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Ratio, proportion & rates of change · Percentages & growth

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Reverse percentages

How to find the original amount before a percentage change, and why you must divide by the multiplier rather than apply the percentage to the new amount.

Ratio, proportion & rates of change · Percentages & growth

Reverse percentages

How to find the original amount before a percentage change, and why you must divide by the multiplier rather than apply the percentage to the new amount.

Why it works

A reverse percentage question gives you the amount after a change and asks for the amount before it. The whole difficulty is a single trap: the percentage in the question is a percentage of the original, which is exactly the number you don't know yet — so you must not apply it to the number you do know.

Think of the forward direction first. If the original is xx and it goes up by 20%20\%, the multiplier is 1.21.2 and the new amount is

1.2×x=new amount.1.2 \times x = \text{new amount}.

That is a simple equation with one unknown. To undo a multiplication you divide, so

x=new amount1.2.x = \frac{\text{new amount}}{1.2}.

That's the entire method: *identify the multiplier, then divide the given amount by it.* Nothing else.

The trap, made concrete. A jacket costs £48\pounds 48 in a 20%20\%-off sale; find the original price. The wrong-but-tempting move is "add the 20%20\% back": 48+0.2×48=£57.6048 + 0.2\times 48 = \pounds 57.60. But the 20%20\% was 20%20\% of the original price, which was bigger than £48\pounds 48 — so 20%20\% of it is more than 20%20\% of 4848. Applying the percentage to the sale price takes a percentage of the wrong number. The multiplier for 20%20\% off is 0.80.8, so

original=480.8=£60.\text{original} = \frac{48}{0.8} = \pounds 60.

Check it forward: 0.8×60=480.8 \times 60 = 48. ✓ (And notice 60≠57.6060 \neq 57.60 — the tempting method is genuinely wrong, not just untidy.)

How to tell reverse from ordinary percentages. The signal is that the number you're given is the one after the change, and the change is stated as a percentage — "after a 5%5\% pay rise she earns £31 500\pounds 31\,500", "a price includes 20%20\% VAT", "reduced by 30%30\% to £21\pounds 21". In every case the amount you have is the multiplier times the amount you want, so you divide.
  • After a p%p\% increase: original =new1+p/100= \dfrac{\text{new}}{1 + p/100}.
  • After a p%p\% decrease: original =new1−p/100= \dfrac{\text{new}}{1 - p/100}.