If a price goes up by 20% to £384, why isn't the original £307.20?

Because the 20% was added to the original, not to £384. The original was multiplied by 1.20 to get £384, so you divide £384 by 1.20 to get back. £384 ÷ 1.20 = £320. Subtracting 20% of £384 takes 20% of the wrong amount.

Why can't I just add two percentages together when something falls twice?

Because the second percentage is taken from the already-reduced amount, not from the original. A 10% then 15% fall isn't a 25% fall — it's a multiplier of 0.90 × 0.85 = 0.765, i.e. a 23.5% fall in total. Adding the percentages overstates the loss.

What's a multiplier and why is it the safest way to do percentage change?

A multiplier is the single number you multiply the original by to get the new value. A 20% increase is ×1.20. A 15% decrease is ×0.85. Multipliers always anchor on the original amount, they chain by multiplying, and they reverse by dividing — so they fix all three percentage failure modes at once.

GCSE Maths Foundation · AQA · Percentages

Percentage change: why % of and % off get swapped

Foundation percentage questions almost always involve a change — an increase, a decrease, a reverse-percent, or a chain of both. AQA examiner reports flag the same cluster of errors every series: applying the percentage to the wrong amount, reversing a change by subtracting forwards, and summing percentages across two stages instead of multiplying multipliers. On 2F NOV24 Q24 (a reverse-percent worth several marks) the examiner noted it was "very rare to see a fully correct solution".

The thirty-second fix is to anchor on the original amount and ask one question before you compute anything: is this change a scale (multiply by a multiplier) or a subtraction (take off a literal pound figure)? In every GCSE percentage question except "deposit of £50 off", the answer is "scale".

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How to spot it in your own work

You answered $\pounds 6450$ on a question asking for the value after a 10% then 15% fall from $\pounds 8600$ .
You answered $\pounds 307{,}200$ on a reverse-percent question by subtracting 20% from the post-rise figure.
You treated a 4%-per-year decrease over 5 years as $5 \times 4\% = 20\%$ instead of multiplying by $0.96$ five times.
You took the percentage of whichever number looked easier (e.g. $64\% \text{ of } 20$ instead of $64\% \text{ of } 50$ ).

An exam question that triggers it

Here is a question almost identical to AQA 2F NOV24 Q24, which the examiner report described as having a "very common" wrong answer of £307.20.

After a 20% price increase, the new price of a coat is $\pounds 384$ .

What was the original price?

The wrong answer almost everyone writes is $\pounds 307.20$ . It comes from computing $20\% \text{ of } \pounds 384 = \pounds 76.80$ and then $\pounds 384 - \pounds 76.80 = \pounds 307.20$ . This takes 20% of the wrong amount — it should be 20% of the original, but the original is the unknown.

The correct answer is $\pounds 320$ , found by recognising that the original was multiplied by $1.20$ to reach £384. Divide by $1.20$ to reverse: $\pounds 384 \div 1.20 = \pounds 320$ . Check: $\pounds 320 \times 1.20 = \pounds 384$ . It works.

Why students fall for this

English reads percentage change as a verb ("20% off", "15% on top") and the brain treats the verb as an instruction to do arithmetic on the most visible number. The most visible number is whatever is in the question stem — usually the post-change figure, because that's the one in the scenario. Once the brain grabs that number, it applies the percentage to it and subtracts. The original amount, which is the actual reference, has been quietly forgotten.

The second root is that primary-school percentage problems are almost all "percent of an amount" questions where there is no change at all. The student arrives at GCSE with a deep habit of computing $p\% \text{ of } n$ and subtracting, with no procedural equipment for chaining or reversing. The multiplier is the equipment they were never given.

The fix — Multiplier method

Anchor on "what's the original amount" and ask "is the change a scale or a subtraction?" In nearly every GCSE percentage problem the change is a scale, which means there is a single number — the multiplier — that turns the original into the new amount. A 20% increase is $\times 1.20$ . A 15% decrease is $\times 0.85$ . To find the new value, multiply. To find the original, divide. To chain two changes, multiply the two multipliers together. One technique, every direction.

The multiplier also keeps the original amount permanently in view. When you write $\text{new} = \text{original} \times 1.20$ , both the original and the multiplier are on the page. You can't accidentally take the percentage of the wrong figure because both figures appear in the same equation. It generalises to simple interest, depreciation, compound growth, tax, and any other Foundation percentage scenario.

Worked example

Reverse the 20% rise on the coat: $\text{new} = \pounds 384$ , find the original.

Identify the multiplier. A 20% rise multiplies the original by $1 + 0.20 = 1.20$ .
Write the relationship.
$\text{new} = \text{original} \times 1.20$
Substitute the known value: $\pounds 384 = \text{original} \times 1.20$ .
Solve by dividing. $\text{original} = \pounds 384 \div 1.20 = \pounds 320$ .
Sanity check. $\pounds 320 \times 1.20 = \pounds 384$ . The check recovers the value given in the question, so the answer is right.

For a two-stage chain (a 10% then 15% fall from £8600) the same logic gives $\pounds 8600 \times 0.90 \times 0.85 = \pounds 6579$ , not $\pounds 6450$ . The additive method is off by £129 because it ignores that the second 15% comes off the already-reduced figure.

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Common questions

How do I turn a percentage into a multiplier in one step?: An increase of $p\%$ is the multiplier $1 + \dfrac{p}{100}$ . A decrease of $p\%$ is $1 - \dfrac{p}{100}$ . So 20% up is $1.20$ , 15% down is $0.85$ , 4% down is $0.96$ . Write the multiplier first, then do the arithmetic — never compute "the percentage of the headline figure" and subtract.
If a price falls by 10% then by 15%, why isn't that the same as falling by 25%?: The second percentage is taken from the already-reduced figure, not from the original. The combined multiplier is $0.90 \times 0.85 = 0.765$ , which is a 23.5% fall. AQA examiners specifically flagged this on 3F NOV24 Q16, where the additive wrong answer $\pounds 6450$ came from subtracting both percentages of the original.
How do I know whether to multiply or divide on a reverse-percent question?: If the question gives you the original and asks for the new, multiply by the multiplier. If it gives you the new and asks for the original, divide by the multiplier. The reverse-percent question is the one where the percentage was already applied — your job is to undo it, and division is the undo operation for multiplication.
What if the percentage is given as a decimal (e.g. "0.7") or a fraction?: Convert it to a percentage first. $0.7 = 70\%$ , not $7\%$ . Mixing up the decimal-to-percent bridge is a separate misconception (see decimal place value), but it commonly piggybacks on this one because both involve interpreting a value before computing.

Related misconceptions

Decimal place value— Longer-is-bigger ordering and the 0.7 ≠ 7% bridge — feeds into every percentage question.
Ratio: additive vs multiplicative scaling— Same multiplicative-vs-additive trap, but for ratios instead of percentages.
Fraction additive— The sister misconception in fractions — treating one number as two streams.

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