Simple Interest
Simple interest is paid only on the amount first invested or borrowed, using \(I = PRN\). In NSW Year 9 Advanced maths this formula is used for savings, loans and buying on terms (NESA outcome MA5‑FIN‑C‑01).
You will find interest for periods given in days, months and years, the total owed on a loan, and the time or rate needed to earn a set amount. You will find the flat interest rate when goods are bought with a deposit and repayments, compare two loan offers, split an investment between two rates, and find when two investments at different rates are worth the same. It is part of Year 9 Advanced Working with Numbers.
Every question with a fully worked solution.
- Simple Interest - Video - Simple Interest Formula Watch
Theory
Simple interest is a fixed percentage of the original amount for each time period: \(I = PRN\). This Year 9 Advanced maths page covers interest for months and days, finding the principal, rate or time, flat-rate loans with a deposit, and comparing investments with equations and straight-line graphs.
Simple interest is interest calculated only on the original amount, the principal \(P\). Each period earns the same interest, so the total grows by equal steps.
The rate \(R\) is the percentage charged or earned per period, written as a decimal in the formula. A rate per annum (p.a.) is per year. \(N\) is the number of periods, in the same unit as the rate.
The amount \(A\) is the principal plus the interest. For a flat-rate loan, interest is charged on the whole amount borrowed for the whole term, and the total is usually repaid in equal instalments. A deposit paid up front is not borrowed, so it does not attract interest.
Because the interest each year is the same, the graph of \(A\) against time is a straight line with gradient \(PR\). Comparing two investments graphically means finding where their lines cross.
Simple interest, where \(P\) is the principal, \(R\) the rate per period as a decimal and \(N\) the number of periods:
Rearranged to find each part:
How to solve simple interest problems
- List \(P\), \(R\) and \(N\). Write \(R\) as a decimal: \(5.25\% = 0.0525\).
- Match the units of \(R\) and \(N\): years with p.a., months with a monthly rate, days as a fraction of \(365\).
- Substitute into \(I = PRN\), or \(A = P(1 + RN)\) if the final amount is involved, and solve for the unknown.
- For a loan, find the amount borrowed (price minus deposit) and the interest (total repaid minus amount borrowed) first.
- Answer the question asked: the interest, the total amount, a whole number of years or months, or the instalment.
(i) \(\$6400\) at \(5.25\%\) p.a. for \(9\) months
(ii) \(\$3650\) at \(4\%\) p.a. for \(146\) days.
The rates are per annum, so write each time in years.
| \(\text{(i) } I\) | \(=\) | \(6400 \times 0.0525 \times \dfrac{9}{12}\) |
| \(\) | \(=\) | \(\mathbf{\$252}\) |
| \(\text{(ii) } I\) | \(=\) | \(3650 \times 0.04 \times \dfrac{146}{365}\) |
| \(\) | \(=\) | \(\mathbf{\$58.40}\) |
Use \(365\) days in a year unless told otherwise.
(i) \(\$8000\) earns \(\$1080\) simple interest in \(3\) years. Find the rate.
(ii) \(\$12\,000\) earns \(4.8\%\) p.a. How many whole months until it has earned at least \(\$1000\)?
(iii) An amount grows to \(\$9450\) in \(5\) years at \(7\%\) p.a. How much was invested?
Substitute what is known into \(I = PRN\) (or \(A = P + PRN\)) and solve.
| \(\text{(i) } 1080\) | \(=\) | \(8000 \times R \times 3\) |
| \(R\) | \(=\) | \(\dfrac{1080}{24\,000} = 0.045 = \mathbf{4.5\%}\ \text{p.a.}\) |
| \(\text{(ii) per month}\) | \(=\) | \(12\,000 \times 0.048 \div 12 = \$48\) |
| \(1000 \div 48\) | \(=\) | \(20.83\ldots\) |
After \(20\) months the interest is only \(\$960\), so it takes \(\mathbf{21}\) months.
| \(\text{(iii) } 9450\) | \(=\) | \(P(1 + 0.07 \times 5) = 1.35P\) |
| \(P\) | \(=\) | \(9450 \div 1.35 = \mathbf{\$7000}\) |
(i) the total she pays
(ii) the interest
(iii) the flat rate of interest p.a., to one decimal place.
| \(\text{deposit}\) | \(=\) | \(0.15 \times 3200 = \$480\) |
| \(\text{(i) total}\) | \(=\) | \(480 + 130 \times 24 = \mathbf{\$3600}\) |
| \(\text{(ii) interest}\) | \(=\) | \(3600 - 3200 = \mathbf{\$400}\) |
(iii) Interest is charged only on the amount borrowed, which is the cash price minus the deposit: \(3200 - 480 = \$2720\).
| \(400\) | \(=\) | \(2720 \times R \times 2\) |
| \(R\) | \(=\) | \(\dfrac{400}{5440} = 0.0735\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{7.4\%}\ \text{p.a.}\) |
Write each value after \(n\) years, then make them equal.
| \(\text{Cai}\) | \(=\) | \(4000 + 280n\) |
| \(\text{Dee}\) | \(=\) | \(5500 + 220n\) |
| \(4000 + 280n\) | \(=\) | \(5500 + 220n\) |
| \(60n\) | \(=\) | \(1500\) |
| \(n\) | \(=\) | \(\mathbf{25}\ \text{years}\) |
Both are then worth \(\$11\,000\). On a graph the two straight lines cross at \(n = 25\).
Common pitfalls
Frequently asked questions
What is the simple interest formula?
\(I = PRN\): interest equals principal times rate per period, as a decimal, times the number of periods. The total amount is \(A = P + I\).
How do you calculate simple interest for months or days?
With a rate per annum, write the time as a fraction of a year: months over \(12\), or days over \(365\). For example, \(9\) months is \(0.75\) years.
How do you find the interest rate from the interest earned?
Rearrange \(I = PRN\) to \(R = \dfrac{I}{PN}\), then write \(R\) as a percentage. For \(\$1080\) on \(\$8000\) over \(3\) years, \(R = \dfrac{1080}{24\,000} = 4.5\%\) p.a.
What is a flat interest rate on a loan?
A flat rate charges simple interest on the full amount borrowed for the whole term, even though the balance goes down as repayments are made. Add the interest to the amount borrowed and divide by the number of payments to get each instalment.
Why is a simple interest graph a straight line?
The same interest is added every year, so the value increases by equal steps. A constant rate of change gives a straight line.