POP

G4.1 Compound Interest and Depreciation

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01 — Explanation

What's the difference between compound interest and simple interest?

Simple interest pays you the same amount every year, always calculated on the original amount. Compound interest instead pays interest on the running total — so the interest itself earns interest, and the value grows faster each year. Depreciation is the same idea in reverse: a value shrinking by a fixed percentage each year (cars, laptops, machinery).

02 — Explanation

How do I build the multiplier?

Every compound change question is solved with one formula:

A = P \times M^n

where $P$ is the starting value, $n$ is the number of time periods, and $M$ is the multiplier — the single decimal that represents "the whole new value as a fraction of the old one":

M = 1 + \frac{r}{100} \ \text{(growth)} \qquad M = 1 - \frac{r}{100} \ \text{(decay)}

A 4% rise gives $M = 1.04$ . An 8% depreciation gives $M = 1 - 0.08 = 0.92$ . Writing $P \times M^n$ explicitly (not just the final number) is what secures the method mark — it proves you're using exponential growth, not just adding the same amount every year.

Exam Tip

Always sanity-check $M$ before you calculate: growth means $M>1$ , decay means $0<M<1$ . If your multiplier doesn't match the scenario, you've mixed up the sign.

03 — Explanation

How do I find the total interest earned (or total value lost)?

This is the single most common way marks are dropped — the question asks for the interest, not the final balance. Calculate $A$ first, then subtract:

\text{Total interest} = A - P \qquad \text{Total depreciation} = P - A

Example: £25 000 invested for 3 years at 2.7% compound interest. $M = 1.027$ , so $A = 25000 \times 1.027^3 = £27\,080.17$ . The interest earned is $27\,080.17 - 25\,000 = £2\,080.17$ — giving just the final balance here would lose the accuracy mark.

04 — Explanation

What if the percentage rate changes each year?

Multiply by a different multiplier for each period instead of raising one multiplier to a power:

A = P \times M_1 \times M_2 \times M_3 \times \dots

Example: a house bought for £350,000 rises 3% in Year 1, rises 2% in Year 2, then depreciates 5% in Year 3: $A = 350000 \times 1.03 \times 1.02 \times 0.95 = £349\,324.50$ . Each year's change always applies to the previous year's value, never the original.

05 — Explanation

How do I find the original amount, given the final value?

Rearrange the formula — this is identical to a reverse-percentage problem:

P = \frac{A}{M^n}

If a car is worth £8,000 after 2 years of 10% depreciation, the original price was $8000 \div 0.90^2 = £9\,876.54$ .

06 — Explanation

How do I find how many years it takes to reach a target value?

Finding $n$ requires logarithms — which are Higher Tier only. At Foundation level you solve this by Trial and Improvement: substitute $n=1,2,3,\dots$ into $P \times M^n$ until the value first reaches or exceeds the target, and the winning $n$ is your answer.

Exam Tip

For currency answers, round only at the very end, to 2 decimal places. The most common arithmetic error is writing the multiplier as $1.2$ instead of $1.02$ for a 2% change — always compute $r \div 100$ explicitly before adding or subtracting from 1.

07 — Worked example

Worked example

Hermione buys a boat for \$26,800. Its value depreciates by 8% per year. Find its value after 3 years, and the total depreciation.

Decay, so $M = 1 - 0.08 = 0.92$ .
$A = 26800 \times 0.92^3 = 26800 \times 0.778688 = \$ 20\,875.26$ (2 d.p.).
Total depreciation $= P - A = 26800 - 20875.26 = \$ 5\,924.74$.