Year 12 Maths Standard 1 (2027) Investment

Simple Interest - Rearranging (P, r, n)

Q: How do you find the principal in simple interest?

Rearrange I equals Prn to P equals I divided by (r times n). For example, to earn $1260 in 4 years at 3.5%, P equals 1260 divided by 0.14, which is $9000.

Q: How do you find the interest rate from I = Prn?

Use r equals I divided by (P times n). This gives a decimal, so multiply by 100 for the percentage. For example, $1152 interest on $6400 over 4 years gives r equals 0.045, which is 4.5% per annum.

Q: How do you find the time in simple interest?

Use n equals I divided by (P times r), with the rate as a decimal. If the rate is per annum, the answer is in years. For example, $1120 interest on $3200 at 5% takes 1120 divided by 160, which is 7 years.

20 practice questions 0 video lessons Theory + worked examples

Rearrange the simple interest formula for NSW Year 12 Mathematics Standard 1 (NESA outcome MST-12-S1-02). The formula $I = Prn$ links four quantities — interest, principal, interest rate and number of periods — so if you know any three, you can find the fourth.

You will find a missing principal, rate or time by rearranging $I = Prn$, write a decimal rate as a percentage, and start from a total value instead of the interest — answering real questions such as how much to invest or how long to wait to reach a savings goal.

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Practice questions

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Theory

When the interest is known but the principal, rate or time is missing, rearrange $I=Prn$ to make the missing quantity the subject, then substitute. This Year 12 Standard 1 (NSW) guide shows the three rearrangements of the simple interest formula, how to turn a decimal rate into a percentage, and what to do when you are given the total value instead of the interest.

The simple interest formula $I=Prn$ links four quantities: the interest $I$, the principal $P$, the rate $r$ per year as a decimal, and the time $n$ in years. If you know any three, you can find the fourth.

To find a missing quantity, rearrange the formula: divide both sides of $I=Prn$ by the two letters you know, which leaves the missing one on its own.

Some questions give the total value $A$ instead of the interest. Find the interest first with $I=A-P$, then rearrange as usual.

Cover the letter you want: $I$ on top is divided by what is left; letters side by side on the bottom row are multiplied.

Given the total value, subtract the principal to get the interest: $I=10{,}620-9000=\$1620$.

Divide both sides of $I=Prn$ by the two letters you know:

\[P = \dfrac{I}{rn}\]

P = \frac{I}{r n}

\[r = \dfrac{I}{Pn}\]

r = \frac{I}{P n}

\[n = \dfrac{I}{Pr}\]

n = \frac{I}{P r}

Given the total value instead of the interest:

\[I = A - P\]

I = A - P

Rate: \(r=\dfrac{I}{Pn}\) gives a decimal — multiply by \(100\) for the percentage (\(0.045=4.5\%\)). Time: \(n=\dfrac{I}{Pr}\) gives the number of years when the rate is per annum. The interest earned each year is \(P\times r\). On a graph of \(I\) against \(n\), a steeper line means a larger \(Pr\).

How to find a missing P, r or n

Find the interest $I$ — if you are given the total value, use $I=A-P$.
Rearrange $I=Prn$ for the missing letter, put the rate in as a decimal, and multiply the bottom row before dividing.
Check by substituting your answer back into $I=Prn$, and write a rate as a percentage.

Example 1 — Find the rate

An investment of $\$6400$ earns $\$1152$ simple interest over $4$ years. What is the annual interest rate?

Solution

Rearrange $I=Prn$ for $r$, substitute, then change to a percentage.

$r$	$=$	$\dfrac{I}{Pn}$
	$=$	$\dfrac{1152}{6400\times4}$
	$=$	$\dfrac{1152}{25{,}600}$
	$=$	$0.045 = 4.5\%$

r = 1152 \div 25600 = 0.045

The rate is $4.5\%$ per annum.

Example 2 — Find the principal

How much must be invested at $3.5\%$ per annum simple interest to earn $\$1260$ in interest over $4$ years?

Solution

Rearrange $I=Prn$ for $P$, with $r=0.035$.

$P$	$=$	$\dfrac{I}{rn}$
	$=$	$\dfrac{1260}{0.035\times4}$
	$=$	$\dfrac{1260}{0.14}$
	$=$	$9000$

P = 1260 \div 0.14 = 9000

Check: $9000\times0.035\times4=1260$. $\$9000$ must be invested.

Example 3 — Find the time

For how many years must $\$3200$ be invested at $5\%$ per annum simple interest to earn $\$1120$?

Solution

Rearrange $I=Prn$ for $n$, then substitute.

$n$	$=$	$\dfrac{I}{Pr}$
	$=$	$\dfrac{1120}{3200\times0.05}$
	$=$	$\dfrac{1120}{160}$
	$=$	$7$

n = 1120 \div 160 = 7

The money must be invested for $7$ years.

Example 4 — Start from the total value

Zoe's $\$9000$ savings bond is worth $\$10{,}620$ after $3$ years of simple interest. What annual interest rate does the bond pay?

Solution

Find the interest from the total value, then use $r=\dfrac{I}{Pn}$.

$I$	$=$	$A-P$
	$=$	$10{,}620-9000 = 1620$
$r$	$=$	$\dfrac{1620}{9000\times3}$
	$=$	$\dfrac{1620}{27{,}000}$
	$=$	$0.06 = 6\%$

r = 1620 \div 27000 = 0.06

The bond pays $6\%$ per annum.

Common pitfalls

Write the rate as a percentage correctly. $r=\dfrac{I}{Pn}$ gives a decimal such as $0.045$ — the rate is $4.5\%$, not $0.045\%$.

Multiply the bottom row before dividing. Put the rate in as a decimal ($6\%=0.06$) and work out the bottom first: $\dfrac{540}{0.06\times3}=\dfrac{540}{0.18}$.

Use the interest, not the total value. If you are given the total value, subtract the principal first. Check your answer by substituting it back into $I=Prn$.

Frequently asked questions

How do you find the principal in simple interest?

Rearrange $I=Prn$ to $P=\dfrac{I}{rn}$: divide the interest by the product $r\times n$. For example, to earn $\$1260$ in $4$ years at $3.5\%$, $P=\dfrac{1260}{0.035\times4}=\dfrac{1260}{0.14}=\$9000$.

How do you find the interest rate from I = Prn?

Rearrange $I=Prn$ to $r=\dfrac{I}{Pn}$: divide the interest by the product $P\times n$. This gives a decimal, so multiply by $100$ for the percentage. For example, $\$1152$ interest on $\$6400$ over $4$ years gives $r=\dfrac{1152}{6400\times4}=0.045$, which is $4.5\%$ per annum.

How do you find the time in simple interest?

Rearrange $I=Prn$ to $n=\dfrac{I}{Pr}$: divide the interest by the product $P\times r$, with the rate as a decimal. If the rate is per annum, the answer is in years. For example, $\$1120$ interest on $\$3200$ at $5\%$ takes $n=\dfrac{1120}{3200\times0.05}=\dfrac{1120}{160}=7$ years.

What if I am given the total value instead of the interest?

Subtract the principal from the total value to find the interest first, $I=A-P$. Then rearrange $I=Prn$ as usual.

Why does my interest rate come out as a decimal?

The formula uses the rate as a decimal, so rearranging it also gives a decimal. Multiply by 100 to write it as a percentage: 0.06 is 6%, not 0.06%.

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Simple Interest Graphs

\(r\)	\(=\)	\(\dfrac{I}{Pn}\)
\(\)	\(=\)	\(\dfrac{1152}{6400\times4}\)
\(\)	\(=\)	\(\dfrac{1152}{25{,}600}\)
\(\)	\(=\)	\(0.045 = 4.5\%\)

\(P\)	\(=\)	\(\dfrac{I}{rn}\)
\(\)	\(=\)	\(\dfrac{1260}{0.035\times4}\)
\(\)	\(=\)	\(\dfrac{1260}{0.14}\)
\(\)	\(=\)	\(9000\)

\(n\)	\(=\)	\(\dfrac{I}{Pr}\)
\(\)	\(=\)	\(\dfrac{1120}{3200\times0.05}\)
\(\)	\(=\)	\(\dfrac{1120}{160}\)
\(\)	\(=\)	\(7\)

\(I\)	\(=\)	\(A-P\)
\(\)	\(=\)	\(10{,}620-9000 = 1620\)
\(r\)	\(=\)	\(\dfrac{1620}{9000\times3}\)
\(\)	\(=\)	\(\dfrac{1620}{27{,}000}\)
\(\)	\(=\)	\(0.06 = 6\%\)

Simple Interest - Rearranging (P, r, n)

📖 Theory

How to find a missing P, r or n

Common pitfalls

Frequently asked questions

Theory