Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Ratio, proportion and rates of change
Practise Compound interest, growth and decay. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Compound interest, growth and decay are repeated percentage changes: each year's change is worked out on the new amount, so you multiply by the same multiplier again and again. Questions ask for the value after a number of years, the interest earned or when a target is passed and, at Higher tier, you may have to work backwards, find the rate or use an iterative formula.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
4
Work out compound interest year by yearAdd each year's interest before working out the next, e.g. £500 at 4% becomes £520, then £540.80.
5
Use a multiplier raised to a powerValue after n years = start × multipliern, e.g. £2500 at 3% for 4 years is 2500 × 1.034.
5
Calculate depreciation and decayUse a multiplier less than 1, e.g. × 0.85n for a 15% fall each year.
5
Find when a value first passes a targetWork out the value year by year and show the values either side of the target.
Notes
Simple and compound interest
Simple interest is worked out on the original amount only, so you get the same interest every year.
Compound interest is added on each year, and the next year's interest is worked out on the new total, so it grows faster.
£1000 at 5% a year: simple interest pays £50 every year. Compound interest pays £50, then £52.50, then £55.13, giving £1157.63 after 3 years.
Using a multiplier and a power
Value after n years = starting value × multipliern.
Growth of r%: the multiplier is \(1 + \frac{r}{100}\), so 3% gives 1.03. Decay or depreciation of r%: the multiplier is \(1 - \frac{r}{100}\), so 12% gives 0.88.
£2500 at 3% compound interest for 4 years: 2500 × 1.034 = £2813.77.
Interest earned = final amount − starting amount.
The same method works for anything that changes by the same percentage in each time period: savings, populations, the value of a car, bacteria.
Check the time period: a rate per month for 2 years means n = 24.
How many years?
Type the starting value and press =, then type × 1.04 and press = again and again: each press is one more year (the calculator uses Ans).
Stop at the first value that passes the target, and write down the values for the year before and the year after.
Cheatsheet
Growth: value after n years = start × \(\left(1 + \frac{r}{100}\right)^n\)
Decay: value after n years = start × \(\left(1 - \frac{r}{100}\right)^n\)
Interest earned = final amount − starting amount
Simple interest is the same every year; compound interest also earns interest on the interest
How to answer each type of question
Work out the value after compound interest
3 marks5
Write the multiplier: 1 + the rate as a decimal (2.5% gives 1.025).
Multiply the amount invested by the multiplier to the power of the number of years.
Round only at the end, to the nearest penny unless told otherwise.
Subtract the amount invested if the question asks for the interest.
Example. Maya invests £4000 in a savings account for 3 years. The account pays compound interest at a rate of 2.5% per year. Work out the value of Maya's investment at the end of the 3 years.
Write the multiplier: 1 − the rate as a decimal (15% gives 0.85).
Multiply the starting value by the multiplier to the power of the number of years.
Round as the question says.
Example. A van cost £18 000 when it was new. The value of the van depreciates by 15% each year. Work out the value of the van after 4 years. Give your answer to the nearest pound.
Work out the value after 1, 2, 3, ... years (use Ans × multiplier on the calculator).
Stop at the first year that passes the target.
Write the values for the year before and the year after the target, then give the number of years.
Example. Josh invests £3000 in an account paying 4% per year compound interest. After how many whole years will the investment first be worth more than £3500? You must show your working.
Show the model answer
3000 × 1.04 = £3120, then £3244.80, ... (M1) After 3 years: £3374.59; after 4 years: £3509.58 (M1) 4 years (A1)
Shortcuts and memory tricks
Calculator trick: type the start value and press =, then × 1.04 =, and every extra press of = moves on one more year.
Sense check: a multiplier above 1 must make the value bigger; a multiplier below 1 must make it smaller.
The power is the number of time periods; the rate only goes into the multiplier.
At the same rate, compound interest earns more than simple interest after the first year, because it earns interest on the interest.
Where marks are lost
Using simple interest (the same amount each year) when the question says compound interest.
Multiplying by 0.15 instead of 0.85 for a 15% decrease: 0.15 is the part that is lost.
Writing the multiplier for 3% as 1.3 instead of 1.03.
Rounding in the middle of the calculation: keep the full calculator value until the end.
Working backwards by multiplying, or by adding the percentage back on: divide by multipliern.
Giving the total when the question asks for the interest, or the other way round.
Exam technique
Write down the calculation you type in, e.g. \(4000 \times 1.025^3\): it earns method marks even if you mistype.
Round money to the nearest penny unless the question says otherwise (e.g. 'to the nearest pound').
'Per annum' means per year.
For 'how many years' questions, show the value just before and just after the target.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
The value of a car is £8000. It decreases by 10% each year. Work out the value after 2 years.[3]
Show the answer and mark scheme
Answer: £6480
M1 for a correct multiplier (0.9) or for finding 10% of £8000 (£800)
M1 for a complete method, e.g. \(8000 \times 0.9^{2}\) or 2 years worked out one at a time
A1 for £6480 cao
Worked solution: Each year multiply by 0.9: after 1 year: \(8000 \times 0.9 = 7200\) after 2 years: \(7200 \times 0.9 = 6480\).
Question 2Medium4 marks
Sanjay has £5000 to invest for 5 years. Account A pays 3% per year compound interest. Account B pays 3.2% per year simple interest.
Which account will give Sanjay more money at the end of the 5 years? You must show your working.[4]
Show the answer and mark scheme
Answer: Account B: A gives £5796.37 and B gives £5800.