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Year 12 Maths Standard 2 (2027) Investment and loans

Simple Interest

20 practice questions 0 video lessons Theory + worked examples

Master simple interest for NSW Year 12 Mathematics Standard 2. You use the formula \(I = Prn\) to find the interest on a principal \(P\) at a rate \(r\) for \(n\) periods, then add it to the principal for the total value \(A = P + I\).

This topic covers finding the interest, the total value, and — by rearranging the formula — a missing rate, time or principal, plus why a simple-interest investment grows in a straight line over time. It is a core Standard 2 skill for comparing savings accounts, term deposits and flat-rate loans.

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Theory

Simple interest is interest charged only on the original principal, found with \(I = Prn\). This Year 12 Standard 2 (NSW) guide shows how to find the interest, add it to the principal for the total value \(A = P + I\), rearrange the formula to find a missing rate, time or principal, and why the value grows in a straight line over time.

Simple interest is interest calculated only on the original principal — the amount first invested or borrowed. Because the principal never changes, the same amount of interest is added every period.

The interest earned is \(I = Prn\), where \(P\) is the principal, \(r\) is the interest rate per period written as a decimal, and \(n\) is the number of periods. The total value (or amount owed) is the principal plus the interest, \(A = P + I\).

Since a fixed amount \(Pr\) is added each period, the value \(A = P + Prn\) grows linearly — its graph against time is a straight line that starts at the principal \(P\) and rises by \(Pr\) each period.

Simple interest grows linearlyValue A=1000+100n is a straight line sitting $1000 above the interest line I=100n n A 1 2 3 4 5 6 500 1000 1500 A=P+Prn I=Prn
The value \(A\) rises in a straight line, always \(\$1000\) above the interest.
Reading a value off a simple-interest lineA=2000+120n; at n=5 the balance is $2600 n A 2 4 6 8 1000 2000 3000 5 2600 A=2000+120n
Read the balance straight off the line: at \(n=5\) years, \(A=\$2600\).

Simple interest is found from the principal, rate and time:

\[I = Prn\]
I=Prn

The total value (future value, or amount owed on a loan) adds the interest to the principal:

\[A = P + I = P + Prn = P(1+rn)\]
A=P+Prn
Rearrange \(I=Prn\) to find a missing quantity: \(r=\dfrac{I}{Pn}\), \(\ n=\dfrac{I}{Pr}\), \(\ P=\dfrac{I}{rn}\). Always write the rate as a decimal and keep the rate and time in the same units.

How to solve a simple-interest problem

  1. List \(P\), \(r\) (as a decimal) and \(n\), making sure \(r\) and \(n\) use the same time unit (convert months to years with \(n=\dfrac{\text{months}}{12}\)).
  2. Substitute into \(I = Prn\) to find the interest.
  3. Add to the principal if the question asks for the total value or amount owed: \(A = P + I\).
  4. Rearrange first for a missing quantity — \(r=\dfrac{I}{Pn}\), \(n=\dfrac{I}{Pr}\) or \(P=\dfrac{I}{rn}\) — then substitute and state the answer with its units.
Example 1 — Find the interest
Find the simple interest earned on \(\$8000\) invested at \(5\%\) per annum for \(4\) years.
Solution

List \(P,\ r,\ n\), then substitute into \(I=Prn\).

\(P=8000,\ r=0.05,\ n\)\(=\)\(4\)
\(I\)\(=\)\(Prn\)
\(\)\(=\)\(8000\times0.05\times4\)
\(\)\(=\)\(\$1600\)
I=1600

The simple interest is \(\$1600\).

Example 2 — Find the total value
\(\$4000\) is invested at \(7.5\%\) per annum simple interest for \(6\) years. Find the total value of the investment.
Solution

Find \(I\), then add it to the principal, \(A=P+I\).

Example 2Value A=4000+300n reaches $5800 after 6 years n A 1 2 3 4 5 6 2000 4000 6 5800 A=4000+300n
\(I\)\(=\)\(Prn = 4000\times0.075\times6\)
\(\)\(=\)\(1800\)
\(A\)\(=\)\(P+I = 4000+1800\)
\(\)\(=\)\(\$5800\)
A=5800

The investment is worth \(\$5800\) after \(6\) years.

Example 3 — Find the rate
An investment of \(\$15{,}000\) earns \(\$2700\) simple interest in \(3\) years. Find the annual interest rate.
Solution

Rearrange \(I=Prn\) to make \(r\) the subject.

\(r\)\(=\)\(\dfrac{I}{Pn} = \dfrac{2700}{15000\times3}\)
\(\)\(=\)\(\dfrac{2700}{45000} = 0.06\)
r=0.06

The rate is \(6\%\) per annum.

Example 4 — Find the time
How long must \(\$2500\) be invested at \(8\%\) per annum simple interest to earn \(\$500\)? Give the answer in months.
Solution

Rearrange \(I=Prn\) for \(n\) (in years), then convert to months.

\(n\)\(=\)\(\dfrac{I}{Pr} = \dfrac{500}{2500\times0.08}\)
\(\)\(=\)\(\dfrac{500}{200} = 2.5\text{ years}\)
\(\)\(=\)\(2.5\times12 = 30\text{ months}\)
n=30

It takes \(30\) months (\(2.5\) years).

Common pitfalls

Make the rate a decimal and match the units. Convert \(7.5\%\) to \(0.075\); if the rate is per annum, put the time in years (\(30\) months \(=2.5\) years).
Interest is not the total. \(I\) is only the interest; the total value or amount owed is \(A=P+I\) — do not stop at \(I\).
Simple is not compound. Simple interest adds the same \(Pr\) each period (a straight line); compound interest adds interest on interest (a curve).

Frequently asked questions

What is simple interest?

Simple interest is interest calculated only on the original principal — the amount first invested or borrowed. The principal never changes, so the same amount of interest is added each period, and the account value grows in a straight line.

What is the simple interest formula?

The interest is I equals P times r times n, where P is the principal, r is the interest rate per period written as a decimal, and n is the number of periods. The total value is the principal plus the interest, A equals P plus I.

How do you find the total amount with simple interest?

First work out the interest with I equals Prn, then add it to the principal: A equals P plus I. For example, $4000 at 7.5% per annum for 6 years earns $1800 interest, so the total value is $5800.

What is the difference between simple and compound interest?

Simple interest is worked out on the original principal only, so the same amount is added each period and the value graph is a straight line. Compound interest is worked out on the growing balance, so it earns interest on interest and its graph is a curve that gets steeper.

How do you convert months to years for the simple interest formula?

If the rate is per annum (per year), the time n must also be in years, so divide the number of months by 12. For example, 30 months is 30 divided by 12, which is 2.5 years.