# 8.03 Representing simple interest graphically

Lesson

Simple interest can be modelled as a linear graph using the simple interest formula $I=Prn$I=Prn.

We can use simple interest graphs to compare different scenarios, such as:

• the interest, $I$I, earned over time, $n$n
• the interest earned for different principals, $P$P
• the total value of a loan or investment, $A$A, over time, $n$n

The graph of simple interest is a straight line, since the interest rate is constant. The gradient of the line indicates how much interest is earned in each time period. If the graph shows interest $I$I against $n$n then the $y$y-intercept will be zero, and if the graph shows total amount $A$A against $n$n then the $y$y-intercept will represent the initial amount invested or borrowed.

When presented with a graph, make sure to read the description of the given graph and the axis labels to understand the context of the question.

#### Worked examples

##### Example 1

Below is the graph showing the amount of interest earned over time at a particular interest rate.

(a) How much interest is earned after $5$5 years?

Think: Find the point on the line that corresponds to $5$5 years.

Do: Here is the point on the line that corresponds to $5$5 years on the horizontal axis:

Reading across, we can see that this point corresponds to $600$600 on the vertical axis.

Therefore the total interest earned after $5$5 years is $\$600$$600. (b) What is the gradient of the line? Think: We can calculate the gradient using \text{gradient }=\frac{\text{rise }}{\text{run }}gradient =rise run . Do: From part (a), we know that \600$$600 of interest is earned in $5$5 years. That is, after a run of $5$5 there is a rise of $600$600. So the gradient will be $\frac{600}{5}=120$6005=120.

This tells us that $\$120$$120 of interest is earned each year. (c) How long will it take to earn \1800$$1800 of interest?

Think: The portion of the graph shown does not extend all of the way to $\$1800$$1800, so we can instead use the gradient that we just found to calculate the time period. Do: We know that \120$$120 is earned each year. So to earn $\$1800$$1800 will take \frac{1800}{120}=151800120=15 years. ##### Example 2 The graph below shows the total amount of an investment over time. a) Find the initial amount of the investment Think: The initial amount of an investment is the amount when time is 00. On the graph, this corresponds to the vertical intercept. Do: Looking at the vertical intercept of the graph, we can see that the initial amount invested was \4000$$4000.

b) What is the interest rate that the investment is earning per year?

Think: The gradient of the line indicates the amount of increase each year. We can work out the interest rate by using the initial value and one other point on the graph.

Do: The value of the investment after $50$50 years is $\$5000$$5000. The interest earned over 5050 years is the difference between \5000$$5000 and the initial investment of $\$4000$$4000 - that is, the interest earned over 5050 years is I=\5000-\4000=\1000I=50004000=1000. We can now use the simple interest formula to find the interest rate rr:  II == PrnPrn 10001000 == 4000\times r\times504000×r×50 (substituting known values) 10001000 == 20000\times r20000×r rr == \frac{1000}{20000}100020000​ (rearranging for rr) == 0.0050.005 == 0.5%0.5% #### Practice questions ##### Question 1 The graph shows the amount of simple interest charged each year by a particular bank, on some 33-year loan. Loading Graph... 1. Find the total amount of simple interest charged on a loan of \5000$$5000.

2. Hence, calculate the simple interest rate per annum, $R$R, charged by the bank on these $3$3-year loans.

Enter each line of working as an equation.

An amount of $\$4000$$4000 is to be invested at a simple interest rate of 6%6% per annum. 1. How much interest is earned each year? 2. Which of the following shows the amount of the investment after xx years?  Loading Graph... y=4000xy=4000x A  Loading Graph... y=4000x+240y=4000x+240 B  Loading Graph... y=240x+4000y=240x+4000 C  Loading Graph... y=240xy=240x D  Loading Graph... y=4000xy=4000x A  Loading Graph... y=4000x+240y=4000x+240 B  Loading Graph... y=240x+4000y=240x+4000 C  Loading Graph... y=240xy=240x D 3. Which feature of the graph represents the interest earned each year? The value of yy when x=1x=1. A The gradient. B The yy-intercept. C The value of yy when x=1x=1. A The gradient. B The yy-intercept. C 4. How would the graph change if the amount was invested at a rate of 3%3% p.a. instead? Select all correct options. The straight line graph would be flatter. A The straight line graph would increase more quickly. B The straight line graph would be steeper. C The graph would shift vertically downwards. D The straight line graph would increase more slowly. E The graph would shift vertically upwards. F The straight line graph would be flatter. A The straight line graph would increase more quickly. B The straight line graph would be steeper. C The graph would shift vertically downwards. D The straight line graph would increase more slowly. E The graph would shift vertically upwards. F ##### Question 3 The graph below shows an amount of \4000$$4000 deposited into a savings account for $10$10 years with simple interest.

1. How much interest has been earned in total over the $10$10 year period?

2. Follow the steps below and fill in the blanks to calculate the interest rate of this investment:

The formula for simple interest is

$I=PRT$I=PRT.

We have just calculated that $I=\editable{}$I=, and we know that $P=\editable{}$P= and $T=\editable{}$T=.

Substituting these values into the formula, we get

$\editable{}=\editable{}\times R\times\editable{}$=×R×.

Rearranging this, we see that

 $R$R $=$= $\frac{\editable{}}{\editable{}\times\editable{}}$×​ $=$= $\editable{}$

The interest rate is therefore $R=\editable{}$R=$%$%.

### Constructing a simple interest graph

In a simple interest graph, the interest $I$I is the dependent variable and the number of time periods $n$n is the independent variable.

To draw a simple interest graph, we can start by constructing a table of values of $I$I for various values of $n$n. It might look something like this:

 $n$n $1$1 $2$2 $3$3 $I$I $\$100$$100 \200$$200 $\$300$$300 #### Worked example ##### Example 3 \5000$$5000 is invested at $6%$6% p.a. over a period of $10$10 years. Construct a table of values and draw the graph of interest over time for this investment.

Think: Simple interest is calculated using $I=Prn$I=Prn and in this question we have that $P=5000$P=5000 and $r=0.06$r=0.06 (as a decimal), with $I$I and $n$n as variables. So the equation of our line is $I=5000\times0.06\times n=300n$I=5000×0.06×n=300n, which we can use to construct a table of values.

Do: Let's substitute some values for $n$n into this equation to find the corresponding values of $I$I:

• When $n=1$n=1, $I=300\times1=300$I=300×1=300
• When $n=2$n=2, $I=300\times2=600$I=300×2=600
• When $n=3$n=3, $I=300\times3=900$I=300×3=900, and so on.

So our table of values will look like this:

### Outcomes

#### MS11-2

represents information in symbolic, graphical and tabular form

#### MS11-5

models relevant financial situations using appropriate tools

#### MS11-6

makes predictions about everyday situations based on simple mathematical models