Simple Interest
Simple interest is interest worked out on the original amount only, so it is the same every year. This NSW Year 9 Core maths subtopic is Stage 5 Core content under NESA outcome MA5‑FIN‑C‑01, built around the formula \(I = Prn\).
You will substitute into \(I = Prn\), writing the rate as a decimal and converting months, weeks and days into years so the rate and time match. You will find the total owed as \(P + I\), work out the cost of buying on terms from a deposit and regular repayments, find the interest paid on a terms purchase, and rearrange the formula to find a flat rate or a number of years. Examples include a coffee shop loan, a car bought on terms and an interest-free phone deal. It is part of Year 9 Core Working with Numbers.
Every question with a fully worked solution.
- Simple Interest - Video - Simple Interest Formula Watch
Theory
Simple interest is interest worked out on the original amount only, so it is the same every period. This Year 9 maths page shows how to use \(I = Prn\), how to change a term in months, weeks or days into years, how to find the total owed and the repayments, and how to work out the interest and the flat rate when something is bought on terms.
Simple interest is extra money paid on a loan, or earned on an investment, that is worked out on the original amount every time. It is the same amount each year, so it never earns interest of its own.
The principal \(P\) is the amount borrowed or invested. The rate \(r\) is the percentage charged for one time period, written as a decimal: \(6\%\) is \(0.06\). The number of periods \(n\) is how long the money is borrowed for, measured in the same unit as the rate. A rate "p.a." (per annum) is a yearly rate, so \(n\) is in years.
The total amount \(A\) is the principal plus the interest. A flat rate loan charges simple interest on the whole amount for the whole time.
Buying on terms means paying a deposit now and then regular instalments. The interest is how much more you pay than the cash price.
The simple interest formula:
The rate and the time must use the same period. For a yearly rate, change the time into years:
The total amount, and the interest on a terms purchase:
To find the rate or the time, substitute what you know into \(I = Prn\) and divide:
How to solve a simple interest problem
- List \(P\), \(r\) and \(n\). Write the rate as a decimal.
- Match the units: a yearly rate needs \(n\) in years, a monthly rate needs \(n\) in months.
- Substitute into \(I = Prn\). If \(r\) or \(n\) is the unknown, divide by the other numbers.
- Read the question again. For the total owed, add \(P\). For a repayment, divide the total by the number of payments.
- On terms, add the deposit to all the instalments and take away the cash price to get the interest.
(i) \(\$6400\) at \(4.5\%\) p.a. for \(10\) months, and
(ii) \(\$3650\) at \(6\%\) p.a. for \(73\) days.
The rate is per annum, so \(n\) must be in years. Divide months by \(12\) and days by \(365\).
(i)
| \(n\) | \(=\) | \(\dfrac{10}{12} \text{ years}\) |
| \(I\) | \(=\) | \(6400 \times 0.045 \times \dfrac{10}{12}\) |
| \(\) | \(=\) | \(\mathbf{\$240.00}\) |
(ii)
| \(n\) | \(=\) | \(\dfrac{73}{365} = 0.2 \text{ years}\) |
| \(I\) | \(=\) | \(3650 \times 0.06 \times 0.2\) |
| \(\) | \(=\) | \(\mathbf{\$43.80}\) |
Putting \(10\) or \(73\) straight in for \(n\) would give an answer \(12\) or \(365\) times too big.
(i) the total he owes and
(ii) each monthly repayment.
(i) Find the interest first, then add the principal back on:
| \(I\) | \(=\) | \(3600 \times 0.08 \times 3\) |
| \(\) | \(=\) | \(\$864\) |
| \(A\) | \(=\) | \(P + I\) |
| \(\) | \(=\) | \(3600 + 864\) |
| \(\) | \(=\) | \(\mathbf{\$4464.00}\) |
(ii) Three years is \(3 \times 12 = 36\) monthly repayments:
| \(\text{monthly}\) | \(=\) | \(4464 \div 36\) |
| \(\) | \(=\) | \(\mathbf{\$124.00}\) |
Divide the TOTAL owed, not just the amount borrowed.
(i) the total paid,
(ii) the interest and
(iii) the flat rate of interest p.a. charged on the balance, to one decimal place.
(i) and (ii) Add the deposit to all \(24\) instalments, then compare with the cash price:
| \(\text{total}\) | \(=\) | \(200 + 68 \times 24\) |
| \(\) | \(=\) | \(\mathbf{\$1832}\) |
| \(I\) | \(=\) | \(1832 - 1600\) |
| \(\) | \(=\) | \(\mathbf{\$232}\) |
(iii) The interest is charged on the balance still owed after the deposit, \(\$1600 - \$200 = \$1400\). Put the known values into \(I = Prn\) and solve for \(r\):
| \(232\) | \(=\) | \(1400 \times r \times 2\) |
| \(232\) | \(=\) | \(2800r\) |
| \(r\) | \(=\) | \(\dfrac{232}{2800}\) |
| \(\) | \(=\) | \(0.08285\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{8.3\%} \text{ p.a.}\) |
This time \(n\) is the unknown. Substitute everything else into \(I = Prn\):
| \(700\) | \(=\) | \(5000 \times 0.04 \times n\) |
| \(700\) | \(=\) | \(200n\) |
| \(n\) | \(=\) | \(700 \div 200\) |
| \(\) | \(=\) | \(\mathbf{3.5} \text{ years}\) |
Check: \(\$200\) a year for \(3.5\) years is \(\$700\). The same steps find the rate \(r\) when the time is known instead.
Common pitfalls
Frequently asked questions
What is the formula for simple interest?
\(I = Prn\), where \(P\) 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 as \(n\). For example \(10\) months is \(\dfrac{10}{12}\) of a year.
How do you find the total amount with simple interest?
Work out the interest with \(I = Prn\), then add the principal: \(A = P + I\).
How do you find the interest when buying on terms?
Add the deposit to all the instalments to get the total paid. Then subtract the cash price. What is left is the interest.
How do you find the interest rate from simple interest?
Substitute the interest, principal and time into \(I = Prn\), then divide: \(r = \dfrac{I}{Pn}\). Multiply by \(100\) to write it as a percentage.