Chapter 14 FM Page 645 Monday, November 13, 2000 3:32 PM
Growth
and
decay
14
VCE coverage
Area of study
Units 3 & 4 • Business
related
mathematics
In this cha
chapter
pter
14A Growth and decay
functions
14B Compound interest
formula
14C Finding time in
compound interest using
trial and error
14D Finding time in
compound interest using
logarithms
14E Flat rate depreciation
14F Reducing balance
depreciation
14G Unit cost depreciation
14H Inflation
Chapter 14 FM Page 646 Monday, November 13, 2000 3:32 PM
646
Further Mathematics
Introduction to growth and decay
Certain quantities in (for example) nature and business may change in a uniform
(constant) way over time. A change may be an increase, as in the case of the value of
an investment such as a house, or it may be a decrease, like the fall in the population of
an endangered species.
Growth and decay can often be modelled by equations or graphs. These in turn can
be used to analyse the situations being modelled and to make predictions about them.
For instance, if we know an equation that relates the falling value of a car to time, the
future value of the car can be determined. (Falling values of this type are called
depreciation.)
This chapter investigates the general principles of growth and decay and takes a
more detailed look at specific examples in the business world, such as compound
interest, depreciation and inflation.
Growth and decay functions
Straight line and exponential growth
If a quantity increases in size over a period of time, it is growing. This growth process
may be straight line growth or exponential growth.
Straight line growth
A quantity may increase by a fixed amount for each time unit; that is, a fixed amount is
added. Hence, the relationship is linear.
A stamp collection worth $5000, for instance, may increase in value by $200 each
year. This growth relationship can be written as the equation
V = 5000 + 200T
where V = value of collection
T = time in years
The general form of linear growth function is
y = a + bx
where y is the dependent variable
x is the independent variable (usually time)
a is the initial or starting value of y
b is the rate of growth.
Value ($)
A graph of the equation could be drawn to represent this situation, like the one
shown.
By using the graph or the equation we can analyse the situation, for example to find
the collection’s value at any future time.
V
6000
5800
5600
5400
5200
5000
0 1 2 3 4 5 T
Time (years)
Chapter 14 FM Page 647 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
647
WORKED Example 1
Write an equation which describes the relationship between the number, N, of bees and
time, T, if the population of 500 bees is known to be increasing by 20 each month. Find:
a the size of the population in a year’s time
b when the population will have doubled.
Using the equation
THINK
WRITE/DISPLAY
a
P = 500 + 20T
1
2
3
b
1
2
Let population be P and
time in months be T.
Substitute T = 12
(12 months = 1 year).
Write your answer.
Substitute P = 1000 and
solve.
Write your answer.
When T = 12,
P = 500 + 20 × 12
P = 740 bees in one year’s time
When P = 1000,
1000 = 500 + 20T
T = 25 months
The population will double after 2 years and 1 month.
Using the graph
Refer to chapter 2 Bivariate data (worked examples 8 and 10) and chapter 3
Introduction to regression (worked example 4) for further details on using a
graphics calculator for plotting relationships between variables.
1
Draw a graph of population
against time by plotting
points.
Set up tables in STAT and
then graph using 2nd
[STAT PLOT].
T
0
10
20
30
P
500
700
900
1100
Continued over page
Chapter 14 FM Page 648 Monday, November 13, 2000 3:32 PM
Further Mathematics
THINK
2
WRITE/DISPLAY
Find T = 12 and read from
the graph.
N
1000
Number of bees
648
800
600
400
0
10 20 30 T
Time (months)
Use STAT , select CALC to
fit a straight line and then
use TRACE to move along
the function to T(x) = 12
and read P(y) = 740.
From graph P = 740 bees
Find N = 1000 and read
from the graph.
Use TRACE to move to
P(y) = 1000 and read
T(x) = 25.
From graph T = 25 months
Exponential growth
Growth is exponential when the quantity present is
multiplied by a constant for each unit time interval.
This constant is called the growth or compounding
factor and is greater than one.
Consider the situation represented in the graph at right
where the changes to a certain population of bacteria
over a period of time are displayed.
It can be seen that the population is increasing but also
that the rate of growth is increasing; that is, the graph is
getting steeper.
N
16
Number of bacteria ('000)
3
12
8
4
0
0
1 2 3 4 5 T
Time (hours)
Chapter 14 FM Page 649 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
649
The readings taken from the graph are:
Time, T
Number, N
0
1
2
3
4
5
500
1000
2000
4000
8000
16 000
WORKED Example 2
The number of bacteria in a group was recorded over a 6-hour
period and a graph of the data is shown at right.
a Determine whether the population growth is exponential.
b If it is exponential, write an approximate equation for the
relationship.
THINK
WRITE
a
1
Find the initial population of bacteria,
N0(T = 0), and population after 1 hour,
N1
N1(T = 1), and evaluate the ratio, ------ .
N0
3
Find N2. For exponential growth, 1.5 is
the growth or compounding factor. That
is, N2 = N1 × 1.5.
Find N3. It should be that N3 = N2 × 1.5.
4
Repeat the process.
5
By using 1.5 as the growth or
compounding factor, the calculated
values compare favourably with values
from the graph.
2
b In the equation the initial value is multiplied
by the growth or compounding factor, 1.5,
raised to the power of unit time interval,
that is, y = ka x.
Number of bacteria ('000)
The population is doubling every hour, so the growth or compounding factor is 2.
The relationship could be written as N = 500 × 2T. This means that as T increases
by 1, the number of bacteria increases by a factor of 2.
In general, the exponential growth function has an equation of the form:
y = ka x, where k and a are constants
a > 1 is the growth or compounding factor
k is the initial value of y (when x = 0).
N
16
14
12
10
8
6
4
2
0
a From the graph,
N0 = 2000,
0 1 2 3 4 5 T
N1 = 3000
Time (hours)
N 1 3000
------ = -----------N 0 2000
= 1.5
N1 = 3000
N2 = 3000 × 1.5
= 4500
N3 = 4500 × 1.5
= 6750
N4 = 6750 × 1.5
= 10 125
N5 = 10 125 × 1.5
= 15 150
Growth or compounding factor = 1.5.
Growth is exponential.
b NT = 2000(1.5)T
Once an equation has been determined for a relationship it can be used to analyse the
situation.
Chapter 14 FM Page 650 Monday, November 13, 2000 3:32 PM
650
Further Mathematics
WORKED Example 3
The cost, C ($), of a deluxe puff pastry after time, T (years), is given by the equation
C = 0.8(1.6)T. Use the equation to complete the table below and plot a graph of cost
against time.
Time, T
0
1
2
3
4
5
Cost, C
THINK
1
Write the equation and substitute T = 0.
2
Substitute T = 1 and evaluate.
3
Substitute T = 2 and evaluate.
4
C = 0.8(1.6)T
When T = 0, C = 0.8(1.6)0
= 0.8 × 1
= 0.8
When T = 1, C = 0.8(1.6)
= 1.28
When T = 2, C = 0.8(1.6)2
= 2.05
T
0
C
0.8
Draw the graph, joining the points with
a smooth curve.
1
2
3
4
6
4
2
0
5
1.28 2.05 3.28 5.24 8.39
C
8
Cost ($)
5
Repeat for T = 3 to 5 and complete the
table of T and C values.
WRITE
0 1 2 3 4 5 T
Time (years)
Note: xy is the power function on a scientific calculator. It may be represented as y x or a x.
On a graphics calculator, use the power button ^ , for example, 5 ^ 3 = 53 = 125.
The other alternative to that outlined above when finding a set of continuous points is
simply to multiply by the growth or compounding factor each time.
Chapter 14 FM Page 651 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
651
Alternative — Using a graphics calculator
Refer to chapter 3 Introduction to regression, worked example 4, for further
details.
THINK
1
Enter the equation in Y= .
2
Set up WINDOW so that the integer values
of x are plotted.
3
GRAPH the function and use TRACE to
investigate values of C.
4
Complete the table or use the TABLE
feature of the calculator.
DISPLAY
Chapter 14 FM Page 652 Monday, November 13, 2000 3:32 PM
652
Further Mathematics
Meaning of growth or compounding factor
The growth or compounding factor takes into account the quantity that we start with as
well as the amount of the increase for the unit time interval.
If the growth or compounding factor is 1.15 then the 1 accounts for the initial
quantity and the 0.15 accounts for the increase.
That is:
1.15 = 1 + 0.15
15
= 1 + -------100
= 100% + 15%
= 115%
So a growth or compounding factor of 1.15 means an increase each unit time interval
of 15% of the previous value.
For example, if the cost of an article increases by 5% per annum (p.a.) then the
growth or compounding factor would be 1.05 (or 100% + 5% = 105%) each year.
WORKED Example 4
The cost, C ($), of a $6.50 cricket ball increases by 3% each year. Write the equation for
the relationship between the cost and time, T (years), and use it to find the cost of the ball
after 8 years.
THINK
WRITE
Find
the
growth
or
compounding
factor
Growth factor = original amount + increase
1
per year.
= 100% + 3%
= 103%
= 1.03
x
C = 6.5(1.03)T
2 Write the equation in the form y = ka ,
where k is the initial value.
When T = 8,
3 Substitute T = 8 and evaluate.
C = 6.5(1.03)8
C = $8.23
Cost of ball after 8 years is $8.23.
4 Write a summary statement.
Decay
If a quantity decreases in size over a period of time, it is decaying. This decay process
may also be linear in nature or exponential, as was the case with growth.
Straight line decay
Number of starfish ($)
In this situation a quantity decreases by a fixed amount for each time unit interval, that
is, a fixed amount is subtracted. Since the quantity is decreasing over time the slope of
the straight line is negative.
Suppose the number of starfish on a reef is 8000 at
N
8000
present but the population is decreasing by 250 each year.
The decay relationship can be written as the equation:
7000
N = 8000 − 250T
where N = number of starfish
T = time in years.
6000
A graph of the equation like the one shown could be
drawn to represent this situation.
50000
The graph or the equation can be used in the same way
0 2 4 6 8 10 T
they were used for growth situations (see worked example 1).
Time (years)
Chapter 14 FM Page 653 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
653
Exponential decay
Decay can be exponential in a similar way to growth. That is, the quantity present can
be multiplied by the growth or compounding factor for each unit time interval. It may
seem to be a contradiction to say that there is a growth or compounding factor associated with decay; however, for decay the growth or compounding factor, a, is less than
one, whereas for exponential growth the growth or compounding factor is greater than
one. In comparison,
the general equation for growth is:
exponential y = ka x where k, a are constants, a >1
the general equation for decay is:
exponential y = ka x where k, a are constants, a <1
The procedure that was adopted to analyse situations of exponential growth can be
applied in the same way to exponential decay.
WORKED Example 5
v
Value ($)
The graph shows how the value of a car decreases
over a 5-year period.
a Determine whether the decay is exponential.
b If it is exponential, write an approximate
equation for the relationship between value, v,
and time, T.
20 000
18 000
16 000
14 000
12 000
10 000
8 000
6 000
4 000
2 000
0
0
THINK
a
1
v1
Find v0 and v1. Evaluate ----- .
v0
3
Find v2. If exponential decay then 0.6 is growth
or compounding factor.
So, v1 × 0.6 = v2.
Find v3. It should be that v2 × 0.6 = v3.
4
Find v4. It should be that v3 × 0.6 = v4.
2
By using 0.6 as the growth or compounding
factor, the calculated values compare favourably
with the values from the graph.
b The decay equation is of the form y = ka x where
a < 1. That is, the initial value, v0, is multiplied by
the growth or compounding factor, 0.6, once for
each unit time interval.
5
1
2
3
4
Time (years)
5
WRITE
a From the graph, v0 = 20 000,
v1 = 12 000
v 1 12000
----- = ---------------v 0 20 000
= 0.6
v1 = 12 000
v2 = 12 000 × 0.6
= 7200
v3 = 7200 × 0.6
= 4320
v4 = 4320 × 0.6
= 2592
Decay factor is 0.6.
Decay is exponential.
b
v = 20 000(0.6)T
T
Chapter 14 FM Page 654 Monday, November 13, 2000 3:32 PM
654
Further Mathematics
Meaning of growth or compounding factor
As previously mentioned, the growth or compounding factor takes into account the
initial quantity and, in the case of decay, also takes into account the decrease for the
unit time interval.
Consider a growth or compounding factor of 0.85.
0.85 = 1 − 0.15
=1−
15
--------100
= 100% − 15%
= 85%
That is, 1 represents the initial quantity and 0.15 represents the actual decrease per
unit time interval.
So a growth or compounding factor of 0.85 means a decrease each unit time interval
of 15% of the previous value.
WORKED Example 6
The number, N, of tigers in a certain population is decreasing by 6% each year from an
initial population of 425. Write an equation for the relationship between the number and
time, T (years), and use it to find how many tigers there will be after 8 years.
THINK
WRITE
Find the compounding factor per year.
Compounding factor
= original amount − decrease
= 100% − 6%
= 0.94
Write the equation in the form y = ka x,
where k is the initial value.
N = 425(0.94)T
3
Substitute T = 8 and evaluate.
When T = 8,
N = 425(0.94)8
N = 259
4
Write a summary statement.
After 8 years there will be 259 tigers.
1
2
Radioactive decay
Radioactive decay is an example of exponential decay. Nuclear radiation is emitted
from many different chemical elements. Uranium and plutonium are probably the best
known radioactive elements.
The original element decays to a different element over a period of time, which means
that the amount of the radioactive element decreases. For example, uranium-238 decays
to thorium-234 and an alpha particle is emitted.
For the purposes of this work on growth and decay, all we need to remember is that,
as with other cases of exponential decay, the rate of decrease (in mass or amount of the
element in this case) is determined by the growth or compounding factor, which, of
course, is less than one. The compounding factor is unique to each radioactive element.
Chapter 14 FM Page 655 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
655
remember
remember
1. Growth and decay can be linear or exponential.
2. Linear growth and decay can be represented by the equation
y = a + bx where y is the dependent variable
x is the independent variable (usually time)
a is the initial or starting value of y
b is the rate of growth or decay.
3. Exponential growth and decay means an initial value multiplied by a growth or
compounding factor for each unit time interval.
4. Exponential growth and decay can be represented by the equation
y = kax,
where a = growth or compounding factor
a > 1 for growth, a < 1 for decay
k = the initial or starting value of y.
14A
Example
1
1 Write an equation that describes the relationship between the variables in each case,
then solve each of the following problems.
a The value of a Beatles record, currently worth $50, will increase by $10 every
year. Assuming this relationship will continue indefinitely:
i what will it be worth in 15 years time?
ii when will its value have doubled?
b From an initial population of 600 ants in a nest, the number grows by 30 each
month indefinitely.
i What will be the population in 2 years’ time?
ii When will the population reach 900?
c If $1200 is invested for 10 years and earns simple interest of $120 each year:
i what will be the amount altogether after 8 years?
ii when will the total amount ($1200 + interest) be $1800?
d A coin collection, currently valued at $1560, will increase in value by 5% of the
current value each year. Assuming the trend will continue:
HEET
14.1
SkillS
WORKED
Growth and decay functions
i what will its value be in 8 1--2- years?
ii when will its value reach $2000?
d
hca
Mat
EXCE
reads
L Sp he
et
2 Verify your answers to question 1 by graphing each relationship.
Function grapher
WORKED
Example
2a
3 Determine whether the situations described below represent exponential growth.
a The price of a certain food item over a 5-year period is detailed below.
Year
Price ($)
1
2
3
4
5
0.85
1.02
1.43
2.00
2.40
Chapter 14 FM Page 656 Monday, November 13, 2000 3:32 PM
656
Further Mathematics
b The value of a coin collection over a period of 5 years is shown below.
Year
Value ($)
c
1
2
3
4
5
500
550
605
665.50
732.05
The number of rabbits in a population over a 3-month period is shown in the graph.
Number of rabbits
N
800
600
400
200
0
0
1
2
3 T
Time (months)
Amount in account, A ($)
d The amount in an investment account after each of 6 years is shown below.
759.38
800
600
506.25
400
200
337.50
100
0
1991
WORKED
Example
2b
150
225
1992 1993 1994
Year
1995
1996
4 Using T to represent time in hours, write an equation to describe the increase in
number, N, of a population of bacteria if initially there are:
a 1000 bacteria and the number increases by a factor of 2 each hour
b 2000 bacteria and the number increases by a factor of 1.4 each hour
c 860 bacteria and the number increases by a factor of 1.25 each hour
d 1250 bacteria and the number increases by 150% each hour
e 2300 bacteria and the number increases by 200% each hour.
5 Using T to represent time in years, write an equation to describe the increase in value,
V ($), of a painting if it was bought for:
a $700 and its value increased by a factor of 1.1 each year
b $1100 and its value increased by a factor of 1.05 each year
c $5000 and its value increased by a factor of 1.16 each year
d $2750 and its value increased to 120% each year
e $380 and its value increased to 108% each year.
6 Using T to represent time in years, write an equation to describe the increase in cost,
C ($), of buying:
a a $25 000 new car if the cost of purchase grew by a factor of 1.07 each year
b a $1.50 loaf of bread if its cost grew by a factor of 1.03 each year
c a $250 bike if its cost grew by a factor of 1.05 each year
d a new $29.95 CD if its cost grew to 106% each year
e a $7.20 packet of cigarettes if its cost grew to 110% each year.
Chapter 14 FM Page 657 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
657
7 Using T to represent time in years, write an equation to describe the increase in the
amount, A ($), in an investment account if it was initially:
a $2000 and it increased by a factor of 1.16 each year
b $850 and it increased by a factor of 1.12 each year
c $1900 and it increased by a factor of 1.06 each year
d $25 000 and it increased to 109% each year
e $12 600 and it increased to 115% each year.
Example
sheet
3
8 a The amount, A ($), in an investment account after time, T (years), is given by the
L Spread
XCE
equation, A = 1500(1.08)T. Using this equation, copy and complete the table below
Plotting
and plot a graph of amount against time.
E
WORKED
relations
T
0
1
2
3
4
5
6
7
Math
cad
A
T
0
1
2
3
4
5
6
7
Growth
and
decay
GC pro
gram
b The value, V ($), of an antique chair over time, T (years), is given by the equation
V = 850(1.06)T. Use the equation to complete the table below and plot a graph of
value against time.
Exponent
V
c
The number, N, of possums in a National Park over time, T (months), is given by
the equation, N = 400(1.02)T. Use the equation to find the values of N for T values
from 0–7 and plot a graph of number against time.
d If the cost, C ($), of a new car is given by C = 17 000(1.1)T, where T is the time in
years, plot a graph of cost against time for 0–5 years.
9 Given the exponential growth equations below, solve the problems provided.
a If N = 650(1.59)T, find N if T = 5.
b If C = 210(1.15)T, find C if T = 9.
c If A = 3600(1.09)T, find A ($) if T = 7 years.
d If V = 1050(1.02)T, find V ($) if T = 20 years.
e If N = 2500(1.85)T, find N (number of bacteria) if T = 12 hours.
Example
11 The value, V ($), of a piece of land bought for $12 000 increases by 4% each year.
a Write an equation for the relationship between the value and time, T (years).
b Use the equation to find how much the land is worth after 10 years.
12 multiple choice
The number, N, of bacteria in a colony increases by 10% per hour from an initial
colony of 1200. After 15 hours the number of bacteria present would be closest to:
A 250
B 1320
C 1350
D 3000
E 5010
HEET
SkillS
10 The amount, A ($), in an investment account is initially $2000 and increases by
8% p.a.
4
a Write the equation for the relationship between the amount and time, T (years).
b Use the equation to find the amount in the account after 6 years.
WORKED
14.2
Chapter 14 FM Page 658 Monday, November 13, 2000 3:32 PM
658
Further Mathematics
13 multiple choice
The cost, C ($), of a can of soft drink, which currently is $1.15, increases by 5.5% p.a.
The cost of the can in 20 years’ time will lie between:
A 30 and 50 cents
B $2.10 and $2.30
C $2.30 and $2.50
D $3.30 and $3.50
E $3.50 and $3.70
14 multiple choice
The equation representing an increase of 9% p.a. in the number of animals, N, in a
certain population which initially numbered 3600 is:
A N = 3600T + 9
B N = 3600T − 9
C N = 3600(1.09)T
T
T
D N = 3600(0.91)
E N = 9(3600)
15 By using the equation that exists between the variables in each case below solve the
given problems.
a The number of elephants in a game reserve was initially 500, but the population is
decreasing by 35 per year. If this trend continues:
i what will be the population in 12 years’ time?
ii when will the population have halved?
b The value of a car, currently worth $12 000, is decreasing by $800 per year. If this
continues:
i what will be its value in 6 years’ time?
ii when will its value be $8800?
c A cylindrical tank, containing water to a height of 3 m, has just been punctured
and water is leaking out so that the water height is falling by 5 cm every minute. If
this trend continues:
i what will be the water height after 15 minutes? (Assume the puncture is below
this height.)
ii when will the height reach 2.65 m?
d The number of wombats on an island is decreasing by a fixed number each year
from an initial population of 366. If this trend continues and the population
reaches 281 after 5 years:
i what is the decay rate per year?
ii what will be the population after 8 years?
iii when will the population reach 162?
e The value of a computer bought for $2500 decreases by a fixed amount each year.
If the value after 4 years is $1740 and this trend continues:
i what is the decay rate?
ii what will be the computer’s value after 7 years?
iii when will its value be $1930?
16 Verify your answers to question 15 by graphing each relationship.
17 Determine whether the situations described below represent exponential decay.
a The value of a car over a 5-year period is detailed below.
5a
WORKED
Example
Year
Value ($)
1
2
3
4
5
30 000
24 000
19 200
15 360
12 288
Chapter 14 FM Page 659 Monday, November 13, 2000 3:32 PM
659
Chapter 14 Growth and decay
b The number in a colony of frogs over a 4-year period is described below.
Year
Number
c
1
2
3
4
2000
1800
1620
1458
The value of a computer over a 6-year period is graphed below.
8000
6000
4800
3375
2531.25
1898.44
1990
1991
1992 1993
Year
1994
1995
d The mass of a radioactive element present over a
4 time-unit interval is given at right. Does this
situation represent exponential decay?
T
Mass of radioactive element (g)
Value ($'000)
V
8000
7000
6000
5000
4000
3000
2000
1000
0
m
800
700
600
500
400
300
200
100
0
0 1
2 3 4 5 t
Time units
18 In each case below, write an equation to represent the decrease in the given variable
over time:
5b
a Amount, A, decreases from 300 by a compounding factor of 1--2- every day.
b Value, V, decreases from $5000 by a compounding factor of 0.75 every year.
c Number, N, decreases from 2500 by a compounding factor of 0.95 each month.
d Mass, m, decreases from 900 g to 80% each minute.
e Value, V, decreases from $850 to 92% each year.
f Number, N, decreases from 15 000 by a compounding factor of 75% each year.
WORKED
Example
19 multiple choice
Which one of the equations below represents a decrease of 17% p.a. in the value,
BV ($), of a boat after time, T (years), if it was bought for $17 800?
A BV = 17 800(1.17)T
B BV = 17 800(0.83)T
C BV = 17 800T 0.83
D BV = 17 800 − 17T
E BV = 17 800 + 17T
20 Given the exponential decay equations below solve the given problems.
a If N = 700(0.85)T, find N if T = 4.
b If A = 10 000(0.92)T, find A if T = 10.
c If V = 1200(0.75)T, find V if T = 7.
d If N = 160(0.96)T, find N if T = 15.
e If A = 185.5(0.52)T, find A if T = 6.
Chapter 14 FM Page 660 Monday, November 13, 2000 3:32 PM
660
Further Mathematics
21 a The mass, m (g), of a radioactive element after time, T (days), is decreasing and is
given by the equation, m = 840(0.76)T. Use this equation to complete the table
below and plot a graph of mass against time.
T
0
1
2
3
4
5
6
7
8
m
b The value, V ($), of a computer after time, T (years), is given by the equation,
V = 2600(0.80)T. Use this equation to complete the table below and plot a graph of
value against time.
T
0
1
2
3
4
5
6
7
8
V
c
The number, N, in a population of seahorses after time, T (months), is given by the
equation, N = 290(0.91)T. Use the equation to find the values of N for T values
from 0–8 and plot a graph of number against time.
d The amount, A (grams), of a radioactive element after time, T (minutes), is
decreasing and is given by the equation, A = 1350(0.70)T. Use the equation to find
the values of A for T values from 0–8 and plot a graph of amount against time.
22 The number, N, of sulphur-crested cockatoos in a certain population is decreasing by
2% each month from an initial number of 1400.
6
a Write an equation for the relationship between the number and time, T (years).
b Use the equation to find the number of cockatoos after 1 year.
WORKED
Example
23 The value, V ($), of a washing machine bought for $899 decreases by 30% p.a.
a Write an equation for the relationship between the value and time, T (years).
b Use the equation to find how much the machine is worth after 5 years.
24 multiple choice
The value, V ($), of a car bought for $27 500 decreases by 20% each year. The value
of the car after 6 years would be closest to:
A $27 300
B $24 400
C $7200
D $5000
E $300
25 multiple choice
The mass, m (g), of a radioactive sample of sodium decreases from an initial mass of
660 g by 5% each hour. The mass of the sodium left after 24 hours would lie within
the range:
A 610–630 grams
B 530–550 grams
C 510–530 grams
D 180–200 grams
E 50–70 grams
26 A 10-year research program is being carried out on two penguin rookery populations.
Rookery A had 2000 penguins at the start of the study and the population has been
decreasing by 5% p.a. Rookery B had 3000 penguins initially and this population has
been decreasing by 10% p.a.
a On the same set of axes draw population against time graphs for the rookeries over
the 10 years.
b From your graphs, estimate when the two populations will be the same.
c When the populations are the same, what is the population?
Chapter 14 FM Page 661 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
661
Compound interest formula
As you have seen in simple interest calculations, the amount present at the start does
not change throughout the life of the investment. Interest is added at the end.
In contrast, in the previous section graphical and algebraic methods were used to
illustrate the concept of exponential growth whereby an amount increases at regular
intervals over a period of time. This increasing factor was called the growth or compounding factor.
Another example of exponential growth is compound interest.
For investments, interest is added to the initial amount (principal) at the end of an
interest-bearing period. Both the interest and the principal then earn further interest
during the next period, which in turn is added to the balance. This process continues for
the life of the investment. The interest is said to be compounded.
The result is that the balance of the account increases at regular intervals and so too
does the interest earned.
Compound interest is illustrated in the next example.
Consider $1000 invested for 4 years at an interest rate of 12% p.a. with interest compounded annually (added on each year). What will be the final balance of this account?
Time
period
Starting principal,
P ($)
Interest ($)
Balance ($)
1
1000
12% of 1000
= 120
1120
2
1120
12% of 1120
= 134.40
1254.40
3
1254.40
12% of 1254.40 = 150.53
1404.93
4
1404.93
12% of 1404.93 = 168.59
1573.52
So the balance after 4 years is $1573.52.
During the total period of an investment, interest may be compounded many times,
so a formula has been derived to make calculations easier.
In the above example the principal is increased by 12% each year. That is, the end of
year balance = 112% or 1.12 of the start of the year balance.
Now let us look at how this growth or compounding factor of 1.12 is applied in the
example.
Time
period
Balance ($)
1
1120 = 1000 × 1.12
= 1000(1.12)1
2
1254.40 = 1120 × 1.12 = 1000 × 1.12 × 1.12
= 1000(1.12)2
3
1404.93 = 1254.40 × 1.12 = 1000 × 1.12 × 1.12 × 1.12
= 1000(1.12)3
4
1573.52 = 1404.93 × 1.12 = 1000 × 1.12 × 1.12 × 1.12 × 1.12 = 1000(1.12)4
If this investment continued for n years the final balance would be:
(
1000(1.12)n = 1000(1 + 0.12)n = 1000 1 +
)
12 n
--------100
Chapter 14 FM Page 662 Monday, November 13, 2000 3:32 PM
662
Further Mathematics
The answer now is only in terms of information that was known at the start of the
investment. From this pattern we are able to write a general formula that can be used to
calculate compound interest.
A = final or total amount ($)
A = PRn where
P = principal ($)
r
R = growth or compounding factor = 1 + --------100
r = interest rate per period
n = number of interest bearing periods
Note that the formula gives the total amount in an account, not just the interest
earned as in the simple interest formula.
To find the total interest compounded, I:
I = A − P where A = final or total amount ($)
P = principal ($)
Now let us consider how the formula is used.
WORKED Example 7
Find the amount in the account (balance) and the interest earned after $5000 is invested
for 4 years at 6.5% p.a., interest compounded annually.
THINK
WRITE
3
What is n?
What is r?
What is P?
4
Write the compound interest (CI) formula.
5
Substitute known values into the formula.
6
8
Simplify.
Evaluate (to 2 decimal places).
Subtract the principal from the balance.
9
Write a summary statement.
1
2
7
n =4
r = 6.5
P = 5000
r
A = P(1 + --------- )n
100
6.5
= 5000(1 + --------- )4
100
= 5000(1.065)4
A = $6432.33
I =A−P
= 6432.33 − 5000
= $1432.33
The amount of interest earned is $1432.33
and the balance is $6432.33.
In the last example interest was compounded annually. However, in many cases the
interest is compounded more often than once a year, for example semi-annually (twice
each year), quarterly (every 3 months), or weekly. In these situations n and r still have
their usual meanings and we calculate them as follows.
Number of interest periods, n = number of years × number of interest periods
per year
nominal interest rate per annum
Interest rate per period, r = ------------------------------------------------------------------------------------------number of interest periods per year
Note: Nominal interest rate per annum is simply the annual interest rate advertised by a
financial institution.
Chapter 14 FM Page 663 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
663
WORKED Example 8
If $3200 is invested for 5 years at 6% p.a., interest compounded quarterly, find:
a the number of interest bearing periods, n
b the interest rate per period, r
c the balance of the account after 5 years.
THINK
WRITE
a Calculate n.
a
b Convert % p.a. to % per quarter to match
time over which interest is calculated.
Divide r% p.a. by the number of
compounding periods per year, namely 4.
Write as a decimal.
6% p.a.
b r% = -----------------4
= 1.5% per quarter
c
c
1
What is P?
2
Write the CI formula.
3
Substitute known values.
4
Simplify.
Evaluate to 2 decimal places.
Write a summary statement.
5
6
n = 5 (years) × 4 (quarters)
= 20
r = 1.5
P = $3200
r
A = P(1 + --------- )n
100
1.5
= 3200(1 + --------- )20
100
= 3200(1.015)20
A = $4309.94
Balance of account after 5 years is $4309.94.
On occasions when interest is compounded monthly or more often, the value of r is a
recurring decimal. Now the accuracy of the calculation should be maintained, so the
value that has been determined for r is not to be approximated or truncated. Therefore,
the order of the calculation on a calculator is changed slightly.
WORKED Example 9
Find the amount that accrues in an account which pays compound interest (compounded
weekly) at a nominal rate of 5.6% p.a. if $2450 is invested for 3 1--2- years.
THINK
1
Calculate n.
2
Calculate r and retain on the calculator for
step 5.
3
What is P?
WRITE
n = 3.5 × 52
= 182
5.6
r = ------52
= 0.107 69
P = $2450
Continued over page
Chapter 14 FM Page 664 Monday, November 13, 2000 3:32 PM
664
Further Mathematics
THINK
WRITE
4
Write the CI formula and substitute.
5
6
Evaluate to 2 decimal places.
Subtract the principal from the balance.
7
Write a summary statement.
r
A = P(1 + --------- )n
100
182
--------------------- )
= 2450(1 + 0.10769
100
A = $2980.18
I =A−P
= 2980.18 − 2450
= $530.18
Amount accrued after 3 1--2- years is $530.18.
Note: The order of calculation used here can be applied in all previous cases when
finding A; see worked examples 7 and 8.
The situation often arises where we require a certain amount of money by a future
date. It may be to pay for a holiday or to finance the purchase of a car. It is then
necessary to know what principal should be invested now in order that it will increase
in value to the desired final balance within the time available.
WORKED Example 10
Find the principal that will grow to $4000 in 6 years, if interest is added quarterly at
6.5% p.a.
THINK
WRITE
1
Calculate n.
2
Calculate r.
3
What is A?
4
Write the CI formula, substitute and
simplify.
5
Transpose to isolate P.
6
Evaluate to 2 decimal places.
Write a summary statement.
7
n = 6 × 4 = 24
6.5
r = ------4
= 1.625
A = $4000
r
A = P(1 + --------- )n
100
1.625
4000 = P(1 + ------------- )24
100
4000 = P(1.01625)24
4000
P = ---------------------------( 1.01625 ) 24
P = $2716.73
$2716.73 would need to be invested.
Sometimes we know how much we can afford to invest as well as the amount we want
to have at a future date. Using the compound interest formula we can calculate the
interest rate that is needed to increase the value of our investment to the amount we
desire. This allows us to ‘shop around’ various financial institutions for an account
which provides the interest rate we want.
Chapter 14 FM Page 665 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
665
We must first find the interest rate per period, r, and convert this to the corresponding
nominal rate per annum.
WORKED Example 11
Find the interest rate per annum (to 2 decimal places) that would enable an investment of
$3000 to grow to $4000 over 2 years if interest is compounded quarterly.
THINK
WRITE
A = $4000
P = $3000
n =2×4
=8
1
What are A, P and n? For this example n
needs to represent quarters of a year and
therefore r will be evaluated in % per
quarter.
2
Write the CI formula and substitute.
A = PRn
4000 = 3000R8
3
Divide A by P.
4000
------------ = R 8
3000
4
Obtain R to the power of 1, that is, raise
 4--- = ( R 8 ) 8 = R
 3
both sides to the power of 1--8- .
1
--8
1
---
1
---
8
r
 4--- = 1 + ------- 3
100
1
---
5
Isolate r and evaluate.
r
4 8
--------- =  --- − 1
100  3
= 0.0366
r = 3.66
r% = 3.66% per quarter
6
Multiply r by the number of interest
periods per year to get the annual rate
(to 2 decimal places).
Annual rate = r% per quarter × 4
= 3.66% per quarter × 4
= 14.65% per annum
7
Write a summary statement.
Interest rate of 14.65% p.a. is required.
remember
remember
1. Compound interest calculations can be made using the formula
A = PRn where A = final amount ($)
P = principal ($)
r
R = growth or compounding factor = 1 + --------100
r = interest rate per period
n = number of interest bearing periods.
2. For compound interest, I = A − P.
Chapter 14 FM Page 666 Monday, November 13, 2000 3:32 PM
666
Further Mathematics
14B
WORKED
Example
7
d
hca
Mat
EXCE
et
reads
L Sp he
WORKED
Example
8a
SkillS
14.3
1 Use the compound interest formula to find the amount, A, when:
a P = $500, n = 2, r = 8
b P = $1000, n = 4, r = 13
c P = $3600, n = 3, r = 7.5
d P = $2915, n = 5, r = 5.25
e P = $850.20, n = 10, r = 1
f P = $1215, n = 24, r = 0.5
2 Find: i the balance, and ii the interest earned (interest compounded annually) after:
a $2000 is invested for 3 years at 8% p.a.
b $7000 is invested for 4 years at 6% p.a.
c $6000 is invested for 2 years at 5% p.a.
d $1900 is invested for 5 years at 10% p.a.
e $3425 is invested for 6 years at 7.5% p.a.
f $962 is invested for 3 years at 6.25% p.a.
g $1650 is invested for 10 years at 4.95% p.a.
h $2505 is invested for 8 years at 7.12% p.a.
Compound interest
HEET
Compound interest formula
WORKED
Example
8b
WORKED
Example
8c
3 Find the number of interest bearing periods, n, if interest is compounded:
a annually for 5 years
b quarterly for 5 years
c semi-annually for 4 years
d monthly for 6 years
e 6-monthly for 4 1--2- years
f quarterly for 3 years and 9 months
g monthly for 2 1--- years
h fortnightly for 3 1--2- years
4
i daily for 4 years
j weekly for 3 1--4- years.
4 Find the interest rate per period, r, if the annual rate is:
a 6% and interest is compounded quarterly
b 4% and interest is compounded half-yearly
c 11% and interest is compounded 6-monthly
d 18% and interest is compounded monthly
e 7% and interest is compounded quarterly
f 7.5% and interest is compounded quarterly
g 8% and interest is compounded monthly
h 5% and interest is compounded fortnightly
i 9% and interest is compounded weekly
j 8.25% and interest is compounded daily.
5 Find the balance of the account after:
a 2 years if $3000 is invested at 8% p.a., interest compounded quarterly
b 5 years if $2000 is invested at 6% p.a., interest compounded 6-monthly
c 4 years if $5000 is invested at 12% p.a., interest compounded monthly
d 7 years if $1500 is invested at 12% p.a., interest compounded quarterly
e 3 years if $2500 is invested at 7% p.a., interest compounded half-yearly
f 6 years if $3400 is invested at 9% p.a., interest compounded quarterly
g 4 years if $840 is invested at 9% p.a. (compounded monthly)
h 3 1--2- years if $1230 is invested at 6% p.a. (compounded quarterly)
i 2 3--4- years if $965 is invested at 7.5% p.a. (compounded monthly)
j 5 years 2 months if $1625 is invested at 15% p.a. (compounded monthly).
Chapter 14 FM Page 667 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
WORKED
Example
9
667
6 Find the amount that accrues in an account which pays compound interest at a nominal
rate of:
a 7% p.a. if $2600 is invested for 3 years (compounded monthly)
b 10% p.a. if $4100 is invested for 2 years (compounded monthly)
c
8% p.a. if $3500 is invested for 4 years (compounded monthly)
d 5% p.a. if $1850 is invested for 3 1--2- years (compounded fortnightly)
e 11% p.a. if $960 is invested for 5 1--2- years (compounded fortnightly)
f
6.6% p.a. if $1290 is invested for 3 years (compounded fortnightly)
g 7.3% p.a. if $2370 is invested for 5 years (compounded weekly)
h 8.2% p.a. if $5020 is invested for 2 3--4- years (compounded weekly)
i
15.25% p.a. if $4605 is invested for 2 years (compounded daily)
j
14.75% p.a. if $8970 is invested for 3 years (compounded daily).
7 multiple choice
The greatest return is likely to be made if interest is compounded:
A annually
B semi-annually
C quarterly
D monthly
E fortnightly.
8 multiple choice
If $12 000 is invested for 4 1--2- years at 6.75% p.a., compounded fortnightly, the amount
of interest that would accrue would be closest to:
A $3600
B $4200
C $5000
D $12 100
E $16 300.
9 multiple choice
Which account would provide the best investment opportunity if $2500 was invested
for one year?
A simple interest at 6% p.a. (see simple interest topic)
B interest compounded annually at 6% p.a.
C interest compounded 6-monthly at 6.2% p.a.
D interest compounded quarterly at 6.1% p.a.
E interest compounded monthly at 6% p.a.
10 Peta wishes to invest $3200 for 5 years. By comparing the interest earned, which of
the following would be Peta’s best investment option?
a 11% p.a. simple interest
b compound interest at 10.5% p.a., compounded annually
c compound interest at 10% p.a., compounded monthly
11 Cyril has just inherited $10 000 and after spending $910 on a holiday he would like to
invest the balance for 4 1--2- years. He is offered the investment opportunities detailed
below. Which option should Cyril choose?
a simple interest at 9.5% p.a.
b compound interest at 9.4% p.a., adjusted semi-annually
c compound interest at 9.3% p.a., adjusted quarterly
12 Agnes invests $2050 for 4 years. Interest is added quarterly. For the first 2 years the
rate is 7% p.a. and for the remaining 2 years the rate rises to 9% p.a. What interest
would accrue during this time?
Chapter 14 FM Page 668 Monday, November 13, 2000 3:32 PM
668
Further Mathematics
13 Alex invests $6185 for 3 years. Interest is added half-yearly. The rate starts at 8.5%
p.a. for the first half of the investment period before it rises to 9.6% p.a. for the
remaining time. What interest would Alex earn from this account?
WORKED
Example
10
14 Use the compound interest formula to find the principal, P, when:
a A = $5000, r = 9, n = 4
c A = $3550, r = 1.5, n = 12
e A = $5495.74, r = 1.2, n = 48
b A = $2600, r = 8.2, n = 3
d A = $6661.15, r = 0.8, n = 36
15 Find the principal that will grow to:
a $3000 in 4 years, if interest is compounded 6-monthly at 9.5% p.a.
b $2000 in 3 years, if interest is compounded quarterly at 9% p.a.
c
$2900 in 3 1--2- years, if interest is compounded quarterly at 10.6% p.a.
d $5600 in 5 1--4- years, if interest is compounded quarterly at 8.7% p.a.
e $10 000 in 4 1--4- years, if interest is compounded monthly at 15% p.a.
f
$28 000 in 2 1--2- years, if interest is compounded monthly at 9% p.a.
g $47 500 in 6 years, if interest is compounded monthly at 10.5% p.a.
h $850 in 2 years, if interest is compounded monthly at 8.25% p.a.
i
$4860 in 2 1--2- years, if interest is compounded fortnightly at 6.5% p.a.
j
$8470 in 2 1--4- years, if interest is compounded weekly at 7.54% p.a.
16 Find the interest accrued in each case in question 15.
17 multiple choice
Lillian wishes to have $24 000 in a bank account after 6 years so that she can buy a
new car. The account pays interest at 15.5% p.a. compounded quarterly. The amount
(to the nearest dollar) that Lillian should deposit in the account now, if she is to reach
her target, is:
A $3720
B $9637
C $10 109
D $12 117
E $22 320
18 multiple choice
Peter has his heart set on a holiday in 2 years’ time and it will cost him $1700. His
bank account pays interest at a rate of 10.25% p.a., compounded 6-monthly. The
amount that Peter will need to deposit now into this account will lie between:
A $1000 and $1050
D $1360 and $1410
B $1150 and $1200
E $1500 and $1550
C $1310 and $1360
19 Sarah needs $1560 for a new stereo system which she is planning to buy in 1 1--2- years’
time. Her bank offers a rate of 9.6% p.a. with interest compounded monthly. How
much should she deposit now?
20 Glen’s credit union offers an account which pays a rate of 8.4% p.a. with interest compounded monthly. His house extension will cost him $15 000 in 4 1--2- years’ time. How
much should Glen invest in this account to be able to pay for his extension?
Chapter 14 FM Page 669 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
669
21 Calculate the interest rate per year (to 2 decimal places), given that the interest rate
per period, r, is:
a 2% and interest is compounded quarterly
b 1.5% and interest is compounded quarterly
c 1% and interest is compounded monthly
d 3.5% and interest is compounded semi-annually
e 0.65% and interest is compounded monthly
f 2.65% and interest is compounded quarterly
g 0.43% and interest is compounded fortnightly
h 0.25% and interest is compounded weekly
i 0.19% and interest is compounded weekly
j 0.025% and interest is compounded daily.
22 Find the interest rates per annum (to 2 decimal places) that would enable investments of:
a $2000 to grow to $3000 over 3 years if interest is compounded 6-monthly
11
b $8000 to grow to $9000 over 2 years (interest compounded quarterly)
c $12 000 to grow to $15 000 over 4 years (interest compounded quarterly)
d $5000 to grow to $7000 over 5 years (compounded semi-annually)
e $2650 to grow to $3750 over 3 1--2- years (compounded quarterly)
f $25 000 to grow to $40 000 over 2 1--2- years (compounded monthly)
g $43 000 to grow to $60 000 over 4 1--2- years (compounded fortnightly)
h $1400 to grow to $1950 over 2 years (compounded weekly)
i $2960 to grow to $3400 over 2 1--4- years (compounded weekly)
j $3250 to grow to $4100 over 2 years (compounded daily).
WORKED
Example
23 If it takes 2 years for $1460 to grow to $2100, find the annual interest rate (compounded
semi-annually).
24 After 3 1--2- years $950 has accumulated to $1300. Find the annual interest rate
(compounded quarterly).
25 multiple choice
What is the minimum interest rate per annum (compounded quarterly) needed for
$2300 to grow to $3200 in 4 years’ time?
A 6% p.a.
B 7% p.a.
C 8% p.a.
D 9% p.a.
E 10% p.a.
26 multiple choice
What will be the minimum annual rate of interest (compounded monthly) needed to
enable $18 500 to accrue $4000 interest in 18 months’ time?
A 12.25% p.a. B 13.25% p.a. C 14.25% p.a. D 15.25% p.a. E 16.25% p.a.
27 Sophie has been told that if you invest $500 at 8% p.a. compounded annually then its
value will double in 9 years. Is this true?
28 At an annual rate of 18.1% (compounded half-yearly) $1000 will double in value in
4 years. Is this true?
ET
SHE
Work
29 a ii If you invested $1000 now at 11% p.a. (compounded quarterly), how much
would you have in 10 years’ time?
ii At that time, what annual interest rate would give you $10 000 in a further
10 years (interest compounded quarterly)?
b How would your answers to part a vary if interest was compounded fortnightly?
14.1
Chapter 14 FM Page 670 Monday, November 13, 2000 3:32 PM
670
Further Mathematics
Finding time in compound interest
In the previous section we studied investments that compound over a period of time by
r
manipulating the compound interest formula, A = PRn, where R = 1 + --------- . Situations
100
where A, P and r were unknown were investigated.
This section deals with two methods that can be used to find n, the number of
interest-bearing periods, that is, to find the time for which an investment runs. The two
methods are:
1. trial and error
2. logarithms.
The value obtained for n may be a whole number, but it is more likely to be a decimal.
That is, the time required will lie somewhere between two consecutive integers. The
smaller of the two integers represents insufficient time for the investment to amount to
the balance desired, while the second represents more than enough time.
In practice, if this is the case, an investor may choose to:
(a) withdraw funds as soon as the final balance is reached, in which case a fee may be
imposed for early withdrawal
(b) withdraw funds at the first integral value of n after the final balance is reached, that
is, when the investment matures.
In this section we will use the second option for calculating the time needed for an
investment to run.
Using trial and error to find time
This process involves substituting reasonable values of n into a simplified form of the
compound interest formula until we find the two integral values mentioned above.
When deciding what values are reasonable we should primarily consider the frequency
at which the interest is compounded. Generally, the greater the frequency, the greater
the value of n. We can also consider by what proportion the principal grows, and the
interest rate per annum. A large proportional increase and a small interest rate may
indicate a large value of n, too. However, in the end the process is guesswork.
WORKED Example 12
Use the compound interest formula to find out how long it will take $2000 to amount to
$3500 at 8% p.a. (interest compounded annually).
THINK
WRITE
1
What are A, P and r?
A = 3500, P = 2000, r = 8
2
Substitute into the CI formula and
simplify.
r
A = P  1 + ---------

100
n
3500 = 2000(1.08)n
1.08n = 1.75
3
Substitute a value for n which you think
may give a value for 1.08n equal to or
greater than 1.75 and compare. Try n = 6.
Need 1.08n ≥ 1.75
1.086 = 1.587 too low
Chapter 14 FM Page 671 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
THINK
4
5
6
671
WRITE
1.086 < 1.75 so repeat the process with a
larger value, say n = 8.
Since 1.088 > 1.75, try n = 7.
n = 7 represents insufficient time while
n = 8 was just in excess of 1.75.
Write your answer. As the compounding
period was annual, n is in years.
1.088 = 1.851 (possibly too high)
1.087 = 1.714 (too small)
Therefore n = 8 is sufficient for 1.08n to
exceed 1.75.
It will take 8 years for $2000 to amount to
$3500.
As we have seen, it may be that interest is compounded more often than once a year
so r will need to be determined in the usual manner. In all these cases n has its usual
meaning and could indicate the number of quarters, months, fortnights, weeks or
days. It is then appropriate for the value of n obtained to be converted to time in more
meaningful terms, for example years and months rather than just months.
A
Note: The decimal value for the ratio --- should be written to 4 decimal places if
P
interest is compounded more often than monthly.
WORKED Example 13
Calculate the number of interest bearing periods, n, and hence the time it will take $3600
to amount to $5100 at a rate of 7% p.a. (interest compounded quarterly).
THINK
1
What are A, P and r? Note that n is the
number of quarters.
WRITE
A = 5100
P = 3600
r =
7
--4
= 1.75
2
Substitute into the formula and simplify.
r n
A = P  1 + ---------

100
5100 = 3600(1.0175)n
1.0175n = 1.417
3
4
5
6
7
Try n = 10 and compare with 1.417.
1.017510 = 1.189
Try a much larger value for n, say n = 20.
(Remember n is in quarters.)
1.017520 = 1.415
1.017520 < 1.417, so try n = 21.
1.017521 = 1.440
So 21 quarters are needed. Change to time
in more meaningful units.
Write a summary statement.
n = 21 quarters
Time = 5 1--4- years
It will take 5 1--4- years for $3600 to amount to
$5100.
Chapter 14 FM Page 672 Monday, November 13, 2000 3:32 PM
672
Further Mathematics
Remember that r may be a recurring decimal (interest compounded monthly or more
often) and although approximations are made to the calculated value of n in a problem,
we should still maintain an accurate r value. This is because n could be very close to an
integer and the accuracy consideration could determine whether n lies above or below
that integer. This is especially important the greater the frequency of compounding.
Once determined, store r in the calculator memory.
The interest accrued, I, may be given rather than the final balance, A. As before, find
A by using A = P + I.
WORKED Example 14
About how long would it take $1800 to accrue $500 interest at 10% p.a., interest credited
fortnightly?
THINK
1
2
WRITE
P and I are given, not A.
A = P + I.
What are A, P and r?
r
Store R = 1 + --------- in your calculator
100
memory.
Substitute and simplify.
A=P+I
= 1800 + 500
= 2300
P = 1800
r =
10
-----26
= 0.384 62
r
A = P  1 + ---------

100
n
2300 = 1800(1.003 8462)n
1.003 8462n = 1.2778
Try n = 60 (remember n is in fortnights)
and compare to 1.2778.
1.003 846260 = 1.2590
4
Try n = 64.
1.003 846264 = 1.2785
5
Try n = 63.
1.003 846263 = 1.2736
3
6
64 fortnights are needed. Change to time
in more meaningful terms.
n = 64 fortnights
Time = 2 years and 12 fortnights
7
Write a summary statement.
2 years and 12 fortnights are needed for
$1800 to accrue $500 interest.
remember
remember
1. Finding time by trial and error involves educated guessing for the value of n.
2. Time = n × number of times interest is compounded each year.
Chapter 14 FM Page 673 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
14C
WORKED
12
Finding time in compound
interest using trial and error
1 Use the compound interest formula to find the amount, n, given:
a A = $4831.53, P = $3000, r = 10
b A = $15 203.25, P = $7000, r = 9
c A = $10 360.01, P = $6500, r = 6
d A = $2754.01, P = $1400, r = 7
e A = $2437.50, P = $950, r = 12.5
f A = $24 472.01, P = $15 000, r = 8.5
g A = $7723.41, P = $5400, r = 1.2
h A = $4439.48, P = $2920, r = 0.5.
Math
cad
Example
673
Compound
interest
E
3 Use your answers from question 2 parts e to h (the number of years needed in each
case) to calculate the actual balance after this time.
WORKED
Example
13
4 For each of the following, calculate the number of interest bearing periods, n, and
hence the time in more meaningful terms.
a A = $2400, P = $1200, r = 3% per 6 months (compounded semi-annually)
b A = $5000, P = $3100, r = 4% per 6 months (compounded half-yearly)
c A = $12 000, P = $8300, r = 2.5% per quarter (compounded quarterly)
d A = $7300, P = $4850, r = 2.25% per quarter
e A = $15 600, P = $9600, r = 1% per month
f A = $1840, P = $990, r = 0.75% per month
g A = $25 000, P = $16 750, r = 0.25% per fortnight
h A = $12 050, P = $9310, r = 0.12% per fortnightly
i A = $4650, P = $2380, r = 0.15% per week
j A = $8810, P = $6950, r = 0.03% per day
5 Calculate the number of interest bearing periods, n, and hence how long it will take for:
a $8900 to amount to $11 000 at a rate of 8% p.a. (compounded quarterly)
b $3400 to amount to $5600 at a rate of 9% p.a. (compounded half-yearly)
c $900 to amount to $2100 at a rate of 11% p.a. (compounded quarterly)
d $2050 to amount to $4800 at a rate of 10% p.a. (compounded 6-monthly)
e $18 200 to amount to $28 600 at 12% p.a. (credited monthly)
f $5640 to amount to $9320 at 15% p.a. (credited monthly)
g $10 500 to amount to $15 000 at 9.5% p.a. (credited quarterly)
h $7950 to amount to $17 100 at 9.6% p.a. (credited monthly).
6 Rachel has $1450 to invest and she would like it to reach $2000 by placing it in an
account at a rate of 10.4% p.a. with interest compounded quarterly. How long will this
take?
7 Billy has invested $6120 at a rate of 8.6% p.a., interest credited semi-annually. How
long will this investment take to accumulate to $8000?
sheet
CEL Spread
X
2 Use the compound interest formula to find out how long it will take (with interest
compounded annually) for:
Compound
interest
a $2000 to amount to $3354.20 at 9% p.a.
b $7500 to amount to $17 284.03 at 11% p.a.
c $9250 to amount to $15 067.28 at 5% p.a.
d $16 000 to amount to $27 031.66 at 6% p.a.
e $750 to amount to $1000 at 8% p.a.
f $925 to amount to $1350 at 12% p.a.
g $1720 to amount to $2600 at 10.5% p.a.
h $11 000 to amount to $20 500 at 14.25% p.a.
Chapter 14 FM Page 674 Monday, November 13, 2000 3:32 PM
674
Further Mathematics
8 Phillipa has invested $4875 in an account which pays interest at a rate of 12.6% p.a.,
interest credited monthly. What time period is needed to enable this investment to
grow to $8000?
9 Michael would like to have $3600 at some time in the near future. He has $2160 that
he can invest now and his bank offers him an account which pays 7.8% p.a., interest
compounded quarterly. How long will it take him to achieve his target?
10 multiple choice
What will be the least number of interest periods, n, required for $7590 to grow to
at least $9000 in an account with interest paid at 6.5% p.a., interest compounded
half-yearly?
A 5
B 6
C 10
D 12
E 14
11 multiple choice
What will be the minimum investment period required for $14 250 to result from an
investment of $11 000 at a rate of 7.5% p.a., interest credited monthly?
A 3 years 5 months
B 3 1--2- years
C 10 years 2 months
1
D 10 --2- years
E 42 years
12 multiple choice
Sonya has $2500 and she wants to be able to take a holiday to Spain, costing $4160.
Her credit union offers an account which pays compound interest at 9.5% p.a.
(credited semi-annually). If Sonya invests her $2500 in this account the first
occasion that she can afford the holiday is after:
A 4 1--2- years B 5 years
C 5 1--2- years
D 6 years
E 11 years
13 About how long would it take for:
a $1500 to accrue $300 interest at 8% p.a., interest compounded monthly?
14
b $3100 to accrue $1500 interest at 15% p.a., interest compounded monthly?
c $9000 to accrue $4400 interest at 9.6% p.a., interest credited fortnightly?
d $14 260 to accrue $6100 interest at 7.4% p.a., interest credited fortnightly?
e $840 to accrue $200 interest at 6.6% p.a., compounded weekly?
f $2875 to accrue $440 interest at 8.3% p.a., compounded weekly?
g $12 120 to accrue $1560 interest at 10.2% p.a., credited daily?
h $19 000 to accrue $11 000 interest at 9.25% p.a., credited daily?
WORKED
Example
14 multiple choice
If, when you were born, your uncle invested $100 at 12% p.a. (credited monthly),
by which of your birthdays would the $100 have amounted to at least $1000?
A 19th
B 20th
C 21st
D 231st!
E 232nd!
15 multiple choice
Consider question 14. If the initial investment that your uncle made was doubled,
the birthday when the balance would reach at least $1000 would be the:
A 14th
B 15th
C 16th
D 161st!
E 162nd!
16 Vivian and Rick both want to save $6000 for a holiday. Vivian has chosen to invest
$5000 with his bank at 5% p.a., compounded monthly. However, Rick has only $4000
but he has found an account which pays interest at 9% p.a., compounded monthly. If
they both invest their money at the same time:
a who will get their holiday first?
b how much longer will the other person have to wait for a holiday?
Chapter 14 FM Page 675 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
675
Finding time in compound interest using logarithms
The logarithm (log) is a special function just as sine and cosine are special functions in
trigonometry. In the same way that we can find the sine of 30° (0.5) on a calculator, we
can find the log of 10 (1). Try this on your calculator so that you are comfortable with
evaluating the log of a number.
In the past you have used a set of rules to simplify indices. There is also a set of rules
which applies to logs. (Your teacher may discuss with you the close relationship that
exists between indices and logs and how the log rules are derived.)
In using logs to find n in the CI formula we use one of the log rules:
log x n = n log x.
That is, the log of a number raised to a power equals the power multiplied by the log of
the number to the power of one.
Now let us see that this rule holds for a numerical example.
Evaluate log 25.
log 25 = log 32
log 25 = 5 log 2 (from log rule)
= 5 × 0.301 03 (from calculator)
= 1.505 15 (from calculator)
= 1.505 15
You could try more of these to satisfy yourself that this log rule works. Now let us
consider finding n in the CI formula.
WORKED Example 15
Using the CI formula, find out how long it will take $2000 to amount to $3500 at 8% p.a.
with interest compounded annually.
THINK
1
What are A, P and r?
2
Substitute into the CI formula and
simplify. Keep 1.75 on the calculator
display for step 6.
3
Take the log of both sides of the equation.
4
Apply the log rule i.e. log xn = n log x.
WRITE
A = 3500
P = 2000
r=8
r n
A = P  1 + ---------

100
3500 = 2000(1.08)n
1.08n = 1.75
log 1.08n = log 1.75
n log 1.08 = log 1.75
5
Isolate n by dividing both sides by
log 1.08
log 1.75
n = -------------------log 1.08
6
Evaluate n.
n = 7.27 years
7
Interest is compounded annually, so n
represents years. Raise n to the next
whole year.
As the interest is compounded annually,
n = 8 years
8
Write a summary statement.
It will take 8 years for $2000 to amount
to $3500.
Chapter 14 FM Page 676 Monday, November 13, 2000 3:32 PM
676
Further Mathematics
WORKED Example 16
Calculate the number of interest bearing periods, n, required and hence the time it will
take $3600 to amount to $5100 at a rate of 7% p.a., with interest compounded quarterly.
THINK
1
What are A, P and r?
2
Substitute into the formula and simplify.
Retain 1.416 66 on your calculator display
for step 6.
3
Take the log of both sides.
Apply the log rule and isolate n.
4
5
6
7
8
Evaluate n.
As n represents quarters, raise n to the
next integer.
Write the time in more meaningful terms.
Write a summary statement.
WRITE
A = 5100
P = 3600
r = 7--4= 1.75
n
r
A = P  1 + ---------

100
5100 = 3600(1.0175)n
1.0175n = 1.416 66
log 1.0175n = log 1.416 66
n log 1.0175 = log 1.416 66
log 1.416 66
n = ------------------------------log 1.0175
n = 20.08 quarters
As the interest is compounded quarterly,
n = 21 quarters.
21
1
- years = 5 --- years.
Time = ----4
4
1
After 5 --4- years, $3600 will amount to $5100.
WORKED Example 17
About how long would it take $1800 to accrue $500 interest at 10% p.a., with interest
credited fortnightly?
THINK
WRITE
P and I are given, not A. A = P + I r
What are A, P and r? Store R = 1 + --------- in
100
the calculator memory.
Substitute and simplify. Retain 1.277 77
on the calculator display for step 6.
A= 2300, P = 1800
------ = 0.384 62
r = 10
26
3
Take the log of both sides.
4
Apply the log rule and isolate n.
5
6
Evaluate n.
Raise n to the next integer and write the
time in more meaningful terms.
7
Write a summary statement.
log (1.003 8462)n = log 1.277 77
n log 1.003 8462 = log 1.277 77
log 1.277 77
n = -----------------------------------------log 1.003 384 62
n = 63.85 fortnights
As interest is credited fortnightly,
n = 64 fortnights
time = 2 years 12 fortnights
2 years 12 fortnights are needed for
$1800 to accrue $500 interest.
1
2
n
r
A = P  1 + ---------

100
2300 = 1800(1.003 846 2)n
(1.003 8462)n = 1.277 77
Chapter 14 FM Page 677 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
677
remember
remember
1. When using logs to find time, use the rule: log xn = n log x.
2. Change the n value to time in more meaningful terms.
1 Use the log rule to solve for n (to 2 decimal places) in each case.
a log 5n = log 3
b log 3n = log 1.5
c log 2n = log 7
d log 1.4n = log 2
e log 1.1n = log 1.9
f log 1.05n = log 1.5
g log 1.05n = log 1.6
h log 1.025n = log 1.513
WORKED
15
14.4
2 Use the compound interest formula to find n, given:
a A = $14 172.49, P = $8000, r = 10
b A = $8687.57, P = $4000, r = 9
Compound
c A = $3759.08, P = $2500, r = 6
d A = $2950.73, P = $1500, r = 7
interest –
e A = $2453.53, P = $850, r = 12.5
f A = $22 840.55, P = $14 000, r = 8.5 finding n
g A = $9374.68, P = $6400, r = 1.2
h A = $2643.75, P = $1960, r = 0.5.
3 Use the CI formula to find n (to 2 decimal places), given:
a A = $6000, P = $4000, r = 8
b A = $7500, P = $5500, r = 10
c A = $1700, P = $1600, r = 6
d A = $2650, P = $1920, r = 9
e A = $16 000, P = $13 300, r = 8.5
f A = $19 500, P = $16 250, r = 12.5
g A = $830, P = $750, r = 2.5
h A = $3680, P = $1320, r = 1.7.
4 Use the compound interest formula to find
out how long it will take (with interest
compounded annually) for:
a $2000 to amount to $3173.75
at 8% p.a.
b $7500 to amount to $18 569.72
at 12% p.a.
c $9250 to amount to $16 565.34
at 6% p.a.
d $16 000 to amount to $34 750.29
at 9% p.a.
e $850 to amount to $1000
at 7% p.a.
f $1025 to amount to $1350
at 11% p.a.
g $1540 to amount to $2600
at 11.5% p.a.
h $12 000 to amount to $20 500
at 13.25% p.a.
5 Having determined the number of years
needed in each case in question 4e–h
calculate the actual balance after this time.
Math
cad
Example
HEET
SkillS
14D
Finding time in compound
interest using logarithms
Chapter 14 FM Page 678 Monday, November 13, 2000 3:32 PM
678
WORKED
Example
16
Further Mathematics
6 Calculate the number of interest bearing periods, n, and hence the time in more
meaningful terms when:
a A = $2100, P = $1200, r = 3% per half-year
b A = $4000, P = $3100, r = 4% per 6-month
c A = $13 500, P = $8300, r = 2.5% per quarter
d A = $8200, P = $4850, r = 2.25% per quarter
e A = $16 900, P = $9600, r = 1% per month
f A = $1920, P = $990, r = 0.75% per month
g A = $24 000, P = $16 750, r = 0.25% per fortnight
h A = $13 050, P = $9310, r = 0.12% per fortnight
i A = $4550, P = $2380, r = 0.15% per week
j A = $9040, P = $6950, r = 0.03% per day
7 Calculate the number of interest bearing periods, n, and hence how long it will take for:
a $7800 to amount to $10 000 at a rate of 8% p.a. (compounded quarterly)
b $2500 to amount to $4600 at a rate of 9% p.a. (compounded half-yearly)
c $800 to amount to $1900 at a rate of 11% p.a. (compounded quarterly)
d $2650 to amount to $4800 at a rate of 10% p.a. (compounded 6-monthly)
e $20 200 to amount to $28 600 at 12% p.a. (credited monthly)
f $6640 to amount to $10 230 at 15% p.a. (credited monthly)
g $11 400 to amount to $16 100 at 9.5% p.a. (credited quarterly)
h $10 500 to amount to $19 280 at 9.6% p.a. (credited monthly).
8 Wanda has invested $1600 in an account at a rate of 10.4% p.a., interest compounded
quarterly. How long will it take to reach $2200?
9 Baden has invested $5680 at a rate of 9.4% p.a. with interest credited semi-annually.
How long will this investment take to accumulate to $7000?
10 Stefan has invested $4400 in an account which pays interest at a rate of 12.6% p.a.,
interest credited monthly. What time period is needed to enable this investment to
grow to $7600?
11 Don would like to have $3800 at some time in the near future. He has $2200 that he
can invest now and his bank offers him an account which pays 8.6% p.a., interest
compounded quarterly. How long will it take him to achieve his goal?
12 multiple choice
What will be the least number of interest periods, n, required for $6470 to grow to at
least $9000 in an account with interest paid at 6.5% p.a. and compounded half-yearly?
A 10
B 11
C 12
D 20
E 22
13 multiple choice
What will be the minimum investment period required for $12 750 to result from an
investment of $8000 at a rate of 6.9% p.a., interest credited monthly?
A 6 months
B 7 months
C 6 years 9 months
D 6 years 10 months
E 7 years
Chapter 14 FM Page 679 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
679
14 multiple choice
Rhiannon has $2200 and she wants to be able to take a holiday to Thailand costing
$3860. Her credit union offers an account which pays compound interest at 9.5% p.a.
(credited semi-annually). If Rhiannon invests her $2200 in this account, the first
occasion that she can afford the holiday is after:
A 3 years
B 2 1--2- years
C 6 years
1
D 6 --2- years
E 13 years
15 About how long would it take for:
a $1400 to accrue $300 interest at 8% p.a., interest compounded monthly?
17
b $2900 to accrue $1500 interest at 15% p.a., interest compounded monthly?
c $8000 to accrue $4400 interest at 9.6% p.a., interest credited fortnightly?
d $13 360 to accrue $6100 interest at 7.4% p.a., interest credited fortnightly?
e $950 to accrue $300 interest at 6.6% p.a., compounded weekly?
f $2725 to accrue $440 interest at 8.3% p.a., compounded weekly?
g $14 260 to accrue $1560 interest at 10.2% p.a., credited daily?
h $18 000 to accrue $11 000 interest at 9.25% p.a., credited daily?
WORKED
Example
16 multiple choice
If, when you were born, your aunt invested $100 at 12% p.a. (credited monthly), by
which one of your birthdays would the $100 have amounted to at least $500?
A 13th
B 14th
C 15th
D 161st
E 162nd
17 multiple choice
Consider question 16. If the initial investment that your aunt made was doubled, the
birthday when the balance would reach at least $500 would be the:
A 8th
B 9th
C 20th
D 92nd
E 93rd
ET
SHE
Work
18 Jennifer and Dawn both want to save $15 000 for a car. Jennifer has $11 000 to invest
in an account with her bank which pays 8% p.a., interest compounded quarterly.
Dawn’s credit union has offered her 11% p.a., interest compounded quarterly.
a How long will it take Jennifer to reach her target?
b How much will Dawn need to invest in order to reach her target at the same time
as Jennifer? Assume their accounts were opened at the same time.
14.2
Chapter 14 FM Page 680 Monday, November 13, 2000 3:32 PM
680
Further Mathematics
Depreciation
Many items such as antiques, jewellery or real estate increase in value (appreciate) with
time. On the other hand, items such as computers, vehicles or machinery decrease in
value (depreciate) with time as a result of wear and tear or a lack of demand for those
specific items.
The estimated loss in value of assets is called depreciation. Each financial year a
business will set aside money equal to the depreciation of an item in order to cover the
cost of the eventual replacement of that item. The estimated value of an item at any
point in time is called its book value.
When the book value becomes zero the item is said to be written off. At the end of an
item’s useful or effective life (as a contributor to a company’s income) its book value is
then called its scrap value.
Book value = cost price − total depreciation to that time
When book value = $0, then the item is said to be written off.
Scrap value is the book value of an item at the end of its useful life.
There are 3 methods by which depreciation can be calculated. They are:
1. flat rate depreciation
2. reducing balance depreciation
3. unit cost depreciation.
Flat rate (straight line) depreciation
If an item depreciates by the flat rate method then its value decreases by a fixed amount
each unit time interval, generally each year. This depreciation value may be expressed
in dollars or as a percentage of the cost price.
This method of depreciation may also be referred to as prime cost depreciation.
Since the depreciation is the same for each unit time interval, the flat rate method is an
example of straight line (linear) decay. The relationship can be represented by the linear
equation:
BV = P − dT
where P = cost price ($)
BV = P − dT
where BVT = book value ($) after time, T
BV = P − dT
where T = time since purchase (years)
BV = P − dT
where d = rate of depreciation ($ per year)
= fixed amount per year or
= percentage of P per year
We can use this relationship to analyse flat rate depreciation or we can use a
depreciation schedule (table) which can then be used to draw a graph of book value
against time. The schedule displays the book value after each unit time interval, that is:
Time, T
Depreciation, d
Book value, V
Chapter 14 FM Page 681 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
681
WORKED Example 18
Fast Word Printing Company bought a new printing press for $15 000 and chose to
depreciate it by the flat rate method. The depreciation was 20% of the prime cost price
each year and its useful life was 4 years.
a Find the annual depreciation.
b Draw a depreciation schedule for the press’ useful life and use it to draw a graph of
book value against time.
c Find the relationship between the book value and time and use it to find the scrap
value.
THINK
WRITE
a
a P = $15 000
d = 20% of $15 000
= $3000 per year
Annual depreciation is $3000.
b
1
2
3
b
1
State the cost price.
Find the depreciation rate as 20% of the
prime cost price.
Write your answer.
Draw a depreciation schedule for
0–4 years, using depreciation of $3000
each year and a starting value of
$15 000.
Time,
T (years)
Depreciation,
d ($)
2
Draw a graph of the tabled values for
book value against time.
Book value ($)
0
1
Set up the equation: BV = P − dT
State d and P.
2
The press is scrapped after 4 years so
substitute T = 4 into the equation.
3
Write your answer.
c
15 000
1
3 000
12 000
2
3 000
9 000
3
3 000
6 000
4
3 000
3 000
V
15 000
12 000
9 000
6 000
3 000
0
0
c
Book
value,
V ($)
1
2
3
4
Time (years)
T
d = 3000
P = 15000
BVT = 15 000 − 3000T
BV4 = 15 000 − 3000(4)
= 15 000 − 12 000
BV4 = $3000
The scrap value is $3000.
Note: If finding time, T, or depreciation, d, simply substitute into the equation and
solve.
The depreciation schedule gives the scrap value, as can be seen in the previous
example. So too does a graph of book value against time, since it is only drawn for the
item’s useful life and its end point is the scrap value.
Chapter 14 FM Page 682 Monday, November 13, 2000 3:32 PM
682
Further Mathematics
Businesses need to keep records of depreciation for tax purposes on a year-to-year
basis. What if an individual wants to investigate the rate at which an item has depreciated
over many years? An example is the rate at which a private car has depreciated. If a
straight line depreciation model is chosen, then the following example demonstrates its
application.
WORKED Example 19
Jarrod bought his car 5 years ago
for $15 000. Its current market value
is $7500. Assuming straight line
depreciation, find:
a the car’s annual depreciation
rate
b the relationship between the
book value and time, and use it
to find when the car will have a
value of $3000.
THINK
a 1 Find the total depreciation over the
5 years and thus the rate of
depreciation.
2
b
1
Write your answer.
Set up the book value equation.
2
Use the equation and substitute
BVT = $3000 and transpose the equation
to find T.
3
Write your answer.
WRITE
a Total depreciation
= cost price − current value
= $15 000 − $7500
= $7500
Rate of depreciation
total depreciation
= -----------------------------------------number of years
$7500
= ----------------5 years
= $1500 per year
The annual depreciation rate is $1500.
b BV = P − dT
BVT = 15 000 − 1500T
When BVT = 3000,
3000 = 15 000 − 1500T
−1500T = 3000 − 15 000
−1500T = −12 000
– 12 000
T = -------------------– 1500
=8
The depreciation equation for the car is
BVT = 15 000 − 1500T. The book value of
$3000 for the car is expected to occur
when it is 8 years old.
Chapter 14 FM Page 683 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
683
remember
remember
1. To make calculations in flat rate depreciation, use the formula
BV = P − dT
where BV = book value after T years ($)
P = purchase or cost price ($)
T = time (years of depreciation)
d = rate of depreciation ($ per year)
= fixed amount per year or
= percentage of P per year.
2. Flat rate depreciation is also known as prime cost depreciation.
3. An item is written off when its book value becomes zero.
4. Scrap value is the item’s book value when it is no longer used.
5. Total depreciation = cost price − current value
total depreciation
6. Rate of depreciation = -----------------------------------------number of years
14E
WORKED
18
1 A mining company bought a vehicle for $25 000 and chose to depreciate it by the flat
rate method. The depreciation was 20% of the cost price each year and its useful life
Flat rate
was 4 years.
depreciation
a Find the annual depreciation.
b Draw a depreciation schedule for the item’s useful life and draw a graph of book
value against time.
c Find the relationship between book value and time. Use it to find the scrap value.
2 All Clean carpet cleaners bought a cleaner for $10 000 and chose to depreciate it by
the flat rate method. The depreciation was 15% of the cost price each year and its
useful life was 5 years.
a Find the annual depreciation.
b Draw a depreciation schedule for the item’s useful life and draw a graph of book
value against time.
For the situations outlined in questions 3 and 4:
a draw a depreciation schedule for the item’s useful life and draw a graph of book
value against time
b find the relationship between book value and time. Use it to find the scrap value.
3 A farming company chose to depreciate a tractor by the prime cost method and the
annual depreciation was $4000. The tractor was purchased for $45 000 and its useful
life was 10 years.
Math
cad
Example
Flat rate depreciation
Chapter 14 FM Page 684 Monday, November 13, 2000 3:32 PM
684
Further Mathematics
4 A winery chose to depreciate a corking machine, costing $13 500 new, by the prime
cost method. The annual depreciation was $2000 and its useful life was 6 years.
For the situations outlined in questions 5 and 6:
a find the annual depreciation
b draw a depreciation schedule for the item’s useful life and draw a graph of book
value against time
c find the relationship between book value and time. Use it to find how long it will
take for the item to reach its scrap value.
5 Machinery is bought for $7750 and depreciated by the flat rate method. The depreciation
is 20% of the cost price each year and its scrap value is $1550.
6 An excavation company buys a digger for $92 000 and depreciates it by the flat rate
method. The depreciation is 15% of the cost price per year and its scrap value is $9200.
For the situations outlined in questions 7 and 8:
a find the annual depreciation
b draw a depreciation schedule for the item’s useful life and draw a graph of book
value against time
c find the relationship between book value and time. Use it to find how long it will
take for the item to be written off.
7 The owner of a rental property chooses to depreciate the carpets, which were purchased for $5000, by the prime cost method. The annual depreciation is 17% of the
cost price per year.
8 A writer buys a set of books for $950 to help him with his work. He opts to depreciate
this ‘library’ by the prime cost method and depreciation is 17% of the cost price each year.
V
3
2
1
0
1500
1000
500
0
0 1 2 3 4 5 T
Time (years)
0 1 2 3 4 T
Time (years)
800
700
600
500
400
300
200
100
0
Book value ($ '000)
10 Each graph below represents the flat rate depreciation of four particular items. In each
case determine:
i the cost price of the item
ii the annual depreciation
iii the time taken for the item to reach its scrap value or to be written off.
a
b V
c V
d
Book value ($)
19
9 For the situations described below, and using a straight line depreciation model, find:
i the annual rate of depreciation
ii the relationship between the book value and time and use it to find at what age the
item will be written off, that is, have a value of $0.
a A car purchased for $50 000 with a current value of $25 000 now it is 5 years old
b A stereo unit bought for $850 seven years ago which has a current value of $150
c A refrigerator with a current value of $285 bought 10 years ago for $1235
d An aeroplane with purchase price of $152 000 sold at market value of $101 250
when it was 7 years old
Book value ($)
Example
Book value ($ '000)
WORKED
0 1 2 3 4 5T
Time (years)
V
18
16
12
8
4
0
0 1 23 4 5 6 7 8T
Time (years)
Chapter 14 FM Page 685 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
685
11 multiple choice
A 1-tonne truck, bought for $31 000, was depreciated using the flat rate method.
If the scrap value of $5000 was reached after 5 years, the annual depreciation
would be:
A $31
B $1000
C $5200
D $6200
E $26 000
12 multiple choice
A half-tonne truck, bought for $27 000, was depreciated using the flat rate method. If
the scrap value of $3600 was reached after 9 years, the annual depreciation would be
closest to:
A $23 400
B $3500
C $3000
D $2500
E $1000
13 A business chose to depreciate its $2600 photocopier by the prime cost method. What
was the annual depreciation if the scrap value of $320 was reached in 6 years?
14 multiple choice
The depreciation of a piece of machinery is given by the equation, V = 6000 − 450T.
The machinery will have a book value of $2400 after:
A 7 years
B 8 years
C 9 years
D 10 years
E 24 years
15 The depreciation of a computer is given by the equation, V = 3450 − 280T. After how
many years will the computer have a book value of $1770?
16 multiple choice
Listed below are the depreciation equations for 5 different items. Which item would
be written off in the least amount of time?
A V = −650T + 7000
B V = −750T + 7000
C V = −650T + 6000
D V = −750T + 6000
E V = −850T + 6000
17 multiple choice
Listed below are the depreciation equations for 5 different items. Which item would
be written off in the greatest amount of time?
A V = 50 000 − 4000T
B V = 50 000 − 5000T
C V = 45 000 − 4000T
D V = 45 000 − 5000T
E V = 40 000 − 5000T
18 A business buys two different photocopiers at the same time. One costs $2200 and is
to be depreciated by $225 per annum. It also has a scrap value of $400. The other costs
$3600 and is to be depreciated by $310 per annum. This one has a scrap value of $500.
a Which machine would need to be replaced first?
b How much later would the other machine need to be replaced?
19 multiple choice
A car valued at $20 000 was bought 5 years ago for $45 000. The straight-line
depreciation model is represented by:
A V = −20 000T + 45 000 B V = −5000T + 45 000 C V = 45 000 − 4000T
D V = 45 000 − 20 000T E V = 45 000 − 25 000T
Chapter 14 FM Page 686 Monday, November 13, 2000 3:32 PM
686
Further Mathematics
Reducing balance depreciation
If an item depreciates by the reducing balance method then its value decreases by a
fixed rate each unit time interval, generally each year. This rate is a percentage of the
previous book value of the item.
Reducing balance depreciation is also known as diminishing value depreciation.
WORKED Example 20
Suppose the new $15 000 printing press considered in worked example 18 was depreciated
by the reducing balance method at a rate of 20% p.a. of the previous book value.
a Draw a depreciation schedule for the first 4 years of work for the press.
b What is the book value after 4 years?
c Draw a graph of book value against time.
THINK
WRITE
a
a d1
= 20% of 15 000
= $3000
1
Find the depreciation for the first year.
2
Find the book value after the first year.
Book value = cost price − depreciation.
BV1 = 15 000 − 3000
= $12 000
3
Calculate year 2 depreciation and book
value after 2 years.
d2
4
Repeat the process for the next 2 years.
d3
5
Draw the depreciation schedule.
= 20% of $12 000
= $2400
BV2 = 12 000 − 2400
= $9600
= 20% of 9600
= $1920
BV3 = 9600 − 1920
= $7680
d4 = 20% of 7680
= $1536
BV4 = 7680 − 1536
BV4 = $6144
Time,
T (years)
Depreciation,
d ($)
Book
value,
V ($)
0
—
15 000
1
3000
12 000
2
2400
9 600
3
1920
7 680
4
1536
6 144
Chapter 14 FM Page 687 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
687
WRITE
b State the book value from the depreciation
schedule.
c Draw a graph of the book value against
time.
b The book value of the press after 4 years
will be $6144.
c
V
Book value ($ '000)
THINK
16
14
12
10
8
6
0
1
2
3
Time (years)
4 T
It is clear from the graph and the schedule that the reducing balance depreciation
results in greater depreciation during the early stages of the asset’s life (the book value
drops more quickly at the start since the annual depreciation falls from $3000 in year 1
to $1536 in year 4).
Now, the Australian Taxation Office allows depreciation of an asset as a tax deduction.
This means that the annual depreciation reduces the amount of tax paid by a business in
that year. The higher the depreciation, the greater the tax benefit. Therefore, depreciating
an asset by the reducing balance method allows a greater tax benefit for a business in
the beginning of an asset’s life rather than towards the end. In contrast, flat rate depreciation remains constant throughout the asset’s life. People have a choice as to whether
they depreciate an item by the flat rate or reducing balance methods, but once a method
is applied to an article it cannot be changed for the life of that article. The percentage
depreciation rates, which are set by the Australian Taxation Office, vary from one item
to another but for each item the rate applied for the reducing balance method is greater
than that for the flat rate method.
Let us compare depreciation for both methods.
WORKED Example 21
A transport business has bought a new bus for $60 000. The business has the choice of
depreciating the bus by a flat rate of 20% of the cost price each year or by 30% of the
previous book value each year.
a Draw depreciation schedules using both methods for a life of 5 years.
b Draw graphs of the book value against time for both methods on the same set of axes.
c After how many years does the reducing balance book value become greater than the
flat rate book value?
THINK
WRITE
a
a d = 20% of $60 000
= $12 000 per year
1
Calculate the flat rate depreciation
per year.
Continued over page
Chapter 14 FM Page 688 Monday, November 13, 2000 3:32 PM
688
Further Mathematics
THINK
3
Draw and complete a flat rate
depreciation schedule for 0–5 years.
Draw and complete a reducing
balance depreciation schedule.
Annual depreciation is 30% of the
previous book value. Subtract this
from the previous book value to
ascertain the present book value.
Continue to calculate the book
value for a period of 5 years.
Time,
T (years)
Depreciation,
d ($)
Book
value,
V ($)
0
—
60 000
1
12 000
48 000
2
12 000
36 000
3
12 000
24 000
4
12 000
12 000
5
12 000
0
Time,
T
(years)
Depreciation,
d ($)
0
—
b Draw graphs using values for V and T from
the schedules.
c Look at the graph to see when the reducing
balance curve lies above the flat rate line.
State the first whole year after this point of
intersection.
Book
value,
V ($)
60 000
1
30% of 60 000 = $18 000
42 000
2
30% of 42 000 = $12 600
29 400
3
30% of 29 400 = $8820
20 580
4
30% of 20 580 = $6174
14 406
5
30% of 14 406 = $4321.80 10 084.20
b
Book value ($ '000)
2
WRITE
V
60
50
40
30
20
10
0
0
1
2 3 4 5 T
Time, (years)
c The book value for the reducing balance
method is greater than that of the flat rate
method after 4 years.
The graph of book value against time in worked example 20 suggests that reducing
balance depreciation may be an example of exponential decay which was discussed
earlier. Let us check this by analysing the schedule.
Chapter 14 FM Page 689 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
689
12 000
Growth or compounding factor = ----------------- = 0.8
15 000
Now
12 000 × 0.8 = 9600
and
9600 × 0.8 = 7680
and
7680 × 0.8 = 6144.
So reducing balance depreciation is an example of exponential decay and in this case
the growth or compounding factor is 0.8.
As we saw in the first section of this chapter an exponential decay situation could be
represented by the equation:
y = kax where k, a are constants, a < 1.
In this situation the growth or compounding factor, a, represents 100% of the previous
book value less the rate of depreciation per unit time interval.
In worked example 20 the annual depreciation is 20% of the previous book value.
Hence, growth or compounding factor = 100% − 20%
= 1 − 0.2
= 0.8
We can write a general formula for reducing balance depreciation which is similar to
the compound interest formula (and can be derived in the same way) except that the
rate is subtracted rather than added to 1.
The reducing balance depreciation formula is:
r T
BVT = book value after time, T
BVT = P  1 – ---------

100
r
= rate of depreciation
P = cost price
T
= time since purchase
That is, given the cost price and depreciation rate we can find the book value
(including scrap value) of an article at any time after purchase.
Let us now see how we can use this formula.
WORKED Example 22
The printing press from worked example 20 was depreciated by the reducing balance
method at 20% p.a. What will be the book value and total depreciation of the press after
4 years if it cost $15 000 new?
THINK
1
State P, r and T.
2
Substitute into the depreciation formula
and simplify.
4
Evaluate.
Total depreciation is:
cost price − current book value.
5
Write a summary statement.
3
WRITE
P = 15 000, r = 20, T = 4
r T
BVT = P  1 – ---------

100
20 4
-)
= 15 000(1 − -------100
= 15 000(0.8)4
BV4 = $6144
Total depreciation = P − BV
= 15 000 − 6144
= $8856
The book value of the press after 4 years
will be $6144 and total depreciation will
be $8856.
Chapter 14 FM Page 690 Monday, November 13, 2000 3:32 PM
690
Further Mathematics
Effective life
The situation may arise where the scrap value is known and we want to know how long
it will be before an item reaches this value, that is, its useful or effective life.
r T
So, in the reducing balance formula BV = P  1 – --------- , T is needed. A similar

100
situation arose in the section ‘Finding time in compound interest’. On that occasion,
two different methods were used: trial and error, and logarithms. We will use both
methods here to find T.
WORKED Example 23
A photocopier purchased for $8000 depreciates by 25% p.a. by the reducing balance
method. If the photocopier has a scrap value of $1200, how long will it be before this value
is reached?
Method 1: Trial and error
THINK
1
2
3
4
5
6
7
WRITE
BV = 1200, r = 25, P = 8000
r T
Substitute into the depreciation formula BV = P  1 – ---------

100
and simplify.
1200 = 8000(0.75)T
1200
0.75T = -----------8000
0.75T = 0.15
Substitute values for T and compare with 0.754 = 0.316
0.15. Since 0.75 < 1, increasing T value
decreases the value of 0.75T. Let T = 4.
0.316 > 0.15. Try T = 6.
0.756 = 0.178
Try T = 7.
0.757 = 0.133
In whole years, the time needed is 7.
T = 7 years
Write a summary statement.
The time needed to reach scrap value is 7 years.
State BV, P and r.
Method 2: Logarithms
1
State BV, P and r.
2
Substitute into the depreciation formula
and simplify.
3
Take the log of both sides.
Apply the log rule and isolate T.
(log xn = n log x)
4
5
6
7
Evaluate T.
State T in whole years.
Write a summary statement.
BV = 1200, r = 25, P = 8000
r T
BV = P  1 – ---------

100
1200 = 8000(0.75)T
1200
0.75T = -----------8000
T
0.75 = 0.15
log 0.75T = log 0.15
T log 0.75 = log 0.15
log 0.15
T = -------------------log 0.75
T = 6.59
T = 7 years
The time needed to reach scrap value is 7 years.
Chapter 14 FM Page 691 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
691
remember
remember
1. To make calculations for reducing balance depreciation, use the formula:
r T
BVT = P  1 – --------- .

100
2. To find time in reducing balance depreciation use either trial and error or
logarithms.
3. Reducing balance depreciation is also known as diminishing value depreciation.
14F
WORKED
20
1 A farming company chose to depreciate its new $60 000 bulldozer by the reducing
balance method at a rate of 20% p.a. of the previous book value.
Reducing
a Draw a depreciation schedule for the first 4 years of the bulldozer’s life.
balance
b What is its book value after 4 years?
depreciation
c Draw a graph of book value against time.
Math
cad
Example
Reducing balance
depreciation
2 A retail store chose to depreciate its new $4000 computer by the reducing balance
method at a rate of 40% p.a. of the previous book value.
a Draw a depreciation schedule for the first 4 years of the computer’s life.
b What is its book value after 4 years?
c Draw a graph of book value against time.
3 A carpenter chose to depreciate her new set of electric hand tools, valued at $1500, by
the diminishing value method at a rate of 40% p.a. of the previous book value.
a Draw a depreciation schedule for the first 4 years of the tool set’s life.
b What is its book value after 4 years?
c Draw a graph of book value against time.
4 An accounting firm chose to depreciate a set of new electronic calculators, valued at
$1000 in total, by the diminishing value method at a rate of 25% p.a. of the previous
book value.
a Draw a depreciation schedule for the first 4 years of the calculators’ life.
b What is the book value after 4 years?
WORKED
21
5 A café buys a cash register for $550. The owner has the choice of depreciating the
register by the flat rate method (at 20% of the cost price each year) or the reducing
Comparing
balance method (at 30% of the previous book value each year).
depreciations
a Draw depreciation schedules for both methods for a life of 5 years.
b Draw graphs of book value against time for both methods on the same set of axes.
c After how many years does the reducing balance book value become greater than
the flat rate book value?
6 Speedy Cabs taxi service has bought a new taxi for $30 000. The company has the
choice of depreciating the taxi by the flat rate method (at 33 1--3- % of the cost price each
year) or the diminishing value method (at 50% of the previous book value each year).
a Draw depreciation schedules for both methods for 3 years.
b Draw graphs of book value against time for both methods on the same set of axes.
c After how many years does the reducing balance book value become greater than
the flat rate book value?
Math
cad
Example
Chapter 14 FM Page 692 Monday, November 13, 2000 3:32 PM
692
Mat
d
hca
WORKED
Example
Reducing
balance
depreciation
formula
22
Further Mathematics
7 Using the reducing balance formula, find BV (to 2 decimal places) given:
a P = 20 000, r = 20, T = 4
b P = 30 000, r = 25, T = 4
c P = 45 000, r = 15, T = 6
d P = 4000, r = 30, T = 5
e P = 1500, r = 20, T = 6
f P = 980, r = 17, T = 5
g P = 2675, r = 22.5, T = 5
h P = 8650, r = 13.5, T = 7
8 multiple choice
A refrigerator costing $1200 new is depreciated by the reducing balance method at
20% a year. After 4 years its book value will be:
A $240
B $491.52
C $960
D $1105
E $2488.32
9 multiple choice
A hot water system, purchased for $2600, was depreciated by the reducing balance
method at 20% p.a. Its book value after 6 years would be closest to:
A $2500
B $2000
C $1000
D $700
E $500
10 The items below are depreciated by the reducing balance method at 25% p.a. What
will be the book value and total depreciation of:
a a TV after 8 years, if it cost $1150 new?
b a photocopier after 4 years, if it cost $3740 new?
c carpets after 6 years, if they cost $7320 new?
d an electric heater after 5 years, if it cost $975 new?
11 The items below are depreciated at 30% p.a. by the diminishing value method. What
will be the book value and total depreciation of:
a a lawn mower after 5 years, if it cost $685 new?
b a truck after 4 years, if it cost $32 500 new?
c a washing machine after 3 years, if it cost $1075 new?
d a bus after 6 years, if it cost $62 000 new?
12 The items below are depreciated at 22.5% p.a. by the diminishing value method. What
will be the book value and total depreciation of:
a a mini bus after 5 years, if it cost $38 400 new?
b a small truck after 4 years, if it cost $26 480 new?
c a car after 6 years, if it cost $42 770 new?
d a machine after 3 years, if it cost $12 300 new?
13 multiple choice
After 7 years, a new $3000 photocopier, which devalues by 25% of its book value
each year, will have depreciated by:
A $400.45
B $750
C $2250
D $2599.55
E $2750.
14 multiple choice
New office furniture valued at $17 500 is subjected to reducing balance depreciation
of 20% p.a. and will reach its scrap value in 15 years. The scrap value will be:
A less than $300
B between $300 and $400 C between $400 and $500
D between $500 and $600 E between $600 and $700.
15 multiple choice
A new chainsaw bought for $1250 has a useful life of only 3 years. If it depreciates
annually at 60% diminishing value rate, its scrap value will be:
A zero
B $60
C $80
D $250
E $270.
Chapter 14 FM Page 693 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
693
16 Use the reducing balance formula to find T (to 2 decimal places), given:
a BV = $3000, P = $40 000, r = 20
b BV = $500, P = $3000, r = 30
23
c BV = $900, P = $4500, r = 25
d BV = $1500, P = $7600, r = 15
e BV = $1250, P = $6100, r = 20
f BV = $2400, P = $10 500, r = 25
g BV = $650, P = $2985, r = 17
h BV = $420, P = $2699, r = 22.5
WORKED
Example
17 Use the reducing balance formula to find the time, T, (to the appropriate integral
value) needed to reach scrap value for an item which has:
a a purchase price $34 000, depreciation of 20% p.a. and a scrap value of $800
b a purchase price $16 000, depreciation of 30% p.a. and a scrap value of $3000
c a purchase price $39 700, depreciation of 25% p.a. and a scrap value of $5000
d a purchase price $920, depreciation of 15% p.a. and a scrap value of $150
e a purchase price $1470, depreciation of 22% p.a. and a scrap value of $250
f a purchase price $2610, depreciation of 17% p.a. and a scrap value of $350
g a purchase price $27 300, depreciation of 33% p.a. and a scrap value of $4500
h a purchase price $54 900, depreciation of 30% p.a. and a scrap value of $5000.
18 A tractor was bought for $42 000 and was depreciated by the diminishing value
method at 22% p.a. Find:
a the book value after 5 years
b the total depreciation after 5 years
c how many years it takes to reach the scrap value of $5000.
19 A car bought for $36 700 was depreciated by the diminishing value method at
25% p.a. Find:
a the book value after 4 years
b the total depreciation after 4 years
c how many years it takes to reach the scrap value of $4000.
20 A boat was bought for $25 670 and depreciated by the reducing balance method at
30% p.a. Find:
a the book value after 4 years
b the total depreciation after 4 years
c how many years it takes to reach the scrap value of $3500.
21 The owner of a bus, bought for $48 900, has the choice of depreciating it by the flat
rate method at 20% p.a. of cost price or the reducing balance method at 30% p.a.
Using the depreciation formulas, find how many years it will take before the reducing
balance book value becomes greater than the flat rate book value.
22 A diving instructor buys new equipment for $4610. She may depreciate the gear at
17% p.a. of cost price or 25% p.a. of the previous book value. Using the depreciation
formulas, find how many years it will take for the diminishing value book value to
exceed the corresponding prime cost value.
23 A business accountant decided to depreciate a $29 000 company vehicle by the
reducing balance method at the rate of 20% p.a. until the car reached its scrap value
of $5000. When the vehicle was bought the business invested money in an account
which paid interest at 10% p.a., compounded monthly.
a How long would it take the vehicle to reach its scrap value?
b How much was invested (to the nearest $100) if the company was able to buy a
new $29 000 vehicle by the time the old one was scrapped?
Chapter 14 FM Page 694 Monday, November 13, 2000 3:32 PM
694
Further Mathematics
Unit cost depreciation
The flat rate and reducing balance depreciations of an item are based on the age of the
item. With the unit cost method, the depreciation is based on the possible maximum
output (units) of the item. For instance, the useful life of a truck could be expressed in
terms of the distance travelled rather than a fixed number of years — for example,
120 000 kilometres rather than 6 years. The actual depreciation of the truck for the
financial year would be a measure of the number of kilometres travelled.
WORKED Example 24
A taxi is bought for $31 000 and it
depreciates by an average of
28.4 cents per kilometre driven.
In one year the car is driven
15 614 km. Find:
a the annual depreciation
for this particular year
b its useful life if its scrap
value is $12 000.
THINK
WRITE
a
a d = 15 614 × 28.4 cents
d = $4434.38
Annual depreciation for the year was
$4434.38.
b Total depreciation = 31 000 − 12 000
= $19 000
19 000
Distance travelled = ----------------0.284
= 66 901 km
1
2
b
1
2
Depreciation amount
= distance travelled × rate
Write a summary statement.
Total depreciation
= cost price − scrap value
total depreciation
Distance travelled = ----------------------------------------------rate of depreciation
where rate of depreciation = 28.4 cents/km
where rate of depreciation = $0.284 per km
State your answer.
The taxi has a useful life of
66 901 km.
Chapter 14 FM Page 695 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
695
WORKED Example 25
A photocopier purchased for $10 800 depreciates at a rate of 20 cents for every 100 copies
made. In its first year of use 500 000 copies were made and in its second year, 550 000.
Find:
a the depreciation each year
b the book value at the end of the second year.
THINK
WRITE
a To find the depreciation, identify the rate
and number of copies made.
Express the rate of 20 cents per 100 copies
in a simpler form of dollars per 100 copies,
0.20
that is $0.20 per 100 copies or --------------------------100 copies
a Depreciation = copies made × rate
b Book value = cost price − total depreciation
0.20
d1 = 500 000 × -------------------------100 copies
= $1000
Depreciation in the first year is $1000.
0.20
d2 = 550 000 × -------------------------100 copies
= $1100
Depreciation in the second year is $1100.
b Total depreciation after 2 years
= 1000 + 1100
= $2100
Book value = 10 800 − 2100
= $8700
WORKED Example 26
The initial cost of a vehicle was $27 850 and its scrap value is $5050. If the vehicle needs to
be replaced after travelling 80 000 km (useful life):
a Find the depreciation rate (depreciation ($) per km).
b Find the amount of depreciation in a year when 16 497 km were travelled.
c Find the book value after it has been used for a total of 60 000 km.
d Set up schedule table listing book value for every 20 000 km.
THINK
WRITE
a
a Total amount of depreciation
= 27 850 − 5050
= $22 800
1
2
To find the depreciation rate, first
find the total depreciation.
Total amount of depreciation
= Cost price − Scrap value
Find the rate of depreciation.
It is common to express rates in
cents per use if less than a dollar.
Depreciation rate
total depreciation
= -----------------------------------------------------total distance travelled
22 800
= ----------------80 000
= $0.285 per km
= 28.5 cents per km
Continued over page
Chapter 14 FM Page 696 Monday, November 13, 2000 3:32 PM
696
Further Mathematics
THINK
b
WRITE
b Amount of depreciation
= amount of use × rate of depreciation
= 16 497 × 28.5
Amount of depreciation is always
= 470 165 cents
expressed in dollars.
= $4701.65
c 1 To find the book value first calculate
c Amount of depreciation
the amount of depreciation for a use
= 60 000 × 28.5
of 60 000 km.
= 1 710 000 cents
= $17 100.00
Book value
2 Calculate the book value. In this case
the 60 000 km has been travelled from
= previous value − amount of depreciation
when the car was new so the previous
= 27 850 − 17 100
book value is $27 850.
= $10 750
Write
your
answer.
Book
value after the car has been used for
3
60 000 km is $10 750.
d Calculate the book value for every
d For every 20 000 km,
20 000 km of use and summarise
amount of depreciation = 20 000 × 0.285
in a table.
= $5700
Find the amount of depreciation
using the rate calculated.
Use (km)
Book value ($)
0
$27 850
20 000
$22 150
40 000
$16 450
60 000
$10 750
80 000
$ 5 050
Alternatively, using a graphics calculator, enter the
equation in Y = .
The depreciation equation is BV = 27 850 – 0.285 km.
Set up TABLE in TABLE SET and display the table.
Start table at 0 km and increment at every 20 000 km.
Chapter 14 FM Page 697 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
697
remember
remember
1. Unit cost depreciation is based on how much an item is used.
2. Current book value ($) = previous book value ($) − amount of depreciation ($)
3. Amount of depreciation ($) = amount of use × rate of depreciation ($ per use)
amount of depreciation ($)
4. Rate of depreciation ($ per use) = ---------------------------------------------------------------amount of use
14G
Example
24
Math
1 Below are depreciation details for 6 vehicles. In each case find:
i the annual depreciation
ii the useful life (km).
Purchase price
($)
cad
WORKED
Unit cost depreciation
Unit cost
depreciation
Scrap value
($)
Average rate of
depreciation
(c/km)
Distance
travelled in first
year (km)
a
25 000
10 000
26
12 600
b
21 400
8 000
21.6
13 700
c
29 600
12 000
28.5
14 000
d
32 000
13 000
32.2
19 600
e
23 900
10 500
24.8
12 656
f
31 440
13 500
22.7
15 144
In each situation in questions 2–4, find:
a the annual depreciation
b the item’s useful life.
2 A company buys a $32 000 car which depreciates at a rate of 23 cents per km driven.
It covers 15 340 km in the first year and has a scrap value of $9500.
3 A new taxi is worth $29 500 and it depreciates at 27.2 cents per km travelled. In its
first year of use it travelled 28 461 km. Its scrap value was $8200.
4 A government car was purchased for $36 949 and depreciated at a rate of 21.5 cents
per km travelled. It travelled 9560 km in its first year of use. It has a scrap value of
$10 350.
WORKED
Example
25
In each situation in questions 5–8, find:
a the depreciation for each year
b the book value at the end of the second year.
5 A photocopier is bought for $8600 and it depreciates at a rate of 22 cents for every
100 copies made. In its first year of use 400 000 copies are made and in its second
year, 480 000 copies are made.
6 A photocopier purchased for $7200 depreciates at a rate of $1.50 per 1000 copies
made. In its first year of use 620 000 copies were made and in its second year,
540 000 were made.
Chapter 14 FM Page 698 Monday, November 13, 2000 3:32 PM
698
Further Mathematics
7 A metal stamping machine purchased for $15 800 depreciates at a rate of 12.5 cents
for every 100 completed stamps. In its first year of use 2 million stamps were completed and in its second year 2.4 million were made.
8 A printing machine was purchased for $38 000 and depreciated at a rate of $1.50 per
million pages printed. In its first year 385 million pages were printed and 496 million
in its second year.
9 A photocopier bought for $11 300 depreciates at a rate of 2.5 cents for every 10 copies
made. Copy and complete the table below.
Time
(years)
Copies made
per year
1
350 000
2
425 000
3
376 200
4
291 040
5
385 620
Annual depreciation
($)
Book value
at end of year ($)
10 A corking machine bought for $14 750 depreciates at a rate of $2.50 for every
100 bottles corked. Copy and complete the table below.
Time
(years)
Bottles corked
per year
1
40 000
2
42 500
3
46 700
4
38 250
5
43 060
Depreciation ($)
Book value
at end of year ($)
11 multiple choice
A vehicle is bought for $25 900 and it depreciates at a rate of 21.6 cents per km
driven. After its first year of use, in which it travelled 13 690 km, the book value of
the vehicle will be closest to:
A $1000
B $3000
C $20 000
D $23 000
E $25 000
In each situation in questions 12 and 13, find:
a the depreciation for each year
b the book value at the end of the second year.
12 A company van is purchased for $32 600 and it depreciates at a rate of 24.8 cents per
km driven. In its first year of use the van travelled 15 620 km and it travelled
16 045 km in its second year.
13 A taxi is bought for $35 099 and it depreciates at a rate of 29.2 cents per km driven. It
travelled 21 216 km in its first year of use and 19 950 km in its second year.
Chapter 14 FM Page 699 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
699
14 A car bought for $28 395 depreciates at a rate of 23.6 cents for every km travelled.
Copy and complete the table below.
Time
(years)
Distance
travelled (km)
1
13 290
2
15 650
3
14 175
4
9 674
5
16 588
Depreciation ($)
Book value
at end of year ($)
15 A farm vehicle bought for $33 949 depreciates at a rate of 32.5 cents per km driven.
Copy and complete the table below.
Time
(years)
Distance
travelled (km)
1
19 150
2
16 490
3
20 565
4
10 986
5
25 782
Depreciation ($)
Book value
at end of year ($)
16 The following information provides depreciation details about various vehicles. In
each case find the depreciation rate (depreciation per km).
26a
a Initial cost = $25 250, scrap value = $5000, useful life = 90 000 km
b Initial cost = $31 040, scrap value = $6000, useful life = 80 000 km
c Initial cost = $36 000, scrap value = $6000, useful life = 80 000 km
d Cost price = $21 005, scrap value = $4000, useful life = 95 000 km
e Cost price = $19 960, scrap value = $4000, useful life = 95 000 km
f Cost price = $29 820, scrap value = $5000, useful life = 85 000 km
WORKED
Example
17 The following information provides depreciation details about various photocopiers.
In each case find the depreciation rate (depreciation per 100 copies).
a Initial cost = $8000, scrap value = $2000, useful life = 4 million copies
b Initial cost = $7800, scrap value = $2000, useful life = 4 million copies
c Initial cost = $5900, scrap value = $1500, useful life = 2 million copies
d Cost price = $10 100, scrap value = $2000, useful life = 3 million copies
e Cost price = $6800, scrap value = $1500, useful life = 2 million copies
f Cost price = $7500, scrap value = $1500, useful life = 4 million copies
18 multiple choice
A machine costing $8500 is estimated to have scrap value of $500 and believed to
have a maximum output of 400 000 units. The depreciation rate (charged per unit) is:
A 20 cents per unit
B 2 dollars per unit
C 2 cents per unit
D 2.125 cents per unit
E 2.125 dollars per unit
Chapter 14 FM Page 700 Monday, November 13, 2000 3:32 PM
700
Further Mathematics
19 multiple choice
An $8500 machine depreciates by 2 cents/unit. By the time the machine had depreciated by $5000, it would have produced:
A 275 000 units
B 250 000 units
C 225 000 units
D 175 000 units
E 150 000 units
20 multiple choice
A machine which was bought for $8500 was depreciated at the rate of 2 cents per unit
produced. By the time the book value had decreased to $2000, the number of units
produced would be:
A 75 000
B 100 000
C 125 000
D 300 000
E 325 000
21 A delivery service purchases a van for $30 000 and it is expected that the van will be
written off after travelling 200 000 km. It is estimated that the van will travel 1600 km
26
each week.
a Find the depreciation rate (charge per km).
b Find how long it will take for the van to be written off.
c Find the distance travelled for the van to depreciate by $13 800.
d Find its book value after it has travelled 160 000 kilometres.
e Set up a schedule table for the value of the van for every 20 000 kilometres.
WORKED
Example
22 A car is bought for $35 000 and a scrap value of $10 000 is set for it. The following
three options for depreciating the car are available:
i flat rate of 10% of the purchase price each year
ii 20% p.a. of the reducing balance
iii 25 cents per km driven (the car travels an average of 10 000 km per year).
a Which method will enable the car to reach its scrap value sooner?
b If the car is used in a business the annual depreciation can be claimed as a tax
deduction. What would the tax deduction be in the first year of use for each of the
depreciation methods?
c How would your answers to part b vary for the 5th year of use?
Chapter 14 FM Page 701 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
701
Inflation
The term inflation is often used when talking about prices. It is a measure of the
average increase in the price of goods and services from one year to the next. The effect
of inflation (price increases) is that money loses its value and therefore its purchasing
power (how much we can buy with it).
Inflation is often expressed as a rate per year and so indicates the annual increase in
the price of a fixed set of goods and services. An average of the annual inflation rates
over several years can be used to make price calculations of items. Since the prices
increase by a fixed rate each year, inflation is another example of exponential growth.
For example, consider an average inflation rate of 8% p.a. and a current price of
$1.65 for a carton of milk. What will be the price in 2 years’ time?
Now, growth factor = 100% + 8%
= 1 + 0.08
= 1.08
Price in 1 year’s time = 1.65 × 1.08
= $1.78
Price in 2 years’ time = 1.78 × 1.08
= $1.92
So, inflation rate works in much the same way as compound interest. In fact we can
use the compound interest formula as follows:
A = PRT
where A = price after time, T
P = original price ($)
T = time in years
r
R = 1 + --------- where r is the inflation rate
100
WORKED Example 27
Assuming an average inflation rate of 5% p.a. find the price for a $7.50 theatre ticket in
6 years’ time.
THINK
1
List P, r and T.
2
Substitute into the formula and evaluate.
WRITE
P = 7.5, r = 5, T = 6
T
r
A = P  1 + ---------

100
= 7.5(1.05)6
= $10.05
The price of a ticket after 6 years is $10.05.
Chapter 14 FM Page 702 Monday, November 13, 2000 3:32 PM
702
Further Mathematics
remember
remember
1. Inflation rate is a measure of the average percentage growth in the cost of
goods and services over a period of years.
2. It is another example of exponential growth and we can use the compound
interest formula as follows:
A = PRT
where A = price after time, T
P = original price ($)
T = time in years
r
R = 1 + --------- where r is the inflation rate
100
14H
WORKED
d
hca
Mat
Example
Inflation
27
Inflation
1 In each case below find the price of the item in:
a 5 years’ time if it now costs $15 and the average inflation is 4% p.a.
b 6 years’ time if it now costs $290 and the average inflation is 5% p.a.
c 4 years’ time if it now costs $42 000 and the average inflation is 4.3% p.a.
d 2 years’ time if it now costs $6870 and the average inflation is 3.8% p.a.
e 7 years’ time if it now costs $8.95 and the average inflation is 2.1% p.a.
f 10 years’ time if it now costs $1.25 and the average inflation is 4.7% p.a.
g 5 years’ time if it now costs $15 600 and the average inflation is 2.9% p.a.
h 4 years’ time if it now costs $105.50 and the average inflation is 5.1% p.a.
2 What could you have expected to pay for an item:
a 5 years ago if it now costs $15 and the average inflation has been 4% p.a.?
b 6 years ago if it now costs $290 and the average inflation has been 5% p.a.?
c 4 years ago if it now costs $42 000 and the average inflation has been 4.3% p.a.?
d 2 years ago if it now costs $6870 and the average inflation has been 3.8% p.a.?
e 7 years ago if it now costs $8.95 and the average inflation has been 2.1% p.a.?
f 10 years ago if it now costs $1.25 and the average inflation has been 4.7% p.a.?
g 5 years ago if it now costs $15 600 and the average inflation has been 2.9% p.a.?
h 4 years ago if it now costs $105.50 and the average inflation has been 5.1% p.a.?
3 How many years (to the nearest year) would it take for the price of an item to go from:
a $25.00 to 34.21 if the average inflation during the time was 4% p.a.?
b $1.30 to $1.74 if the average inflation during the time was 5% p.a.?
c $500.00 to $623.09 if the average inflation during the time was 4.5% p.a.?
d $12 000 to $17 796 if the average inflation during the time was 8.2% p.a.?
e 80 cents to $1.08 if the average inflation during the time was 1.5% p.a.?
f $2695 to $3270 if the average inflation during the time was 2.8% p.a.?
g $945 to $1089 if the average inflation during the time was 3.6% p.a.?
h $89.00 to $142.87 if the average inflation during the time was 5.4% p.a.?
4 The price of a standard cricket bat changes from $140 to $190 over a 6-year period.
Find the average annual inflation rate during this time.
Chapter 14 FM Page 703 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
703
5 The price of 100W light bulbs changes from 85 cents to $1.03 over a 5-year period.
Find the average annual inflation rate during this time (to 2 decimal places).
6 Over a 12-month period the price of a certain cereal changes from $4.85 to $5.25.
Find the annual inflation rate during the year (to 2 decimal places).
7 Calculate the average inflation rate for the 9-year period 1989–1998 of the items in
the table (to 1 decimal place):
Item
a
Video
b
1989 Price ($)
1998 Price ($)
23.95
29.95
Pen
0.97
1.20
c
Can of coke
1.67
2.10
d
Council rates
950.00
1100.00
e
Cough medicine
6.40
9.95
f
Freddo frog
0.50
0.65
g
Pair of socks
7.70
8.50
h
Toaster
64.50
69.90
8 At the start of 1992 the price of a tennis racquet was $110.50. Using an average
inflation rate of 14% per annum, find the price of this racquet at the start of 1998.
9 In 1985 the cost of a house was $75 000. Find the 1995 cost of the house (to the
nearest $100) if the inflation rate has been 5.2% per annum.
10 multiple choice
In a year when the inflation rate was 12.4% and the cost of a book at the start of the
year was $8.95, the cost at the end of the year would be closest to:
A $6
B $8
C $10
D $12
E $13
11 multiple choice
In a year when the price of a cubic metre of firewood was $35, the inflation rate over
the past three years when its cost rose from $25 would be closest to:
A 9%
B 10%
C 11%
D 12%
E 13%
Chapter 14 FM Page 704 Monday, November 13, 2000 3:32 PM
704
Further Mathematics
summary
Growth and decay functions
• Growth and decay can be linear or exponential.
• Linear growth and decay can be represented by the equation
y = a + bx
where y is the dependent variable
x is the independent variable (usually time)
a is the initial or starting value of y
and b is the rate of growth or decay.
• Exponential growth and decay means an initial value multiplied by a growth or
compounding factor for each unit time interval.
• Exponential growth and decay can be represented by the equation
y = kax
where a = growth or compounding factor
a > 1 for growth, a < 1 for decay
and k = the initial or starting value of y.
Compound interest formula
• Compound interest is an example of exponential growth and is calculated using the
formula:
where A = final amount ($)
P = principal ($)
r
R = growth or compounding factor = 1 + --------100
r = interest rate per period
n = number of interest bearing periods
• For compound interest, I = A − P
A = PRn
Finding time in compound interest using trial and error
• Finding time by trial and error involves educated guessing for the value of n.
• Time = n × number of times interest is compounded each year.
Finding time in compound interest using logarithms
• When using logs to find time, use the rule: log xn = n log x.
• Change the n value to time in more meaningful terms.
Depreciation
• There are 3 methods by which depreciation can be calculated:
1. flat rate depreciation
2. reducing balance depreciation
3. unit cost depreciation.
• An item is written off when its book value becomes zero.
• Scrap value is an item’s book value when it is no longer used.
Chapter 14 FM Page 705 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
705
Flat rate depreciation
• To make calculations in flat rate depreciation, use the formula
BVT = P − dT where BVT = book value after T years ($)
P
= purchase or cost price ($)
T
= time (years of depreciation)
d
= rate of depreciation ($ per year)
= fixed amount per year or
= percentage of P per year
• Flat rate depreciation is also known as prime cost depreciation.
• Total depreciation = cost price − current value
total depreciation
• Rate of depreciation = -----------------------------------------number of years
Reducing balance depreciation
• Reducing balance depreciation is an example of exponential decay and is calculated
using the formula:
r T
BVT = P  1 – ---------

100
where BVT is the book value after time, T
P is the cost price
r is the rate of depreciation
T is the time since purchase
Unit cost depreciation
• Unit cost depreciation is based on how much an item is used
• Current book value ($) = previous book value ($) − amount of depreciation ($)
• Amount of depreciation ($) = amount of use × rate of depreciation ($ per use)
amount of depreciation ($)
• Rate of depreciation ($ per use) = ---------------------------------------------------------------amount of use
Inflation
• Inflation rate is a measure of the average percentage growth in the cost of goods
and services over a period of years.
• It is another example of exponential growth and we can use the compound interest
formula as follows:
A = PRT
where A = price after time, T
P = original price ($)
T = time in years
r
R = 1 + --------- where r is the inflation rate
100
Chapter 14 FM Page 706 Monday, November 13, 2000 3:32 PM
706
Further Mathematics
CHAPTER
review
Multiple choice
14A
1 What type of growth or decay does the graph at right display?
A Linear growth
B Linear decay
C Exponential growth
D Exponential decay
E Steady/no change
14A
2 Which of the following graphs displays an exponential decay?
A
B
C
D
E
FM 17 44
The following information refers to questions 3 and 4. An investment of $4500 earns compound
interest at a rate of 6.4% p.a. and is made for 5 years.
14B
3 The balance in the account at the end of the investment period, if interest is compounded
quarterly, is:
A $6181.40
B $4871.71
C $6136.50
D $15 561.27 E $8592.20
14B
4 The principal plus interest accrued during the investment, if interest is credited
weekly, is:
A $6136.50
B $15 561.27 C $7985.47
D $6195.86
E $4527.76
14B
5 After 4 1--2- years $1200 has grown to $1750 in an account where interest is compounded
14C,D
monthly. The annual interest rate is:
A 7.0%
B 0.7%
C 8.4%
D 3.2%
E 38%
6 A sum of $850 is invested at 8% p.a. compound interest, credited fortnightly. For the
balance to grow to $1200 the investment should be left for a minimum of:
A 112 years
B 113 years
C 4 years 8 fortnights
D 4 years 9 fortnights
E 5 years
Chapter 14 FM Page 707 Monday, November 13, 2000 3:32 PM
Chapter 14 Growth and decay
707
7 In an account which pays compound interest at 12% p.a., credited daily, an investment of
$13 000 will accrue $4000 interest in:
A 2 years 86 days
B 2 years 87 days
C 3585 days
D 9 years 301 days
E 10 years
14D
8 A machine bought for $8500 is depreciated by the flat rate method. If its useful life is
6 years and its scrap value is $1000 then the annual depreciation will be:
A $1417
B $1583
C $7500
D $7494
E $1250
14E
The following information refers to questions 9–11. A refrigerator valued at $1250
is depreciated at 25% p.a. of the previous book value.
9 The book value of the refrigerator after 4 years will be:
A $395.51
B zero
C $1150
D $4.88
10 The total depreciation after 4 years will be:
A $2150
B $1245.12
C $854.49
D $4.88
E $1245.12
14F
E $100
14F
11 If the refrigerator has a scrap value of $200 then its useful life will be closest to:
A 5 years
B 6 years
C 7 years
D 4 years
E 3 years
14F
The following information refers to questions 12 and 13. A car was purchased for $32 600 and
depreciated at a rate of 25.6 cents per km driven.
12 The depreciation in a year when the car travelled 15 780 km would be:
A $31 983.59 B $28 560.32 C $5020
D $4039.68
E $616.41
14G
13 If the car travelled an average of 15 000 km each year then it would be written off after:
A 2 years
B 3 years
C 4 years
D 8 years
E 9 years
14G
14 The inflation rate for the past 6 years has averaged 3.75% per annum. If the salary of a
teacher 6 years ago was $38 000 then the teacher, to maintain the standard of earnings at the
present time, should be receiving close to:
A $44 000
B $45 000
C $46 000
D $47 000
E $48 000
14H
Chapter 14 FM Page 708 Monday, November 13, 2000 3:32 PM
708
Further Mathematics
Short answer
14A
1 If the average inflation rate during a 5-year period had been 3.5% p.a., what would be the
cost of a jar of peanut butter at the end of the period if it cost $2.60 at the start?
14B
2 If $5400 is to be invested for 5 years, which of the options below would be the most
productive to use?
a 12% p.a. simple interest
b compound interest at 11.8% p.a., credited quarterly
c compound interest at 11.7% p.a., credited monthly.
14B
3 What amount must be invested at 9.25% p.a., interest compounded 6-monthly, if it is to
grow to $5000 over 4 years?
14B
4 How much interest would $950 earn if it was invested for 3 years at 12% p.a., interest
credited daily?
14C,D
5 How long would it take for $2000 to amount to $3450 by earning interest at 6.8% p.a.,
compounded monthly?
14E
6 A computer is depreciated by the prime cost method at 15% p.a. If it was bought for $4900
how many years elapsed before it was written off?
14F
7 Furniture is bought for $15 000 and depreciated at 18% p.a. by the reducing
balance method. What would its value be in 6 years’ time?
14G
8 A taxi depreciates at 29.5 cents per km driven. If it was bought for $29 600, how far would
it have travelled for it to be valued at $10 000?
14G
9 A taxi was purchased for $38 000 and it depreciates at an average rate of 30 cents per km
driven. During its first year the taxi travelled 21 650 km and during its second it travelled
19 880 km. Find:
a the depreciation in each of the first 2 years
b how far the car had travelled if its total depreciation was $20 000
c how far the car had travelled when it reached its scrap value of $5000.
14H
10 A stamp collection was purchased 12 years ago for $3350. It has increased in value at the
same rate as inflation. It is currently valued at $5680.
a Find the inflation rate.
b How long will it take for the stamp collection to double in value?
c How long ago did it have a value of $4980?
d Graph its value over the first 10 years on a graphics calculator or spreadsheet.
Chapter 14 FM Page 709 Monday, November 13, 2000 3:32 PM
709
Chapter 14 Growth and decay
Analysis
1 A small number of rabbits have migrated to Mr Smith’s farm. Mr Smith has brought in
experts to make some investigations.
a The small number of rabbits is estimated to be 80. From experience, the experts know that
the rabbit population increases by 20% on each previous month’s population.
i Copy and complete the table of the expected population of the rabbits for the first three
months.
Month
0
Rabbit population
80
1
2
3
ii A population of 1000 or more is considered to be harmful to the ecosystem. How soon
(in months) will this occur on Mr Smith’s farm?
b The area of land used (and devastated) by a rabbit population is given in the table below.
Number of rabbits
100
200
300
400
500
600
Area of land used (hectares)
1.25
2.5
3.75
5.0
6.25
7.5
i State whether it is an exponential or straight line growth.
ii Give evidence to justify your response in part i.
c The cost (C in dollars) to the farmer for every hectare of land (H) lost is given by the
equation C = 1000 × 1.4H.
i What is the cost to the farmer if there are 400 rabbits on the farm?
ii Mr Smith can only sustain about $2000 loss. How much land is this equivalent to (to
the nearest hectare)?
d At the completion of the investigation, the experts suggest Mr Smith budget $12 400 for a
rabbit eradication program. Mr Smith needs to borrow this amount and he has 2 loan options:
• 10% p.a. simple interest with full payment in 2 years
• 7% p.a. compound interest (adjusted monthly) with full payment in 3 years’ time.
i Calculate the total payment to be made at the end of the term for each of the loan
options.
ii Which is the best option? Explain why.
Another option is to use the cash set aside for a tractor and get the tractor on hire-purchase.
iii On hire-purchase he can get the $13 000 tractor
with a deposit of
$600 and monthly
instalments at
6% p.a. over 3 years.
Find the monthly
instalments.
Chapter 14 FM Page 710 Monday, November 13, 2000 3:32 PM
710
Further Mathematics
2 Alisha has saved a sum of money and has several options as she turns 18. She chooses to
investigate these options.
a Option 1 is for her to use the sum of $10 000 towards a new 4WD car valued at
$27 000. She can get it on a hire-purchase agreement with $10 000 deposit and
monthly instalments over 4 years at 6% p.a.
i Calculate the monthly instalments.
ii What is the total cost of the car?
iii What is the effective interest rate?
b Another option is to buy a second-hand car for $12 000. Alisha can borrow the extra
$2000 at 7% p.a. compounded annually for 3 years with a single payment at the end
of the term.
i Find the amount of interest Alisha will have to pay at the end of the 3 years for
the $2000 loan.
ii What will be the total cost of the car?
c
The third option is to buy a reliable car for $5000 and use the other $5000 to buy a
collectable painting, the value of which will increase at the same rate as inflation.
i What is the expected value of the painting in 4 years’ time if inflation is running
at 5% p.a.?
ii The $5000 car is expected to depreciate at 10% p.a. using the straight line
depreciation model. What is its book value in 4 years’ time?
d The other 2 cars, being up-market models, depreciate at 15% p.a. on reducing value.
i What will be the book value in 4 years’ time for the $27 000 4WD car?
ii What will be the book value in 4 years’ time for the $12 000 car?
iii Find which of the three options loses the most money.
3 Complete analysis task 2 and use the information to fill in a copy of the following table.
Description
Cost of loan (interest charged only)
Depreciation of goods
Total cost
Less benefits, e.g. increase in values
CHAPTER
test
yyourself
ourself
14
Total cost
$27 000
4WD car
$12 000
car
$5000 car
and painting