IGCSE Compound Interest and Depreciation: Why the Multiplier Matters

Primary audience and intent: Cambridge IGCSE Mathematics students and parents looking for a concept-first explanation of repeated percentage change; not a general overview of IGCSE tiers or a simple ratio lesson.

Original illustration brief: A two-panel vector infographic. Left: an £8,000 balance rising by 4% annually over three years, with labels £8,000 → £8,320 → £8,652.80 → £8,998.91. Right: a £15,000 asset falling by 18% annually over two years, £15,000 → £12,300 → £10,086. Add the multipliers ×1.04 and ×0.82 over the arrows. Alt text: “Compound growth and depreciation shown as repeated percentage multipliers over successive years.” Copyright/licensing: Draw the graphic from these original figures using self-created shapes and type; no copied textbook diagrams, exam questions, third-party logos or unlicensed stock artwork.

Why does the same percentage produce a different change each year?

Suppose a savings account starts with £8,000 and increases by 4% per year. The first year's increase is 4% of £8,000, or £320. The new balance is £8,320.

The second year's increase is not another £320. It is 4% of the new £8,320 balance: £332.80. That is the central idea of compound growth: each percentage is calculated on the amount remaining at the start of that period.

A useful habit is to ask: “What is the new base amount?” If a student cannot answer that, memorising a formula will not fix the misconception.

Step 1: Turn a percentage into a multiplier

For a 4% increase, keep 100% of the original amount and add 4%:

$100\%+4\%=104\%=1.04$.

Multiply the current amount by 1.04 for each 4% increase. For an 18% decrease, keep 82% of the current amount:

$100\%-18\%=82\%=0.82$.

Multiply by 0.82 for each decrease.

This method works because percentages can be written as decimals. For example, 104% is 104 ÷ 100 = 1.04.

Worked example 1: Compound growth over three years

Question: An investment of £8,000 grows by 4% at the end of each year. What is its value after three years?

Year 1: £8,000 × 1.04 = £8,320.

Year 2: £8,320 × 1.04 = £8,652.80.

Year 3: £8,652.80 × 1.04 = £8,998.912.

To the nearest penny, the final value is £8,998.91.

Instead of writing each year separately, use the repeated multiplier:

$$A=P(1+r/100)^n$$

Here $P=8000$, $r=4$ and $n=3$, so $A=8000(1.04)^3$.

The exponent 3 counts the number of growth periods, not the percentage rate. This formula assumes the rate is applied once per year, as the question states. Different compounding frequencies need a different period rate and number of periods.

Worked example 2: Depreciation of equipment

Question: A machine is worth £15,000. Its value falls by 18% each year. Find its value after two years.

A decrease of 18% leaves 82%, so the multiplier is 0.82.

Year 1: £15,000 × 0.82 = £12,300.

Year 2: £12,300 × 0.82 = £10,086.

Or calculate £15,000 × (0.82)² = £10,086.

Common error: Subtracting 36% of the original £15,000 gives £9,600. That assumes the same original base in both years and therefore does not model compound depreciation.

Worked example 3: Find the starting amount (reverse growth)

Question: After two annual increases of 5%, an amount is £5,292. What was it before the increases?

The total multiplier is $(1.05)^2=1.1025$. To reverse a multiplication, divide:

$$\text{Original amount}=5292\div1.1025=4800.$$

Answer: £4,800. Check by moving forwards: £4,800 × 1.05 = £5,040; £5,040 × 1.05 = £5,292.

Reverse percentage questions are not solved by subtracting the stated percentage from the final amount. The final amount is already measured on a different base.

Compound interest versus simple interest

The distinction becomes visible even over two years. On £2,500 at 6% per year:

• Simple interest: £150 each year, giving £2,500 + £300 = £2,800 after two years.

• Compound interest: £2,500 × 1.06² = £2,809 after two years.

The £9 difference is the interest earned on the first year's £150 increase. Both methods are mathematically valid, but the wording of the question determines which model applies.

A subtle trap: up 12%, then down 12%

A shop raises a price by 12%, then reduces the new price by 12%. Does the price return to where it started?

Take a £100 item. After the increase: £100 × 1.12 = £112. After the decrease: £112 × 0.88 = £98.56.

The final price is 1.44% below the original. The two percentages do not cancel because they are applied to different bases. Multipliers reveal the change directly: $1.12\times0.88=0.9856$.

This idea also helps students reason about discounts, population change and repeated adjustments without assuming that equal percentage changes reverse one another.

How to decide which calculation to use

Read for three details before calculating: the starting value, the percentage change per period, and the number of periods. Then decide whether the question asks for the final amount or the original amount.

If the value increases, use a multiplier greater than 1. If it decreases, use one between 0 and 1. If the final amount is given and the starting amount is unknown, divide by the total multiplier. Finally, estimate whether the result makes sense: growth should produce a larger positive amount, and depreciation should produce a smaller one.

For financial examples, retain full precision during calculation and round to currency units at the end unless the question gives different instructions.

Four original practice questions

1. A balance of £1,200 grows by 3% each year for two years. Find the final amount to the nearest penny.

2. A vehicle worth £9,000 loses 10% of its value annually. What is it worth after three years?

3. After two increases of 5%, a sum becomes £6,615. What was the original sum?

4. A product costs £3,000. Its price rises by 8% in one year and 5% in the next. Find its final price.

Answers and reasoning

1. £1,200 × 1.03² = £1,273.08.

2. £9,000 × 0.9³ = £6,561.

3. £6,615 ÷ 1.05² = £6,000. Check: £6,000 × 1.1025 = £6,615.

4. £3,000 × 1.08 × 1.05 = £3,402. The rates differ, so use two distinct multipliers rather than squaring one.

What parents can ask instead of “Did you get the answer?”

Ask the student to identify the percentage remaining after each change, explain why the second year's base is different, and estimate whether the answer should rise or fall. Then ask them to reverse the calculation as a check.

A focused diagnostic exercise can separate three difficulties: converting percentages to multipliers, using powers correctly, and deciding whether to multiply or divide. Those are more useful targets for practice than simply repeating twenty similar questions.

Curriculum and source note (checked 27 October 2026): Cambridge IGCSE Mathematics (0580) syllabus for examinations 2025–2027 includes exponential growth and decay, such as depreciation and population change, under Extended E1.17. The worked examples above are original and not Cambridge examination questions. See Cambridge official syllabus. For a separate UK GCSE comparison, see AQA GCSE Mathematics formulae: compound interest.

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