When we talk about making an investment we're talking about finding something to purchase or put our money into that will increase in value, so that we can make a profit. Some options include investing in a house, investing in shares, or putting your money in the bank.
When deciding what to do with our money, the most important factor is usually how much profit stands to be made. If we lend money to the bank, this profit comes in the form of interest. This is what we will focus on here.
Interest
Interest is a fee that is charged on money that is borrowed.
When you make an investment by putting your money in a bank account, the bank is borrowing money from you, and so pays you interest.
When you borrow money from a bank (or another lending agency), you have to pay the bank interest for taking out a loan with them.
There are several different types of interest, but this lesson focuses on simple interest.
What is simple interest?
Simple, or straight line interest is a method where the interest amount is fixed (i.e. it doesn't change). The interest charged is always based on the original principal, so interest on interest is not included.
It is calculated using the formula:
$I=PRT$I=PRT
where $P$P is the principal (the initial amount borrowed)
$R$R is the interest rate, expressed as a decimal or fraction
$T$T is the number of time periods (the duration of the loan)
If we are given an interest rate of $r$r$%$% per annum, then to convert it to the $R$R used in the formula above, we need to divide it by $100$100. So:
$R=\frac{r}{100}$R=r100
Examples
Question 1
Calculate the simple interest on a loan of $\$8580$$8580 at $2%$2% p.a. for $10$10 months. Write your answer to the nearest cent.
Think: We can substitute in the values for the principal, interest rate and time periods into our simple interest formula. Remember here, we'll need to change our time, 10 months, to years. To do this we need to divide by 12.
Do:
| $I$I | $=$= | $PRT$PRT |
| $=$= | $8580\times0.02\times\frac{10}{12}$8580×0.02×1012 | |
| $=$= | $\$143$$143 |
Question 2
The interest on an investment of $\$3600$$3600 over $10$10 years is $\$2520.00$$2520.00. If the annual interest rate is $R$R, find $R$R as a percentage correct to $1$1 decimal place.
Think: What values do we know that we can sub in?
Do:
| $I$I | $=$= | $PRT$PRT | |
| $2520$2520 | $=$= | $3600\times R\times10$3600×R×10 | Substitute in the values we know. |
| $2520$2520 | $=$= | $36000R$36000R | |
| $R$R | $=$= | $\frac{2520}{36000}$252036000 | |
| $R$R | $=$= | $0.07$0.07 | Turn this into a percentage. |
| $R$R | $=$= | $7%$7% |
QUESTION 3
Sally made loan repayments adding up to $\$4320$$4320 on a loan of $\$4000$$4000 over $4$4 years.
Calculate $I$I, the total simple interest charged on the loan.
Calculate $r$r, the annual simple interest rate.
Express the rate as a percentage value.