The compound interest formula for GCSE Maths
The compound interest formula used in GCSE Maths is A = P(1 + r/100)ⁿ, where P is the amount you start with, r is the percentage interest rate per year, n is the number of years and A is the amount at the end. Each year's interest is added on, and then earns interest itself, so the amount is multiplied by the same number, the multiplier 1 + r/100, every year. £2000 at 3% compound interest for 5 years grows to 2000 × 1.03⁵ = £2318.55.
The compound interest formula
| Symbol | Meaning |
|---|---|
| A | Final amount |
| P | Principal: the starting amount |
| r | Percentage rate per period (usually per year) |
| n | Number of periods (usually years) |
AQA's GCSE Maths specification asks students to set up, solve and interpret growth and decay problems, including compound interest. Understanding where the formula comes from means you can rebuild it, and adapt it for decay, in any question.
Why it works: the multiplier
Increasing an amount by 3% is the same as multiplying it by 1.03, because 100% + 3% = 103%. After one year £2000 becomes 2000 × 1.03. After two years, that new amount is multiplied by 1.03 again: 2000 × 1.03 × 1.03 = 2000 × 1.03². After n years it is 2000 × 1.03ⁿ.
| Change | Multiplier |
|---|---|
| Increase by 3% | 1.03 |
| Increase by 12% | 1.12 |
| Decrease by 12% | 0.88 |
| Decrease by 5% | 0.95 |
Compound interest examples
Example 1: the amount after n years
£2000 is invested at 3% compound interest per year. How much is there after 5 years, and how much interest has been earned?
After 5 years there is £2318.55 (to the nearest penny), so the interest earned is 2318.55 − 2000 = £318.55.
Example 2: compound vs simple interest
With simple interest, you earn interest only on the original amount, the same every year: 3% of £2000 is £60, so after 5 years there is 2000 + 5 × 60 = £2300. Compound interest gives £18.55 more, because each year's interest also earns interest. The gap grows the longer the money is invested.
Example 3: depreciation
A car bought for £15 000 loses 12% of its value each year. What is it worth after 3 years?
A decrease of 12% uses the multiplier 1 − 0.12 = 0.88:
The car is worth £10 222.08. The same formula with a minus sign, A = P(1 − r/100)ⁿ, works for any decay.
Example 4: interest added more often
An account pays 2% interest every 6 months. How much does £1000 become after 2 years?
2 years contains four 6-month periods, so n = 4 and the rate per period is 2%:
Example 5: finding the rate
£5000 grows to £5832 in 2 years with compound interest. Find the yearly rate.
The multiplier squared is 5832 ÷ 5000 = 1.1664, so the multiplier is √1.1664 = 1.08. The rate is 8% per year.
Example 6: how many years?
£1000 is invested at 4% per year. After how many whole years will it first be worth more than £1200?
Try values of n: 1.04⁴ = 1.1699, so after 4 years there is £1169.86, not enough. 1.04⁵ = 1.2167, so after 5 years there is £1216.65. The answer is 5 years.
Example 7: working backwards
After 3 years at 5% compound interest, an investment is worth £11 576.25. How much was invested?
Divide by the multiplier as many times as it was applied:
The original investment was £10 000. Dividing is the key step: taking off 3 × 5% = 15% gives the wrong answer.
Here is a quick check.
Compound interest beyond GCSE
The same idea is exponential growth. At A-level you can solve 1.04ⁿ = 1.2 directly with logarithms, n = log 1.2 ÷ log 1.04 = 4.65, instead of trying values. Interest added more and more often leads to the exponential function eˣ. Our guide to log rules covers solving equations like this.
Common mistakes
- Adding the percentage each year (simple interest) when the question says compound.
- Using the wrong multiplier. A 3% increase is × 1.03, not × 1.3 or × 0.03.
- Decrease multipliers. A 12% decrease is × 0.88, not × 1.12 or × 0.12.
- Reverse percentages. To go back, divide by the multiplier; do not subtract the percentage of the final amount.
- Giving the interest when asked for the total, or the other way round. Read which one the question wants.
- Rounding during the working. Round money to the nearest penny only at the end.
Compound interest questions with answers
1. £500 is invested at 6% compound interest. How much is there after 2 years?
2. A van worth £8000 depreciates by 15% a year. What is it worth after 2 years?
3. What is the multiplier for a 7% decrease?
4. A population of 300 grows by 10% a year. What is it after 3 years?
5. An amount grows by 3.5% a year. By roughly what factor has it grown after 10 years?
Practise percentages in a short timed paper
GCSE Maths Higher and A-level Maths: Algebra & Functions each give you a short timed paper every day, with 8 questions worth 25 to 35 marks across the course, so growth and decay turn up regularly but not in every paper.
GCSE Maths Higher
Start today's paperA-level Maths: Algebra & Functions
Start today's paperPapers reset in
Common questions
What is the compound interest formula for GCSE?
What is the difference between simple and compound interest?
How do you work out depreciation?
How do you find the original amount?
What if interest is added monthly or every 6 months?
How do you find how many years it takes?
Revision cards
Say the answer first, then flip the card to check it.
Where to go next
- Test it
GCSE Maths Higher
A new exam-style paper every day, with instant marking and a mark scheme
Start today's paper - Test it
A-level Maths: Algebra & Functions
A new exam-style paper every day, with instant marking and a mark scheme
Start today's paper