Compound interest, explained with examples

By The STEM-Tutor team · Updated October 2026

Compound interest is interest on interest: each year's interest is added to the balance, and the next year's interest is calculated on the larger amount. The same method describes depreciation, population growth and many other real-world changes.

Simple versus compound interest

With simple interest, the interest is the same every year, because it is always calculated on the original amount. £1000 at 5% simple interest earns £50 a year, so after 3 years it is worth £1150.

With compound interest, each year's interest is added to the balance, and the next year's interest is calculated on the new, larger balance. £1000 at 5% compound interest becomes £1050 after one year, then £1102.50, then £1157.63. The extra £7.63 compared with simple interest is interest earned on interest. The gap grows quickly over long periods: after 30 years, the compound account would hold about £4322, compared with £2500 from simple interest.

The multiplier method

To increase an amount by a percentage, multiply by a multiplier:

• increase by 5%: multiply by 1.05;
• increase by 12%: multiply by 1.12;
• decrease by 20%: multiply by 0.80;
• decrease by 3.5%: multiply by 0.965.

For compound change over n years, multiply by the multiplier n times, which is the same as multiplying by the multiplier to the power n.

The formula

Final amount = P × (1 + r/100)ⁿ

P is the starting amount (the principal), r is the percentage rate per year, and n is the number of years. For a decrease, such as depreciation, use (1 − r/100)ⁿ. Even when a formulae sheet is provided, as it is for GCSE exams in 2025 to 2027, understanding multipliers means you can always rebuild the formula yourself.

Worked examples

Example 1: compound interest

£2000 is invested at 4% compound interest per year. Find the value after 5 years.

• Multiplier: 1.04. Value = 2000 × 1.04⁵.
• £2433.31 to the nearest penny.

Example 2: interest earned

How much interest was earned in Example 1?

• Interest = final amount − original = 2433.31 − 2000 = £433.31.
• Read the question: “value” and “interest” are different answers.

Example 3: depreciation

A car bought for £18,000 loses 15% of its value each year. Find its value after 3 years.

• Multiplier: 0.85. Value = 18,000 × 0.85³.
• £11,054.25.

Example 4: comparing accounts

Account A pays 3% compound interest. Account B pays 3.2% simple interest. Which gives more on £5000 after 10 years?

• A: 5000 × 1.03¹⁰ ≈ £6719.58.
• B: interest 5000 × 0.032 × 10 = £1600, total £6600.
• Account A gives more, even though its rate is lower. Over long periods, compounding wins.

Example 5: how many years?

£1000 is invested at 6% compound interest. After how many whole years will it first be worth more than £1500?

• Try values: 1.06⁶ ≈ 1.419, so £1418.52 after 6 years. 1.06⁷ ≈ 1.504, so £1503.63 after 7 years.
• 7 years. Trial and improvement is the expected GCSE method; at A level you would use logarithms.

Example 6: reverse percentage

After 2 years of 5% compound interest, an account holds £2205. How much was invested?

• Original × 1.05² = 2205, so original = 2205 ÷ 1.1025.
• £2000. To go backwards, divide by the multiplier; never subtract the percentage.

Example 7: different rates each year

An investment grows by 4% in year 1, 6% in year 2 and falls by 2% in year 3. Starting with £3000, find its value.

• Use each multiplier in turn: 3000 × 1.04 × 1.06 × 0.98.
• £3241.06.

Example 8: finding the rate

£1000 grows to £1210 in 2 years with compound interest. Find the annual interest rate.

• 1000 × m² = 1210, so m² = 1.21 and m = 1.1.
• The rate is 10% per year.

Borrowing works the same way

Compound interest also applies to money you owe. Loans and credit card debts grow by a percentage each period if they are not paid off, so a debt can increase surprisingly quickly. A £1000 balance at 2% per month, left unpaid, grows to 1000 × 1.02¹² ≈ £1268 in a year. Understanding the multiplier method helps you compare savings accounts, loans and finance offers, which is one reason this topic appears on every GCSE paper.

Interest paid more often (Higher)

Some accounts pay interest monthly. If the annual rate is 6% paid monthly, the monthly rate is 6/12 = 0.5%, and there are 12 compounding periods each year. £1000 for one year becomes 1000 × 1.005¹² ≈ £1061.68, slightly more than the £1060 from yearly compounding. More frequent compounding always gives a little more, and this is why banks quote an AER (annual equivalent rate) to compare accounts fairly.

Growth and decay beyond money

The same multiplier method models many real situations:

• Population growth: a town of 20,000 growing by 2% a year has 20,000 × 1.02ⁿ people after n years.
• Depreciation: cars, phones and machinery lose a percentage of their value each year.
• Radioactive decay and drug levels: a fixed percentage disappears in each time period.
• Bacteria: a population that doubles every hour has multiplier 2 per hour.

Whenever a quantity changes by the same percentage in each time period, it is exponential growth or decay, and the multiplier method applies. A multiplier above 1 means growth; a multiplier between 0 and 1 means decay.

Common mistakes

• Multiplying by 1.5 instead of 1.05 for a 5% increase.
• Using simple interest when the question says compound.
• Subtracting the percentage to reverse a change, instead of dividing by the multiplier.
• Giving the total value when the question asks for the interest, or the other way round.
• Rounding during the calculation instead of only at the end.

Practice questions

Try each one before opening the answer.

Frequently asked questions

Is compound interest on the Foundation paper?

Yes. Simple and compound interest, and depreciation, are on both Foundation and Higher papers. Higher papers add reverse problems and finding the number of years.

How is compound interest used at A level?

A level Maths models growth and decay with exponentials such as P = P₀e^kt, and uses logarithms to find the time taken, which replaces trial and improvement.

Key takeaways

• Compound interest earns interest on interest.
• Use a multiplier: 1 + r/100 for increases, 1 − r/100 for decreases.
• Final amount = P × multiplierⁿ.
• Reverse a change by dividing by the multiplier.
• The same method models depreciation, population growth and decay.

Related guides

• Logarithms
• Standard form
• GCSE Maths topics