Number and ratio · GCSE Maths
Compound interest and depreciation
GCSE Maths compound interest: use multiplier to the power n, not simple interest, and apply the same method to depreciation, with a full line-by-line calculation.
Compound interest repeats the multiplier. After n years at r%, multiply by (1 + r/100)^n. Simple interest adds the same pound amount each year; compound does not.
The important bits
What you need to know
- 1
Simple interest: the extra is always a percentage of the original principal. £2000 at 5% simple for 3 years earns 3 × 0.05 × 2000 = £300, so £2300 in total.
- 2
Compound interest: each year the extra is a percentage of the new amount. The compact method is original × (multiplier)^n, where n is the number of time periods.
- 3
Growth multiplier is 1 + r/100. 4% interest is × 1.04 each year. Depreciation or decay uses 1 − r/100: 15% depreciation is × 0.85 each year.
- 4
After 3 years at 4% compound, multiply by 1.04³, not by 1.12. Adding 12% is simple interest, and it understates compound growth.
- 5
Write the formula before the calculator: value = P × (1.04)^3. Keep the power on the multiplier, not on P, and not on 4%.
- 6
If the question asks for the interest earned, subtract the original from the final value. If it asks for the final value, do not subtract.
- 7
Compound depreciation of a car, population decay, or a 3% annual fall in value all use the same structure with a multiplier less than 1.
- 8
To find how many years until a target, you may test n = 1, 2, 3… on the calculator, or take logs on Higher if the specification allows. Show at least one evaluated year so the method is visible.
Quotations worth analysing
Short evidence. Real method.
“Multiply by (1 + r/100)^n”
The power n is the number of compounding periods. Leaving the power off, or putting it on the original amount, loses the accuracy mark even when 1.04 is correct.
“This is not simple interest.”
Simple interest of 4% for 3 years is × 1.12. Compound is × 1.04³ ≈ 1.124864. The difference is small for one year and obvious after several.
“Value after n years = original × (0.85)^n”
A 15% fall each year leaves 85%, so the multiplier is 0.85, not 0.15, and not −0.15. The value stays positive; it does not change sign.
Go deeper
Why the power sits on the multiplier
£2000 at 4% compound for 3 years means: after year 1, 2000 × 1.04; after year 2, that new amount × 1.04 again; after year 3, × 1.04 again. That is 2000 × 1.04 × 1.04 × 1.04 = 2000 × 1.04³. Students who do 2000 × 1.04 × 3 have invented simple interest with a different story, and 2000 × 1.12 is the same error in one step. Calculate 1.04³ first if you like, then multiply: 1.04³ = 1.124864, then 2000 × 1.124864 = £2249.73 to the nearest penny. The interest earned is 2249.73 − 2000 = £249.73, not £249.73 + 2000 written twice. Write “compound, so power n” in the margin when the word compound is in the question.
Go deeper
Depreciation is compound with a multiplier below 1
A machine costs £8000 and loses 15% of its value each year. After 1 year it is worth 8000 × 0.85 = £6800. After 4 years it is worth 8000 × 0.85⁴. 0.85⁴ = 0.52200625, so value = £4176.05 to the nearest penny. Do not subtract 15% of 8000 four times: that would be simple depreciation, 8000 − 4 × 1200 = £3200, which is a different (and smaller) number. A multiplier of 0.15 would destroy the value in one year and is the classic “I used the percentage that left” error. Check the size: after several 15% falls the value should still be more than half of £8000 after four years, which 0.85⁴ confirms. If your answer is a few hundred pounds, the multiplier was 0.15.
Go deeper
One year is the same; many years are not
Simple and compound interest agree after exactly one period, because there is nothing yet to compound. After that they diverge. Exam questions exploit this by asking you to compare, or by giving a table of year-by-year values. If year 1 is £1040 on £1000, both models fit; if year 2 is £1081.60 rather than £1080, it is compound. When a question says “compounded annually” or “value falls by 8% each year”, use a power. When it says “simple interest” or “£40 interest per year”, add a constant. Mixed questions appear: a savings account at 3% compound for 5 years, then a withdrawal. Do the 1.03⁵ step first, then subtract the withdrawal from the result, not from the original £P.
See the idea in action
£2500 is invested at 3% compound interest per year. Find the value after 4 years, to the nearest penny, and the interest earned. Step 1: Multiplier = 1 + 3/100 = 1.03. Number of years n = 4. Step 2: Value = 2500 × 1.03⁴. Step 3: 1.03⁴ = 1.12550881. Step 4: 2500 × 1.12550881 = 2813.772025, so £2813.77 to the nearest penny. Step 5: Interest earned = 2813.77 − 2500 = £313.77. Check against simple interest: simple would be 2500 × 0.03 × 4 = £300, total £2800. Compound is slightly more, as expected.
Exam technique
Turn knowledge into marks
Write value = P × (multiplier)^n before you touch the calculator. If the question asks for interest, subtract P at the end; if it asks for the value, stop after the product.
Common mistakes
Do not give these marks away
- 01
Using 1 + nr/100 (simple interest) when the question says compound, or adding r% of the original n times.
- 02
Putting the power on the original amount, such as 2500⁴ × 1.03, which is dimensionally nonsense.
- 03
Using 0.15 as a depreciation multiplier instead of 0.85, so the value collapses in a single year.
£2000 is invested at 4% compound interest for 3 years. The value after 3 years is
A£2240
B£2249.73
C£2400
D£2080
Show the answer
£2249.73. Value = 2000 × 1.04³ = 2000 × 1.124864 = £2249.73 to the nearest penny. £2240 is simple interest (2000 × 1.12). £2400 is 2000 × 1.04 × 3, multiplying by 3 instead of using a power. £2080 is one year of 4% on £2000.
Quick questions
If this is the bit you searched
What is the difference between simple and compound interest?
Simple interest is always a percentage of the original principal, so you add the same amount each year. Compound interest is a percentage of the current value, so you multiply by the same multiplier each year.
How do you calculate compound interest GCSE?
Value = original × (1 + r/100)^n, where r is the percentage rate per period and n is the number of periods. Subtract the original if you need the interest earned.
How do you calculate depreciation?
Use a multiplier less than 1: after n years at r% depreciation, value = original × (1 − r/100)^n. Do not subtract r% of the original n times unless the question says simple depreciation.
Why is 3 years at 4% not a 12% increase when it is compound?
Each 4% is 4% of a larger amount than the year before. The overall multiplier is 1.04³ ≈ 1.1249, which is about 12.49%, not 12%.