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When the same percentage change is applied again and again over several time periods, you get compound interest (for growth, like savings) or depreciation (for loss, like a car losing value). These are central to the Rates of Change strand of OCR GCSE Mathematics (J560) and appear mainly on the calculator papers. This lesson builds directly on the multiplier idea from the percentage-change lesson, covering simple interest for contrast, compound interest, depreciation, and general repeated percentage change.
This lesson develops AO1 fluency with the compound multiplier (1±100r)n, AO2 reasoning when you compare simple and compound interest, and AO3 problem-solving when you must find a number of years or work back to an original amount.
| Term | Meaning |
|---|---|
| Principal | The starting amount invested or borrowed |
| Simple interest | Interest paid only on the original principal each year |
| Compound interest | Interest paid on the principal plus previous interest |
| Depreciation | A decrease in value over time (e.g. cars, machinery) |
| Multiplier | The decimal applied each period, e.g. 1.05 for +5% |
| Per annum (p.a.) | Per year |
It is worth meeting simple interest first, because contrasting it with compound interest makes the key idea of compounding much clearer. With simple interest, the same amount of interest is added every year, calculated only on the original principal — the interest never itself earns interest. This means the total grows in equal steps, like a straight line, rather than accelerating.
interest=P×100r×n,
where P is the principal, r the rate per year, and n the number of years.
£2,000 is invested at 4% simple interest per year. Work out the total interest after 5 years.
Interest each year =2,000×1004=80, i.e. £80 per year.
Over 5 years: 80×5=400, i.e. £400 total interest.
Answer: £400. Notice the interest is the same £80 every year, because it is always 4% of the original £2,000 — this is the defining feature of simple interest.
Work out the total amount in an account after £5,000 is invested at 3% simple interest per year for 4 years.
Interest each year =5,000×1003=150, i.e. £150 per year.
Total interest =150×4=600, i.e. £600.
Total amount =5,000+600=5,600, i.e. £5,600.
Answer: £5,600.
Common error: giving £600 (the interest) when the question asks for the total amount. Read whether the question wants the interest or the final balance.
Compound interest is the form used by almost all real savings accounts and loans, and it is the one examined most often. The crucial difference from simple interest is that each year's interest is added to the running total, so the next year's interest is calculated on a larger amount — you earn "interest on your interest". This is why a compound account grows faster and faster over time, and why the total cannot be found just by multiplying one year's interest by the number of years. Instead, you apply the multiplier once for each year, which is the same as raising the multiplier to a power:
A=P(1+100r)n,
where A is the final amount, P is the principal, r is the annual rate and n is the number of years. The power n is what makes the growth compound rather than simple, so make sure it appears in your working.
£3,000 is invested at 5% compound interest per year. Work out the value after 3 years.
The multiplier is 1.05, applied 3 times:
A=3,000×1.053=3,000×1.157625=3,472.875.
Rounding to the nearest penny, the value is £3,472.88.
Answer: £3,472.88.
Show the year-by-year build-up of £2,000 at 5% compound interest for 3 years.
| Year | Start (£) | Interest at 5% (£) | End (£) |
|---|---|---|---|
| 1 | 2,000.00 | 100.00 | 2,100.00 |
| 2 | 2,100.00 | 105.00 | 2,205.00 |
| 3 | 2,205.00 | 110.25 | 2,315.25 |
Using the formula: 2,000×1.053=2,315.25, matching the table. The final value is £2,315.25.
Common error: calculating 5% of the original each year (that is simple interest). With compound interest, the interest grows because the base grows.
£1,500 is invested at 2% compound interest per year. Work out the total interest earned after 3 years, to the nearest penny.
First find the final amount using the multiplier 1.02:
A=1,500×1.023=1,500×1.061208=1,591.812.
So the final amount is £1,591.81 (nearest penny). The question asks for the interest, not the total, so subtract the principal:
interest=1,591.81−1,500=91.81.
Answer: £91.81 interest.
Common error: giving the final amount (£1,591.81) when the question asks for the interest earned. Always check whether the question wants the total value or just the interest.
A favourite exam question asks you to compare the two kinds of interest, or to work out the difference between them over a number of years. The difference is small at first but grows steadily, because the compound account keeps building on an ever-larger balance while the simple account adds the same fixed amount each year. The reliable approach is to work out each total separately and then subtract — and always be clear about whether the question wants the final amount or just the interest.
£4,000 is invested for 3 years at 6% per year. Work out how much more is earned with compound interest than with simple interest.
Simple: interest =4,000×0.06×3=720, i.e. £720.
Compound: A=4,000×1.063=4,000×1.191016=4,764.064, so interest =764.06, i.e. £764.06.
Difference =764.06−720=44.06, i.e. £44.06 more with compound interest.
Answer: £44.06 more.
£3,000 is invested for 2 years. Work out the difference between the interest earned at 5% compound interest and at 5% simple interest.
Simple interest: 3,000×0.05×2=300, i.e. £300.
Compound: A=3,000×1.052=3,000×1.1025=3,307.50, so the compound interest is 3,307.50−3,000=307.50, i.e. £307.50.
Difference =307.50−300=7.50, i.e. £7.50.
Answer: £7.50 more with compound interest. The gap is small over just two years but would widen each year, because compounding keeps building on a larger balance.
Depreciation is what happens when something loses value over time — most famously a car, which is worth noticeably less each year. Mathematically it is just compound interest in reverse: instead of the amount growing by a fixed percentage each year, it shrinks by a fixed percentage. The method is identical, except the multiplier is below 1. A 20% loss each year, for instance, leaves 80% of the value, so the multiplier is 0.8.
A=P(1−100r)n.
A car is bought for £18,000 and depreciates by 20% each year. Work out its value after 3 years.
The multiplier is 1−0.20=0.8:
A=18,000×0.83=18,000×0.512=9,216.
So the value is £9,216.
Answer: £9,216. Notice the car loses nearly half its value in just three years — depreciation is steepest early on, which is why a brand-new car is often a poor investment compared with a nearly-new one.
A machine costing £25,000 depreciates by 15% per year. Work out its value after 2 years.
The multiplier is 0.85:
A=25,000×0.852=25,000×0.7225=18,062.50.
So the value is £18,062.50.
Answer: £18,062.50.
A motorbike is bought for £6,000 and loses 30% of its value in the first year and 30% again in the second year. Work out its value after 2 years.
A 30% loss has multiplier 0.7, applied twice:
A=6,000×0.72=6,000×0.49=2,940.
So the value is £2,940.
Answer: £2,940.
Common error: treating "30% then 30%" as a 60% loss, which would give 6,000×0.4=2,400 (£2,400). Because the second year's loss is taken from the already-reduced value, the correct answer of £2,940 is higher.
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