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.

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`=`P``R``T`

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)

Calculate the simple interest on a loan of $\$8580$$8580 at $2%$2% p.a. for $10$10 years. Write your answer to the nearest cent.

**Think:** We can sub in the values for the principal, interest rate and time periods.

**Do:**

$I$I |
$=$= | $PRT$PRT |

$=$= | $8580\times0.02\times10$8580×0.02×10 | |

$=$= | $\$1716$$1716 |

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 |

$2520$2520 | $=$= | $36000R$36000R |

$R$R |
$=$= | $\frac{2520}{36000}$252036000 |

$R$R |
$=$= | $0.07$0.07 |

$R$R |
$=$= | $7%$7% |

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.

Apply everyday compounding rates

Apply numeric reasoning in solving problems