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C H A P T E R
6Financial 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 changeIn 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:
Discount = 10% of original price
= 0.10 × original price
and 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:
Discount = r
100× original price
New price = original price − discount = (100 − r )
100× original price
209
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210 Essential Standard General Mathematics
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:� subtracting the discount from
the original price, or� directly calculating 90% of
the original price.
Discount = 10
100× $50.00 = $5.00
New price = original price − discount
= $50.00 − $5.00 = $45.00
New price = $90
100× $50.00 = $45.00
Sometimes, prices are increased, or marked up. If a price is increased by 10%:
Increase = 10% of original price
= 0.10 × original price
and 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:
Increase = r
100× original price
New price = original price + increase = (100 + r )
100× original price
Example 2 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:� adding the increase to the
original price, or� directly calculating 110%
of the original price.
Increase = 10
100× 50.00 = $5.00
New price = original price + increase
= 50.00 + 5.00 = $55.00
New price = 110
100× 50.00 = $55.00
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Chapter 6 — Financial arithmetic 211
Calculating the percentage changeGiven 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
original price× 100
1%
Percentage increase = increase
original price× 100
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 = 5.00
50.00× 100
1= 10%
Increase = new price − original price
= 55.00 − 50.00 = $5.00
Percentage increase = 5.00
50.00× 100
1= 10%
Calculating the original priceSometimes we are given the new price and the percentage increase or decrease (r%), and asked
to determine the original price. Since we know that:
for a discount: New price = (100 − r )
100× original price
for an increase: New price = (100 + r )
100× original price
we can rearrange these formulas to give rules for determining the original price, as follows:
When r % discount has been applied: Original price = new price × 100
(100 − r )
When r % increase has been applied: Original price = new price × 100
(100 + r )
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212 Essential Standard General Mathematics
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.
Original price = 50 × 100
90= $55.56
Original price = 50 × 100
110= $45.45
Exercise 6A
1 Calculate the amount of the discount in each of the following, to the nearest whole cent.
a 24% discount on $360 b 72% discount on $250
c 6% discount on $9.60 d 9% discount on $812
e 12% discount on $86.90 f 19% discount on $38
g 2.6% discount on $900 h 14.2% discount on $1650
2 Calculate the amount of the increase in each of the following, to the nearest whole cent.
a 1.08% increase on $26 000 b 15.9% increase on $4760
c 3 14 % increase on $1520 d 12 1
2 % increase on $9460
e 9 12 % increase on $18 650 f 2.8% increase on $1 000 000
g 0.2% increase on $10 000
3 Calculate the following as percentages, correct to 2 decimal places.
a $19.56 of $400 b 60c/ of $2 c $6.82 of $24
d $24 of $2600 e 30c/ of 90c/ f $1.50 of $13.50
g 38c/ of $10.50 h 25c/ of $186
4 Calculate the new increased price for each of the following.
a $260 marked up by 12% b $580 marked up by 8%
c $42.50 marked up by 60% d $5400 marked up by 17%
e $42.80 marked up by 35% f $9850 marked up by 8%
g $258 marked up by 14.2% h $1960 marked up by 17.4%
5 Calculate the new discounted price for each of the following.
a $2050 discounted by 9% b $11.60 discounted by 4%
c $154 discounted by 82% d $10 600 discounted by 3%
e $980 discounted by 13.5% f $2860 discounted by 8%
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Chapter 6 — Financial arithmetic 213
g $15 700 discounted by 22.7% h $147 discounted by 2.8%
i $674 discounted by 12%
6 Find the original prices of the items that have been marked down as follows.
a Marked down by 10%, now priced $54.00
b Marked down by 25%, now priced $37.50
c Marked down by 30%, now priced $50.00
d 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 Marked up by 20%, now priced $15.96
b Marked up by 12.5%, now priced $70.00
c Marked up by 5%, now priced $109.73
d 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 interestWhen 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 × 5
100= $50
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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214 Essential Standard General Mathematics
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.
Interest = 1000 × 5
100= $50
Interest = 1000 × 5
100= $50
Interest = 1000 × 5
100= $50
Interest for 3 years = 50 + 50 + 50
= $150
The same principles apply when the simple interest is applied to a loan rather than an
investment.
The simple interest formulaSince the amount of simple interest earned is the same every year, we can apply a general rule.
Interest = amount invested × interest rate (per annum) × length of time (in years)
100
or I = P × r × t
100= Prt
100
where the amount invested or borrowed ($P) is known as the principal, r% is the interest
rate per annum and t is the time in years.
How does this relationship look graphically?
Suppose we were to borrow $1000 at 5%
per annum simple interest for a period of
years. A plot of interest against time is shown.
0
100
200
300
400
500
1 2 3 4 5Year
6 7 8 9 10
Inte
rest
($)
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
A = P + I = P + Prt
100
where $P is the amount invested or borrowed, r% is the interest rate per annum and t is the
time in years.
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Chapter 6 — Financial arithmetic 215
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).
I = Prt
100= 5000 × 10
100× 3.5
= $1750
Example 7 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.
I = Prt
100= 16 000 × 8
100× 2
= $2560
2 Find the total owed by adding the
interest to the principal.
A = P + I = 16 000 + 2 560
= $18 560
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?
Steps1 Substitute P = $10 000 and r = 8.25%
in the formula for interest rate. I = Prt
100= 10 000 × 8.25 × t
100
= 825t
2 Press o to display the Y= editor.
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216 Essential Standard General Mathematics
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 accountsOne 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 Transaction Debit Credit Total
30 June Pay 400.00 400.00
3 July Cash 50.00 350.00
15 July Cash 100.00 450.00
1 August 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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Chapter 6 — Financial arithmetic 217
Solution
1 Determine the minimum monthly
balance for July.
2 Determine the interest payable
for July.
The minimum balance in the account
for July was $350.00
I = Prt
100= 350 × 3
100× 1
12
= $0.88 or 88 cents
Exercise 6B
1 Calculate the amount of simple interest for each of the following:
a
b
c
d
e
f
g
h
i
Principal Interest rate Time
$400 5% 4 years
$750 8% 5 years
$1000 7 12 % 8 years
$1250 10 14 % 3 years
$2400 12 34 % 15 years
$865 15% 2 12 years
$599 10% 6 months
$85.50 22.5% 9 months
$15000 33 13 % 1 1
4 years
2 Calculate the amount to be repaid for each of the following:
a
b
c
d
e
Principal Interest rate Time
$500 5% 4 years
$780 6 12 % 3 years
$1200 7 14 % 6 months
$2250 10 34 % 8 months
$2400 12% 250 days
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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218 Essential Standard General Mathematics
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 b $125 000 for 6 months
c $37 750 for 18 months d $200 000 for 2 years
e $74 386 for 8 months 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 $1000 withdrawn
12 October $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 Transaction Debit Credit Balance
1 March 500.00
15 March Cash 250.00 750.00
31 March Cash 250.00 1000.00
1 April 1000.00
How much interest was earned in March?
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Chapter 6 — Financial arithmetic 219
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 Transaction Debit Credit Balance
1 March 650.72
8 April Cash 250.00 900.72
21 May Cash 250.00 1150.72
1 June 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 formulaThe 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.
2 Apply the formula with P = $8000, I = $3040
and t = 4 to find the value of r.
3 Since the unit of time was years, the interest
rate can be written as the interest per annum.
Interest, I = 11 040 − 8000
= $3040
r = 100IP t
= 100 × 3040
8000 × 4
= 9.5%
Interest rate = 9.5% per annum
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:
t = 100I
Pr
where $P is the principal, $I is the amount of interest and r% is the interest rate per annum.
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220 Essential Standard General Mathematics
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 = 100I
Pr= 100 × 1 800
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
r t
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:
P = A(1 + r t
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 = 100Irt
= 100 × 1500
5 × 3
= $10 000
P = A(1 + rt
100
) = 15 600(1 + 5 × 4
100
)
= 15 600
1.2= $13 000
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Chapter 6 — Financial arithmetic 221
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 Rate Time Simple interest Total investment
$600 6% 5 years a b
$880 6 12 % c $171.60 d
$1290 e 6 months $45.15 f
g 10% 4 months $150 h
$3600 i 200 days $98.63 j
$980 7 12 % k l $1200.50
m 7 14 % 6 months $52.50 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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222 Essential Standard General Mathematics
6.4 Compound interestWe 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 = $332.75
And so on.
Continuing in this way, we can build a table of values. For comparison, the amount of the
money if invested at 10% per annum simple interest is also shown.
Amount of investment ($)
Year (n) 10% simple interest 10% compound interest
0 250 250
1 275 275
2 300 302.50
3 325 332.75
4 350 366.03
5 375 402.63
6 400 442.89
7 425 487.18
8 450 535.90
9 475 589.49
10 500 648.44
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Chapter 6 — Financial arithmetic 223
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.
250
300
350
400
450
500
550
600
650
1 2 3 4 5
Year
Am
ount
of i
nves
tmen
t ($)
6 7 8 9 10
Compound interest at 10%per year
Simple interest at10% per 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 formulaCalculation 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% at the end of each year.
The multiplying factor to increase a quantity by 10% is
(1 + 10
100
)= 1.1.
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
A = P(
1 + r
100
)t
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 $A.
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224 Essential Standard General Mathematics
To find the amount of interest earned, we need to subtract the initial investment from the final
amount.
The interest ($I ) that would result from investing $P at r% per annum, compounded
annually for t years, is given by:
I = P − A = P − P ×(
1 + r
100
)t
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.
b Subtract the principal from this amount
to determine the interest earned.
A = P ×(
1 + r
100
)t
= 2000 ×(
1 + 8
100
)3
= $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.
Steps1 Substitute P = $2000 and r = 8 into the
formula for compound interest.
A = 2000 ×(
1 + 8
100
)t
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.
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Chapter 6 — Financial arithmetic 225
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 = $519
5 Note that a graph of the compound interest
function can also be constructed using the
graphics calculator. Select p and
set Xmin to 0 and Ymin to 2000. Then
press q, followed by 0:ZoomFit, to construct the
graph shown here. Select r to find values from the graph.
How would our answer to Example 12 change if, instead of the interest being added to the
account at the end of each year (called compounded annually), the interest is added every
3 months (called compounded quarterly)? Compounding 8% quarterly means that the
interest rate for the 3-month period is reduced to 2%. However, the number of times the
interest is added increases from three to 12, since the interest is added 4 times each year.
This situation is shown in the following table.
Amount of investment ($)
Year Compounding annually Compounding quarterly
0.25 2000 2040
0.5 2000 2080.80
0.75 2000 2122.42
1 2160 2164.86
1.25 2160 2208.16
1.5 2160 2252.32
1.75 2160 2297.37
2 2332.80 2343.32
2.25 2332.80 2390.19
2.5 2332.80 2437.99
2.75 2332.80 2486.75
3 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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226 Essential Standard General Mathematics
In general:
Amount of the investment, A = P ×(
1 + r/n
100
)nt
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.
A = P ×(
1 + r/n
100
)nt
= 5000 ×(
1 + 12/12
100
)12×2
= 5000 × (1.01)24
= $6349 to the nearest dollar
b Subtract the principal from this final
amount to determine the interest
earned.
I = A − P = $6349 − $5000
= $1349
Calculating the principalSometimes 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
P = A(1 + r/n
100
)nt
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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Chapter 6 — Financial arithmetic 227
Solution
Substitute A = $5000, r = 8, n = 12
and t = 2 into the formula giving the
principle.
P = A(1 + r /n
100
)n t = $5000(1 + 8/12
100
)12×2
= $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 b Compound interest calculated annually
c Compound interest calculated quarterly d Compound interest calculated monthly
e Compound interest calculated daily
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 Sue’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 depreciationDepreciation and book valueIn 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 − 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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Chapter 6 — Financial arithmetic 229
Flat rate depreciationUsing 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.
Depreciation = purchase price × depreciation rate (per annum) × length of time (in years)
100
or D = Prt
100
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
V = P − D = P − 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.
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.
Annual depreciation = 24000 × 10
100= $2400
b Substitute P = $24 000, r = 10 and t = 4
into the formula for flat rate depreciation.D = Prt
100= 24000 × 10 × 4
100= $9600
c The book value at the end of 4 years is the
cost of the item less the depreciation.
After 4 years
V = 24000 − 9600 = $14400
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230 Essential Standard General Mathematics
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?
Steps1 Substitute P = $24 000 and r = 10 into
the formulas for depreciation and book
value under flat rate depreciation.
D = 24000 × 10 × t
100
V = 24000 − 24000 × 10 × t
100
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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Chapter 6 — Financial arithmetic 231
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
10= $2400
b To find the flat rate of depreciation,
the amount of the annual depreciation
is expressed as a percentage of the
original cost.
r = 2 400
30 000× 100
1= 8%
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.
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.
v = 30 000 − 30 000 × 8 × t
100
= 30 000 − 2400 t
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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232 Essential Standard General Mathematics
2 Geoffrey’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 Calculate the annual depreciation of the equipment.
b Calculate the book value of the equipment after 5 years.
c Draw a graph to show book value against time.
d 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 $122 870, and its value depreciates by 12.5% per year by the flat rate method.
a Calculate the annual depreciation of the system.
b Calculate the book value of the system after 4 years.
c Draw a graph to show book value against time.
d Estimate the number of years it would take for the book value of the system to be zero.
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Chapter 6 — Financial arithmetic 233
6.6 Reducing balance depreciationIn reducing balance depreciation, the annual depreciation is not a fixed amount but is
calculated as a percentage of the book value at the beginning of each year. The amount of
depreciation is largest in the first year, becoming smaller in each subsequent year. Reducing
balance depreciation is the equivalent, but opposite, situation to compound interest, where an
investment increases by a constant percentage each year. We can apply some of our knowledge
of compound interest to this situation. Earlier we established the following formula:
The amount of money ($A) that would result from investing $P at r% per annum,
compounded annually for a period of t years, is:
A = P ×(
1 + r
100
)t
Extending this to reducing balance depreciation we can say
Book value is given by
V = P ×(
1 − r
100
)t
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
D = P − V = P − P ×(
1 − r
100
)t
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.
V = P ×(
1 − r
100
)t
= $30 000 ×(
1 − 15
100
)4
= $15 660 to the nearest dollar
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234 Essential Standard General Mathematics
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 agraphics 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?
Steps1 Substitute P = $30 000 and r = 15 into the
formulas for book value and depreciation
under reducing balance depreciation.
V = 30000 ×(
1 − 15
100
)t
D = 30000 − 30000 ×(
1 − 15
100
)t
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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Chapter 6 — Financial arithmetic 235
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.
0
5000
10000
15000
20000
25000
30000
2 4 6 8 10Years
Boo
k va
lue
($)
12
Straight line depreciation at 8% per year
Reducing balance depreciationat 15% per year
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 year’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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236 Essential Standard General Mathematics
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-purchaseOne 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 rateIf 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:
r = 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:
Flat rate of interest per annum, r f = 100I
Ptwhere $I is the total interest paid, $P is the principal owing after the deposit has been
deducted and t is the number of years.
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Chapter 6 — Financial arithmetic 237
Example 18 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.
Total paid = deposit + repayments
= 370 + 12 × 141.50
= $2068
2 Determine the interest paid. This is the
difference between the total paid and
the purchase price.
Interest paid = 2068 − 1850
= $218
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.
r f = 100IP t
= 100 × 218
1480 × 1
= 14.73
2 Since the unit of time is years, the
interest rate can be written as the
interest rate per annum.
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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238 Essential Standard General Mathematics
2 Determine the interest paid. This
is the difference between the total
paid and the purchase price.
Interest paid = 1510 − 1400
= $110
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.
r f = 100IP t
= 100 × 110
1000 × 0.5= 22%
2 Since the unit of time is years, the interest
rate can be written as the interest rate per
annum.
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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Chapter 6 — Financial arithmetic 239
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 b the total amount to be repaid
c the amount of each instalment.
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 $112.50 over 2 years. Calculate:
a the total interest paid b the flat rate of annual interest.
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240 Essential Standard General Mathematics
Key ideas and chapter summary
Percentage increase Percentage incresase or decrease is the amount of the increase or
or decrease decrease expressed as a percentage of the original value.
Percentage change = amount of change
original value× 100
1
Simple interest Simple interest is paid on an investment or loan on the basis of the
original amount invested or borrowed, called the principal (P). The
amount of simple interest is constant from year to year, and thus is
linearly related to the term of the investment. The amount of interest
earned ($I ) when a principal ($P) is invested at r% per annum for
t years is given by:
I = Prt
100Amount of the The total amount owed or invested ($A) after simple interest has been
investment or loan added to a principal ($P) for t years at r% per annum is given by:
A = P + I = P + Prt
100where $I is the amount of interest earned.
(simple interest)
Minimum monthly The lowest amount an account contains in each calendar month is its
minimum monthly balance.balance
Compound interest Under compound interest, the interest paid on a loan or investment is
credited or debited to the account at the end of each period, and the
interest for the next period is based on the sum of the principal and
previous interest. The amount of the compound interest increases each
year, and thus there is a non-linear relationship between compound
interest and the term of the investment.
The amount of interest earned ($I ) when a principal ($P) is invested at
r% per annum for t years, compounded n times a year, is given by:
I = A − P = P ×(
1 + r/n
100
)nt
− P
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 ($P) for t years at r% per annum,
compounded n times a year, given by:
A = P ×(
1 + r/n
100
)nt
investment or loan
(compound interest)
Depreciation Depreciation is the amount by which the value of an item decreases over
time.
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Chapter 6 — Financial arithmetic 241
Book value The book value of an item is its depreciated value.
Scrap value The scrap value of an item is the value at which it is no longer of value
to the business, so it is replaced.
Flat rate Flat rate depreciation is when the value of an item is reduced by the
same percentage of the purchase price, or amount, for each year the item
is in use. It is equivalent, but opposite, to simple interest.
The depreciation ($D) of an item that has a purchase price of $P and
is depreciating at a flat rate of r% per annum for t years is given by:
D = Prt
100
depreciation
Book value The book value ($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:
V = P − D = P − Prt
100where $D is the amount of depreciation.
(flat rate
depreciation)
Reducing balance Reducing balance depreciation is when the value of an item is reduced
by a constant percentage for each year it is in use. It is the equivalent,
but opposite, situation to compound interest.
The depreciation ($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:
D = P − V = P − P ×(
1 − r
100
)t
where $V is the book value of an item after t years.
depreciation
Book value The book value ($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:
V = P ×(
1 − r
100
)t
(reducing balance
depreciation)
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.
Flat rate of interest per annum, rf = 100I
Ptwhere $I is the total interest paid, $P is the principal owing after the
deposit has been deducted and t is the number of years.
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242 Essential Standard General Mathematics
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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Chapter 6 — Financial arithmetic 243
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
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):
A 3600(0.154) B 3600(0.854) C 3600/(1.154)
D 3600 × 0.6 E (3600 × 0.85)4
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 15To 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%
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244 Essential Standard General Mathematics
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 JD’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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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.
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.