Growth and Decay
From compound interest to the depreciation of a car’s value, growth and decay describe how quantities change by a percentage over time. This guide will show you how to master these concepts using the powerful shortcut of multipliers.
The Power of Multipliers
Multipliers combine the original 100% and the percentage change into a single decimal, making multi-year calculations a one-step operation.
Growth (e.g., Compound Interest)
An increase of $X\%$ means you retain $100\% + X\%$. The multiplier is $1 + \frac{X}{100}$.
Multiplier $= 1 + 0.03 = 1.03$
Decay (e.g., Depreciation)
A decrease of $X\%$ means you retain $100\% – X\%$. The multiplier is $1 – \frac{X}{100}$.
Multiplier $= 1 – 0.15 = 0.85$
The General Formula
To apply a percentage change over multiple periods, raise the multiplier to the power of the number of periods.
where $n$ is the number of periods (e.g., years).
Worked Examples
Example 1: Compound Interest (Growth)
£4,000 is invested at 3% compound interest per year. Find the total amount after 3 years.
- Growth of 3% → multiplier $= 1.03$.
- Number of periods: $n = 3$ years.
- Calculate: $£4{,}000 \times (1.03)^3 = £4{,}370.91$.
Answer: £4,370.91
Example 2: Depreciation (Decay)
A car costs £18,000 and depreciates by 15% each year. Find its value after 2 years.
- Decay of 15% → multiplier $= 0.85$.
- Number of periods: $n = 2$ years.
- Calculate: $£18{,}000 \times (0.85)^2 = £13{,}005$.
Answer: £13,005
Tutor Insights
🤔 Common Misunderstandings
- Confusing simple and compound interest. Compound interest is recalculated on the growing total each year — that’s why we use powers, not simple multiplication.
- Multiplier mistakes. Using $0.05$ for a 5% increase instead of $1.05$. Growth multipliers are always $> 1$; decay multipliers are always $< 1$.
📝 Common Exam Mistakes
- Rounding too early. Only round your final answer — for money, to 2 decimal places.
- Forgetting the power. Calculating for one year and failing to raise the multiplier to the correct power $n$.
- Not showing the setup. Always write out e.g. $£4{,}000 \times (1.03)^3$ for method marks.
Practice Questions
- Elara puts £600 into a savings account paying 4% compound interest per year. How much will be in her account after 2 years?
- A village has a population of 2,500. It is predicted to increase by 2% each year. What will the population be after 3 years (to the nearest whole number)?
- A new laptop costs £850. Its value depreciates by 20% each year. What is its value after 2 years?
- A pond contains 400 fish. The number of fish decreases by 5% each month. How many fish will be in the pond after 3 months (to the nearest whole number)?
Show Answers
- $£600 \times (1.04)^2 = £648.96$. Answer: £648.96.
- $2{,}500 \times (1.02)^3 \approx 2{,}653.02$. Answer: 2,653.
- $£850 \times (0.80)^2 = £544$. Answer: £544.
- $400 \times (0.95)^3 \approx 342.95$. Answer: 343 fish.
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