Reverse Percentages

Grade 5

What even is a reverse percentage?

Normal percentage questions give you the starting amount and ask for the new one. Reverse percentage questions flip that: you're given the amount after a change, and you have to find where it started.

The classic trap is to just take the percentage off the number you're given. Don't. That number isn't the original — so a percentage of it isn't the right size. We need a proper method.

The one idea that makes it click

Every reverse percentage question comes down to this: the amount you're given represents a certain percentage of the original. Your job is to get back to 100%.

Find what 1% is worth first. Once you know 1%, you can build up to 100% — which is the original amount.

To find the percentage your number represents:

A rise of 20% means your number is 100 + 20 = 120% of the original. A fall of 20% means it's 100 − 20 = 80%.

Worked example — an increase

EXAMPLE

A jacket costs £60 in the sale after a 20% price rise from the shop. What was the original price?

The £60 is after a 20% rise, so it's 120% of the original. Find 1%: 60 ÷ 120 = £0.50. Now scale up to 100%: 0.50 × 100 =£50. Quick check: 20% of £50 is £10, and £50 + £10 = £60. ✓

Worked example — a decrease

EXAMPLE

After a 15% discount, a game costs £34. What was the price before the discount?

A 15% discount means £34 is 85% of the original. Find 1%: 34 ÷ 85 = £0.40. Scale up: 0.40 × 100 =£40. Check: 15% of £40 is £6, and £40 − £6 = £34. ✓

Don't get caught out

If a question says a price "increased by 20% to £60", the £60 is the new price — that's a reverse percentage. But if it says "increased by 20%, what is the new price" and gives you the original, that's a normal forward one. Read which amount they've actually handed you.

Try this one

A car depreciates in value by 20% each year.

At the end of three years, the car is worth £5120.

(a) Show that the original price of the car was £10 000. [3]

(b) By what overall percentage did the car depreciate over the three years? [2]

Nova's hint:

Part (a) uses reverse compound percentage: undo each year's 20% decrease by dividing by the multiplier three times (or divide by the multiplier cubed).

Ask yourself: 'What single multiplier represents one year's 20% decrease, and how do I raise a multiplier to a power of 3?' For example, two 10% decreases in succession use multiplier \( 0.9^2 \).

Part (b) compares the total change to the original using \( \frac{\text{change}}{\text{original}} \times 100 \).

Try to Solve it with Nova?