GCSE Maths mock exam Free full practice test
A full-length practice test built from our exam-style questions on every topic of GCSE Maths. Each answer is marked as you go, and every question has a worked solution to open afterwards.
Before you start
- Set a timer for about 85 minutes and work without notes. A formulae sheet is provided in the 2025, 2026 and 2027 GCSE Maths exams. Answers within 2.5% are marked correct, and you can type fractions such as 23/20.
- There are no hints in the mock. Answer every question before you open its worked solution: opening it first counts the question as missed.
- Write the exam-style questions on paper, then mark them yourself with the scoring guide.
Section I: 30 exam-style questions
Hint
Divide the increase by the original price.
Worked solution
- Increase = 92 − 80 = 12.
- 12 ÷ 80 × 100 = 15%.
Hint
Add the powers when multiplying, subtract when dividing.
Worked solution
- 23 + 5 − 4 = 24.
- = 16.
Hint
A 20% increase means multiplying by 1.2. Undo it.
Worked solution
- Original × 1.2 = 96.
- Original = 96 ÷ 1.2 = 80, so £80.
Hint
Find square factors: 50 = 25 × 2 and 18 = 9 × 2.
Worked solution
- √50 = 5√2 and √18 = 3√2.
- 5√2 + 3√2 = 8√2, so a = 8.
Hint
The upper bounds are half a unit above each measurement.
Worked solution
- Upper bounds: 12.5 cm and 5.5 cm.
- Upper bound of area = 12.5 × 5.5 = 68.75 cm².
Hint
Expand the bracket first.
Worked solution
- 5x − 10 = 2x + 11.
- 3x = 21.
- x = 7.
Hint
Add the equations to eliminate y.
Worked solution
- Adding: 3x = 12, so x = 4.
- Then y = 3. So x = 4.
Hint
Change in y divided by change in x.
Worked solution
- (12 − 3)/(4 − 1) = 9/3.
- Gradient = 3.
Hint
It does not factorise. Use the quadratic formula with a = 1, b = −4, c = −7.
Worked solution
- x = (4 ± √(16 + 28))/2 = (4 ± √44)/2.
- x = (4 + 6.633)/2 = 5.32 (the other solution is −1.32).
Hint
Multiply every term by 6 to clear the fractions.
Worked solution
- 3(x + 3) + 2(x − 1) = 24.
- 3x + 9 + 2x − 2 = 24, so 5x = 17.
- x = 3.4.
Hint
Find the flour for one biscuit, then scale up.
Worked solution
- 250 ÷ 10 = 25 g per biscuit.
- 25 × 24 = 600 g.
Hint
Multiply by 1.03 once for each year.
Worked solution
- 2000 × 1.034 = 2000 × 1.12551.
- = 2251.02, so £2251.02.
Hint
Fewer workers take longer: this is inverse proportion.
Worked solution
- Total work = 6 × 10 = 60 worker-days.
- 60 ÷ 4 = 15 days.
Hint
Density = mass ÷ volume.
Worked solution
- 540 ÷ 200 = 2.7 g/cm³.
Hint
Make the b parts match: write b : c as 4 : 10.
Worked solution
- b : c = 2 : 5 = 4 : 10, so a : b : c = 3 : 4 : 10.
- a = 12 means 1 part = 4, so c = 10 × 4 = 40.
Hint
Use Pythagoras' theorem.
Worked solution
- c2 = 92 + 122 = 225.
- c = 15 cm.
Hint
Opposite and hypotenuse: use sine.
Worked solution
- sin 35° = opp/10.
- opp = 10 × sin 35° = 5.74 cm.
Hint
V = πr2h.
Worked solution
- V = π × 9 × 10 = 90π.
- V = 282.7 cm³.
Hint
Each exterior angle is 360° ÷ 6.
Worked solution
- Exterior angle = 60°.
- Interior angle = 180 − 60 = 120°.
Hint
Two sides and the angle between them: use the cosine rule.
Worked solution
- a2 = 72 + 92 − 2 × 7 × 9 × cos 60° = 49 + 81 − 63 = 67.
- a = √67 = 8.19 cm.
Hint
Multiply for independent events.
Worked solution
- 0.3 × 0.6 = 0.18.
Hint
After the first red is taken, there are 2 red left out of 7.
Worked solution
- P = 3/8 × 2/7 = 6/56.
- = 3/28 (about 0.107).
Hint
Draw a Venn diagram, starting with the 5 in the overlap.
Worked solution
- Football only 13, tennis only 7, both 5: total 25.
- Neither = 30 − 25 = 5, so P = 5/30 = 1/6.
Hint
Relative frequency = frequency ÷ trials.
Worked solution
- 46 ÷ 200 = 0.23.
Hint
Given they play football, only the 18 football players count.
Worked solution
- 5 of the 18 football players also play tennis.
- P = 5/18 (about 0.278).
Hint
Multiply each score by its frequency, add, then divide by the total frequency.
Worked solution
- Σfx = 3 + 10 + 24 + 16 = 53, and Σf = 20.
- Mean = 53 ÷ 20 = 2.65.
Hint
Find the total of all five first.
Worked solution
- Total = 5 × 12 = 60.
- Fifth = 60 − (10 + 11 + 13 + 15) = 11.
Hint
IQR = UQ − LQ.
Worked solution
- 26 − 14 = 12.
Hint
Use the midpoint of each class.
Worked solution
- Midpoints 5, 15, 25: Σfx = 20 + 90 + 250 = 360.
- Estimated mean = 360 ÷ 20 = 18.
Hint
Frequency = frequency density × class width.
Worked solution
- Class width = 25 − 10 = 15.
- Frequency = 2.4 × 15 = 36.
Section II: 4 exam-style questions
Exam-style question 1: Number
Exam-style question style · 5 points · about 6 minutes
Sam earns £1800 a month. He spends 1/3 of it on rent. He then spends 15% of what is left on travel.
- (a) How much does Sam spend on rent?
- (b) How much does Sam spend on travel?
- (c) What fraction of his earnings does Sam have left after rent and travel? Give it in its simplest form.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
1800 ÷ 3 = £600.
Part (b)
What is left: 1800 − 600 = 1200. 15% of 1200 = £180.
Part (c)
1800 − 600 − 180 = 1020. 1020/1800 = 17/30.
Your mark: 0 of 5
Exam-style question 2: Ratio, proportion and rates of change
Exam-style question style · 6 points · about 6 minutes
A car costs £15 000. Its value falls by 12% each year.
- (a) Work out its value after 1 year.
- (b) Work out its value after 3 years.
- (c) Work out the total percentage decrease over the 3 years, to 1 decimal place.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
15 000 × 0.88 = £13 200.
Part (b)
15 000 × 0.883 = £10 222.08.
Part (c)
0.883 = 0.681472, so the value has fallen by 31.9%.
Your mark: 0 of 6
Exam-style question 3: Geometry and measures
Exam-style question style · 6 points · about 6 minutes
A ladder 6.5 m long leans against a vertical wall. The top of the ladder is 6 m above the ground, which is horizontal.
- (a) How far is the foot of the ladder from the wall?
- (b) Work out the angle between the ladder and the ground, to 1 decimal place.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
d2 = 6.52 − 62 = 42.25 − 36 = 6.25, so d = 2.5 m.
Part (b)
sin θ = 6/6.5, so θ = 67.4°.
Your mark: 0 of 6
Exam-style question 4: Statistics
Exam-style question style · 5 points · about 5 minutes
Two classes sat the same test. Class A had median 64 and interquartile range 12. Class B had median 58 and interquartile range 20.
- (a) Compare the scores of the two classes.
- (b) Why is the IQR a better measure of spread than the range here?
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
Class A scored higher on average, since its median (64) is greater than Class B's (58). Class A's scores were more consistent, since its IQR (12) is smaller than Class B's (20).
Part (b)
The IQR is not affected by extreme scores, because it uses only the middle 50% of the data.
Your mark: 0 of 5
After the mock
Note every question you missed and its topic, then go back to that topic's practice page: all GCSE Maths topics. Then try the real thing with official GCSE Maths past papers.