Sample paper · GCSE Foundation · grades 1–5
GCSE Foundation sample Paper 1 (non-calculator)
The real Paper 1 is 1.5 hour 30 minutes and 80 marks, non-calculator, and all three papers carry equal weight. This sample is 20 original questions in the same content proportions as the Foundation qualification — number 25%, algebra 20%, ratio, proportion and rates of change 25%, geometry and measures 15%, probability 7.5%, statistics 7.5% — with no calculator-only items. AO1 / AO2 / AO3 at this tier: 50% / 25% / 25%.
Non-calculatorGCSE Foundation
GCSE Foundation sample Paper 1 (non-calculator)
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- 1.A train leaves Leeds at 14:35. The journey takes 1 hour 50 minutes. Work out the time the train arrives. Give your answer using the 24-hour clock.
- 2.A straight line passes through the points (1, 3) and (2, 5). Work out the equation of the line.
- 3.In a class, 50% of the students study French, 30% study Spanish and the rest study German. Write down the ratio of French : Spanish : German students, in its simplest form.
- 4.A treasure map has a scale of 1 cm to 4 m. A path from the start to a rock is drawn as two straight sections, measuring 3 cm and 2.5 cm. Work out the total real distance from the start to the rock, in metres.
- 5.A red fair dice and a blue fair dice are rolled together. Every outcome is listed as an ordered pair, red score then blue score. Work out the probability that the red score is less than the blue score.
- 6.A scatter graph of the number of ice creams sold at a seaside kiosk in Bournemouth and the number of sunburn cases treated at a nearby pharmacy, recorded on the same 30 days, shows strong positive correlation. Which statement about this correlation is correct?
- 7.The attendance at a football match is 23,000, correct to the nearest thousand. Which of these attendances is not possible for the match?
- 8.The area of a circle is given by the formula A = πr², where r is the radius. Rearrange the formula to make r the subject.
- 9.A cyclist rides at an average speed of 24 km/h for 1 hour 15 minutes. Work out the distance travelled, in km.
- 10.Triangle T has a vertex at (2, 1). It is enlarged by a scale factor of 3, centre (0, 0). Work out the coordinates of the image of this vertex.
- 11.Two ordinary fair dice are rolled, one after the other. Work out the probability that the first dice shows a 6 and the second dice shows an even number.
- 12.On a number line, point A is at −1. Point B is 4 units from point A. Write down the two possible positions of point B.
- 13.Solve the inequality 2(x − 1) ≤ 8.
- 14.x is 2/3 of y. Write the ratio x : y in its simplest form.
- 15.A robotic arm's tip starts at the point (12.5, −4.25) on a grid measured in centimetres. It moves by the vector to pick up a component, then by the vector to place it. Work out the coordinates of the tip after both moves.
- 16.A car travels 180 km using 6 litres of fuel. Work out the car's fuel consumption in kilometres per litre, then work out how many kilometres it can travel on a full tank of 12 litres at this rate.
- 17.A ball's height, h metres, t seconds after being thrown follows h = (t − 1)(9 − t). Given that the ball is at ground level at t = 1 and t = 9, work out at what time t the ball reaches its maximum height, using symmetry.
- 18.Triangle A and triangle B are mathematically similar. A side of triangle A is 4 cm long and the corresponding side of triangle B is 10 cm long. Write the ratio of the length in triangle A to the length in triangle B in its simplest form.
- 19.Estimate the value of √45, giving your answer to the nearest whole number.
- 20.The density of a liquid is 0.92 g/cm³. Work out the density of the liquid in kg/m³.
Answer key
- (c) 16:25 — Add the whole hour first: 14:35 plus 1 hour is 15:35. Then add the 50 minutes. From 15:35, 25 minutes reaches 16:00 and the remaining 25 minutes gives 16:25. 15:85 adds 35 and 50 to get 85 minutes and never exchanges 60 of them for an hour, 16:05 treats 1 hour 50 minutes as 1.5 hours, and 12:45 subtracts the journey time instead of adding it.
- (c) y = 2x + 1 — Method: the gradient of the line through two points is the change in y divided by the change in x, and the constant is then found by substituting one of the points into y = mx + c. Working: m = (5 − 3) ÷ (2 − 1) = 2 ÷ 1 = 2, so the line is y = 2x + c; substituting x = 1 and y = 3 gives 3 = 2 × 1 + c, so c = 3 − 2 = 1. Answer: y = 2x + 1. The distractors: y = x + 2 comes from taking the gradient as the change in x, 2 − 1 = 1, and then substituting (1, 3) to reach a constant of 2; y = 2x − 1 comes from working out the constant as mx − y, 2 × 1 − 3 = −1, instead of y − mx; y = 2x + 3 comes from using the y-coordinate of (1, 3) as the constant without substituting at all.
- (d) 5:3:2 — German = 100% − 50% − 30% = 20%. The ratio 50 : 30 : 20 simplifies by dividing every part by 10 to give 5 : 3 : 2.
- (b) 22 — Method: add the two drawn lengths together first, then apply the scale to the total. Working: 3 cm + 2.5 cm = 5.5 cm; 5.5 cm × 4 = 22 m. A student who answers 10 has only converted one of the two sections (2.5 cm × 4) and forgotten the other. A student who answers 5.5 has added the two drawn lengths but forgotten to apply the scale at all. A student who answers 44 has doubled the correct answer, effectively applying the scale twice. Answer: 22 m.
- (b) 5/12 — There are 36 equally likely ordered pairs. Listing the pairs where the red score is smaller than the blue score gives 15 outcomes, so the probability is 15/36 = 5/12. Choosing 1/2 comes from assuming the 36 outcomes split evenly between 'red smaller' and 'red bigger', which ignores the 6 outcomes where the two dice show the same score. Choosing 7/12 comes from including the outcomes where the two scores are equal, 15 + 6 = 21, giving 21/36 = 7/12, instead of using 'less than' strictly. Choosing 5/36 comes from counting only the outcomes where the two scores differ by exactly 1, such as (1, 2) and (2, 3), which is 5 outcomes, and missing the pairs that differ by more than 1.
- (c) Neither causes the other; sunshine links both. — Both ice cream sales and sunburn cases tend to rise on hot, sunny days, so the amount of sunshine is a third factor linked to both — neither variable causes the other. Saying ice cream sales cause the sunburn assumes a causal link in one direction that the correlation alone cannot establish. Saying sunburn cases cause the ice cream sales assumes the reverse causal link, which is no more justified. Saying a strong correlation always means causation is the general error this question is testing: correlation, however strong, does not by itself prove that one variable causes the other.
- (d) 23,600 — 23,000 correct to the nearest thousand means the true attendance, a, satisfies 22,500 ≤ a < 23,500. 23,600 is outside this range, so it is not possible. 22,500 is the lower bound of the range, so it is possible. 23,499 is just below the upper bound, so it is possible. 23,000 is the given value itself, so it is possible.
- (c) r = √(A/π) — A = πr² means r has been squared and then multiplied by π. To make r the subject, first divide both sides by π to get A/π = r², then take the square root of both sides: r = √(A/π). Writing r = A/π stops after dividing by π and forgets that r is still squared — it never undoes the square. Writing r = √A/π takes the square root before dividing by π, which square-roots only the A and not the whole of A/π. Writing r = (A/π)² squares A/π instead of taking its square root — the opposite of what is needed to undo r². The correct rearrangement is r = √(A/π).
- (c) 30 km — First convert 1 hour 15 minutes to hours: 15 minutes is a quarter of an hour, so the time is 1.25 hours. Then multiply by the speed: 24 × 1.25 = 30 km. Reading 15 minutes as 0.15 hours (writing the minutes after the decimal point instead of as a fraction of 60) gives 24 × 1.15 = 27.6 km. Working out 24 × 75 = 1800 multiplies by the number of minutes instead of converting to hours first. Working out 24 ÷ 1.25 = 19.2 divides by the time instead of multiplying. The cyclist travels 30 km.
- (d) (6, 3) — For an enlargement centred at the origin, each coordinate is multiplied by the scale factor: (2, 1) → (2 × 3, 1 × 3) = (6, 3). ((5, 4) comes from adding the scale factor to each coordinate instead of multiplying; (6, 1) comes from multiplying only the x-coordinate by 3 and leaving the y-coordinate unchanged; (2, 3) comes from multiplying only the y-coordinate by 3 and leaving the x-coordinate unchanged.)
- (a) 1/12 — The probability that the first dice shows a 6 is 1/6. The probability that the second dice shows an even number, 2, 4 or 6, is 3/6 = 1/2. Since the two dice are independent, multiply the probabilities: 1/6 × 1/2 = 1/12. A candidate who answers 1/6 has considered only the first dice and forgotten the condition on the second dice. A candidate who answers 1/2 has considered only the second dice and forgotten the condition on the first dice. A candidate who answers 1/36 has treated 'an even number' as a single specific value rather than three possible values, using 1/6 × 1/6.
- (c) 3 or −5 — Method: a point a fixed distance from another can lie on either side of it, so move the given distance in each direction from the starting point. Working: moving 4 units to the right gives −1 + 4 = 3, and moving 4 units to the left gives −1 − 4 = −5. Answer: 3 or −5. The distractors: 5 or −3 comes from starting at 1 instead of −1, giving 1 + 4 and 1 − 4; 3 only comes from moving to the right and forgetting that the point could lie to the left as well; 4 or −4 comes from measuring the distance from zero instead of from point A, which just repeats the given distance.
- (b) x ≤ 5 — Method: divide out the bracket first, then undo the number term; the inequality sign turns round only if both sides are multiplied or divided by a negative number. Working: dividing both sides of 2(x − 1) ≤ 8 by 2 gives x − 1 ≤ 4, and 2 is positive so the ≤ is unchanged; adding 1 to both sides gives x ≤ 5. Answer: x ≤ 5. The distractors: x ≤ 3 comes from subtracting 1 from 4 instead of adding 1 to both sides; x ≤ 4 comes from stopping at 8 ÷ 2 = 4 and never undoing the −1 inside the bracket; x < 5 comes from reading ≤ as a strict inequality, which wrongly leaves the boundary value out of the solution set.
- (d) 2 : 3 — x is 2/3 of y means for every 3 parts of y, x is 2 parts, so x : y = 2 : 3. 3 : 2 comes from writing the ratio the wrong way round. 2 : 5 comes from comparing x with the total of x and y (2 parts out of 5), instead of with y alone. 3 : 5 comes from comparing y with the total of x and y (3 parts out of 5), instead of with x.
- (d) (11, 0.75) — Add the moves to the starting point one component at a time. x: 12.5 + (−3.75) + 2.25 = 11; y: −4.25 + 6.5 + (−1.5) = 0.75, giving (11, 0.75). (8.75, 2.25) stops after the first move only and never applies the second vector. (6.5, 3.75) comes from subtracting the second vector instead of adding it. (0.75, 11) comes from swapping the final x-coordinate and y-coordinate.
- (d) 360 km — Method: first find the kilometres per litre by dividing distance by fuel used, then multiply this rate by the new tank size. Working: 180 ÷ 6 = 30 km per litre; 30 × 12 = 360 km. Answer: 360 km. 30 km comes from finding the correct fuel consumption but stopping there, without scaling it up to the full tank. 2160 km comes from multiplying the original distance (180) by the tank size (12) directly, skipping the unit rate. 90 km comes from pairing the numbers the wrong way round: dividing the distance by the new tank size, 180 ÷ 12 = 15, and then multiplying by the original 6 litres, 15 × 6 = 90.
- (d) t = 5 — The maximum height occurs halfway between the two times when the ball is at ground level: the midpoint of t = 1 and t = 9 is (1 + 9) ÷ 2 = 5, so the ball reaches its maximum height at t = 5 seconds. Getting t = 4 comes from halving the DIFFERENCE between the times, 9 − 1 = 8, then 8 ÷ 2 = 4, instead of finding their midpoint. Getting t = 8 uses that difference, 9 − 1 = 8, as if the gap between the two times were itself the time of the maximum. Getting t = 10 adds the two times, 1 + 9 = 10, but forgets to divide by 2.
- (c) 2 : 5 — A ratio is written in the order the question names the two shapes, so triangle A's length comes first: 4 : 10. Both parts divide by 2: 4 ÷ 2 = 2 and 10 ÷ 2 = 5, giving 2 : 5. Lengths are compared using the lengths themselves, so nothing is squared here; squaring both parts would give the ratio of the areas instead.
- (b) 7 — Method: trap the number between the two square numbers on either side of it, then decide which of them it is nearer to. Working: 6² = 36 and 7² = 49, so √45 lies between 6 and 7; 49 − 45 = 4 while 45 − 36 = 9, so 45 is nearer to 49. Answer: 7. The distractors: 6 comes from taking the square number below 45 and stopping there, without checking which of 36 and 49 is nearer; 22.5 comes from halving 45 instead of looking for the number that multiplies by itself to give 45; 2,025 comes from squaring 45 instead of taking its square root.
- (c) 920 kg/m³ — Convert each unit in turn. Mass: 1 g = 0.001 kg. Volume: 1 m³ = 100 × 100 × 100 = 1 000 000 cm³. So a density of 0.92 g per cm³ is 0.92 × 1 000 000 = 920 000 g in every cubic metre, and 920 000 g = 920 000 × 0.001 = 920 kg. The two conversions leave a single factor of 1 000 000 × 0.001 = 1000, so in one step multiply g/cm³ by 1000: 0.92 × 1000 = 920 kg/m³. Multiplying by 100 instead of 1000 gives 92 kg/m³, using the factor for 1 m² rather than 1 m³ of volume. Multiplying by 10 instead of 1000 gives 9.2 kg/m³, moving the decimal point one place for a conversion that moves it three. Dividing by 1000 instead of multiplying gives 0.00092 kg/m³, going the wrong way between the units — a kilogram is heavier than a gram, but a cubic metre is a million times bigger than a cubic centimetre, so the number must get larger, not smaller. The liquid's density is 920 kg/m³.
How the 20 questions are shared out
- Number — 5 questions (25% of the qualification)
- Algebra — 4 questions (20% of the qualification)
- Ratio, proportion and rates of change — 5 questions (25% of the qualification)
- Geometry and measures — 3 questions (15% of the qualification)
- Probability — 2 questions (7.5% of the qualification)
- Statistics — 1 question (7.5% of the qualification)
Where an area has fewer printable questions than its share, the shortfall is filled from the other areas. These are original questions, not past papers.