buy a car costing $23 000, Janet pays $5000 deposit and then payments of $440 per month for the next 5 years. 14 How much interest does Janet pay under this scheme? A $3400 B $8400 C $3120 D $1600 E $1250 15 What flat rate of interest per annum does this amount to? A 46.7% B 9.3% C 36.5% D 7.3%
E 14.6%
Review
5 The total value of an investment of $1000 after 3 years if simple interest is paid at the rate of 5.5% per annum is: A $55 B $1055 C $1165 D $3165
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Short-answer questions 1 Rabbit Easter Eggs were reduced in price from $2.99 to $2.37 as they were not selling quickly. Bilby Easter Eggs were discounted to $3.83 from $4.79. a Which type of Easter Egg had the larger percentage reduction? b Calculate the difference in the percentage rates. 2 After Christmas, all stock in JDâ&#x20AC;&#x2122;s was discounted by 20%. The sale price of a pair of cross-trainers was $110. Calculate the original marked price. 3 How much additional interest is earned if $8000 is invested for 7 years at 6.5% when interest is compounded annually, as compared with simple interest paid at the same rate? 4 A person wishes to have $20 000 available for a world trip in 10 years time. Calculate the principal to be invested over 10 years at 6% per annum for this to be achieved, if the interest is compounded monthly. 5 A computer system costing $4400 depreciates at a reducing balance rate of 20% per annum. a What is the value of the system after 4 years? b By what constant yearly amount would the system be reduced to give the same value at the end of 4 years? 6 A television set, which normally costs $880, can be bought through hire-purchase with a $200 deposit and a payment of $30 a month for 30 months. Calculate: a the amount of interest being charged b the flat rate of interest per annum.
Extended-response questions 1 a The wholesale price of a digital camera is $350. The maximum profit that a retailer is allowed to make when selling this particular camera is 75% of the wholesale price. Calculate the maximum retail price of the camera. b Suppose that the wholesale price of the camera increases at 5% per annum simple interest for the next 5 years. i By how much will the wholesale price have increased at the end of 5 years? ii What is the new wholesale price of the camera? iii What is the new retail price of the camera (with 75% profit)? iv What percentage increase is this in the retail price determined in part a? 2 Peter has $10 000 that he wishes to invest for 5 years. a Aussie Bank offers him 6% per annum simple interest. How much will he have at the end of the 5 years under this plan? b Bonza Bank offers him 5.5% per annum compound interest, compounding monthly. How much will he have at the end of the 5 years under this plan? c Find, correct to 1 decimal place, the simple interest rate that Aussie Bank should offer if the two investments are to be equal after 5 years.
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Plan B offers 5.0% per annum compound interest, compounding annually. a Use your graphics calculator to construct a graph of the interest earned under Plan A against time. b On the same axes, use your graphics calculator to construct a graph of the interest earned under Plan B against time. c Which of the plans would you choose, A or B, if the investment is for: i 3 years? ii 6 years? 4 The Smiths bought a new car priced at $34 800 and paid a deposit of $5000 cash. They borrowed the balance of the purchase price at simple interest. They then agreed to repay the loan, plus interest, in equal monthly payments of $850 over 4 years. a Calculate the total amount of interest to be repaid over the term of this loan. b Calculate, correct to 1 decimal place, the annual simple interest rate charged on this loan. c The value of the car was depreciated using a reducing balance method at a rate of 15% per year. Calculate, to the nearest dollar, the depreciated value of the car after 3 years. 5 Phil has had his present van for 6 years, and it cost him $28 500. a What is the current value of his van if it has depreciated at a flat rate of 10% per annum? b What is the current value of the van if it has depreciated at a rate of 13.5% per annum on its reducing value? Give your answer to the nearest dollar. c i Use your graphics calculator to construct graphs on the same axes of the depreciation of the van over time under each of the terms given in parts a and b. ii When is the van worth more under reducing balance depreciation than under flat rate depreciation? Give your answer in years correct to 1 decimal place. 6 The DVD player that Emily wants to buy usually costs $400, but is on sale for $350. a What percentage discount does this amount to? b Emily considers entering into an agreement to buy the DVD player where she pays no deposit and 24 monthly payments of $22.50. i How much interest would Emily pay under this agreement on the purchase price of $350? ii What is the flat rate of interest per annum that this represents? Express your answer as a percentage correct to 1 decimal place.
Review
3 Suppose that you have $30 000 to invest, and there are two alternative plans for investment: Plan A offers 5.3% per annum simple interest.