A radioactive substance has a half-life of 20 years. Which statement is correct?

After 40 years a quarter of the substance remains

After 40 years none of the substance remains

After 40 years a quarter of the substance remains

After 40 years half of the substance remains

After 10 years a quarter of the substance remains

At the end of each of four successive years a quantity measures 300, 360, 432 and 518.4. Which statement describes the change?

An exponential increase of 20% each year

A linear increase of 60 each year

An exponential increase of 20% each year

An exponential increase of 60% each year

A linear increase of 20 each year

Is exponential growth and decay a Higher tier only topic?

No. Spec reference R16 is common content, so Foundation candidates also set up and solve growth and decay problems, usually over two or three years with a friendly multiplier. Higher adds the general iterative processes clause and the harder question types: finding k and a in y = ka^x from two points, working the annual rate back out of a start and end value, and finding the number of steps by trial.

Do I get the growth and decay formula on the Edexcel formulae sheet?

Partly. The Exam Aid sheet issued with 1MA1 papers for the 2025 to 2027 windows carries the compound interest formula, written for an amount invested at a percentage rate over a number of years. Nothing on the sheet mentions decay, depreciation or half-life, so you build those multipliers yourself: a fall of r percent per step means multiplying by 1 - (r)/(100) each step.

How do you find the number of years without using logarithms?

By trial, which is the method Edexcel expects and marks. Work out the value for a sensible number of steps, then adjust up or down until you find the step where the quantity crosses the target. Both trials need to be on the page, because the process marks are awarded for the pair, and the answer is the number of steps or the calendar year, not the value you calculated.

What does half-life mean in a GCSE maths question?

It is the time taken for the mass of a substance to fall to half of what it was, and for exponential decay it stays the same however much you start with. Count how many half-lives fit into the given time, then halve that many times: after n half-lives the fraction left is (1)/(2^n), so 3 half-lives leave an eighth. Half-life questions turn up on Paper 1 as well, because the arithmetic is repeated halving.

Exponential Growth and Decay - Edexcel GCSE Higher Maths

Typically 3-5 marks per paper, usually one question in two or three parts · Spec 1MA1

Growth and decay is one percentage change applied over and over. Build the multiplier for a single time step, raise it to the power of the number of steps, and the whole model is $y = ka^n$ , where $k$ is the starting amount, $a$ is the multiplier and $n$ counts the steps. Edexcel files this under spec reference R16, which is common content across both tiers, with the clause about general iterative processes reserved for Higher. The arithmetic is the same as compound interest; what changes is the thing being counted, which might be bacteria in a dish, the population of a town, the number of trees left standing after a disease, the mass of a radioactive sample, or the trade-in value of a four year old car.

Most of these questions land on Papers 2 and 3, where a calculator turns $0.86^4$ into three keystrokes. The shapes recur: give the amount after a stated number of steps to 3 significant figures; find the year in which a quantity first passes a target; work out the percentage change per step from a start value and an end value; find $k$ and $a$ for a curve $y = ka^x$ that passes through two named points. Paper 1 is non-calculator, so there growth and decay arrives as doubling or halving, where a half-life of 6 hours across 30 hours is five halvings and the answer is a division by 32.

The first line of working decides most of the marks. Writing 0.08 for an 8 percent rise, or 0.12 for a 12 percent fall, loses everything that follows, and so does dividing by 1.04 when a value drops by 4 percent. After that comes the power: multiplying by 0.85 once and then by 3, or taking 15 percent off the original three times over, answers a linear question that nobody asked. On a how many years question the mark scheme wants two trials written down, the one that falls short and the one that clears the target, then the number of steps rather than the amount you calculated on the way.

Worked example

At the start of a study a colony contains 3200 insects. The number of insects increases exponentially at a rate of 15% per week. (a) Work out the number of insects after 4 weeks. Give your answer to the nearest whole number. (b) Work out the smallest number of complete weeks after which the colony contains more than 10000 insects.

[5 marks]

An increase of 15% multiplies the count by 1.15 each week, so after $n$ weeks there are $3200 \times 1.15^n$ insects.
M1: a correct multiplier raised to a power; 0.15 or 1.5 here scores nothing
(a) $3200 \times 1.15^4 = 5596.8\ldots$ , so there are 5597 insects to the nearest whole number.
A1: the full calculator value rounded once, at the end
(b) Try $n = 8$ : $3200 \times 1.15^8 = 9788.8\ldots$ , which is still below 10000.
M1: a trial below the target, written on the script rather than done on the calculator and abandoned
Try $n = 9$ : $3200 \times 1.15^9 = 11257.2\ldots$ , which is above 10000.
M1: the trial that clears the target; both trials are needed for the method to be complete
The colony first passes 10000 during week 9, so the answer is 9 weeks.
A1: the number of weeks, not the population 11257; an answer of 8 is the classic off by one

Practice questions

These are original questions in Edexcel 1MA1 Higher style, covering population and bacterial growth, decay of value and of radioactive mass, half-life, finding the rate from a start and end figure, and the trial method for the number of steps. The doubling and halving ones work without a calculator, the way Paper 1 sets them; use a calculator for anything with an awkward multiplier. Where a question states a degree of rounding, type the rounded value.

No account needed: the questions below are always free to check, plus 8 extra questions today. Your score so far updates as you go.

Question 1 Exam pace [3 marks]

A radioactive sample has a mass of 240 g. The half-life of the sample is 6 hours. Work out the mass of the sample, in grams, after 30 hours.

Your answer (fractions like 1/2 are fine)
Question 2 Exam pace [2 marks]

A radioactive substance has a half-life of 20 years. Which statement is correct?

Answer options for question 2
After 40 years none of the substance remains After 40 years a quarter of the substance remains After 40 years half of the substance remains After 10 years a quarter of the substance remains
Question 3 Exam pace [3 marks]

A colony of bacteria grows exponentially, increasing by 50% each hour. After 3 hours there are 6750 bacteria. Work out the number of bacteria at the start.

Your answer (fractions like 1/2 are fine)
Question 4 Exam pace [2 marks]

At the end of each of four successive years a quantity measures 300, 360, 432 and 518.4. Which statement describes the change?

Answer options for question 4
A linear increase of 60 each year An exponential increase of 20% each year An exponential increase of 60% each year A linear increase of 20 each year
Question 5 Stretch [4 marks]

A car costs £24000 when new. Its value falls by 15% during the first year and by 10% during each year after that. Work out the value of the car after 4 years, in pounds, correct to the nearest pound.

Your answer (fractions like 1/2 are fine)
Question 6 Stretch [4 marks]

Colony A starts with 500 bacteria and increases by 40% each hour. Colony B starts with 4000 bacteria and increases by 15% each hour. Work out the smallest number of complete hours after which colony A contains more bacteria than colony B.

Your answer (fractions like 1/2 are fine)

Common mistakes examiners see

Using the percentage itself as the multiplier: 8 percent growth entered as $\times 0.08$ , or 12 percent decay as $\times 0.12$ . A population of 9000 growing at 8 percent for 3 years comes out as 4.6 rather than 11337.

Say what one step does to 100: a rise of 8 percent takes 100 to 108, so the multiplier is 1.08; a fall of 12 percent takes 100 to 88, so it is 0.88. Write the multiplier on its own line before any power goes near it.
Dividing by $1.04$ to model a 4 percent decrease instead of multiplying by $0.96$ . The two are close enough that the answer looks plausible, which is why it slips through, and on a depreciation question it scores nothing at all.

Division reverses a change that has already happened; that is a reverse percentage question. Going forwards, always multiply. A 4 percent drop is $\times 0.96$ , and the check is that $0.96 < 1$ , so the value falls.
Treating the change as linear: taking 15 percent of the original value away three times, so £20000 depreciating at 15 percent a year becomes $20000 - 3 \times 3000 = 11000$ instead of $20000 \times 0.85^3 = 12282.50$ .

Each step is a percentage of what is there at the time, not of the original, so the multiplier gets a power. One expression, $20000 \times 0.85^3$ , replaces three subtractions and is what the process mark is for.
Handling half-life by dividing rather than halving repeatedly. With 640 g and a half-life of 3 hours, dividing 640 by the 4 half-lives in 12 hours gives 160 g when the correct answer is 40 g.

Count the half-lives first, then halve that many times: $640 \div 2 \div 2 \div 2 \div 2$ , or $640 \times \frac{1}{2^4}$ . Four half-lives leave one sixteenth, not one quarter.
Answering a how many years question with the amount rather than the count, or stopping at the first trial and being one out. A trial showing 11348 at 6 years is evidence that 6 is too few, not that 6 is the answer.

Write both trials with their values, then underline the step number that first clears the target. If the question names a starting year, add the number of steps to it and give a calendar year.

Frequently asked questions

Is exponential growth and decay a Higher tier only topic?: No. Spec reference R16 is common content, so Foundation candidates also set up and solve growth and decay problems, usually over two or three years with a friendly multiplier. Higher adds the general iterative processes clause and the harder question types: finding $k$ and $a$ in $y = ka^x$ from two points, working the annual rate back out of a start and end value, and finding the number of steps by trial.
Do I get the growth and decay formula on the Edexcel formulae sheet?: Partly. The Exam Aid sheet issued with 1MA1 papers for the 2025 to 2027 windows carries the compound interest formula, written for an amount invested at a percentage rate over a number of years. Nothing on the sheet mentions decay, depreciation or half-life, so you build those multipliers yourself: a fall of $r$ percent per step means multiplying by $1 - \frac{r}{100}$ each step.
How do you find the number of years without using logarithms?: By trial, which is the method Edexcel expects and marks. Work out the value for a sensible number of steps, then adjust up or down until you find the step where the quantity crosses the target. Both trials need to be on the page, because the process marks are awarded for the pair, and the answer is the number of steps or the calendar year, not the value you calculated.
What does half-life mean in a GCSE maths question?: It is the time taken for the mass of a substance to fall to half of what it was, and for exponential decay it stays the same however much you start with. Count how many half-lives fit into the given time, then halve that many times: after $n$ half-lives the fraction left is $\frac{1}{2^n}$ , so 3 half-lives leave an eighth. Half-life questions turn up on Paper 1 as well, because the arithmetic is repeated halving.

Exponential Growth and Decay - Edexcel GCSE Higher Maths

Worked example

Practice questions

Common mistakes examiners see

Frequently asked questions

Related topics