Simple Interest
Simple Interest in NSW Year 10 Core maths uses the formula \(I = Prn\), where interest is earned or charged only on the original amount (NESA outcome MA5‑FIN‑C‑01). Here \(P\) is the principal, \(r\) the rate per period and \(n\) the number of periods, so the interest grows by the same amount each year.
You will find interest and total amounts for loans and investments, convert months to years, and rearrange the formula to find the rate, the time or the principal. You will also read simple interest from straight-line graphs, compare a yearly rate with a quarterly rate, work out how long money takes to double, and split savings between two accounts. It is part of Year 10 Core Earning Money.
Every question with a fully worked solution.
- Simple Interest - Video - Simple Interest Formula Watch
Theory
Simple interest is worked out on the original amount only, using \(I = Prn\). This Year 10 maths page shows how to find the interest and the total amount, change months into years, solve for the principal, rate or time, and read simple interest from a straight-line graph.
The principal (\(P\)) is the amount invested or borrowed. Interest (\(I\)) is the extra money earned on an investment or paid on a loan.
With simple interest, the interest is worked out on the principal only, so the same amount is added every period. The rate is usually per annum (p.a.), which means per year.
The amount or future value at the end is the principal plus the interest: \(A = P + I\). A flat-rate loan charges simple interest on the amount borrowed.
\(r\) is the rate per period as a decimal (\(4.5\% = 0.045\)) and \(n\) is the number of periods.
How to solve a simple interest problem
- List \(P\), \(r\) and \(n\), with the rate as a decimal and the time in the same period as the rate.
- Substitute into \(I = Prn\). If \(I\) is known, solve the equation for the unknown.
- Add the interest to the principal if the total amount, value or amount owed is asked for.
Other questions you will meet
- Value known, principal unknown: find the total percentage added, rate \(\times\) time, then divide the value by \(1 +\) that decimal (Example 3).
- Value known, time unknown: interest \(=\) value \(-\) principal, then divide by one year's interest.
- Comparing offers: change rates to the same period. \(1.2\%\) per quarter is \(4 \times 1.2\% = 4.8\%\) p.a.
- Short loans: to give a rate for a few months as a yearly rate, multiply by the number of those periods in a year.
- Doubling: the money has doubled when the interest equals the principal, so \(n = 1 \div r\).
- Money split between two accounts: let \(x\) be the amount in one account and the rest be (total \(- x\)), then write total interest as an equation.
- Graphs: the interest or value grows by the same amount each year. Read values from the graph; the rate is one year's interest \(\div\) principal. Where two value lines cross, the investments are worth the same.
(i) the interest earned
(ii) the value of the investment at the end.
The rate is per year, so write the time in years: \(30\) months \(= \dfrac{30}{12} = 2.5\) years.
| \(I\) | \(=\) | \(Prn\) |
| \(\) | \(=\) | \(8400 \times 0.045 \times 2.5\) |
| \(\) | \(=\) | \(\mathbf{\$945.00}\) |
(ii) The value at the end is the principal plus the interest.
| \(A\) | \(=\) | \(P + I\) |
| \(\) | \(=\) | \(8400 + 945 = \mathbf{\$9345.00}\) |
(i) \(\$2500\) earns \(\$337.50\) simple interest in \(3\) years. Find the interest rate per annum.
(ii) How long will it take \(\$6200\) at \(3.5\%\) p.a. to earn \(\$542.50\) simple interest? Answer in months.
Substitute what you know into \(I = Prn\) and solve for the unknown.
(i)
| \(337.50\) | \(=\) | \(2500 \times r \times 3\) |
| \(337.50\) | \(=\) | \(7500r\) |
| \(r\) | \(=\) | \(337.50 \div 7500 = 0.045\) |
| \(\) | \(=\) | \(\mathbf{4.5\%} \text{ p.a.}\) |
(ii) One year's interest is \(6200 \times 0.035 = \$217.00\).
| \(542.50\) | \(=\) | \(217 \times n\) |
| \(n\) | \(=\) | \(542.50 \div 217 = 2.5 \text{ years}\) |
| \(\) | \(=\) | \(2.5 \times 12 = \mathbf{30} \text{ months}\) |
Find the total percentage added over the whole loan, then undo it.
| \(n\) | \(=\) | \(\dfrac{15}{12} = 1.25 \text{ years}\) |
| \(\text{total rate}\) | \(=\) | \(8.4\% \times 1.25 = 10.5\%\) |
So \(\$7735\) is \(110.5\%\) of the amount borrowed.
| \(1.105 \times P\) | \(=\) | \(7735\) |
| \(P\) | \(=\) | \(7735 \div 1.105\) |
| \(\) | \(=\) | \(\mathbf{\$7000.00}\) |
Check: \(7000 \times 0.084 \times 1.25 = 735\), and \(7000 + 735 = 7735\).
(i) Write a formula for the interest \(I\) after \(n\) years.
(ii) After how many years will the interest reach \(\$650\)?
(iii) After how many years will his money have doubled?
(i) Substitute \(P\) and \(r\) into \(I = Prn\).
| \(I\) | \(=\) | \(2500 \times 0.04 \times n\) |
| \(\) | \(=\) | \(\mathbf{100}\boldsymbol{n}\) |
(ii) Solve \(100n = 650\): \(n = \mathbf{6.5}\) years. On the graph, read across from \(650\) to the line, then down.
(iii) Doubled means the interest equals the principal, \(\$2500\).
| \(100n\) | \(=\) | \(2500\) |
| \(n\) | \(=\) | \(\mathbf{25} \text{ years}\) |
The graph of simple interest against time is a straight line through the origin; its gradient is the interest for one year.
Common pitfalls
Frequently asked questions
What is the formula for simple interest?
The formula is \(I = Prn\), where \(I\) is the interest, \(P\) is the principal, \(r\) is the rate per period as a decimal and \(n\) is the number of periods.
How do you calculate simple interest for months?
Write the months as a fraction of a year by dividing by \(12\), then use that value for \(n\) with the yearly rate. For example, \(21\) months is \(1.75\) years.
How do you find the interest rate in simple interest?
Rearrange \(I = Prn\) to get \(r = \dfrac{I}{Pn}\). Multiply the decimal by \(100\) to write it as a percentage per annum.
What does p.a. mean?
p.a. stands for per annum, which means per year. A rate of \(5\%\) p.a. adds \(5\%\) of the principal each year with simple interest.
Why is a simple interest graph a straight line?
Simple interest adds the same amount every year, because it is always worked out on the original principal. Equal steps each year make a straight line.