Simple Interest
Simple interest is calculated only on the amount first invested or borrowed, so it is the same every year. It uses the formula \(I=Prn\), where \(P\) is the principal, \(r\) the rate per period and \(n\) the number of periods. In NSW Year 10 Advanced maths the formula is rearranged to find any unknown and applied to loans, bonds and savings (NESA outcome MA5‑FIN‑C‑01).
You will find interest for days, months and years, then work backwards to the principal, the rate or the time. You will read a simple-interest investment from a linear rule, compare a broken term deposit with a full one, find the cost of store finance, and calculate interest on a daily balance. Harder questions form equations for two investments or a part-repaid loan, and show how long money takes to triple. It is part of Year 10 Advanced Earning Money.
Every question with a fully worked solution.
- Simple Interest - Video - Simple Interest Formula Watch
Theory
Simple interest is a fixed percentage of the principal for each year: \(I = Prn\). This Year 10 Advanced maths page covers time in days and months, finding the principal, rate or time, the straight-line growth of an investment, rounding up to whole months and forming equations.
The principal \(P\) is the amount invested or borrowed. Simple interest is calculated on the principal only, at a rate \(r\) per year (per annum, p.a.). So the same amount of interest is added every year, and \(I = Prn\), where \(n\) is the time in years.
The rate goes into the formula as a decimal: \(4.5\%\) is \(0.045\). The time must be in years: divide months by \(12\) and days by \(365\). If a rate is given per month, it can be used with the time in months instead, or multiplied by \(12\) to give a rate p.a.
The total value of an investment, or the total owed on a loan, is \(V = P+I = P+Prn\). Because the interest each year is the same, \(V\) graphs as a straight line against time, starting at \(P\).
Flat-rate loans, government bonds, term deposits and some store finance use simple interest. Any of \(I\), \(P\), \(r\) or \(n\) can be found when the other three are known.
| To find | Rearranged formula |
|---|---|
| Interest | \(I = Prn\) |
| Principal | \(P = \dfrac{I}{rn}\) |
| Rate | \(r = \dfrac{I}{Pn}\), then \(\times 100\%\) |
| Time | \(n = \dfrac{I}{Pr}\) years |
How to solve a simple interest problem
- List what you know: \(P\), \(r\), \(n\), \(I\) or the final value \(V\).
- Convert the rate to a decimal and the time to years.
- Find the interest from a total if needed: \(I = V-P\) (amount repaid minus amount borrowed).
- Substitute into \(I = Prn\) and solve for the unknown.
- Answer in the units asked (dollars and cents, % p.a., months), rounding only at the end.
Other question types
- Several parts: if the rate changes, money is added part-way, or the balance changes during a month, find the interest for each part separately and add.
- Payments during the year: half-yearly payments are half the yearly interest; count the number of payments.
- Least whole months: if the time is not a whole number, round up and check.
- Penalties and fees: breaking a term deposit may mean a lower rate and a fee; subtract the fee from what is received.
- Equations: when a principal or rate is unknown, write each interest in terms of the pronumeral and use the fact given (equal interest, total interest), as in Example 4.
- Average rate: the single rate giving the same total interest is \(r = \dfrac{\text{total } I}{P\times\text{total years}}\).
(i) \(\$14\,600\) at \(5.5\%\) p.a. for \(219\) days (use \(365\) days in a year)
(ii) \(\$3800\) at \(6.3\%\) p.a. for \(10\) months.
(i) Convert the time to years first.
| \(n\) | \(=\) | \(\dfrac{219}{365} = 0.6\) |
| \(I\) | \(=\) | \(14\,600\times 0.055\times 0.6\) |
| \(\) | \(=\) | \(\mathbf{\$481.80}\) |
(ii) \(10\) months is \(\dfrac{10}{12}\) of a year.
| \(I\) | \(=\) | \(3800\times 0.063\times\dfrac{10}{12}\) |
| \(\) | \(=\) | \(\mathbf{\$199.50}\) |
(i) Ari borrowed \(\$5200\) and repaid \(\$5642\) after \(15\) months. Find the simple interest rate p.a.
(ii) How long does \(\$4500\) at \(3.6\%\) p.a. take to earn \(\$405\) interest?
(i) The interest is the extra repaid: \(5642-5200 = \$442\), and \(n = \dfrac{15}{12} = 1.25\).
| \(442\) | \(=\) | \(5200\times r\times 1.25\) |
| \(442\) | \(=\) | \(6500r\) |
| \(r\) | \(=\) | \(0.068 = \mathbf{6.8\%}\text{ p.a.}\) |
(ii) The interest each year is \(4500\times 0.036 = \$162\).
| \(n\) | \(=\) | \(\dfrac{405}{162} = 2.5\text{ years}\) |
| \(\) | \(=\) | \(\mathbf{2}\text{ years }\mathbf{6}\text{ months}\) |
Give the time in the units the question asks for: \(2.5\) years is \(30\) months.
(i) the interest each year
(ii) the principal
(iii) the rate
(iv) the least whole number of months until it is worth at least \(\$8600\).
(i) Simple interest adds the same amount each year: \(7800-6960 = \$840\) in \(3\) years, so \(\mathbf{\$280}\) a year.
| \(\text{(ii) } P\) | \(=\) | \(6960-2\times 280 = \mathbf{\$6400}\) |
| \(\text{(iii) } r\) | \(=\) | \(\dfrac{280}{6400} = 0.04375 = \mathbf{4.375\%}\text{ p.a.}\) |
(iv) It needs \(8600-6400 = \$2200\) of interest.
| \(n\) | \(=\) | \(\dfrac{2200}{280} = 7.857\ldots\text{ years}\) |
| \(\) | \(=\) | \(94.28\ldots\text{ months}\) |
After \(94\) months it is still just under \(\$8600\), so round up: \(\mathbf{95}\) months.
The rest is \(\$(15\,000-P)\). Write the interest from each account for \(2\) years and add.
| \(P\times 0.05\times 2+(15\,000-P)\times 0.035\times 2\) | \(=\) | \(1245\) |
| \(0.1P+1050-0.07P\) | \(=\) | \(1245\) |
| \(0.03P\) | \(=\) | \(195\) |
| \(P\) | \(=\) | \(\boldsymbol{6500}\) |
Check: \(6500\times 0.05\times 2 = 650\) and \(8500\times 0.035\times 2 = 595\); \(650+595 = 1245\).
Common pitfalls
Frequently asked questions
What is the formula for simple interest?
\(I = Prn\), where \(P\) is the principal, \(r\) is the interest rate per year as a decimal and \(n\) is the time in years. The total value is \(P+I\).
How do you calculate simple interest for months or days?
Change the time to years first: divide the number of months by \(12\), or the number of days by \(365\). Then use \(I = Prn\).
How do you find the interest rate from simple interest?
Rearrange \(I = Prn\) to \(r = \dfrac{I}{Pn}\). Then multiply by \(100\) to write it as a percentage per annum.
Why is a simple interest graph a straight line?
Because the same amount of interest is added every year, the value increases by equal steps. Its graph against time is a straight line that starts at the principal.
What is the difference between simple and compound interest?
Simple interest is always calculated on the original principal, so it is the same each year. Compound interest is calculated on the growing balance, so it gets larger each year.