Growth and decay problems involve quantities that increase or decrease by a fixed percentage repeatedly over time. The formula uses a multiplier raised to a power.
$$\text{Amount} = P \times r^n$$
where $P$ is the initial value, $r$ is the multiplier (e.g. $1.05$ for 5% growth, $0.92$ for 8% decay), and $n$ is the number of time periods.
Compound interest example: £2000 invested at 3% per year for 5 years. $$A = 2000 \times 1.03^5 \approx 2000 \times 1.1593 = £2318.55$$
Depreciation example: A car worth £12,000 depreciates at 15% per year. Value after 4 years: $$V = 12000 \times 0.85^4 \approx 12000 \times 0.5220 = £6264$$
This is not the same as simple interest/decrease (which adds the same amount each time) — compound growth is multiplicative.
Common error: adding the percentage change each year instead of multiplying by the power. Also: using the wrong multiplier (e.g. 0.15 instead of 0.85 for 15% decrease).
Percentage change Basic indices
New to Bow Tie Maths? It generates questions on this topic, marks them instantly, and tracks what you've mastered. Free to sign up.
Use the $x^n$ key: enter $1.03$ then $x^5$ to compute $1.03^5$ in one step.
2025 Jun 2H GCSE Q18 (1 mark) 2025 Jun 3H GCSE Q12 (3 marks) 2024 Jun 3H GCSE Q4 (3 marks) 2024 Jun 3H GCSE Q14 (2 marks) 2023 Jun 2H GCSE Q8 (4 marks) 2023 Jun 3H GCSE Q9 (2 marks) 2022 Jun 2H GCSE Q6 (3 marks) 2022 Jun 3H GCSE Q10 (1 mark) 2022 Jun 3H GCSE Q20 (2 marks) 2019 Nov 2H GCSE Q13 (1 mark) 2019 Jun 3H GCSE Q2 (3 marks) 2018 Nov 2H GCSE Q4 (3 marks) 2018 Jun 2H GCSE Q9 (2 marks) 2017 Jun 2H GCSE Q6 (2 marks) 2017 Nov 2H GCSE Q13 (4 marks) 2017 Nov 3H GCSE Q9 (2 marks) 2017 Jun 3H GCSE Q10 (1 mark)