GCSE Maths · Number

Percentages GCSE Questions, Worked Examples and Answers

Percentage questions become easier when you identify the original amount, the amount of change and the new result. Learn which value is the 100% base, choose a multiplier or fraction method, then practise original questions for Edexcel, AQA and OCR.

Edexcel · AQA · OCRFoundation & Higher15 original questions

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Keep the three amounts separate

What you need to know about GCSE percentages

If 25 of 100 equal squares are shaded, 25% are shaded: the percent sign means “out of 100”. So 25% = 25/100 = 0.25, or one quarter. The same share applies to any whole: 25% of £80 is one quarter of £80, which is £20. The whole amount is always 100%; we call it the base.

A first example

See the base, change and result

A £100 price increases by 15%. The original amount is £100, the amount of change is £15, and the new result is £115.

Original · 100%£100the base

Change · 15%£15the increase

New · 115%£115the result

The blue bar is the original 100%; the gold extension is the extra 15%.

From problem to method

Typical percentage problems you need to be able to solve

Each problem below states what you may be asked to find, followed by the calculation that solves it.

Percentage of an amount

What the problem asks: “Find 18% of £250.” The whole is given and a part is wanted.

How to solve it: Multiply the whole by the percentage written as a decimal.

One amount as a percentage

What the problem asks: “45 out of 60” or “45 as a percentage of 60”.

How to solve it: Compare part with whole, then multiply by 100.

Increase or decrease

What the problem asks: The original is given and the new result is wanted.

How to solve it: A multiplier includes the original 100% and the change.

Percentage change

What the problem asks: Old and new values are given; the percentage change is wanted.

How to solve it: Find the difference, divide by the original value, then multiply by 100 to express the change as a percentage.

Reverse percentage

What the problem asks: The value after a change is given; the original is missing.

How to solve it: Divide by the forward multiplier to undo it.

Repeated changes

What the problem asks: Look for “each year”, repeated periods or successive changes.

How to solve it: Each stage has a new base, so multiply the multipliers.

Simple interest

What the problem asks: The same percentage of the original is added for each period.

How to solve it: Find one period's interest, multiply by the number of periods, then add it to the original.

Profit, loss or best buy

What the problem asks: A buying price and selling price, or two different offers, must be compared.

How to solve it: Find each actual change or final price before calculating a percentage or choosing the cheaper option.

Build percentages from known parts

Work without a calculator

Start with a percentage you can find exactly, then combine parts. This keeps the meaning visible and gives a quick size check.

Use place value

$\begin{aligned}10\%\text{ of }A&=A\div10\\1\%\text{ of }A&=A\div100\end{aligned}$

Then halve 10% to get 5%, or halve 5% to get 2.5%. For example, 17.5% is 10% + 5% + 2.5%.

Use fraction equivalents

$25\%=\frac14\qquad50\%=\frac12\qquad75\%=\frac34\qquad125\%=\frac54$

A percentage can exceed 100%. For example, 125% of an amount is the whole amount plus one quarter.

The multiplier keeps the 100%

Percentage increase and decrease in one step

Increase by

r\%

\boxed{1+\frac{r}{100}}

A 7% increase uses 1.07: the result is 107% of the original.

Decrease by

r\%

\boxed{1-\frac{r}{100}}

A 7% decrease uses 0.93: 93% of the original remains.

Direction check: a positive increase uses a multiplier above 1. A decrease of less than 100% uses a multiplier between 0 and 1; a 100% decrease uses 0 and leaves nothing.

Three dependable relationships

The formulas describe the base

\text{percentage amount}=\text{original}\times\frac{\text{percentage}}{100}

\text{percentage change}=\frac{\text{amount of change}}{\text{original amount}}\times100\%

\text{new result}=\text{original amount}\times\text{multiplier}

For a reverse percentage, rearrange the last relationship: $\text{original}=\text{new}\div\text{multiplier}$ .

A new base at every stage

Repeated changes and compound interest

Apply each multiplier to the latest value. If the same change repeats $n$ times, a power abbreviates the multiplication.

Start500

After 1520

After 2540.80

500\times1.04\times1.04=500\times1.04^2=540.80

Compound interest is repeated percentage increase on a changing balance. The same structure models depreciation and population growth or decay.

Fully worked

Percentages GCSE worked examples

Each example names the clue, the 100% base and the reason the method fits.

Example 1

Find a percentage of an amount

2 marks

Question

Find $17.5\%$ of $240$ .

Recognise it: “of” asks for a part of the given whole. Without a calculator, build $17.5\%$ from percentages that are easy to find.

$10\%\text{ of }240=24$

$5\%\text{ of }240=12$

$2.5\%\text{ of }240=6$

$17.5\%\text{ of }240=24+12+6=42$

\boxed{42}

Sense check: $17.5\%$ is less than one fifth, so the answer should be a little below $48$ .

Example 2

Express one quantity as a percentage of another

2 marks

Question

A learner answers $54$ of $72$ questions correctly. Express this as a percentage.

Recognise it: there are two quantities and the wording asks for one as a percentage of the other. The total $72$ is the whole, so it belongs underneath the fraction.

\frac{\text{part}}{\text{whole}}=\frac{54}{72}

\frac{54}{72}\times100=75

\boxed{75\%}

Base check: “54 out of 72” means $54\div72$ , not $72\div54$ .

Example 3

Separate a discount from the sale price

3 marks

Question

A coat costs £160 and is reduced by $15\%$ . Find the amount of the discount and the sale price.

Recognise it: the original price is known. The question asks for both the change and the new result, so label them separately.

The amount of change is

160\times0.15=24

\boxed{\text{Discount}=\text{£}24}

The new result is

160\times0.85=136

\boxed{\text{Sale price}=\text{£}136}

“Discount” means the £24 removed; “sale price” means the £136 left.

Example 4

Find percentage change

3 marks

Question

A ticket price rises from £84 to £105. Calculate the percentage increase.

Recognise it: old and new values are both given. Find their difference, then compare that difference with the original £84.

\text{change}=105-84=21

\text{percentage increase}=\frac{21}{84}\times100

\boxed{25\%\text{ increase}}

The percentage describes how much £84 changed, so £84 — not £105 — is the base.

Example 5

Use a multiplier to add tax

3 marks

Question

A service costs £250 before tax. Tax is $20\%$ . Find the price after tax.

Recognise it: the original is given and the result after an increase is wanted. A multiplier gives the result in one step.

100\%+20\%=120\%

120\%=1.20

250\times1.20=300

\boxed{\text{Price after tax}=\text{£}300}

Multiplier $0.20$ would find only the £50 tax; $1.20$ finds the £300 total.

Example 6

Reverse a percentage decrease

Harder3 marks

Question

A bicycle costs £204 after a $15\%$ discount. Find its original price.

Recognise it: £204 is the result after a change and the original is missing. The result is $85\%$ of the original, so undo multiplication by $0.85$ .

100\%-15\%=85\%

204=\text{original}\times0.85

\text{original}=204\div0.85

\boxed{\text{Original price}=\text{£}240}

Check: $15\%$ of £240 is £36, and $240-36=204$ .

Example 7

Combine different percentage changes

Harder4 marks

Question

A £1,800 asset rises by $4\%$ , then falls by $6\%$ . Find its final value and overall percentage change.

Recognise it: there are two successive changes. The $6\%$ fall is based on the value after the $4\%$ rise, so multiply the two multipliers.

\text{final}=1800\times1.04\times0.94

\text{final}=1759.68

\text{change}=1800-1759.68=40.32

\frac{40.32}{1800}\times100=2.24

\boxed{\text{£}1{,}759.68\text{, an overall }2.24\%\text{ decrease}}

Do not combine $4\%$ and $-6\%$ as $-2\%$ : their bases are different.

Example 8

Connect repeated change to compound interest

3 marks

Question

£950 is invested at $3.2\%$ compound interest per year for 3 years. Find the balance.

Recognise it: “per year for 3 years” means the same increase happens three times. Use power $3$ to repeat multiplier $1.032$ .

\text{balance}=950\times1.032^3

\text{balance}=1044.1495296

\boxed{\text{Balance}=\text{£}1{,}044.15}

Keep the full calculator value until all three increases have been applied.

15 original questions · total 44 marks

Percentages GCSE exam-style questions

Try each question before opening its fully worked answer. The set moves from direct skills to reverse and multi-step contexts.

Before you startAllow about 45 minutes · show the 100% base and each multiplier · answers start collapsed

Find a percentage without a calculator

2 marks

Find $17.5\%$ of $360$ without a calculator.

Show worked answer

Split $17.5\%$ into $10\%+5\%+2.5\%$ .

$10\%=36,\qquad5\%=18,\qquad2.5\%=9$

$36+18+9=\boxed{63}$

Express one amount as a percentage

2 marks

Express $48$ as a percentage of $160$ .

Show worked answer

The amount being compared is $48$ and the whole, or base, is $160$ .

\frac{48}{160}\times100=30

\boxed{30\%}

Use a fraction equivalent for a percentage

2 marks

Work out $125\%$ of $64$ without a calculator.

Show worked answer

The percentage is greater than $100\%$ , so the answer must be greater than $64$ . Use $125\%=\frac54$ .

$64\times\frac54=16\times5$

$\boxed{80}$

Calculate a sale price

2 marks

A bag costs £85. It is reduced by $30\%$ . Work out the sale price.

Show worked answer

A $30\%$ reduction leaves $70\%$ of the original price.

85\times0.70=59.5

Write money with two decimal places.

\boxed{\text{Sale price}=\text{£}59.50}

Find a percentage increase

3 marks

The number of members in a club rises from $320$ to $368$ . Calculate the percentage increase.

Show worked answer

First find the amount of change.

368-320=48

The original $320$ is the base.

\frac{48}{320}\times100=15

\boxed{15\%\text{ increase}}

Use a quantity as the base

2 marks

A water tank holds $84$ litres when full. It currently contains $63$ litres. What percentage full is the tank?

Show worked answer

The full capacity, $84$ litres, represents $100\%$ .

\frac{63}{84}\times100=75

\boxed{75\%\text{ full}}

Calculate simple interest

3 marks

£800 is invested at $4\%$ simple interest per year for 3 years. Find the final balance.

Show worked answer

Simple interest is calculated from the original £800 every year.

$\text{interest each year}=800\times0.04=32$

$\text{interest for 3 years}=32\times3=96$

$\text{final balance}=800+96=\boxed{\text{£}896}$

Reverse a discount

Harder3 marks

A pair of trainers costs £93 after a $25\%$ discount. Find the original price.

Show worked answer

The £93 is the new result. It represents $75\%$ of the original price.

75\%=0.75

\text{original}=93\div0.75

\boxed{\text{Original price}=\text{£}124}

Check: $124\times0.75=93$ .

Reverse an increase

Harder3 marks

After an $8\%$ increase, a fee is £286.20. Work out the fee before the increase.

Show worked answer

The final £286.20 represents $108\%$ of the original fee.

\text{original}=286.20\div1.08

\boxed{\text{Original fee}=\text{£}265}

Dividing reverses the earlier multiplication. Subtracting $8\%$ of £286.20 would use the wrong base.

Apply repeated percentage growth

3 marks

A population of bacteria starts at $1{,}250$ and grows by $6\%$ each hour. Find the population after 3 hours. Give your answer to the nearest whole number.

Show worked answer

Each hour uses the new population as its base, so apply $1.06$ three times.

1250\times1.06^3=1488.77

Round only after all three changes.

\boxed{1489\text{ bacteria}}

Combine an increase and a decrease

Harder4 marks

A machine worth £800 falls in value by $15\%$ , then rises in value by $10\%$ . Find its final value and the overall percentage change from £800.

Show worked answer

Use a separate multiplier for each change.

\text{final}=800\times0.85\times1.10=748

\text{decrease}=800-748=52

\frac{52}{800}\times100=6.5

\boxed{\text{Final value}=\text{£}748}

\boxed{\text{Overall change}=6.5\%\text{ decrease}}

Compare two offers

4 marks

Headphones have a marked price of £240. Shop A offers $25\%$ off. Shop B offers £45 off. Which shop is cheaper, and by how much?

Show worked answer

Compare the final prices, not the labels on the offers.

$\text{Shop A}=240\times0.75=180$

$\text{Shop B}=240-45=195$

$195-180=15$

$\boxed{\text{Shop A is cheaper by £15}}$

Remove VAT from a total

Harder3 marks

A repair bill is £75.60 including VAT at $5\%$ . Find the price before VAT.

Show worked answer

The total includes a $5\%$ increase, so £75.60 represents $105\%$ .

105\%=1.05

\text{price before VAT}=75.60\div1.05

\boxed{\text{Price before VAT}=\text{£}72}

Calculate percentage profit

3 marks

A trader buys an item for £48 and sells it for £60. Calculate the percentage profit.

Show worked answer

Profit is measured against the buying price, because £48 is the original amount.

$\text{profit}=60-48=12$

$\frac{12}{48}\times100=25$

$\boxed{25\%\text{ profit}}$

Interpret two different changes

Harder5 marks

A reservoir contains $640$ m³ of water. Rain increases the amount by $12.5\%$ . Then $90$ m³ is released. Calculate the overall percentage change from the original $640$ m³. Give your answer to $1$ decimal place.

Show worked answer

The first change is a percentage, so use a multiplier. The second is a fixed amount, so subtract $90$ .

\text{after rain}=640\times1.125=720

\text{after release}=720-90=630

The final amount is $10$ m³ below the original.

\frac{10}{640}\times100=1.5625

\boxed{1.6\%\text{ decrease (to 1 d.p.)}}

Examiner-style feedback

Common percentage mistakes

Using the wrong base

Percentage change is measured against the original amount. Put the change over the old value, not the new result.

Confusing change with result

For £80 reduced by 25%, £20 is the amount removed and £60 is the new result. Match your answer to the noun.

Subtracting in a reverse question

If £72 is the price after 10% off, £72 represents 90%. Divide by 0.90; do not add 10% of £72.

Building the wrong multiplier

A 7% increase uses 1.07 and a 7% decrease uses 0.93. Using 0.07 gives only the change.

Rounding too early

Keep the full calculator display through repeated changes. Round only after the final operation.

Adding repeated rates

A 10% fall followed by a 10% rise does not return to the start because the second 10% has a different base.

30-SECOND RECAP

Base, operation, answer

Before using the calculator, name the original amount, the change and the new result.

Circle what the question gives and underline what it asks for.
Decide which amount represents the original 100%.
Multiply to move forwards; divide by the multiplier to reverse.
State whether the answer is a change, an original or a new result.

Quick answers

Percentages FAQ

How do I find a percentage of an amount?

Convert the percentage to a decimal by dividing by 100, then multiply by the whole amount.

How do I express one quantity as a percentage of another?

Divide the quantity being compared by the original or whole quantity, then multiply by 100.

What multiplier gives a percentage increase or decrease?

For an increase of r%, use 1 + r/100. For a decrease of r%, use 1 − r/100.

How do reverse percentages work?

Identify what percentage the final amount represents, convert it to a multiplier, then divide by that multiplier.

Why do repeated percentage changes not usually add together?

Each percentage is applied to the latest value, so the base can change at every step.

How can I find a percentage without a calculator?

Build it from easy percentages: 10% means divide by 10, 5% is half of 10%, and 1% means divide by 100. Useful equivalents include 25% = 1/4, 50% = 1/2 and 75% = 3/4.

What is the difference between simple and compound interest?

Simple interest adds the same percentage of the original amount each period. Compound interest applies the percentage to the changing balance, so the amount added can change.

Build connected skills

What to revise next

Strengthen the base

Ratio

Use ratios and fractions to compare parts with a whole.

Revise ratio

Use repeated growth

Compound interest

Apply the same multiplier to a changing balance over several periods.

Revise compound interest

Generalise the model

Growth and decay

Model repeated change, then Higher iterative processes with deposits and repayments.

Revise growth and decay →

Content standards

Curriculum and rights review

Curriculum references checked 4 September 2026. Percentage calculations, percentage change and original-value problems are included in the shared GCSE Maths content used by Edexcel, AQA and OCR. All questions, values, contexts and solution wording are original Pass an Exam content.

Official specification references

Percentages GCSE Questions, Worked Examples and Answers

Practise GCSE percentages for free with an AI tutor

What you need to know about GCSE percentages

See the base, change and result

Typical percentage problems you need to be able to solve

Work without a calculator

Percentage increase and decrease in one step

The formulas describe the base

Repeated changes and compound interest

Percentages GCSE worked examples

Percentages GCSE exam-style questions

Common percentage mistakes

Base, operation, answer

Percentages FAQ

What to revise next

Ratio

Compound interest

Growth and decay

Continue with Higher exam preparation

Curriculum and rights review