FR10, FR16, HR10, HR17 - Interest, growth and decay | Eduqas GCSE Mathematics

Distinguish simple interest from compound interest, then use repeated percentage multipliers to model growth and depreciation. FR10 and HR10 cover simple interest; FR16 and HR17 connect repeated changes with equations, time periods and sensible interpretation of answers.

Simple interest

£ $1000$ earns $5\%$ interest each year for three years. Must the interest be exactly £ $150$ ? The answer depends on whether each year's interest uses the original amount or the growing balance.

Principal and simple interest

The principal is the original amount saved or borrowed. Simple interest is calculated on that unchanged principal in every period, so equal periods earn equal amounts of interest.

If £ $1000$ earns $5\%$ simple interest per year, the annual interest is $1000\times0.05=50$ pounds. Three years earn $3\times50=150$ pounds. Interest is the extra amount; the total balance also includes the principal.

Simple interest

I=P\times\frac r{100}\times t,\qquad A=P+I

Here $P$ is the principal, $r\%$ is the simple interest rate per period, $t$ is the number of those periods, and $A$ is the total.

Not in the formula booklet: learn it.

Match the time to the rate. A yearly rate uses years; a monthly rate uses months. For a fractional period, use proportional interest only when that is the stated model.

Worked example 1: several whole years

Question. £ $1200$ is invested at $4\%$ simple interest per year for $3$ years. Find the interest and the total balance.

Plan. Calculate one year's interest on £ $1200$ , then multiply by the number of years.

\begin{aligned} \text{annual interest}&=1200\times0.04=48\\ I&=48\times3=144\\ A&=1200+144=1344 \end{aligned}

Answer. Interest £ $144$ ; total balance £ $1344$ .

Check. Over three years the interest is $12\%$ of the original: $0.12\times1200=144$ .

Keep the same principal and rate, but give the duration in months. Convert it before using the yearly rate.

Worked example 2: converting the duration

Question. £ $1200$ earns $4\%$ simple interest per year for $18$ months. Interest accrues proportionally with time. Find the interest and total balance.

Plan. Convert $18$ months to years, then use the same unchanged principal.

\begin{aligned} t&=18\div12=1.5\text{ years}\\ I&=1200\times0.04\times1.5=72\\ A&=1200+72=1272 \end{aligned}

Answer. Interest £ $72$ ; total balance £ $1272$ .

Check. A year earns £ $48$ , and half a year earns £ $24$ . Together they give £ $72$ .

For interest credited only at each year end, a question asking for the first balance greater than a target needs a whole-year answer. For example, £ $800$ at $5\%$ simple interest gains £ $40$ a year. After $5$ years the balance is exactly £ $1000$ ; after $6$ it is £ $1040$ . The first year above £ $1000$ is therefore year $6$ , not year $5$ .

Simple interest adds the same amount each period because the reference principal stays fixed. If interest itself also earns interest, the reference balance changes: repeated multiplication is needed instead.

Watch out

Using $18$ with an annual rate in the second example would mean eighteen years. First make the rate period and the duration agree.

Convert a fresh duration before finding the extra interest.

Compound change

With compound interest, each period's interest is calculated on the current balance, including earlier interest. A constant percentage therefore gives a changing cash increase.

For example, £ $1000$ growing by $5\%$ becomes £ $1050$ . The following $5\%$ is calculated on £ $1050$ , so it adds £ $52.50$ , not £ $50$ .

Repeated multiplier

If each period multiplies an amount by the same number $m$ , then after $n$ periods the amount is $Pm^n$ , starting from $P$ . The power counts how many multiplications occur.

After one period the amount is $Pm$ . After two it is $(Pm)m=Pm^2$ ; after three it is $Pm^3$ . This explains the compound-interest formula rather than treating it as a separate rule.

Compound interest

A=P\left(1+\frac r{100}\right)^n

Here $r\%$ is the interest rate per compounding period, $n$ is the number of those periods, and $P$ is the principal. This form assumes a constant rate and no deposits or withdrawals.

In the supplied additional formulae sheet.

Repeated percentage loss uses the same idea with multiplier $1-\frac r{100}$ . For ordinary gradual decay, $0<r<100$ , so this multiplier lies between $0$ and $1$ .

Worked example 3: compound growth

Question. £ $1000$ earns $5\%$ compound interest annually for $3$ years. Find the final balance to the nearest penny.

Plan. Multiply by $1.05$ three times, keeping full intermediate values.

\begin{aligned} A&=1000(1.05)^3\\ &=1157.625 \end{aligned}

Answer. £ $1157.63$ .

Check. The successive unrounded balances are $1050$ , $1102.5$ , $1157.625$ . Simple interest would give £ $1150$ ; compound interest is a little larger because earlier interest also earns interest.

Change only the direction of the repeated percentage change. The base multiplier is now below one.

Worked example 4: compound depreciation

Question. Equipment worth £ $1000$ loses $5\%$ of its current value each year. Find its value after $3$ years, to the nearest penny.

Plan. Retain $95\%$ of the current value each year, using $0.95^3$ .

\begin{aligned} A&=1000(0.95)^3\\ &=857.375 \end{aligned}

Answer. £ $857.38$ .

Check. The successive values are $950$ , $902.5$ , $857.375$ , all decreasing. Subtracting $3\times50$ would incorrectly use the original value every year.

The power and the rate must use the same period. A rate of $2\%$ every six months for three years uses $m=1.02$ and $n=6$ . If rates change between periods, use their different multipliers: growth of $4\%$ then $3\%$ uses $1.04\times1.03$ , not one unchanged rate.

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FR10, FR16, HR10, HR17 - Interest, growth and decay

Simple interest

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