Compound interest, growth and decay Cambridge IGCSE Mathematics revision
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In plain words
With compound interest you earn interest on your interest, so the amount grows faster every year. The same idea, a fixed percentage change repeated again and again, describes a car losing value, a population growing, and bacteria multiplying.
Three things to know
- Compound interest: amount = P × (1 + r/100)ⁿ, where P is the starting amount, r the percentage rate and n the number of years.
- For something losing value (depreciation), the multiplier is less than 1: a 10% fall each year is × 0.9 each year.
- Exponential growth means multiplying by the same number in each equal period of time.
Worked example
$2000 is invested at 5% per year compound interest. Find its value after 3 years.
- The multiplier for a 5% increase is 1.05.
- After 3 years: 2000 × 1.05³.
- = $2315.25.
Tips and tricks
- Use a power, not three separate steps: it is quicker and there is less to go wrong.
- "How much interest?" means take the starting amount off at the end.
It lands in your notebook with its questions as flashcards.
Compound interest, growth and decay: 5 questions and answers
These are the quiz’s questions. Do the quiz first, then come back here for the ones that got you.
Which calculation gives the value of $300 after 4 years at 3% compound interest?
The multiplier 1.03 is applied four times.
$100 is invested at 10% per year compound interest. What is it worth after 2 years?
100 × 1.1 × 1.1.
A phone worth $1000 loses 20% of its value each year. What is it worth after 2 years?
1000 × 0.8 × 0.8.
How does compound interest differ from simple interest?
That is why the amount grows faster each year.
A colony of 10 bacteria triples every hour. How many are there after 3 hours?
10 × 3³.
Quiz
5 questions
Tap an answer and you’ll see straight away whether it’s right, and why.
Worksheet
4 questions, 9 marks. Write your answers on paper, then check them.
Compound interest, growth and decay
Cambridge IGCSE Mathematics 0580 · 9 marks · papermunch.org
Name ______________________________ Date ______________
$500 is invested at 4% per year compound interest. Find its value after 2 years.[2]
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$540.80. 500 × 1.04².
A car worth $12 000 loses 10% of its value each year. Find its value after 2 years.[2]
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$9720. 12 000 × 0.9².
A population of 800 bacteria doubles every hour. Find the population after 3 hours.[2]
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6400. 800 × 2³.
Find the interest earned when $1000 is invested for 2 years at 10% per year compound interest.[3]
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$210. 1000 × 1.1² = 1210; 1210 − 1000.
Answers: Compound interest, growth and decay
- 1. $540.80. 500 × 1.04².
- 2. $9720. 12 000 × 0.9².
- 3. 6400. 800 × 2³.
- 4. $210. 1000 × 1.1² = 1210; 1210 − 1000.



