Is £2000 growing by 10% a year for 2 years worth £2400?

No. After year 1, £2000 grows to £2200. Year 2's 10% is taken on £2200, not on the original £2000, so it adds £220, not £200. The end value is 2000 × 1.1 × 1.1 = £2420. The £2400 answer adds 10% of the original twice and misses the £20 of growth on year 1's growth.

A £8600 laptop is reduced by 15%, then by a further 10%. What is the final price?

£6579. The 10% comes off the already-reduced price, not the £8600, so you multiply the multipliers: 8600 × 0.85 × 0.90 = £6579. Adding the cuts to a flat 25% off gives £6450, which is wrong by £129.

If something grows by 5.1% a year, does it double after 14 years?

Yes — it at least doubles. The multiplier after 14 years is 1.051 to the power 14 = 2.0065, which is at least 2. The simple figure 14 × 5.1 = 71.4% under-counts the growth, because each year's rise is on a bigger amount than the last.

GCSE Maths Foundation · AQA · Percentages

Compound and repeated percentages: why you multiply the multipliers, not add the percentages

Repeated and staged percentage questions trip students who treat the percentages as if they add up. Told £2000 grows by 10% a year for 2 years, they answer £2400 — $2000 + 20\%$ — instead of $2000 \times 1.1 \times 1.1 = 2420$ . Year 2’s 10% is on the bigger £2200, not the original £2000.

The thirty-second fix: each change acts on the previous total, so a repeated or staged percentage change multiplies the multipliers — × 1.1 × 1.1, or × 0.85 × 0.90, or the multiplier to a power. Never just add the percentages.

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How to spot it in your own work

You read two years of 10% growth as a flat +20% — writing $2000 + 20\% = 2400$ instead of $2000 \times 1.1 \times 1.1 = 2420$ .
You added two stage-cuts into one — treating a 15% then 10% reduction as $25\%$ off, rather than $\times\, 0.85 \times 0.90$ .
You multiplied a percentage by the number of years — claiming a 4% yearly fall for 5 years is $20\%$ off, instead of $\times\, 0.96^5$ .
You gave the simple added-percentage value rather than the compound one — £5300 instead of $5000 \times 1.03^2 = 5304.50$ .

An exam question that triggers it

Here is a canonical AQA Foundation trigger (Nov24 P3 Q16 shape, calculator):

A laptop priced at £8600 is reduced by 15% in a sale. The next day the sale price is reduced by a further 10%. Work out the final price.

The misconception answer is $\pounds 6450$ — adding the cuts to 25% and taking 25% off the original £8600.

But the 10% comes off the already-reduced price, so multiply the multipliers: $8600 \times 0.85 \times 0.90 = 6579$ .

Why students fall for this

A percentage change is a multiplication. “Up 10%” means × 1.1; “down 15%” means × 0.85. When a change happens more than once, the second multiplication acts on the output of the first, not on the original. So the multipliers multiply together — they do not collapse into a single added percentage.

The trap hides a scale error. 10% of the grown £2200 is more than 10% of the original £2000, because £2200 is bigger. Adding 10% of the original twice therefore under-counts growth (and, for a fall, over- or under-counts depending on direction). The gap is the change-on-the-change, which compounding captures and adding does not.

AQA Foundation calculator papers exploit every face of this: two different stage-changes (Nov24 P3 Q16), the same change repeated over several years (Jun23 P3 Q18: a 4% yearly fall for 5 years), and “show it at least doubles” growth (Jun24 P3 Q26: 5.1% a year for 14 years).

The fix: Each change acts on the previous total: write one multiplier per stage, then multiply

Two different stages: multiply the two multipliers. A 15% then 10% reduction is $8600 \times 0.85 \times 0.90 = 6579$ . The flat 25%-off answer, £6450, takes both cuts on the original.

The same change repeated: raise the multiplier to a power. A 4% yearly fall for 5 years is $\times\, 0.96^5 = 0.8153726976$ , so a town of 50 000 becomes $50\,000 \times 0.96^5 = 40\,769$ . The simple 20%-off answer is 40 000.

Use the power to test “does it double?” A 5.1% yearly rise for 14 years is $\times\, 1.051^{14} = 2.0065$ , which is at least 2 — so the price has at least doubled, even though $14 \times 5.1 = 71.4\%$ would wrongly suggest it has not.

Worked example

A town has a population of 50 000. Each year the population falls by 4%. What is the population after 5 years?

Write the multiplier for one year. A 4% fall is $100\% - 4\% = 96\%$ , so $\times\, 0.96$ .
$\text{after 1 year} = 50\,000 \times 0.96$
Repeat it 5 times — raise to the power 5.
$50\,000 \times 0.96^5 = 50\,000 \times 0.8153726976$
Compute and round.
$= 40\,768.63\ldots \approx 40\,769$

The trap answer $40\,000$ treats the fall as $4\% \times 5 = 20\%$ off — but that adds the percentages. Each year’s 4% is on a smaller population than the year before, so the change compounds: you raise the multiplier to a power, never multiply the percentage by the number of years.

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Common questions

Why can’t I just add the percentages together?: Because each change acts on the running total, not the original. After £2000 grows 10% to £2200, the next 10% is of £2200 (£220), not of £2000 (£200). Multiplying the multipliers captures that: $2000 \times 1.1 \times 1.1 = 2420$ , while adding to a flat +20% gives £2400 and misses the £20.
What do I do when the same change repeats every year?: Raise the one-year multiplier to the power of the number of years. A 4% fall is × 0.96, so after 5 years it is $\times\, 0.96^5$ ; a 5.1% rise for 14 years is $\times\, 1.051^{14}$ . Use the power key on your calculator.
How do I show a price “at least doubles”?: Work out the multiplier over the whole period and compare it with 2. For 5.1% a year over 14 years, $1.051^{14} = 2.0065$ , which is at least 2 — so the price has at least doubled. The simple $14 \times 5.1 = 71.4\%$ is an under-estimate and would give the wrong conclusion.
Is the simple value ever close enough to the compound value?: They are close for small percentages over short periods — $5000 \times 1.03^2 = 5304.50$ versus the simple $5000 \times 1.06 = 5300$ — but the exam wants the compound value, and the gap grows fast over more stages. Always multiply the multipliers.

Related misconceptions

Percentage change: why the percentage is of the original, not the new valueThe foundational percentages misconception — the same 'percentage is of which figure' confusion, met on a single change rather than a repeated one.
Reverse percentages: why you divide by the multiplier, not take the percentage offThe sibling percentages step — once you trust the multiplier going forwards, you divide by it to go back, and multiply several to compound.

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