9.6Simple Interest

Simple interest is charged on the original principal every year, so it grows by the same amount each year rather than compounding. The total is the principal plus its interest.

Simple interest formula

\displaystyle I = \dfrac{P \times R \times T}{100}

$P$ = principal, the original amount deposited or borrowed
$R$ = annual interest rate as a percentage (e.g. $3.5\%$ means $R = 3.5$ )
$T$ = time in years, always convert months to years before substituting

The example below, year by year.

\$6000

3.5\%

earns

\$210

each year, so after 4 years the interest is

\$840

Finding interest on a savings account

Wei Ling deposits $\displaystyle \$6000$ at $\displaystyle 3.5\%$ simple interest per annum for 4 years. Find the interest and the total amount.

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\displaystyle I

\displaystyle = \dfrac{6000 \times 3.5 \times 4}{100}

\displaystyle = \dfrac{84\,000}{100}

\displaystyle = \$840

\displaystyle A

\displaystyle = P + I

\displaystyle = \$6000 + \$840

\displaystyle = \$6840

Interest over a number of months

Ahmad borrows $\displaystyle \$4500$ at $\displaystyle 6\%$ simple interest per annum for 10 months. How much interest does he owe?

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\displaystyle \text{Time in years}

\displaystyle = \dfrac{10}{12}

\displaystyle = \dfrac{5}{6} \text{ years}

\displaystyle I

\displaystyle = \dfrac{4500 \times 6 \times \frac{5}{6}}{100}

\displaystyle = \dfrac{4500 \times 5}{100}

\displaystyle = \dfrac{22\,500}{100}

\displaystyle = \$225

Now you try

Find the simple interest on $\$1000$ at $2\%$ per annum for $5$ years.

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\displaystyle I

\displaystyle = \frac{P \times R \times T}{100}

\displaystyle = \frac{1000 \times 2 \times 5}{100}

\displaystyle = \$100

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