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Middle Years

9.02 Simple interest

Lesson

It costs money to borrow money from financial institutions (like banks). The extra money that these lenders charge is called interest. Interest can also refer to money earned from investing money, such as in a savings accounts.

The starting amount, either borrowed or invested, is called the principal. The interest is usually described as a rate (percentage) per annum. For example, an investment of $\$100$$100 at a rate of $3%$3% per annum. $3%$3% of $100$100 is $3$3 , so this investment produces $\$3$$3 every year.

Calculating simple interest

Simple, or straight line interest is a method where the interest amount is fixed (i.e. it doesn't change). The interest is based on the original principal.

It is calculated using the formula:

$I=\frac{PrT}{100}$I=PrT100 ,

where $I$I is the total interest earned

$P$P is the principal (the initial amount borrowed/invested)

$r$r% is the interest rate, expressed as a whole number

$T$T is the number of time periods (the duration of the loan/investment)

Alternatively, when the interest rate is expressed as a decimal or fraction, the following formula can be used

$I=PRT$I=PRT  , where $R=\frac{r}{100}$R=r100$%$%

Remember to be careful with percentages!

The percentage symbol $%$% means per cent, which is a number out of $100$100. For example, $10%=\frac{10}{100}$10%=10100. Great care must be taken when substituting interest rates into either formula above. The first formula, the interest rate can be entered as a whole number. For example, for $10%$10% interest, using formula $I=\frac{PrT}{100}$I=PrT100, the interest rate substitution would be $r=10$r=10.

Whereas the second formula, the interest rate must be entered as a fraction or decimal. For example, for $10%$10% interest, using the formula $I=PRT$I=PRT, the interest rate substitution would be $R=10%=\frac{10}{100}=\frac{1}{10}=0.10$R=10%=10100=110=0.10.

The interest rate is often given as a percentage per time period. For example $4.5%$4.5% p.a. where "p.a." is an abbreviation of per annum, which means every year. However, the unit of time used for $T$T must match the unit of time used in the interest rate $r$r. Sometimes this means a conversion is required. For example, the time of a loan / investment may be given as a number of days, which would then need to be converted into a number (or fraction) of years.

Calculating principal, rate or time

Using any three known pieces of information from the simple interest formula it is possible to find the remaining unknown variable.

The formula can be rearranged to calculate the principle, the rate or the time. As long as you know the simple interest rule you don't need to remember all the variations but they are useful to see.

Rearranging the simple interest formula

To find the interest rate, $r$r%:

$r=\frac{100I}{PT}$r=100IPT

To find the time, $T$T:

$T=\frac{100I}{Pr}$T=100IPr

To find the principal amount, $P$P:

$P=\frac{100I}{rT}$P=100IrT

Calculating the amount of the loan / investment

For simple interest, once the interest is calculated, it can then be added to the principal amount to calculate the total amount of the loan / investment, represented by $A$A. This is found using the following formula:

$A=P+I$A=P+I

In some cases you may know the final amount of the investment $A$A, and not $I$I, the interest that has been earned. In this case use the formula:

$P=\frac{A}{(1+RT)}$P=A(1+RT) , where $R$R is the interest rate expressed as a decimal or fraction.

Practice questions

Question 1

Calculate the simple interest on a loan of $\$8000$$8000 at $8%$8% p.a. for $6$6 years.

Give the answer to the nearest dollar.

Question 2

Question 3

For a simple interest rate of $6%$6% p.a. , calculate the number of years $T$T needed for an interest of  $\$1174.20$$1174.20 to be earned on the investment $\$1957$$1957.

Give your answer as a whole number of years.

Enter each line of working as an equation.

Graphing simple interest

Simple interest can be modelled as a linear graph.

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

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

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 $t$t then the $y$y-intercept will be zero, and if the graph shows total amount $A$A against $t$t 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.

Exploration

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

We can find how much interest is earned after $5$5 years by finding the point on the line that corresponds to $5$5 years.

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.

we can find the gradient of the line by calculate the gradient using $\text{gradient }=\frac{\text{rise }}{\text{run }}$gradient =rise run .

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.

We can calculate how long will it take to earn $\$1800$$1800 of interest by considering the portion of the graph shown does not extend all of the way to $\$1800$$1800, so we can instead use the gradient that we just found to calculate the time period. We know that $\$120$$120 is earned each year. So to earn $\$1800$$1800 will take $\frac{1800}{120}=15$1800120=15 years.

Practice questions

Question 4

The graph shows the amount of simple interest charged per year by a particular bank.

Loading Graph...

  1. Find the total amount of simple interest charged on a loan of $\$5000$$5000 over a $3$3 year period.

  2. Determine the simple interest rate per year, $r$r, charged by the bank on these loans.

    Enter each line of working as an equation, and give your answer as a percentage.

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