What formula is used for growth and decay at GCSE?

Use $A=P\times m^n$, where $P$ is the starting value, $m$ is the multiplier and $n$ is the number of time periods. For growth at $r\%$, $m=1+\frac{r}{100}$. For decay, $m=1-\frac{r}{100}$.

How do I know whether to use a growth or decay multiplier?

Use a multiplier greater than $1$ when the quantity increases. Use a multiplier between $0$ and $1$ when it decreases. For example, a $12\%$ increase uses $1.12$, while a $12\%$ decrease uses $0.88$.

Is compound interest a growth and decay topic?

Yes. Compound interest is a form of repeated growth because interest is added to the current balance each period. Depreciation is a common form of compound decay.

How do I find an original value before compound growth?

Divide the final value by the complete growth multiplier. After $n$ periods at multiplier $m$, use $P=A\div m^n$. Do not simply subtract the stated percentage from the final value.

Should I round after every year in a growth and decay calculation?

Normally, no. Keep the full calculator value throughout and round only the final answer. Round money to the nearest penny unless the question gives a different instruction.

Why is repeated percentage change exponential rather than linear?

Each change is calculated from the latest value, so the numerical amount added or removed changes every period. The ratio between consecutive values stays constant, which produces an exponential model.

Growth and Decay GCSE Maths: Higher Revision Guide

Growth and decay GCSE Maths questions involve a quantity changing by a percentage over time. If the same percentage change is applied repeatedly, the quantity changes exponentially, so each percentage is calculated from the latest value rather than the original value.

This Higher-tier guide explains multipliers, compound interest, depreciation, reverse growth and changing rates. It also includes worked examples, common mistakes and an exam-style question with a full answer.

What are growth and decay in GCSE Maths?

Growth occurs when a quantity increases by a percentage during each time period. Examples include money earning compound interest, an increasing population and the spread of bacteria.

Decay occurs when a quantity decreases by a percentage during each time period. Examples include depreciation, falling populations and the amount of a substance decreasing over time.

The GCSE Mathematics subject content requires students to set up, solve and interpret growth and decay problems, including compound interest. At Higher tier, students may also need to connect repeated change with exponential functions and general iterative processes.

The central model is:

\text{new value}=\text{original value}\times\text{multiplier}^{\text{number of periods}}

This formula works when the multiplier remains constant. If the percentage changes, each multiplier must be applied separately.

Finding growth and decay multipliers

A multiplier represents the proportion of a quantity remaining after a percentage change.

To increase a quantity by $r%r\%$ , use:

1+\frac{r}{100}

To decrease a quantity by $r%r\%$ , use:

1-\frac{r}{100}

Percentage change	Calculation	Multiplier
Increase by $7%7\%$	$1 + 0.07$	$1.07$
Increase by $25%25\%$	$1 + 0.25$	$1.25$
Decrease by $7%7\%$	$1 - 0.07$	$0.93$
Decrease by $25%25\%$	$1 - 0.25$	$0.75$

A growth multiplier is greater than $1$ . A decay multiplier is between $0$ and $1$ .

The multiplier for a $17%17\%$ decrease is therefore $0.83$ , not $0.17$ . Multiplying by $0.17$ finds the amount lost, whereas multiplying by $0.83$ finds the amount remaining.

The growth and decay formula

For a starting quantity $P$ , multiplier $m$ and number of periods $n$ , the final quantity $A$ is:

A=P\times m^n

For compound growth at $r%r\%$ per period:

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

For compound decay at $r%r\%$ per period:

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

The time period attached to the rate must match the number of periods. A rate of $4%4\%$ per year applied for $6$ years has $n = 6$ . A monthly rate applied for $2$ years has $n = 24$ .

Growth and decay worked examples

Worked example 1: compound growth

£ $1200$ is invested at $4.5%4.5\%$ compound interest per year. Find its value after $3$ years.

The growth multiplier is:

1+\frac{4.5}{100}=1.045

Apply it three times:

1200\times1.045^3=1369.39935

Money should be rounded to the nearest penny:

\boxed{£1369.40}

The account did not gain the same cash amount each year. Each year's interest was calculated from the new balance, including interest already added.

Worked example 2: compound decay

A car is worth £ $50018\,500$ . Its value depreciates by $17%17\%$ each year. Find its value after $4$ years.

A $17%17\%$ decrease leaves $83%83\%$ , so the decay multiplier is $0.83$ .

18\,500\times0.83^4=8779.789385

Therefore, the car is worth:

\boxed{£8779.79}

Do not calculate $17%17\%$ of the original price and subtract that same amount four times. Depreciation is normally compound because each decrease is based on the car's current value.

Worked example 3: reverse growth

An investment is worth £ $1728$ after increasing by $20%20\%$ per year for $2$ years. Find its original value.

The multiplier over two years is:

1.2^2=1.44

The final value was produced by multiplying the original value by $1.44$ , so reverse the process by dividing:

1728\div1.44=1200

The original investment was:

\boxed{£1200}

This is a reverse percentage problem. Subtracting $40%40\%$ from £ $1728$ would not reverse two successive increases of $20%20\%$ .

Worked example 4: different rates in different years

A machine costs £ $00024\,000$ . It depreciates by $22%22\%$ in its first year and by $9%9\%$ in each of the next two years. Find its value after three years.

The first-year multiplier is $0.78$ . The multiplier for each later year is $0.91$ .

24\,000\times0.78\times0.91^2=15\,502.032

The machine is worth:

\boxed{£15\,502.03}

A single power cannot be used for all three years because the percentage rate is not constant throughout.

Worked example 5: finding the percentage rate

An investment increases from £ $5000$ to £ $6050$ in two years at a constant annual rate. Find the rate.

First find the overall multiplier:

\frac{6050}{5000}=1.21

If the annual multiplier is $m$ , then:

m^2=1.21

Therefore:

m=\sqrt{1.21}=1.1

The multiplier $1.1$ represents an annual increase of $10%10\%$ , so the rate is:

\boxed{10\%}

Only the positive square root is meaningful because a multiplier representing a financial value cannot be negative.

Simple change compared with compound change

Repeated growth and decay should not be confused with simple interest or a fixed yearly change.

Simple change	Compound change
The same amount is added or removed each period	The same percentage is applied each period
Produces linear change	Produces exponential change
Calculated from the original value	Calculated from the latest value
Has a constant numerical difference	Has a constant multiplier

For example, simple interest of $5%5\%$ on £ $1000$ adds £ $50$ every year. Compound interest of $5%5\%$ adds £ $50$ in the first year, but the second year's interest is calculated from £ $1050$ .

Solving growth and decay exam questions

Use the following method for most growth and decay exam questions:

Identify the starting value.
Identify whether the value is increasing or decreasing.
Convert the percentage into the correct multiplier.
Count the number of time periods carefully.
Write one complete calculation before using the calculator.
Round only the final answer unless the question instructs otherwise.
Include the correct unit and interpret the result in context.

Writing the full calculation is important because method marks may be available even if the final arithmetic or rounding is incorrect. The MathsGenie compound interest and depreciation lesson provides further revision and worked solutions.

Embedded exam-style question

A town has a population of $00032\,000$ . Its population grows by $3.5%3.5\%$ each year for $5$ years. At the end of the fifth year, $8%8\%$ of the population moves away.

Calculate the population remaining after these changes. Give your answer to the nearest whole person.

The annual growth multiplier is $1.035$ . The final decrease has multiplier $0.92$ .

32\,000\times1.035^5\times0.92=34\,965.484\ldots

To the nearest whole person, the population is:

\boxed{34\,965}

The order of the multipliers does not affect this calculation, but both changes must be included. The power of $5$ applies only to the multiplier representing five years of growth.

Try 4 more growth and decay exam questions, using the worked solutions and mark schemes to check each stage of your method.

Common growth and decay mistakes

Using the percentage as the multiplier

For an $8%8\%$ increase, the multiplier is $1.08$ , not $0.08$ . The value must retain its original $100%100\%$ and gain another $8%8\%$ .

For an $8%8\%$ decrease, the multiplier is $0.92$ , not $0.08$ and not $1.08$ .

Multiplying the percentage by the number of years

An increase of $6%6\%$ for $4$ years is not generally a $24%24\%$ increase. The correct model is:

\text{original value}\times1.06^4

Multiplying $6%6\%$ by $4$ ignores compounding.

Applying a power to the starting value

The power belongs on the multiplier:

P\times m^n

It should not be written as $Pn×mP^n\times m$ .

Rounding during the calculation

Early rounding changes the value used in later stages and may produce an inaccurate final answer. Keep the calculator's full displayed value and round once at the end.

Miscounting time periods

A value given at the start of 2025 and required at the start of 2029 changes four times, not five. A timeline can help when the wording is less direct.

Reversing by subtracting the percentage

If a price after a $15%15\%$ increase is known, divide by $1.15$ to find the original. Decreasing the new price by $15%15\%$ does not undo the increase because it calculates $15%15\%$ of a different amount.

How to revise growth and decay effectively

Begin by practising percentage multipliers until you can form them reliably. Then work on compound growth, compound decay, changing rates and reverse problems in that order.

Use the MathsGenie revision guide to review the method, followed by the printable exam booklet. Marking your answers against worked solutions will reveal whether an error came from choosing the multiplier, counting periods or using the calculator.

For a focused explanation of financial applications and reverse percentages, read GCSE compound interest questions made easy. You can also use the GCSE Maths topics by exam board and tier checklist to place repeated percentage change within your wider revision plan.

Conclusion

Growth and decay GCSE Maths questions are based on repeated multiplication. Convert the percentage to a multiplier, raise a constant multiplier to the number of periods and apply each different rate separately when the conditions change.

For reverse questions, divide by the overall multiplier. Keep full calculator accuracy until the end, show the complete mathematical model and interpret the result using the required units. MathsGenie's revision lesson, practice questions, booklet and mark schemes provide a structured way to develop these skills before attempting full past papers.

Sources and referenced URLs