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C H A P T E R
6
E
Financial arithmetic
How do we determine the new price when discounts or increases are applied?
M
PL
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:
SA
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
and
In general, if r% discount is applied:
r
× original price
Discount =
100
New price = original price − discount =
224
(100 − r )
× original price
100
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Chapter 6 — Financial arithmetic
Example 1
225
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
10
× $50.00 = $5.00
100
E
Discount =
New price = original price − discount
= $50.00 − $5.00 = $45.00
90
× $50.00 = $45.00
New price = $
100
M
PL
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.
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 =
Calculating the increase and the new price
SA
Example 2
(100 + r )
× original price
100
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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226
Essential Standard General Mathematics
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.
100
discount
×
%
original price
1
Percentage increase =
100
increase
×
%
original price
1
Example 3
E
Percentage discount =
Calculating the percentage discount or increase
M
PL
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.
SA
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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Chapter 6 — Financial arithmetic
Example 4
227
Calculating the original price
Solution
100
= $55.56
90
100
= $45.45
Original price = 50 ×
110
Original price = 50 ×
M
PL
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.
E
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?
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.
1.08% increase on $26 000
3 14 % increase on $1520
9 12 % increase on $18 650
0.2% increase on $10 000
SA
a
c
e
g
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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228
Essential Standard General Mathematics
g $15 700 discounted by 22.7%
i $674 discounted by 12%
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
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
M
PL
a
b
c
d
E
7 Find the original prices of the items that have been marked up as follows.
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
SA
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 ×
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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Chapter 6 — Financial arithmetic
Example 5
229
Calculating simple interest
How much interest will be earned if we invest $1000 at 5% interest per annum for 3 years?
Solution
2 Calculate the interest for the second
year.
3 Calculate the interest for the third year.
5
= $50
100
Interest for 3 years = 50 + 50 + 50
= $150
Interest = 1000 ×
M
PL
4 Calculate the total interest.
5
= $50
100
5
= $50
Interest = 1000 ×
100
Interest = 1000 ×
E
1 Calculate the interest for the first year.
The same principles apply when the simple interest is applied to a loan rather than an
investment.
The simple interest formula
Since the amount of simple interest earned is the same every year, we can apply a general rule.
amount invested × interest rate (per annum) × length of time (in years)
100
Prt
P ×r ×t
=
or
I =
100
100
where the amount invested or borrowed ($P) is known as the principal, r% is the interest
rate per annum and t is the time in years.
Interest =
500
Interest ($)
SA
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.
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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230
Essential Standard General Mathematics
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.
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=
10
Prt
= 5000 ×
× 3.5
100
100
= $1750
Calculating the total amount borrowed or invested
M
PL
Example 7
E
Solution
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.
SA
How to solve simple interest problems using the TI-Nspire CAS
How much interest is earned if $10 000 is invested at 8.25% simple interest for
10 years? Show that the graph of simple interest earned is linear.
Steps
1 Substitute P = $10 000 and r = 8.25%
in the formula for simple interest.
I=
Prt
10 000 × 8.25 × t
=
100
100
= 825t
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Chapter 6 — Financial arithmetic
231
Note: You can also use the sequence command
to do this.
3 Place the cursor in the grey formula
cell in the list named interest and type
in = 825 × time.
key and paste
M
PL
Note: You can also use the
time from the variable list.
E
2 Start a new document (by pressing
/ + N ) and select 3:Add Lists &
Spreadsheet.
Name the lists time (to represent time in
years) and interest.
Enter the data values 1–10 into the list
named time, as shown.
SA
Press enter to display the values.
By scrolling down the table (using )
we can see that the interest amount of
$8250 will be earned after 10 years.
4 Press
/5:Add Data & Statistics and
plot the graph as shown.
a To connect the data points
Move the cursor to the graphing area
and press / + b . Select
2:Connect Data Points.
b To display a value
Place the cursor on the data point
before
and hold . Press
moving to another data point.
From the plot we can see that the
graph of the amount of simple
interest earned is linear. The slope of
the graph is equal to the interest paid
each year.
Note: You can also graph this example in the
Graphs & Geometry application.
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232
Essential Standard General Mathematics
How to solve simple interest problems using the ClassPad
Steps
1 Substitute P = $10 000 and
r = 8.25%
in the formula for simple
interest.
I=
10 000 × 8.25 × t
Prt
=
100
100
= 825t
M
PL
2 To form a table of values,
E
How much interest is earned if $10 000 is invested at 8.25% simple interest for
10 years? Show that the graph of simple interest earned is linear.
open the Sequence (
)
application.
Select the Explicit tab and
move the cursor to the box
opposite a n E: and type
825n. The n is found in the
toolbar ( ).
Press E to confirm your
entry (which is indicated by
a tick in the square to the
left of a n E:).
SA
3 To display the terms of the
sequence in table format,
tap the # icon.
By scrolling down the table,
it can be seen that the
interest amount of $8250
will have been earned after
10 years.
Note: Tap the Sequence
TableInput (8) icon in the
toolbar and adjust the Start and
End values if more values are
required.
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Chapter 6 — Financial arithmetic
233
Note: To define the graph
window scale, select the View
Window (6) icon from the
toolbar and set the values as
shown. Tap OK to confirm your
settings. Leave the xdot and
ydot settings as they are. These
values control the trace
increment of the cursor.
SA
M
PL
5 From the menu bar, select
Analysis and then Trace.
This will place a marker on
the graph at the first value
in the table. Use the cursor
arrow keys ( and ) to
move from one table value
to the next.
From the plot we can see
that the graph of the amount
of simple interest earned is
linear. The slope of the
graph is equal to the interest
paid each year.
E
4 To graph the sequence of
simple interest values,
select the Sequence Grapher
( ) icon from the toolbar.
Interest paid to 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 next example.
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234
Essential Standard General Mathematics
Example 8
Calculating interest paid to a bank account
The table shows the entries in Tom’s bank account.
Transaction
Pay
Cash
Cash
Debit
Credit
400.00
50.00
100.00
Total
400.00
350.00
450.00
450.00
E
Date
30 June
3 July
15 July
1 August
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.
M
PL
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
Prt
3
1
I=
= 350 ×
×
100
100 12
= $0.88 or 88 cents
Exercise 6B
1 Calculate the amount of simple interest for each of the following:
Principal
Interest rate
Time
$400
5%
4 years
b
$750
8%
5 years
c
$1000
7 12 %
8 years
d
$1250
3 years
e
$2400
10 14 %
12 34 %
15 years
f
$865
15%
2 12 years
g
$599
10%
6 months
h
$85.50
22.5%
9 months
$15000
33 13 %
1 14 years
SA
a
i
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Chapter 6 — Financial arithmetic
235
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
E
2 Calculate the amount to be repaid for each of the following:
3 A sum of $10 000 was invested in a fixed term account for 3 years. Calculate:
M
PL
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.
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.
SA
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
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236
Essential Standard General Mathematics
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 three consecutive months are:
$240.00
$350.50
$478.95
E
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.
Transaction
Debit
Credit
Balance
500.00
750.00
1000.00
1000.00
M
PL
Date
1 March
15 March
31 March
1 April
Cash
Cash
250.00
250.00
How much interest was earned in March?
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
Cash
Cash
Debit
Credit
250.00
250.00
Balance
650.72
900.72
1150.72
1150.72
SA
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.
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Chapter 6 — Financial arithmetic
Example 9
237
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 =
E
Interest rate = 9.5% per annum
M
PL
3 Since the unit of time was years, the interest
rate can be written as the interest per annum.
100 × 3040
100I
=
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.
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
SA
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:
100I
P=
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.
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238
Essential Standard General Mathematics
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
100 × 1500
100I
=
rt
5×3
= $10 000
E
P=
A
P=
rt
1+
100
15 600
5×4
1+
100
15 600
= $13 000
=
1.2
= M
PL
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.
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.
SA
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?
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Chapter 6 — Financial arithmetic
239
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?
E
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?
6.4
M
PL
13 How long does it take for $2400 invested at 12% per annum simple interest to earn
$360 interest?
Compound interest
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
SA
$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 = $332.75
And so on.
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240
Essential Standard General Mathematics
Continuing in this way, we can build a table of values. For comparison, the amount of the
money if invested at 10% per annum simple interest is also shown.
Amount of investment ($)
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
M
PL
E
Year (n)
0
1
2
3
4
5
6
7
8
9
10
Compound interest at 10%
per year
Simple interest at
10% per year
650
600
550
500
450
400
350
300
250
SA
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.
1 2 3 4 5 6 7 8 9 10
Year
From the graph we see that the growth in the investment over time when interest is
compounded is clearly not linear, rather, it curves upwards and away from the simple interest
graph at an ever increasing rate.
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 previous example, the principal was increased by 10% at theend of each year.
10
= 1.1.
The multiplying factor to increase a quantity by 10% is 1 +
100
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Chapter 6 — Financial arithmetic
241
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
E
It can be seen that a formula for calculating the amount of an investment earning compound
interest can be written as follows.
M
PL
In general, the amount of the investment is given by
r t
A = P 1+
100
$A = final amount
where
$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 $A.
To find the amount of interest earned, we need to subtract the initial investment from the final
amount.
SA
The interest ($I ) that would result from investing $P at r% per annum, compounded
annually for t years, is given by:
r t
−P
I = 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.
b Subtract the principal from this amount
to determine the interest earned.
8 3
r t
= 2000 × 1 +
A = P × 1+
100
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 both the tables and graphing facility
of the calculator.
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242
Essential Standard General Mathematics
How to investigate compound interest problems using the TI-Nspire CAS
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.
c Show that the graph of the amount of money accumulated curves up wards.
8
A = 2000 × 1 +
100
t
E
Steps
1 Substitute P = $2000 and r = 8 into the
formula for compound interest.
M
PL
2 Start a new document (by pressing / + N )
and select 3:Add Lists & Spreadsheet.
Name the lists time (to represent time in years)
and amount.
Enter the data values 1–10 into the list named
time, as shown.
Note: You can also use the sequence command
to do this.
3 Place the cursor in the grey formula cell in the
list named amount and type in
= 2000 × (1 + 8 ÷ 100)∧ time
Note: You can also use the
from the variable list.
key and paste time
SA
Press enter to display the values as shown.
By scrolling down the table we can see
that the
a amount of money accumulated after
3 years is $2519.42
b interest earned = $2519.42 − $2000
= $519.42
4 Press
/5:Add Data & Statistics and plot the
graph as shown.
Notes:
1 To connect the data points: Press / + b
and select 2:Connect Data Points.
2 To display a value: Place the cursor on the data
point and hold . Press
before moving to
another data point.
3 You can use / + b and select
1:Zoom/1:Window Settings and set the
Ymin to 0, if you prefer.
c From the plot we see that, for compound
interest, the graph of amount of money
accumulated curves upwards with time.
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Chapter 6 — Financial arithmetic
243
How to investigate compound interest problems using the ClassPad
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.
c Show that the graph of the amount of money accumulated curves up wards.
2 To form a table of values, open the
M
PL
Sequence (
) application.
Select the Explicit tab and move the
cursor to the box opposite an E: and
type 2000 × (1 + 8/100)∧ n. The n
is found in the toolbar ( ).
Press E to confirm your entry
(which is indicated by a tick in the
square to the left of an E:).
8 t
A = 2000 × 1 +
100
E
Steps
1 Substitute P = $2000 and r = 8
into the formula for compound
interest.
SA
3 Tap # from the toolbar to view a
table of values.
By scrolling down the table we
can see that the
a amount of money accumulated
after 3 years is $2519.42
b interest earned =
$2519.42 − $2000 = $519.42
4 To graph the sequence of simple
interest values, select the Sequence
Grapher ( ) icon from the
toolbar.
Note: To define the graph window
scale, select the View Window (6)
icon from the toolbar and set the
values as shown. Tap OK to confirm
your settings.
c From the plot we see that, for
compound interest, the graph of
amount of money accumulated
curves upwards with time.
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244
Essential Standard General Mathematics
How would our answer to Example 12 change if, instead of the interest being added to the
account at the end of each year (called compounded annually), the interest is added every
3 months (called compounded quarterly)? Compounding 8% quarterly means that the
interest rate for the 3-month period is reduced to 2%. However, the number of times the
interest is added increases from three to 12, since the interest is added four times each year.
This situation is shown in the table below.
E
M
PL
Year
0.25
0.5
0.75
1
1.25
1.5
1.75
2
2.25
2.5
2.75
3
Amount of investment ($)
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 stated previously to take into account situations in which
interest is compounded, or adjusted, other than annually.
In general:
SA
r/n
Amount of the investment, A = P × 1 +
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.
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Chapter 6 — Financial arithmetic
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.
r /n
A = P × 1+
100
n t
12/12 12×2
= 5000 × 1 +
100
= 5000 × (1.01)24
= $6349 to the nearest dollar
I = A − P = $6349 − $5000
= $1349
E
Solution
245
Calculating the principal
M
PL
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
SA
How much money should be invested at 8% per annum compound interest, compounding
monthly, if $5000 is needed in 2 years time?
Solution
Substitute A = $5000, r = 8, n = 12
and t = 2 into the formula giving the
principle.
$5000
A
n t = r /n
8/12 12×2
1+
1+
100
100
= $4263 to the nearest dollar
P =
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.
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246
Essential Standard General Mathematics
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.
E
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.
b Compound interest calculated annually
d Compound interest calculated monthly
M
PL
a Simple interest
c Compound interest calculated quarterly
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?
SA
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?
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.
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Chapter 6 — Financial arithmetic
247
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.
Flat rate depreciation
Depreciation and book value
E
6.5
M
PL
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 − depreciation
SA
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.
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.
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248
Essential Standard General Mathematics
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.
E
V =P−D=P−
M
PL
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.
SA
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 =
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.
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Chapter 6 — Financial arithmetic
249
How to determine flat rate depreciation and book value using the TI-Nspire CAS
24 000 × 10 × t
100
= 2400t
24 000 × 10 × t
V = 24 000 −
100
D =
M
PL
Steps
1 Substitute P = $24 000 and r = 10 into
the formulae for depreciation and book
value under flat rate depreciation.
E
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?
2 Start a new document (by pressing / +
N ) and select 3:Add Lists &Spreadsheet.
Name the lists time, depreciation, and
book value.
Hint: Use / +
for the underscore or just
write as bookvalue.
3 Enter the data values 1–10 into the list time.
SA
4 Move the cursor to the grey formula cell of
the list depreciation and type in
= (24 000 × 10 × time)/100
Press enter to calculate the values for
depreciation.
Move cursor to the grey formula cell of the
list book value and type in
= 24 000 − (24 000 × 10 × time)/100
Press enter to calculate the values for
book value.
Note: An alternative formula to use to
calculate the list book value would be
= 24 000 −depreciation
Hint: You can use the
key to display the
variable list rather than retyping the list names.
5 Write your answers.
a After 1 year, depreciation = $ 2400
b After 4 years, depreciation = $ 9600
c After 1 year, book value = $ 14400
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250
Essential Standard General Mathematics
How to determine flat rate depreciation and book value using the ClassPad
24 000 × 10 × t
100
= 2400t
24 000 × 10 × t
V = 24 000 −
100
D =
M
PL
Steps
1 Substitute P = $24 000 and
r = 10 into the formulae for
depreciation and book value
under flat rate depreciation.
E
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?
2 To form a table of values, open
the Sequence (
) application.
Select the Explicit tab.
Opposite
r a E: type in 2400n
n
r b E: type in 24 000 – 2400n
n
The n is found in the toolbar
( ).
Press E to confirm your entries
(which is indicated by ticks in
the squares to the left of an E:
and bn E:).
SA
3 Tap # from the toolbar to view
the table of values.
4 Write your answers.
Example 16
a After 1 year, depreciation = $ 2400
b After 4 years, depreciation = $ 9600
c After 1 year, book value = $ 14400
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 against time.
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Chapter 6 — Financial arithmetic
251
Solution
a
b To find the flat rate of depreciation,
the amount of the annual depreciation
is expressed as a percentage of the
original cost.
Depreciation per year =
r =
24 000
= $2400
10
100
2 400
×
= 8%
30 000
1
30 000 × 8 × t
100
= 30 000 − 2400 t
V = 30 000 −
M
PL
c To draw a graph of the book value
against time:
1 Substitute P = $30 000 and
r = 8 into the formula for flat
rate depreciation (book value (V)).
2 Use a graphics calculator to plot
V (book value) against t (time).
Total depreciation = 30 000 − 6000 = $24 000
E
1 Work out the total depreciation
over the 10-year period.
2 Determine how much this is
per year.
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
SA
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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252
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?
E
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.
M
PL
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.
SA
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 $122 870, and its value depreciates by 12.5% per year by the flat rate method.
a
b
c
d
Calculate the annual depreciation of the system.
Calculate the book value of the system after 4 years.
Draw a graph to show book value against time.
Estimate the number of years it would take for the book value of the system to be zero.
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Chapter 6 — Financial arithmetic
6.6
253
Reducing balance depreciation
E
In reducing balance depreciation, the annual depreciation is not a fixed amount but is
calculated as a percentage of the book value at the beginning of each year. The amount of
depreciation is largest in the first year, becoming smaller in each subsequent year. Reducing
balance depreciation is the equivalent, but opposite, situation to compound interest, where an
investment increases by a constant percentage each year. We can apply some of our knowledge
of compound interest to this situation. Earlier we established the following formula:
The amount of money ($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
M
PL
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.
SA
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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254
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 a TI-Nspire CAS
15
V = 30 000 × 1 −
100
t
15
D = 30 000 − 30 000 × 1 −
100
t
M
PL
Steps
1 Substitute P = $30 000 and r = 15 into the
formulae for book value and depreciation
under reducing balance depreciation.
E
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? By how much has it depreciated in
value? Draw a graph of book value against time for 10 years.
2 Start a new document (by pressing / + N )
/3:Add Lists & Spreadsheet.
and select
Name the lists time, book value, and
depreciation.
Hint: Use / +
as bookvalue.
for the underscore or just write
3 Enter the data values 1–10 into the list time.
SA
4 To calculate book value
a Move the cursor to the grey formula cell of
the list book value and type in
= 30 000 × (1 − 15 ÷ 100)∧ time
b Press enter to list the values for book value.
From this list, we see that, after 4 years, the
book value of the machine is $15 660, to the
nearest dollar.
5 To determine depreciation
a Move the cursor to the grey formula cell of
the list depreciation and type in
= 30 000 − 30 000 ÷ (1 − 15/100)∧ time
Note: An alternative formula to use to calculate
the list depreciation would be
= 24 000 − book value.
b Press enter to list the values for depreciation.
From this list, we see that, after 4 years, the
machine depreciates in value by $14 340, to
the nearest dollar.
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Chapter 6 — Financial arithmetic
255
E
6 Use the Data & Statistics application to
construct a graph of book value against time.
The graph shows that book value decreases
in a non-linear way with time.
How to determine reducing balance depreciation and book value using a ClassPad
M
PL
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? By how much has it depreciated in
value? Draw a graph of book value against time for 10 years.
Steps
1 Substitute
P = $30 000 and r = 15 into
the formulae for book value and
depreciation under reducing
balance depreciation.
2 To form a table of values, open
15
V = 30 000 × 1 −
100
t
15
D = 30 000 − 30 000 × 1 −
100
t
SA
the Sequence (
) application.
Select the Explicit tab.
Opposite
r a E: type in
n
30 000 × (1 − 15/100)∧ n
and press E.
r b E: type in 30 000 − a E
n
n
and press E.
Note: To obtain an E, tap n,an in the
menu bar and select an E (i.e. the
depreciation value).
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256
Essential Standard General Mathematics
3 Tap # from the toolbar to view
the table of values.
Note: Tap table values for a higher
degree of accuracy. A more accurate
value is then displayed at the bottom
of the screen.
E
From the table we see that, after
4 years, the
a book value of the machine is
$15 660, to the nearest dollar
b machine depreciates in value
by $14 340, to the nearest
dollar
M
PL
4 To graph the sequences of book
and depreciation values, select
the Sequence Grapher ( ) icon
from the toolbar.
Note: To define the graph window
scale select the View Window (6)
icon from the toolbar and set the
values as shown. Tap OK to confirm
your settings.
The graph shows that book value
decreases in a non-linear way
with time.
SA
How do the two methods of depreciation compare? The graph below shows book value plotted
against time for both the flat rate method and the reducing balance method, to allow
comparison, using the values from Examples 16 and 17. It can be clearly seen that reducing
balance depreciation is not linear. This is not unexpected, since reducing balance depreciation
is the equivalent but opposite situation to compound interest.
30000
Book value ($)
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
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Chapter 6 — Financial arithmetic
257
depreciation reduces from year to year until eventually the two lines cross. The points have
been joined to show this.
Exercise 6F
E
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.
M
PL
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.
SA
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.
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258
Essential Standard General Mathematics
6.7
Hire-purchase
Flat interest rate
E
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.
M
PL
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 =
Calculating the interest and flat rate when t = 1
SA
Example 18
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.
Total paid = deposit + repayments
= 370 + 12 × 141.50
= $2068
Interest paid = 2068 − 1850
= $218
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Chapter 6 — Financial arithmetic
2 Since the unit of time is years, the
interest rate can be written as the
interest rate per annum.
rf =
100I
100 × 218
=
Pt
1480 × 1
= 14.73 . . .
Interest rate = 14.73% per annum
E
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 .
259
M
PL
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 is also the flat
rate of interest per annum. The distinction should be clear in the next example, in which 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.
SA
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.
Total paid = deposit + repayments
= 400 + 6 × 185
= $1510
Interest paid = 1510 − 1400
= $110
rf =
100 × 110
100I
=
Pt
1000 × 0.5
= 22%
Interest rate = 22% per annum
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260
Essential Standard General Mathematics
Exercise 6G
a the total cost of buying the racquet by
hire-purchase
b the extra cost of buying by hire-purchase.
E
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:
M
PL
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.
SA
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.
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Chapter 6 — Financial arithmetic
261
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 $112.50 over 2 years. Calculate:
E
b the flat rate of annual interest.
SA
M
PL
a the total interest paid
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Essential Standard General Mathematics
Key ideas and chapter summary
Percentage incresase or decrease is the amount of the increase or
decrease expressed as a percentage of the original value.
amount of change 100
×
Percentage change =
original value
1
Simple interest
Simple interest is paid on an investment or loan on the basis of the
original amount invested or borrowed, called the principal (P). The
amount of simple interest is constant from year to year, and thus is
linearly related to the term of the investment. The amount of interest
earned ($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 $I is the amount of interest earned.
M
PL
E
Percentage increase
or decrease
Amount of the
investment or loan
(simple interest)
Minimum monthly
balance
The lowest amount an account contains in each calendar month is its
minimum monthly balance.
Compound interest
Under compound interest, the interest paid on a loan or investment is
credited or debited to the account at the end of each period, and the
interest for the next period is based on the sum of the principal and
previous interest. The amount of the compound interest increases each
year, and thus there is a non-linear relationship between compound
interest and the term of the investment.
The amount of interest earned ($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
SA
Review
262
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 ($P) for t years at r % per annum,
investment or loan
(compound interest) compounded n times a year, given by:
r/n nt
A = P × 1+
100
Depreciation
Depreciation is the amount by which the value of an item decreases over
time.
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Chapter 6 — Financial arithmetic
263
Scrap value
The scrap value of an item is the value at which it is no longer of value
to the business, so it is replaced.
Flat rate
depreciation
Flat rate depreciation is when the value of an item is reduced by the
same percentage of the purchase price, or amount, for each year the item
is in use. It is equivalent, but opposite, to simple interest.
The depreciation ($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 $D is the amount of depreciation.
M
PL
Book value
(flat rate
depreciation)
E
The book value of an item is its depreciated value.
Reducing balance depreciation is when the value of an item is reduced
by a constant percentage for each year it is in use. It is the equivalent,
but opposite, situation to compound interest.
The depreciation ($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.
SA
Reducing balance
depreciation
Cambridge University Press • Uncorrected Sample Pages • 978-0-521-74049-4
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Book value
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Essential Standard General Mathematics
Skills check
M
PL
E
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.
SA
Review
264
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
Cambridge University Press • Uncorrected Sample Pages • 978-0-521-74049-4
2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
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Chapter 6 — Financial arithmetic
265
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%
E
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
M
PL
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
SA
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%
Cambridge University Press • Uncorrected Sample Pages • 978-0-521-74049-4
2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
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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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.
E
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?
M
PL
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?
SA
Review
266
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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Chapter 6 — Financial arithmetic
267
E
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?
M
PL
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.
SA
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.
Cambridge University Press • Uncorrected Sample Pages • 978-0-521-74049-4
2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
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.