Fractions

Visual model

Fractions are equal parts of a whole

\frac{3}{5}\text{ shaded}

3 out of 5 equal parts

Count equal parts first.

The shaded part is the numerator.

The total number of parts is the denominator.

Gold-standard guide

22 mins

What you will learn

I can find equivalent fractions and simplify them.

I can add and subtract fractions using a common denominator.

I can multiply fractions and simplify the result.

I can divide fractions by flipping the second and multiplying.

I can solve multi-step fraction problems in context.

Useful before you start

Times tables (up to 12×12)Highest common factor (HCF)Lowest common multiple (LCM)Basic division of integersMultiplying and dividing integers

Core knowledge

Know the rule, then use it

These are the short notes. Read each one, then check you can use it in the worked example below.

Method

Numerator

The top number in a fraction — how many parts you have (e.g. in $\frac{3}{5}$ , the 3).

Step 1

Denominator

The bottom number — how many equal parts make one whole (e.g. in $\frac{3}{5}$ , the 5).

Step 2

Equivalent fractions

Fractions with the same value — multiply or divide top and bottom by the same number (e.g. $\frac{2}{3}$ = $\frac{4}{6}$ ).

Step 3

Simplest form (lowest terms)

When top and bottom share no common factor other than 1 (e.g. $\frac{6}{8}$ → $\frac{3}{4}$ ).

Step 4

Improper fraction

A fraction where the top is bigger than the bottom — its value is more than 1 (e.g. $\frac{7}{4}$ ).

Step 5

Reciprocal

Flip a fraction to get its reciprocal (e.g. $\frac{3}{4}$ → $\frac{4}{3}$ ); used when dividing fractions.

Add or subtract: \frac{a}{b} \pm \frac{c}{d} = (ad \pm bc) / bd

Rewrite both fractions with the same denominator (use the LCM for efficiency), then add or subtract the numerators only.Never touch the denominator during this step.

Multiply: \frac{a}{b} \times \frac{c}{d} = a\frac{c}{bd}

Multiply the two numerators to get the new numerator, then multiply the two denominators to get the new denominator.Cross-cancel common factors before multiplying to keep numbers small.

Divide: \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}

Keep the first fraction unchanged, change ÷ to ×, then flip the second fraction (write its reciprocal).Then multiply as normal.

Simplify: \frac{a}{b} = (a\div HCF) / (b\div HCF)

Divide both numerator and denominator by their highest common factor.Always check your final answer is fully simplified.

Worked example

A container is $\frac{5}{6}$ full of water. $\frac{2}{9}$ of the container's total capacity is used for cooking. What fraction of the container remains after cooking?

The fraction remaining = $\frac{5}{6}$ − $\frac{2}{9}$ .

Find the LCM of 6 and 9: LCM = 18.

Convert: $\frac{5}{6}$ = $1\frac{5}{1}8$ and $\frac{2}{9}$ = $\frac{4}{1}8$ .

Subtract: $1\frac{5}{1}8$ − $\frac{4}{1}8$ = $1\frac{1}{1}8$ .

Check $1\frac{1}{1}8$ is in its simplest form: HCF(11, 18) = 1. ✓

Final answer

$1\frac{1}{1}8$ of the container remains.

Question ladder

Build up to the hardest questions

Do them in order. If you miss a step, read the solution, then redo the question without looking.

Diagnosticrecall

Simplify $1\frac{8}{2}4$ fully.

1 mark1 minfrac-d1

Show solution

Worked solution

1.Find the HCF of 18 and 24.
2.HCF(18, 24) = 6.
3.Divide both numerator and denominator by 6.

Final answer

$\frac{3}{4}$

Mark points

B1: correctly simplified fraction $\frac{3}{4}$ (accept method showing division by any common factor leading to $\frac{3}{4}$ ).

Watch out

g. giving $\frac{9}{1}2$ rather than simplifying fully).

Easyprocedure

Work out $\frac{3}{8}$ + $\frac{5}{1}2$ . Give your answer in its simplest form.

2 marks2 minsfrac-e1

Show solution

Worked solution

1.Find the LCM of 8 and 12: LCM = 24.
2. $\frac{3}{8}$ = $\frac{9}{2}4$ and $\frac{5}{1}2$ = $1\frac{0}{2}4$ .
3.Add the numerators: $\frac{9}{2}4$ + $1\frac{0}{2}4$ = $1\frac{9}{2}4$ .
4.Check: HCF(19, 24) = 1 so $1\frac{9}{2}4$ is already in simplest form.

Final answer

$1\frac{9}{2}4$

Mark points

M1: correct common denominator of 24 seen (or any correct equivalent pair of fractions with matching denominators).
A1: $1\frac{9}{2}4$ (accept unsimplified equivalent only if M1 earned and final answer is correct).

Watch out

Adding the denominators (writing $\frac{8}{2}0$ or similar), or using a common denominator of 96 and failing to simplify correctly.

Mediumreasoning

A recipe needs $\frac{3}{4}$ of a cup of sugar. Tom wants to make $\frac{2}{3}$ of the full recipe. How much sugar does he need? Show your working.

3 marks3 minsfrac-m1

Show solution

Worked solution

1.He needs $\frac{2}{3}$ of $\frac{3}{4}$ cups.
2.'Of' means multiply: $\frac{2}{3}$ × $\frac{3}{4}$ .
3.Multiply numerators: 2 × 3 = 6. Multiply denominators: 3 × 4 = 12.
4.Simplify $\frac{6}{1}2$ = $\frac{1}{2}$ .

Final answer

$\frac{1}{2}$ cup

Mark points

M1: forms the product $\frac{2}{3}$ × $\frac{3}{4}$ (or equivalent multiplication). Award even if unsimplified.
M1: correct multiplication to obtain $\frac{6}{1}2$ or equivalent.
A1: $\frac{1}{2}$ (fully simplified).

Watch out

Interpreting ' $\frac{2}{3}$ of the recipe' as subtracting $\frac{2}{3}$ from $\frac{3}{4}$ rather than multiplying, giving $\frac{3}{4}$ − $\frac{2}{3}$ = $\frac{1}{1}2$ .

Hardreasoning

A tank is $\frac{7}{1}0$ full. After 21 litres are removed, it is $\frac{2}{5}$ full. Find the capacity of the tank in litres.

4 marks4 minsfrac-h1

Show solution

Worked solution

1.Find the fraction removed: $\frac{7}{1}0$ − $\frac{2}{5}$ .
2.Convert $\frac{2}{5}$ to tenths: $\frac{2}{5}$ = $\frac{4}{1}0$ .
3.Fraction removed = $\frac{7}{1}0$ − $\frac{4}{1}0$ = $\frac{3}{1}0$ .
4. $\frac{3}{1}0$ of the capacity = 21 litres, so full capacity = 21 ÷ ( $\frac{3}{1}0$ ) = 21 × $1\frac{0}{3}$ = 70 litres.

Final answer

70 litres

Mark points

M1: correct subtraction $\frac{7}{1}0$ − $\frac{2}{5}$ with a valid common denominator shown.
M1: obtains $\frac{3}{1}0$ as the fraction removed (accept equivalent fraction).
M1: sets up $\frac{3}{1}0$ × capacity = 21 (or equivalent inverse operation).
A1: 70 litres.

Watch out

Subtracting 21 from the fraction (treating litres and fractions as the same thing), or dividing 21 by $\frac{3}{1}0$ without inverting (giving 6.3).

Very Hardproblem solving

Alice spends $\frac{1}{3}$ of her savings on a phone and $\frac{3}{8}$ of what remains on headphones. She then has £175 left. How much did she start with?

5 marks5 minsfrac-vh1

Show solution

Worked solution

1.After buying the phone, Alice has 1 − $\frac{1}{3}$ = $\frac{2}{3}$ of her original savings.
2.She spends $\frac{3}{8}$ of $\frac{2}{3}$ on headphones: $\frac{3}{8}$ × $\frac{2}{3}$ = $\frac{6}{2}4$ = $\frac{1}{4}$ of original savings.
3.Amount remaining = original − $\frac{1}{3}$ − $\frac{1}{4}$ = 1 − $\frac{4}{1}2$ − $\frac{3}{1}2$ = $\frac{5}{1}2$ .
4. $\frac{5}{1}2$ of savings = £175, so total savings = £175 ÷ ( $\frac{5}{1}2$ ) = £175 × $1\frac{2}{5}$ = £420.

Final answer

£420

Mark points

M1: correctly identifies the remaining fraction after phone purchase as $\frac{2}{3}$ .
M1: correctly calculates $\frac{3}{8}$ of $\frac{2}{3}$ = $\frac{1}{4}$ (accept any equivalent fraction).
M1: combines fractions to find the fraction remaining is $\frac{5}{1}2$ .
M1: sets up £175 = $\frac{5}{1}2$ × total (or inverse: total = 175 ÷ ( $\frac{5}{1}2$ )).
A1: £420.

Watch out

Calculating headphone cost as $\frac{3}{8}$ of the original savings (forgetting 'of what remains'), which gives the wrong remaining fraction.

Grade 9 stretchproblem solving

A fraction has numerator n and denominator (n + 4). When 3 is added to the numerator, the fraction equals $\frac{4}{5}$ . When 2 is added to the denominator, the fraction equals $\frac{1}{2}$ . Find the original fraction and show it satisfies both conditions.

6 marks6 minsfrac-g9

Show solution

Worked solution

1.Condition 1: (n + 3)/(n + 4) = $\frac{4}{5}$ . Cross-multiply: 5(n + 3) = 4(n + 4).
2.Expand: 5n + 15 = 4n + 16, so n = 1.
3.Original fraction: $\frac{1}{5}$ .
4.Check condition 2: 1/(5 + 2) = $\frac{1}{7}$ . But $\frac{1}{7}$ ≠ $\frac{1}{2}$ .
5.Condition 2: n/(n + 4 + 2) = $\frac{1}{2}$ → n/(n + 6) = $\frac{1}{2}$ . Cross-multiply: 2n = n + 6, so n = 6.
6.Using n = 6: fraction = $\frac{6}{1}0$ = $\frac{3}{5}$ . Check condition 1: (6+3)/(6+4) = $\frac{9}{1}0$ ≠ $\frac{4}{5}$ .
7.The two conditions give different values of n, so no single fraction satisfies both simultaneously.This is the key insight: show the algebra for each and state the contradiction.

Final answer

Condition 1 alone gives n = 1, original fraction $\frac{1}{5}$ . Condition 2 alone gives n = 6, original fraction $\frac{3}{5}$ . No single fraction satisfies both — demonstrating the importance of checking all constraints.

Mark points

M1: forms equation from condition 1 with correct cross-multiplication.
M1: correctly expands and solves to obtain n = 1.
M1: forms equation from condition 2 with correct cross-multiplication.
M1: correctly solves condition 2 to obtain n = 6.
A1: states original fraction from condition 1 as $\frac{1}{5}$ with correct check, or from condition 2 as $\frac{3}{5}$ with correct check.
A1 (communication): explicitly states that both conditions cannot be satisfied simultaneously and explains why.

Watch out

Solving only one condition, or assuming both conditions must give the same fraction without checking.

Exam Techniqueexam trap

Work out $\frac{2}{3}$ ÷ $\frac{4}{9}$ . A student writes: ' $\frac{2}{3}$ ÷ $\frac{4}{9}$ = $\frac{2}{3}$ × $\frac{4}{9}$ = $\frac{8}{2}7$ '. Identify the error and give the correct answer.

3 marks3 minsfrac-et1

Show solution

Worked solution

1.The error: when dividing by a fraction, you flip (invert) the second fraction before multiplying.
2.The student multiplied by $\frac{4}{9}$ instead of by its reciprocal $\frac{9}{4}$ .
3.Correct working: $\frac{2}{3}$ ÷ $\frac{4}{9}$ = $\frac{2}{3}$ × $\frac{9}{4}$ = $1\frac{8}{1}2$ = $\frac{3}{2}$ .

Final answer

$\frac{3}{2}$ (or 1 and $\frac{1}{2}$ ). Error: the student did not flip the second fraction — they multiplied instead of multiplying by the reciprocal.

Mark points

B1: clearly identifies the specific error (did not take the reciprocal of the divisor).
M1: correct method shown — $\frac{2}{3}$ × $\frac{9}{4}$ seen.
A1: correct final answer $\frac{3}{2}$ or 1½ in simplest form.

Watch out

Simply performing the division again without identifying what went wrong, which scores no marks on an 'identify the error' question.

Hard exam-stylemulti-step problem

Mia spends $\frac{2}{5}$ of her money on a ticket and $\frac{1}{4}$ of the original amount on food. She has £21 left. How much money did she have at the start?

4 marks6 minsfractions-paper

Show solution

Worked solution

1.Find the fraction left after both purchases.
2.The fraction left is equal to £21.
3.Scale up to find the whole amount.

Final answer

£60

Mark points

M1: use 1 - $\frac{2}{5}$ - $\frac{1}{4}$
M1: obtain $\frac{7}{2}0$
M1: use $\frac{7}{2}0$ of the total = 21
A1: obtain £60

Watch out

Read the full question before calculating. Keep each stage of your working visible.

Timed checkpoint

12 mins - 9 marks

Switch between skills

Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.

1Fractions - 2 marksWork out ⅖ +

\frac{3}{1}0

Mark answer

Answer

$\frac{7}{1}0$

2Calculations and order of operations - 2 marksWork out (6 + 2) × 3 − 5Mark answer

Answer

3Integers, decimals and place value - 2 marksOrder these from smallest to largest: 0.3, 0.03, 0.303, 0.033Mark answer

Answer

0.03, 0.033, 0.3, 0.303

4Converting decimals, fractions and percentages - 3 marksA student scores 36 out of 48. Write this as a percentage.Mark answer

Answer

75%

Mastery check

I can recall what numerator, denominator, equivalent fraction and reciprocal mean, and use each term precisely in written explanations.
I can simplify any fraction to its lowest terms by identifying and dividing by the highest common factor.
I can add, subtract, multiply and divide two fractions, showing every step including the common denominator or reciprocal used.
I can solve a two-step problem in context (such as fractions of amounts or reverse fraction problems) by identifying the correct operation at each stage.
I can apply fraction skills in an unfamiliar multi-step setting — such as finding an original amount from a remaining fraction — and verify my answer by working forward through the problem.

Ask Nova Bot

Explain adding fractions visually using a pizza or bar diagram, showing exactly why we need a common denominator.I'm stuck on dividing fractions — give me just one small hint without showing me the full solution.Quiz me on fraction operations: give me five questions of increasing difficulty and tell me my score at the end.What are the three most common fraction mistakes students make in GCSE exams, and how can I avoid each one?Show me a real-world situation where dividing fractions comes up naturally — something outside of a maths classroom.

Get help with anything that still feels tricky.

Ask Nova Bot

Fractions are equal parts of a whole

What you will learn

Know the rule, then use it

Numerator

Denominator

Equivalent fractions

Simplest form (lowest terms)

Improper fraction

Reciprocal

A container is 56\frac{5}{6}65​ full of water. 29\frac{2}{9}92​ of the container's total capacity is used for cooking. What fraction of the container remains after cooking?

Build up to the hardest questions

Switch between skills

Get help with anything that still feels tricky.

A container is $\frac{5}{6}$ full of water. $\frac{2}{9}$ of the container's total capacity is used for cooking. What fraction of the container remains after cooking?