R16 - Compound interest, growth and decay

Repeated percentage changes use a changing amount, so their effects build up by multiplication. Calculate growth, depreciation and compound interest, reverse changes to find a starting amount or rate, and decide what a model's predictions mean.

Repeated percentage change and compound interest

If £1000 earns $10\%$ interest each year, will it become £1200 after two years? That depends on whether the second year's interest uses the original deposit or the new balance.

You already know

Increasing by $10\%$ means multiplying by $1.1$ . Decreasing by $20\%$ means multiplying by $0.8$ . A power records repeated multiplication: $1.1^3=1.1\times1.1\times1.1$ .

Compound interest adds each period's interest to the balance before the next period's interest is calculated. £1000 becomes £1100 after one year at $10\%$ . The next year's interest is $10\%$ of £1100, or £110, giving £1210. You earn interest on earlier interest.

Compound growth

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

$P$ is the starting amount, $r$ the percentage increase per period and $A$ the amount after $n$ complete periods. The rate must stay constant and no other changes are included. Here $n$ is a non-negative integer.

On the formula sheet (2026 version; the 2027 sheet will have the same content).

For decay, replace the growth multiplier by the fraction retained. A $20\%$ decrease leaves $80\%$ ; for example, a £5000 machine becomes £4000, £3200 and £2560 after three annual changes. The losses shrink because the base shrinks.

Repeated percentage decay

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

For an exact repeated decrease of $r\%$ per complete period, with $0<r<100$ , the multiplier lies between $0$ and $1$ .

Not on the formula sheet: derive it from the compound-growth formula by using the percentage retained.

The final amount is $A$ . Interest earned or value lost requires a subtraction from the starting amount.

Worked example 1: repeated growth

Question. £1000 grows by $10\%$ each year for three years. Find the final balance and the interest earned.

Plan. Apply the multiplier three times, then subtract the original deposit.

\begin{aligned} A&=1000(1.1)^3 &&\text{three complete changes}\\ &=1331 &&\text{evaluate the power}\\ \text{interest}&=1331-1000=331 &&\text{remove the original deposit} \end{aligned}

Why a power? Each new balance becomes the base for the next change.

Answer. Balance £1331; interest £331.

Check. The yearly balances are £1100, £1210 and £1331.

Change only growth to a $10\%$ decrease, keeping the starting amount and duration.

Worked example 2: repeated decay

Question. A £1000 item loses $10\%$ of its current value each year for three years. Find its value and total loss.

Plan. Use the retained multiplier $0.9$ three times, then subtract from the starting value.

\begin{aligned} V&=1000(0.9)^3 &&\text{retain }90\%\text{ each year}\\ &=729 &&\text{evaluate the power}\\ \text{loss}&=1000-729=271 &&\text{compare with the start} \end{aligned}

Why not subtract £100 each year? Each loss is $10\%$ of a different current value.

Answer. Value £729; total loss £271.

Check. £1000 becomes £900, £810 and £729.

Even equal percentage changes in opposite directions do not cancel: $1.1(0.9)=0.99$ , so a $10\%$ rise followed by a $10\%$ fall leaves $99\%$ of the start.

Watch out

Wrong: $1000(1-3\times0.1)=700$ for three $10\%$ decreases. That removes $10\%$ of the original amount each time. Correct: $1000(0.9)^3=729$ , because each decrease uses the current value.

Change just the starting value in the decay example.

Periods, growth and depreciation

Count changes using the same time unit as the rate: a quarterly rate needs a number of quarters.

One year has four quarters, so $2\%$ per quarter for two years means eight changes. A monthly rate for six months means six changes. You cannot simply replace a quarterly rate by four times that rate and expect the same compounded result.

Worked example 3: matching rate and duration

Question. An account starts with £1200 and grows by $2\%$ at the end of each quarter. Find its balance after one year, to the nearest penny.

Plan. Convert years to quarters, then use the quarterly multiplier.

\begin{aligned} n&=1\times4=4 &&\text{four quarters per year}\\ A&=1200(1.02)^4 &&\text{four quarterly changes}\\ &=1298.918592 &&\text{retain calculator precision} \end{aligned}

Why use $4$ rather than $1$ ? The interest is added quarterly, so it acts four times during the year.

Answer. £1298.92.

Check. Four lots of $2\%$ simple growth would give £1296; the compound balance is slightly larger.

Change only the duration to two years.

Worked example 4: eight quarterly changes

Question. Find the same account's balance after two years, to the nearest penny.

Plan. Use eight quarterly multipliers on the original deposit.

\begin{aligned} n&=2\times4=8 &&\text{convert the duration}\\ A&=1200(1.02)^8 &&\text{eight complete changes}\\ &\approx1405.991257 &&\text{evaluate without early rounding} \end{aligned}

Why start with £1200 again? The power already includes every quarter from the start; applying eight changes to the one-year balance would count some quarters twice.

Answer. £1405.99.

Check. $1200(1.02)^4(1.02)^4=1200(1.02)^8$ .

The same period-counting rule applies to depreciation. Keep the stored calculator value until the end unless a question explicitly says to round each posted payment or interest amount.

Watch out

Wrong: $1200(1.02)^2$ for two years at $2\%$ per quarter. This includes only two quarters. Correct: two years contain $8$ quarters, giving $1200(1.02)^8$ .

Change only the duration once more.

Recovering initial values

To recover a starting amount, undo the whole repeated multiplier by division.

Subtracting the same percentage from the final amount uses the wrong base. If a $5\%$ increase changes £100 into £105, subtracting $5\%$ of £105 gives £99.75, rather than £100. Dividing $105$ by $1.05$ undoes the actual multiplication.

Worked example 5: reversing growth

Question. A balance is £1764 after two years of $5\%$ annual compound growth. Find the initial investment.

Plan. Write the forward equation, then divide by its two-year multiplier.

\begin{aligned} 1764&=P(1.05)^2 &&\text{forward model}\\ P&=\frac{1764}{1.05^2} &&\text{undo both multiplications}\\ &=1600 &&\text{evaluate} \end{aligned}

Why divide? Multiplying the unknown start by $1.05^2$ produced the final balance.

Answer. £1600.

Check. $1600(1.05)=1680$ , then $1680(1.05)=1764$ .

Change only the direction of the annual change to a $5\%$ decrease. The final value and duration stay fixed.

Worked example 6: reversing decay

Question. An item is worth £1764 after two annual $5\%$ decreases. Find its original value to the nearest penny.

Plan. Divide by the retained multiplier twice.

\begin{aligned} 1764&=P(0.95)^2 &&\text{retain }95\%\text{ each year}\\ P&=\frac{1764}{0.95^2} &&\text{undo the decay}\\ &\approx1954.570637 &&\text{evaluate before rounding} \end{aligned}

Why is the original larger? Dividing by a positive factor below $1$ increases the amount.

Answer. £1954.57.

Check. The unrounded starting value multiplied by $0.9025$ returns £1764.

A convenient exact decay example is £3200 after two $20\%$ decreases: $3200/0.8^2=5000$ .

Watch out

Wrong: $1764(1.05)^2$ to recover a value before two $5\%$ decreases. A $5\%$ increase is not the inverse of a $5\%$ decrease. Correct: divide by $0.95^2$ .

Change only the final value in Worked example 6.

Finding a rate from repeated change

Equal repeated changes have an equal multiplier, which you can recover using a root.

If a value grows from £1000 to £1210 in two years, its overall factor is $1.21$ . Two equal annual factors must multiply to $1.21$ , so each is $\sqrt{1.21}=1.1$ . The annual increase is $10\%$ , although the total increase is $21\%$ .

For three equal periods, take a cube root instead. For example, an overall factor $0.729$ gives $\sqrt[3]{0.729}=0.9$ , a $10\%$ decrease per period. Use the positive root for a positive multiplier. This root method is not on the formula sheet: derive it by reversing the power in the compound-change model.

Worked example 7: finding an annual increase

Question. £800 becomes £968 after two equal annual compound increases. Find the annual percentage rate.

Plan. Find the total factor, take its positive square root, then remove the original $100\%$ .

\begin{aligned} m^2&=968/800=1.21 &&\text{final divided by initial}\\ m&=\sqrt{1.21}=1.1 &&\text{two equal positive factors}\\ r&=(1.1-1)100\%=10\% &&\text{convert the extra fraction} \end{aligned}

Why a square root? The same multiplier acted twice.

Answer. $10\%$ per year.

Check. $800(1.1)^2=968$ .

Last Accessed

Papers 1, 2 and 3

R16 - Compound interest, growth and decay

Repeated percentage change and compound interest

Periods, growth and depreciation

Recovering initial values

Finding a rate from repeated change

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