Reverse percentages Revision Notes for Edexcel GCSE Maths

Reverse percentages

What are reverse percentages?

Reverse percentages are used when you know the final amount after a percentage increase or decrease, but need to find the original amount. This is the opposite of normal percentage calculations.

infoNote

You can identify reverse percentage problems when:

You're given the amount after a percentage change
You need to find the original amount before the change
The question asks "what was the original price/amount?"

Understanding multipliers

To solve reverse percentage problems, you need to understand multipliers.

For percentage decreases

Subtract the percentage from 100%
Convert to a decimal by dividing by 100
Example: 20% decrease → 100% - 20% = 80% → $80 \div 100 = 0.8$

For percentage increases

Add the percentage to 100%
Convert to a decimal by dividing by 100
Example: 5% increase → 100% + 5% = 105% → $105 \div 100 = 1.05$

The reverse percentage method

When you have the final amount and need the original amount:

chatImportant

Divide the final amount by the multiplier

$\text{Original Amount} = \frac{\text{Final Amount}}{\text{Multiplier}}$

infoNote

Step-by-step process:

Identify whether it's an increase or decrease
Calculate the multiplier
Divide the given amount by the multiplier
Check your answer by working forwards

Worked examples

lightbulbExample

Worked Example 1: Price reduction

A sweater's sale price is £40 after a 20% reduction.

Solution:

This is a 20% decrease
Multiplier: 100% - 20% = 80% = 0.8
Original price: $£40 \div 0.8 = £50$

Check: $£50 \times 0.8 = £40$ ✓

lightbulbExample

Worked Example 2: Temperature increase

The temperature after a 5% increase is 23.1°C.

Solution:

This is a 5% increase
Multiplier: 100% + 5% = 105% = 1.05
Original temperature: $23.1°C \div 1.05 = 22°C$

lightbulbExample

Worked Example 3: Trainer sale

Trainers cost £75.65 in a 15% off sale.

Solution:

Calculate multiplier: 100% - 15% = 85% = 0.85
Find original price: $£75.65 \div 0.85 = £89$
Check: $£89 \times 0.15 = £13.35$ reduction, $£89 - £13.35 = £75.65$ ✓

lightbulbExample

Worked Example 4: Height comparison

Paul's height in 2014 was 154cm after a 10% increase since 2013. Amy's height in 2014 was 150.8cm after a 4% increase since 2013.

Solutions:

Paul's 2013 height: $154 \div 1.1 = 140cm$
Amy's 2013 height: $150.8 \div 1.04 = 145cm$
Conclusion: Amy was taller than Paul by 5cm in 2013

Practice problems

lightbulbExample

Practice Problem 1

Hannah buys shoes in a sale marked "40% off" for £27. What was the original price?

Method:

40% decrease → multiplier = 0.6
Original price = $£27 \div 0.6 = £45$

lightbulbExample

Practice Problem 2

Jared's house increased in value by 8% from 2010 to 2012. In 2012, it was worth £237,600. What did he originally pay?

Method:

8% increase → multiplier = 1.08
Original price = $£237,600 \div 1.08 = £220,000$

Exam tips

chatImportant

Essential Exam Tips:

Always identify whether the change is an increase or decrease
Remember to convert percentages to decimals for multipliers
Check your answer by working forwards
Show clear working in exam questions
Read the question carefully to identify what you're finding

bookmarkSummary

Key Points to Remember:

Reverse percentages help you find original amounts when given the final amount after a percentage change
Calculate the multiplier by adding to or subtracting from 100%, then converting to a decimal
Divide by the multiplier to find the original amount - this reverses the percentage operation
Always check your answer by working forwards to verify your solution
Practice identifying whether you're dealing with an increase or decrease from the question context

Reverse percentages (Edexcel GCSE Maths): Revision Notes