Year 9 Maths Number and Financial Mathematics

Simple Interest

Q: What is the formula for simple interest?

I equals P times r times n, where P is the principal, r is the rate per period as a decimal and n is the number of periods.

Q: 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 10 twelfths of a year.

Q: How do you find the total amount with simple interest?

Work out the interest with I equals Prn, then add the principal: A equals P plus I.

Q: How do you find the interest rate from simple interest?

Substitute the interest, principal and time into I equals Prn, then divide the interest by P times n. Multiply by 100 to write it as a percentage.

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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 $I = Prn$, change a term in months, weeks or days into years, find the total owed and the repayments, work out the interest and the flat rate when something is bought on terms, and find the rate or the time.

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.

Simple interest adds the same amount every period, so the total grows along a straight line. Here $I = 1000 \times 0.05 \times 1 = \$50$ each year.

On terms you pay the deposit plus every instalment. The part past the cash price is the interest: $\$1980 - \$1800 = \$180$.

The simple interest formula:

\[I = Prn\]

I = P r n

The rate and the time must use the same period. For a yearly rate, change the time into years:

\[n = \dfrac{\text{months}}{12}\]\[n = \dfrac{\text{weeks}}{52}\]\[n = \dfrac{\text{days}}{365}\]

n = \frac{months}{12}

The total amount, and the interest on a terms purchase:

\[A = P + I\]\[\text{total paid} = \text{deposit} + \text{instalment} \times \text{number of instalments}\]

A = P + I

To find the rate or the time, substitute what you know into $I = Prn$ and divide:

\[r = \dfrac{I}{Pn} \qquad n = \dfrac{I}{Pr}\]

r = \frac{I}{P n}

A rate per month. If the rate is given per month, measure \(n\) in months. \(1\%\) per month for \(2\) years is \(r = 0.01\) and \(n = 24\).

A different balance each month. When a bank pays interest on each month's minimum balance, each month is \(\dfrac{1}{12}\) of a year. Add the balances, then multiply by the rate and by \(\dfrac{1}{12}\).

At least a target. If a question asks how many whole years until the interest is at least a target, work out \(n\) and round up. At \(\$240\) a year, \(\$1000\) needs \(1000 \div 240 = 4.17\ldots\) years, so \(5\) whole years: after \(4\) years there is only \(\$960\).

Reading a graph. A graph of simple interest is a straight line through the origin. Read the interest for one year (or divide, such as \(\$450\) in \(3\) years is \(\$150\) a year), then \(r = \dfrac{I}{P}\) for one year: on \(\$5000\), \(\dfrac{150}{5000} = 3\%\) p.a.

Same interest. To compare two investments, work out the interest for one of them first, then use \(I = Prn\) for the other with that \(I\).

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.

Example 1 — A term in months or days

Find the simple interest on
(i) $\$6400$ at $4.5\%$ p.a. for $10$ months, and
(ii) $\$3650$ at $6\%$ p.a. for $73$ days.

Solution

The rate is per annum, so $n$ must be in years. Divide months by $12$ and days by $365$.

(i)

$n$	$=$	$\dfrac{10}{12} \text{ years}$
$I$	$=$	$6400 \times 0.045 \times \dfrac{10}{12}$
	$=$	$\mathbf{\$240.00}$

I = 6400 \times 0.045 \times \frac{10}{12} = 240

(ii)

$n$	$=$	$\dfrac{73}{365} = 0.2 \text{ years}$
$I$	$=$	$3650 \times 0.06 \times 0.2$
	$=$	$\mathbf{\$43.80}$

I = 3650 \times 0.06 \times 0.2 = 43.80

Putting $10$ or $73$ straight in for $n$ would give an answer $12$ or $365$ times too big.

Example 2 — Total owed and the monthly repayment

Ravi borrows $\$3600$ at a flat rate of $8\%$ p.a. for $3$ years. He repays the loan and the interest in equal monthly amounts. Find
(i) the total he owes and
(ii) each monthly repayment.

Solution

(i) Find the interest first, then add the principal back on:

$I$	$=$	$3600 \times 0.08 \times 3$
	$=$	$\$864$
$A$	$=$	$P + I$
	$=$	$3600 + 864$
	$=$	$\mathbf{\$4464.00}$

A = 3600 + 864 = 4464

(ii) Three years is $3 \times 12 = 36$ monthly repayments:

$\text{monthly}$	$=$	$4464 \div 36$
	$=$	$\mathbf{\$124.00}$

4464 \div 36 = 124

Divide the TOTAL owed, not just the amount borrowed.

Example 3 — Buying on terms, then the flat rate

A fridge has a cash price of $\$1600$. On terms it costs a $\$200$ deposit and $\$68$ a month for $2$ years. Find
(i) the total paid,
(ii) the interest and
(iii) the flat rate of interest p.a. charged on the balance, to one decimal place.

Solution

(i) and (ii) Add the deposit to all $24$ instalments, then compare with the cash price:

$\text{total}$	$=$	$200 + 68 \times 24$
	$=$	$\mathbf{\$1832}$
$I$	$=$	$1832 - 1600$
	$=$	$\mathbf{\$232}$

I = 1832 - 1600 = 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 $r$:

$232$	$=$	$1400 \times r \times 2$
$232$	$=$	$2800r$
$r$	$=$	$\dfrac{232}{2800}$
	$=$	$0.08285\ldots$
	$\approx$	$\mathbf{8.3\%} \text{ p.a.}$

r = \frac{232}{2800} \approx 8.3 %

Example 4 — Finding the time

Lena invests $\$5000$ at $4\%$ p.a. simple interest. How long will it take to earn $\$700$ in interest?

Solution

This time $n$ is the unknown. Substitute everything else into $I = Prn$:

$700$	$=$	$5000 \times 0.04 \times n$
$700$	$=$	$200n$
$n$	$=$	$700 \div 200$
	$=$	$\mathbf{3.5} \text{ years}$

700 = 200 n, n = 3.5

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

Change the time to match the rate. Watch out for a term in months, weeks or days with a rate per annum. $18$ months is $n = 1.5$, not $n = 18$.

Interest is not the total. The interest is only the extra. The amount owed is $A = P + I$.

Repayments come from the total. To find a monthly repayment, divide $P + I$ by the number of months, not just $P$.

Count the deposit. On a terms purchase the deposit is part of the total paid. A payment-free period means fewer instalments, so count only the months you actually pay.

Round time up to reach a target. If $n = 4.17$ years and the question asks for whole years until at least a target is reached, the answer is $5$, not $4$.

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.

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\(n\)	\(=\)	\(\dfrac{10}{12} \text{ years}\)
\(I\)	\(=\)	\(6400 \times 0.045 \times \dfrac{10}{12}\)
\(\)	\(=\)	\(\mathbf{\$240.00}\)

\(n\)	\(=\)	\(\dfrac{73}{365} = 0.2 \text{ years}\)
\(I\)	\(=\)	\(3650 \times 0.06 \times 0.2\)
\(\)	\(=\)	\(\mathbf{\$43.80}\)

\(I\)	\(=\)	\(3600 \times 0.08 \times 3\)
\(\)	\(=\)	\(\$864\)
\(A\)	\(=\)	\(P + I\)
\(\)	\(=\)	\(3600 + 864\)
\(\)	\(=\)	\(\mathbf{\$4464.00}\)

\(\text{monthly}\)	\(=\)	\(4464 \div 36\)
\(\)	\(=\)	\(\mathbf{\$124.00}\)

\(\text{total}\)	\(=\)	\(200 + 68 \times 24\)
\(\)	\(=\)	\(\mathbf{\$1832}\)
\(I\)	\(=\)	\(1832 - 1600\)
\(\)	\(=\)	\(\mathbf{\$232}\)

\(232\)	\(=\)	\(1400 \times r \times 2\)
\(232\)	\(=\)	\(2800r\)
\(r\)	\(=\)	\(\dfrac{232}{2800}\)
\(\)	\(=\)	\(0.08285\ldots\)
\(\)	\(\approx\)	\(\mathbf{8.3\%} \text{ p.a.}\)

\(700\)	\(=\)	\(5000 \times 0.04 \times n\)
\(700\)	\(=\)	\(200n\)
\(n\)	\(=\)	\(700 \div 200\)
\(\)	\(=\)	\(\mathbf{3.5} \text{ years}\)

Simple Interest

📖 Theory

How to solve a simple interest problem

Common pitfalls

Frequently asked questions

Theory