Growth and Decay at GCSE: What the Question Is Really Asking
Growth and decay is one of those GCSE maths topics that sounds harder than it is. Strip away the wording about savings accounts, house prices, populations and radioactive substances, and every single growth and decay question comes down to the same idea: something changes by the same percentage over and over again.
That word compound is the key. It means the percentage is applied to the new value each time, not the original, so your interest earns interest and a depreciating car loses value on its already-reduced price. Get comfortable with one method and you can answer every version of this question, on both Foundation and Higher papers.
This guide covers the one formula you need, worked examples for both growth and decay, and the mistakes examiners see most often.
The One Formula for Growth and Decay
You do not need a separate rule for growth, decay, interest, depreciation and population. They all use a single multiplier.
First, turn the percentage into a multiplier:
- For growth (things getting bigger): $\text{multiplier} = 1 + \dfrac{r}{100}$
- For decay (things getting smaller): $\text{multiplier} = 1 - \dfrac{r}{100}$
where $r$ is the percentage rate. So a 5% increase is a multiplier of $1.05$, and a 5% decrease is a multiplier of $0.95$.
Then, to find the value after $n$ time periods, raise the multiplier to the power of $n$:
$$\text{Final amount} = \text{Starting amount} \times \left(\text{multiplier}\right)^{n}$$
That's it. The whole topic is choosing the right multiplier and counting the number of years (or hours, or days) correctly. The power is what makes it compound, applying the change again and again.
Compound Growth: Worked Examples
Example 1: compound interest. You invest £2,000 in an account paying 3% compound interest per year. How much is in the account after 4 years?
The value is growing, so the multiplier is $1 + \frac{3}{100} = 1.03$. There are 4 years, so:
$$2000 \times 1.03^{4} = 2000 \times 1.12550881 = \pounds 2251.02$$
Rounded to the nearest penny, the account holds £2,251.02. Notice you raise the multiplier to the power 4. You do not work out one year and multiply by 4, because that would be simple interest and it gives the wrong answer.
Example 2: population growth. A town has 45,000 people and its population grows by 2% each year. What will the population be after 6 years, to the nearest hundred?
Multiplier $= 1.02$, and $n = 6$:
$$45000 \times 1.02^{6} = 45000 \times 1.12616\ldots = 50677.4\ldots$$
So the population is about 50,700 people. The same method handles money, people or anything else that grows by a fixed percentage.
Compound Decay: Worked Examples (Depreciation)
Decay works in exactly the same way. You just subtract the percentage to build the multiplier. Decay is often called depreciation when it refers to something losing value.
Example 3: car depreciation. A car is bought for £18,000 and loses 15% of its value each year. What is it worth after 3 years?
The value is falling by 15%, so the multiplier is $1 - \frac{15}{100} = 0.85$. With $n = 3$:
$$18000 \times 0.85^{3} = 18000 \times 0.614125 = \pounds 11054.25$$
The car is worth £11,054.25 after 3 years. A common slip here is to subtract $15% \times 3 = 45%$ in one go, ignoring the fact that each year's loss is calculated on a smaller amount. The multiplier method deals with that automatically.
Example 4: working backwards. A different car depreciates by 10% a year and is worth £7,290 after 3 years. What was its original price?
Here the final amount is known and the starting amount is missing. Set up the equation with the multiplier $0.9$:
$$\text{Starting amount} \times 0.9^{3} = 7290$$
$$\text{Starting amount} = \frac{7290}{0.729} = \pounds 10000$$
The car cost £10,000 when new. Reverse problems like this are a favourite on Higher papers, so divide by the multiplier raised to the power rather than multiply.
Simple vs Compound: Where the Marks Are Lost
The slip examiners see most often on growth and decay is confusing simple change with compound change.
- Simple interest applies the percentage to the original amount every year. £2,000 at 3% simple interest earns exactly £60 every year, forever.
- Compound interest applies the percentage to the current amount, so the yearly gain grows. £2,000 at 3% compound interest earns £60 in year one, but £61.80 in year two.
If a question says "compound", you must use powers. If it genuinely says "simple", you multiply the yearly amount by the number of years. Read the wording carefully, because the exam will tell you which one it wants.
What Examiners Look For
A few habits will protect your marks on this topic:
- Convert the percentage properly. A 4% rise is a multiplier of $1.04$, not $0.4$ or $1.4$. This is the most common careless error of all.
- Count the time periods carefully. "After 5 years" means the power is 5. Watch for questions that switch to months, or give you the rate per month.
- Never add the percentages together. Losing 20% for two years is not a 40% loss. It is $0.8^2 = 0.64$, a 36% loss overall.
- Show the multiplier and the power. Even if your final rounding slips, writing $18000 \times 0.85^{3}$ earns method marks. Examiners award marks for the correct setup, so always write it down.
- Round only at the end. Keep the full figure on your calculator until the final step, then round money to the nearest penny and people to a sensible whole number.
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Summary
Every growth and decay question at GCSE uses one formula: find the multiplier ($1 + \frac{r}{100}$ for growth, $1 - \frac{r}{100}$ for decay), then raise it to the power of the number of time periods. The compound part, applying the change again and again, is exactly what those powers capture, and it is why you must never simply add or multiply the percentages. Practise a few of each type, always writing out the multiplier and the power, and this becomes a dependable source of marks on the paper.
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