NZ Level 6 (NZC) Level 1 (NCEA) Simple Interest
Lesson

It costs money to borrow money. The extra money that banks and other lenders charge us to borrow money is called interest. However, interest may also refer to additional money you earn from investing money, such as in a savings accounts. There are different types of interest and today we are going to learn about 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 charge 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)

#### Examples

##### Question 1

Calculate the simple interest on a loan of $\$8580$$8580 at 2%2% p.a. for 1010 years. Write your answer to the nearest cent. Think: We can sub in the values for the principal, interest rate and time periods. Do:  II == PRTPRT == 8580\times0.02\times108580×0.02×10 == \1716$$1716

##### Question 2

The interest on an investment of $\$3600$$3600 over 1010 years is \2520.00$$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 $2520$2520 $=$= $36000R$36000R $R$R $=$= $\frac{2520}{36000}$252036000​ $R$R $=$= $0.07$0.07 $R$R $=$= $7%$7%

##### 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.

Enter each line of working as an equation.

### Outcomes

#### NA6-3

Apply everyday compounding rates

#### 91026

Apply numeric reasoning in solving problems