Compound Interest and Growth
A percentage change can happen more than once. Each time, the new amount becomes the starting point for the next change. This is the key to compound interest, growth and decay.
Foundation & Higher — with a first-principles start: begin with two simple yearly changes, then use multipliers and powers. The final two worked examples offer further practice in total change and working backwards.
The new idea: use the current amount
In Percentage Increase and Decrease, you used a multiplier for one change. A multiplier is the number you multiply the starting amount by.
A 5% increase keeps 105% of the amount, so the multiplier is 1.05. A 5% decrease leaves 95%, so the multiplier is 0.95.
Now let the change happen more than once. After each change, use the new amount as the starting point for the next.
Compound growth means repeated percentage increases based on the current amount. Compound decay means repeated percentage decreases based on the current amount.
We begin with a constant percentage at each step, then consider changing rates. Even with a constant rate, the amount added or removed can change.
Example Question:
£100 grows by 10% each year for 2 years. What is it worth at the end?
Would the second increase be £10 again? Follow both years in Example 1.
Exam Tip
Underline the rate, how often it applies and the number of changes.
“Per year” means each year. Three annual changes need three multiplications.
Check whether the question asks for the final amount or just the interest earned. Keep full calculator values and round money only at the end.
Method — One change at a time
1. Identify the original amount. This is the amount before any changes.
2. Choose the multiplier. For growth, add the percentage to 100%. For decay, subtract it from 100%. Write the result as a decimal.
3. Apply it once for each change. Each result becomes the starting amount for the next step.
4. Answer the question asked. State the final amount, or subtract the original if asked for interest or total gain.
Start with the year-by-year method below. We will introduce powers after seeing why the amount changes each year.
Worked Examples
Example 1 — Follow two years of growth
£100 grows by 10% each year for 2 years.
Year 1: 10% of £100 is £10. Add it to the starting amount:
£100 + £10 = £110
The same step using the 110% multiplier is:
£100 × 1.10 = £110
Year 2 starts at £110. This time 10% is £11:
£110 + £11 = £121
£110 × 1.10 = £121
The final amount is £121. The second increase was based on £110, not the original £100.
| Time | Amount |
|---|---|
| Start (year 0) | £100 |
| After year 1 | £110 |
| After year 2 | £121 |
The total increase is £121 − £100 = £21. Adding two lots of £10 would miss the extra £1.
Example 2 — Write repeated multiplication as a power
£500 grows by 4% each year for 3 years. Find the final amount.
100% + 4% = 104%
104% = 1.04
Three changes mean three factors of 1.04:
Final amount = £500 × 1.04 × 1.04 × 1.04
As in Powers and Roots, a raised number counts repeated factors:
Final amount = £500 × 1.043
The raised 3 is the exponent. It tells us to multiply by 1.04 three times, not multiply 1.04 by 3.
500 × 1.043 = 562.432
The final amount is £562.43 to the nearest penny.
The same structure for any fixed rate
Final amount = original amount × multipliernumber of changes
The original amount is where you start. The multiplier gives one percentage change. The exponent counts how many changes happen.
For example, 1.053 means 1.05 × 1.05 × 1.05. Use the number of changes, which need not always be a number of years.
Example 3 — Compound interest: balance and interest
£600 is saved at 3% compound interest per year for 4 years. No money is added or withdrawn. Find the final balance and the interest earned.
Interest is money added to the savings. The balance is the total in the account. With compound interest, later interest is calculated on the balance including earlier interest.
100% + 3% = 103%
103% = 1.03
Final balance = £600 × 1.034
600 × 1.034 = 675.305286
Final balance: £675.31 to the nearest penny.
For the interest alone, subtract the original £600 from the unrounded balance:
675.305286 − 600 = 75.305286
Interest earned: £75.31 to the nearest penny.
Example 4 — Growth outside a money question
A model starts with 2000 bacteria. The number increases by 10% each hour for 3 hours. Find the number predicted by the model.
100% + 10% = 110%
110% = 1.10
2000 × 1.103 = 2662
The model predicts 2662 bacteria. Here, the exponent counts hours because the change happens every hour.
This is a model with a fixed growth rate, not a claim that every real population follows this rule.
Repeated decreases
Depreciation means a fall in value. When the same percentage of the current value is lost each year, it is compound decay.
Use the fraction that remains. Losing 15% leaves 85%, so the multiplier is 0.85.
Example 5 — Repeated depreciation
A car is worth £20,000. It loses 15% of its current value each year for 3 years. Find its value after 3 years.
100% − 15% = 85%
85% = 0.85
20,000 × 0.853 = 12,282.5
The car is worth £12,282.50.
| Time | Value |
|---|---|
| Start | £20,000 |
| After year 1 | £17,000 |
| After year 2 | £14,450 |
| After year 3 | £12,282.50 |
Each year removes 15% of a smaller value. The amounts lost get smaller too. It is not a single 45% decrease.
Example 6 — Why decreases do not simply add
An amount of 100 decreases by 20% twice. Is that the same as a 40% decrease?
A 20% decrease leaves 80%, so use 0.80 at each step:
100 × 0.80 = 80
80 × 0.80 = 64
The repeated decrease leaves 64. A single 40% decrease would leave 60.
The actual loss is 100 − 64 = 36. The second 20% removes 16, not another 20.
Example 7 — A GCSE savings question with rounding
£750 is saved at 2.5% compound interest per year for 3 years, with no deposits or withdrawals. How much interest is earned?
100% + 2.5% = 102.5%
102.5% = 1.025
750 × 1.0253 = 807.66796875
Subtract the original before rounding:
807.66796875 − 750 = 57.66796875
The interest earned is £57.67 to the nearest penny.
£807.67 would be the final balance. It answers a different question.
Further practice — the whole change and working backwards
Try these after the forward calculations feel secure. They use the same multipliers but ask for a different result.
Example 8 — Further practice: find the total percentage increase
A quantity grows by 10% twice. Find the overall percentage increase.
The combined multiplier is:
1.102 = 1.21
This means the final amount is 121% of the original.
121% − 100% = 21%
The overall increase is 21%. Keep the original 100% separate from the extra 21%.
Example 9 — Further practice: work backwards
A value is £1210 after two annual increases of 10%. Find the original value.
Write the forward calculation first:
Original × 1.102 = £1210
Divide by the whole combined multiplier to undo both changes:
Original = £1210 ÷ 1.102
1.102 = 1.21
1210 ÷ 1.21 = 1000
The original value was £1000.
Check by applying both changes forwards:
1000 × 1.102 = 1210
This recovers the given final value. Subtracting 20% would not undo the two multiplications.
Changing rates — use a multiplier for each stage
A single repeated power works only when the multiplier stays the same. If the rate changes, write the multipliers in their time order and apply each to the current amount.
Worked example: £800 increases by 5% in year 1, then decreases by 10% in year 2. After year 1: £800 × 1.05 = £840. After year 2: £840 × 0.90 = £756. Equivalently, £800 × 1.05 × 0.90 = £756. The combined multiplier is 0.945, a 5.5% decrease overall, not 5%.
Quick practice — £900 grows by 4% in year 1 and 2% in year 2. Find the final amount.
£900 × 1.04 × 1.02 = £954.72. Do not use 1.04² or add the percentage rates.
Higher — construct a general growth or decay model
For a fixed multiplier m, starting amount P and number of changes n, the model is A = Pmn. The equivalent step rule is “next amount = current amount × m”. Both assume the rate stays fixed and there are no other additions or withdrawals.
A car starts at £12 000 and loses 10% of its current value each year. After n years, its modelled value is V = 12 000 × 0.9n. At n = 2 this gives £9720. Repeated numerical percentage calculations above belong to both tiers; constructing and interpreting a general symbolic model is the Higher extension.
General iterative methods are developed in Numerical Methods and Iteration.
Quick practice — Higher: £500 grows by 4% each year with no deposits or withdrawals. Write a model for its balance B after n years.
The multiplier is 1.04, so B = 500 × 1.04n. When n = 0 the starting balance is £500.
Calculator and interest checks
Use a calculator carefully
For 500 × 1.043, enter 500 multiplied by 1.04 raised to power 3. The power applies to 1.04 only, not to 500 × 1.04.
The general power function may look like xy or a raised box. Its label and entry sequence depend on the calculator; x and y on the key are placeholders.
Check the expression on the display. Keep the full result and round the final money answer to two decimal places when needed.
Simple interest is different
Simple interest adds the same percentage of the original amount each time. Compound interest uses the current amount.
For £100 at 10% per year over 2 years, simple interest adds £10 each year and gives £120. Compound interest gives £121.
Quick practice — £480 earns 4% simple interest each year for 3 years. Find the interest and the final balance.
Each year adds 4% of the original £480: 480 × 0.04 = £19.20.
Three years of interest: 3 × £19.20 = £57.60.
Final balance: £480 + £57.60 = £537.60. For simple interest, do not raise 1.04 to a power.
Read whether the question specifies simple or compound interest before choosing the method.
These examples assume a fixed rate and no deposits or withdrawals. Follow any different instructions in the question.
Check that your answer is sensible
For positive starting amounts, a growth multiplier is greater than 1. A decay multiplier is between 0 and 1 when the decrease is between 0% and 100%.
Repeated growth should give a larger amount; repeated decay should leave a smaller, positive amount.
At 4% growth for 3 years, £500 should grow moderately. £562.43 is sensible; a result of several thousand pounds would need checking.
Compare the final answer with a rough estimate. For the same fixed positive rate over several periods, compound growth gives more than simple interest because the increases are based on growing amounts.
Check the exponent, units and what the question asks for. For a reverse calculation, apply the forward changes again.
Think Like a Mathematician
Why do two 10% increases give 21%, not 20%?
The amounts are 100, then 110, then 121. The second increase includes 10% of the first increase.
10% of the first extra 10 is 1. That extra 1 explains the difference between 120 and 121.
With a fixed positive percentage, a larger current amount gives a larger increase. That is why the amounts added by compound growth get larger.
Equal percentage increases and decreases do not cancel either: 100 × 1.10 = 110, then 110 × 0.90 = 99. The percentages use different starting amounts.
Common mistakes to avoid
Adding the percentages. Two 10% increases use 1.10 twice. They are not a single 20% increase.
Choosing the share instead of the growth multiplier. 0.05 finds just 5%. A 5% increase keeps the whole and adds 5%, so use 1.05.
Using a growth multiplier for decay. Losing 15% leaves 85%. Use 0.85, not 1.15.
Returning to the original amount each year. Start the next step from the latest amount.
Using the wrong power. Count the changes. Three years at an annual rate means three factors, not two or four.
Confusing balance and interest. Interest earned is the final balance minus the original savings.
Rounding each year. Keep the full calculator value until the final answer unless the question explicitly instructs otherwise.
Trying to undo growth with an equal percentage decrease. Work backwards by dividing by the forward multiplier.
Try it yourself
Start with multipliers, then work through growth, interest and decay. Show your working. Questions 11–12 practise total change and working backwards. Assume fixed rates, with no deposits or withdrawals in savings questions.
1. Write the multiplier for a 6% increase.
100% + 6% = 106%
106 ÷ 100 = 1.06
The multiplier is 1.06. It includes the original amount.
2. Write the multiplier for a 12% decrease.
100% − 12% = 88%
88 ÷ 100 = 0.88
The multiplier is 0.88. It represents what remains.
3. £200 grows by 10% each year for 2 years. Work year by year.
A 10% increase uses 1.10.
Year 1: £200 × 1.10 = £220
Year 2: £220 × 1.10 = £242
The final amount is £242. The second increase is £22.
4. £400 grows by 5% each year for 3 years. Write a calculation using a power and find the final amount.
100% + 5% = 105%, so use 1.05.
Three changes need power 3:
400 × 1.053 = 463.05
The final amount is £463.05.
5. £800 earns 4% compound interest per year for 3 years. Find the final balance and the interest earned, to the nearest penny.
Use multiplier 1.04 and power 3:
800 × 1.043 = 899.8912
Final balance: £899.89.
Use the unrounded value for the interest:
899.8912 − 800 = 99.8912
Interest earned: £99.89.
6. A car worth £12,000 loses 10% of its current value each year. Find its value after 2 years.
A 10% loss leaves 90%, so use 0.90.
12,000 × 0.902 = 9720
The value is £9720.
Check year by year: £12,000 × 0.90 = £10,800, then £10,800 × 0.90 = £9720.
7. A model starts with 500 bacteria and predicts 20% growth each hour. How many bacteria are predicted after 2 hours?
120% = 1.20. Two hourly changes need power 2.
500 × 1.202 = 720
The model predicts 720 bacteria. The units are bacteria, not pounds.
8. £350 earns 3.5% compound interest per year for 4 years. Find the final balance to the nearest penny.
103.5% = 1.035. Use four factors:
350 × 1.0354 = 401.63305021875
The third decimal digit is 3, so the hundredths digit stays the same.
The final balance is £401.63.
9. A student says two 10% decreases take 100 down to 80. Explain the mistake and find the correct result.
Each decrease acts on the current amount.
100 × 0.90 = 90
90 × 0.90 = 81
The result is 81. The decreases remove 10 and then 9, not 10 twice.
10. Equipment worth £2500 loses 20% of its current value each year for 3 years. Find its final value and the total value lost.
A 20% decrease leaves 80%, so use 0.80.
2500 × 0.803 = 1280
Final value: £1280.
Subtract the final value from the original to find the loss:
2500 − 1280 = 1220
Total value lost: £1220.
11. Further practice: a quantity increases by 5% twice. Find the overall percentage increase.
Find the combined multiplier:
1.052 = 1.1025
The final amount is 110.25% of the original.
110.25% − 100% = 10.25%
The overall increase is 10.25%.
12. Further practice: a value is £968 after two annual increases of 10%. Find the original value.
Original × 1.102 = £968
Divide by the combined multiplier:
1.102 = 1.21
968 ÷ 1.21 = 800
The original value was £800.
Check forwards:
800 × 1.102 = 968
Frequently Asked Questions
What does compound mean?
Each percentage change uses the latest amount. Earlier changes become part of the amount used for the next calculation.
Why does the multiplier have a power?
The power is a short way to write repeated multiplication. Power 3 means use the multiplier three times.
Does the exponent always count years?
No. It counts changes. If growth happens every hour, use the number of hourly changes.
Is interest the same as the final balance?
No. The final balance includes the original savings. Subtract the original to find just the interest earned.
Should I round after each year?
For these GCSE calculations, retain full precision and round only the final answer. Follow a question’s instructions if it explicitly gives a different rounding rule.
Can I use one power if the percentage rate changes?
One multiplier raised to a power assumes the same rate each time. Different rates need their own multipliers in the correct sequence.
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