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.Maya buys 4 plants at £3.20 each and a bag of compost for £6.75. She pays with a £20 note. Work out her change.
- 2.The nth term of a sequence is 4n − 3. Work out the 7th term of the sequence.
- 3.The rent on a flat increases by 10% one year and by a further 10% the following year. Work out the overall percentage increase over the two years.
- 4.A quadrilateral has vertices A(1, 1), B(5, 1), C(6, 4) and D(2, 4). By comparing the y-coordinates of A, B and of D, C, write down whether the sides AB and DC are parallel, and give a reason for your answer.
- 5.A fair coin is flipped and a fair five-sided spinner, labelled 1 to 5, is spun. All the outcomes are listed as pairs, such as (H, 3). Work out the probability that the outcome is a tail and an odd number.
- 6.In a scatter graph of the age, in years, and the wingspan, in cm, of 20 birds of the same species, all the points lie close to a rising line of best fit except one, which lies a long way below the line. That bird was later found to have a damaged wing. Give a reason why this point should not be used when drawing the line of best fit.
- 7.Work out the reciprocal of (2 + 3)
- 8.A delivery van's distance-time graph shows the following: it travels 9 km in the first 15 minutes at a steady speed, stops for 45 minutes to unload, then travels a further 21 km in the next 30 minutes, also at a steady speed. Work out the average speed, in km/h, for the whole journey.
- 9.A fruit squash is made by mixing squash and water. In a jug, 2/9 of the total volume is squash. A caterer makes up 4.5 litres of the squash mixture. Work out how many litres of squash are in the mixture.
- 10.A courier drone delivers a parcel. It starts at the point (2, 1) on a map grid measured in kilometres. It flies by the vector to a warehouse, then by the vector to the delivery address. Work out the coordinates of the delivery address.
- 11.Two cards are dealt one after the other from an ordinary pack of 52 playing cards. The first card is not put back before the second is dealt. The pack contains 4 aces. Work out the probability that neither card is an ace. Give your answer as a product of two fractions.
- 12.Insert one pair of brackets into 2 + 3 × 5 − 1 so that the calculation is equal to 24. Which calculation is correct?
- 13.Solve 2x − 2 = 2
- 14.A laptop is bought for £600. Its value decreases by 20% after 1 year. Work out the value of the laptop after 1 year.
- 15.A ship's radar shows a lighthouse at the point (12, −4) on a grid measured in nautical miles. The ship is at (2, 5). The ship sails along the vector that takes it directly to the lighthouse, then sails along that same vector again. Work out the ship's final position.
- 16.A circle has a diameter of 10 cm. Work out the exact circumference of the circle, in terms of π.
- 17.Here are the first five terms of an arithmetic sequence: 4, 7, 10, 13, 16. Work out the 9th term.
- 18.A smoothie recipe combines 350 cm³ of mango juice with 650 cm³ of orange juice. Work out the total volume of the smoothie in litres.
- 19.Jack buys 3 books, each costing £4.25, and pays with a £20 note. Work out how much change he receives.
- 20.The number of tickets a group can afford is inversely proportional to the price per ticket. At £4 per ticket, the group can afford 12 tickets. Work out how many tickets the group can afford at £6 per ticket.
Answer key
- (c) £0.45 — Method: find the total cost, then subtract from £20. Working: 4 × £3.20 = £12.80. £12.80 + £6.75 = £19.55. Change = £20.00 − £19.55 = £0.45. Answer: £0.45. (£7.20 comes from forgetting to include the compost and subtracting only the plants' cost from £20. £1.45 comes from dropping the carry when adding the pence: 80p + 75p = £1.55, but only the 55p is written down, giving £18.55 instead of £19.55. £10.05 comes from buying only one plant instead of four, using £3.20 + £6.75.)
- (b) 25 — Substitute n = 7: 4 × 7 − 3 = 28 − 3 = 25. Forgetting to subtract 3 gives 4 × 7 = 28. Subtracting 3 from 7 before multiplying by 4, 4 × (7 − 3) = 16, applies the operations in the wrong order. Substituting n = 8 by miscounting the position gives 4 × 8 − 3 = 29.
- (c) 21% — Method: an increase of 10% is a multiplier of 1.1, and two successive increases are found by multiplying the multipliers. Working: 1.1 × 1.1 = 1.21, so the rent is 121% of the original, which is an increase of 21%. Answer: 21%. The distractors: 20% comes from adding the two percentages, which ignores that the second 10% is taken of a larger amount; 121% is the multiplier written as the change rather than the change itself; 11% comes from slipping in the multiplication and getting 1.11 instead of 1.21.
- (b) Parallel, since AB and DC are both horizontal lines — A and B both have y-coordinate 1, so AB is horizontal; D and C both have y-coordinate 4, so DC is horizontal too. Two horizontal lines are always parallel, so AB and DC are parallel. "Not parallel, since AB and DC have different lengths" is wrong twice over: AB runs from x = 1 to x = 5 and DC from x = 2 to x = 6, so both are in fact 4 units long, and in any case length has no bearing on whether two lines are parallel — only direction does. "Not parallel, since AB is horizontal and DC is vertical" misreads the coordinates of D and C, which share the y-coordinate 4 and so give a horizontal line, not a vertical one. "Parallel, since A and D have the same x-coordinate" rests on a false claim: A has x-coordinate 1 and D has x-coordinate 2, so they do not share an x-coordinate — and even if they did, it would say nothing about whether AB is parallel to DC.
- (d) 3/10 — Method: list the full possibility space as pairs of coin and spinner results, then count how many pairs satisfy both conditions and divide by the size of the whole space. Working: the coin gives 2 outcomes and the spinner gives 5, so the full space has 2 × 5 = 10 equally likely pairs. The pairs with a tail and an odd number are (T,1), (T,3) and (T,5), which is 3 out of 10. Answer: 3/10. Watch out: writing down 1/2 uses only the coin's own chance of a tail and ignores that the spinner also has to land on an odd number. Writing down 3/5 uses only the spinner's chance of landing on an odd number and ignores the coin altogether. And writing down 1/10 counts just one matching outcome, such as (T,1), instead of all three pairs that satisfy both conditions.
- (d) An outlier from the damaged wing, not the trend. — That bird's point lies a long way from the rising trend followed by every other bird, and its low wingspan is explained by the damaged wing rather than by its age — it is an outlier caused by an unusual factor, not part of the general relationship between age and wingspan, so it should not be used when drawing the line of best fit. Saying every point must be used ignores that an outlier caused by a separate, identifiable factor can rightly be set aside. Saying it shows no correlation ignores that the other 19 points do show a clear rising trend; one outlier does not remove that. Saying it proves the line is inaccurate confuses one unusual bird with a fault in the line itself, when the line correctly describes the trend followed by the rest of the data.
- (a) 1/5 — Work out the bracket first: 2 + 3 = 5. The reciprocal of 5 is 1/5. A candidate who forgot to take the reciprocal and just gave the value of the bracket wrote 5. A candidate who took the reciprocal but made a sign error wrote −1/5. A candidate who found the reciprocal of each number separately and added them, treating reciprocal as if it distributes over addition, worked out 1/2 + 1/3 = 5/6.
- (d) 20 km/h — Method: total distance = 9 + 21 = 30 km. Total time = 15 + 45 + 30 = 90 minutes = 1.5 hours. Average speed = total distance ÷ total time = 30 ÷ 1.5 = 20 km/h. Distractor origins: 40 km/h ignores the 45-minute stop, dividing the total distance by the travel time only (30 km ÷ 45 minutes = 40 km/h); 39 km/h averages the two section speeds, 36 km/h and 42 km/h, instead of using total distance over total time; 36 km/h uses only the first section's speed (9 km in 15 minutes), ignoring the second section and the stop.
- (d) 1.00 litres — Squash is 2/9 of the mixture, so the squash volume is 4.5 × 2/9 = 1.00 litres. Using the water's fraction, 7/9, instead of squash's gives 4.5 × 7/9 = 3.50 litres — the volume of water, not squash. Dividing 4.5 by 9 but forgetting to multiply by the numerator 2 gives 4.5 ÷ 9 = 0.50 litres, which is only 1/9 of the mixture. Halving the total volume instead of applying the fraction 2/9 gives 4.5 ÷ 2 = 2.25 litres, which assumes the mixture is half squash.
- (d) (6, 2) — Applying the first vector: (2, 1) + (5, −3) = (7, −2), which is the warehouse. Applying the second vector: (7, −2) + (−1, 4) = (6, 2), the delivery address. '(7, −2)' stops at the warehouse and forgets the second flight. '(8, −6)' comes from adding (1, −4) instead of (−1, 4) for the second vector, getting both signs wrong. '(11, −3)' comes from swapping the components of the second vector to (4, −1) before adding.
- (b) (48/52) × (47/51) — Method: for two deals one after the other with nothing put back, multiply the probability of the first by the probability of the second worked out from the cards that are left. Working: 52 − 4 = 48 cards are not aces, so the first card is not an ace with probability 48/52. One card has now gone and it was not an ace, so 51 cards remain and 47 of them are not aces, giving 47/51. Answer: the probability is (48/52) × (47/51). The distractors: (48/52) × (48/52) comes from leaving the pack at 52 cards for the second deal, which is only true if the first card is replaced; (4/52) × (3/51) comes from working out the probability that both cards ARE aces instead of neither; (4/52) × (4/51) comes from the same misreading with the ace count left at 4 while the total is reduced, adjusting only half of the second fraction.
- (b) (2 + 3) × 5 − 1 — 2 + 3 = 5, then 5 × 5 = 25, then 25 − 1 = 24, so the brackets belong around 2 + 3. Placing them around 5 − 1 instead gives 5 − 1 = 4, then 3 × 4 = 12, then 2 + 12 = 14. Leaving the multiplication bracketed instead changes nothing, because it already had priority: 3 × 5 = 15, then 2 + 15 = 17, then 17 − 1 = 16. Bracketing both 2 + 3 and 5 − 1 uses two pairs instead of the one asked for: 2 + 3 = 5, 5 − 1 = 4, then 5 × 4 = 20.
- (a) x = 2 — Method: undo the subtraction first, then undo the multiplication, doing the same to both sides each time. Working: adding 2 to both sides gives 2x = 4, and dividing both sides by 2 gives x = 2. Answer: x = 2. The distractors: x = 4 comes from stopping at 2x = 4 and writing 4 as the value of x; x = 0 comes from subtracting 2 from both sides instead of adding it, giving 2x = 0; x = 8 comes from multiplying 4 by 2 instead of dividing by 2.
- (c) £480.00 — To decrease by 20%, multiply by 0.80 (100% − 20%). £600 × 0.80 = £480.00. £120.00 comes from working out only the decrease (£600 × 0.20) and forgetting to subtract it from the original value. £580.00 comes from subtracting 20 directly instead of 20% of £600. £720.00 comes from multiplying by 1.20, adding the percentage instead of subtracting it.
- (b) (22, −13) — The vector from the ship to the lighthouse is (12, −4) − (2, 5) = (10, −9). Sailing along this vector twice from the start gives (2, 5) + 2 × (10, −9) = (2 + 20, 5 − 18) = (22, −13). '(12, −4)' stops after the ship reaches the lighthouse and ignores the second identical leg. '(−18, 23)' comes from finding the vector the wrong way round, as (2, 5) − (12, −4) = (−10, 9), and then doubling that. '(2, 5)' comes from adding the vector and then subtracting it again, wrongly cancelling the two legs instead of adding them.
- (d) 10π cm — Method: for a circle, circumference = π × diameter. Working: circumference = π × 10 = 10π cm. Answer: 10π cm. 20π cm comes from using the formula 2 × π × radius but plugging in the diameter as if it were the radius: 2 × π × 10 = 20π. 25π cm comes from confusing circumference with area, using π × radius² with radius 5: π × 5² = 25π. 5π cm comes from halving the diameter and then multiplying by π, as if circumference were π × radius: π × 5 = 5π.
- (b) 28 — Method: find the common difference, then use the position-to-term rule (or extend the sequence) to reach the 9th term. Working: the common difference is 7 − 4 = 3, so the nth term is 4 + 3(n − 1). For n = 9: 4 + 3 × 8 = 4 + 24 = 28. Answer: 28. 27 comes from using 3n instead of 3n + 1, dropping the constant term from the rule, 3 × 9 = 27. 31 comes from extending from the 5th term by one step too many, adding 3 five times instead of four, 16 + 3 × 5. 25 comes from extending by one step too few, adding 3 three times instead of four, 16 + 3 × 3.
- (b) 1.00 litres — Total volume = 350 + 650 = 1000 cm³. Since 1000 cm³ = 1 litre, the smoothie is 1.00 litre. Using only the orange juice's 650 cm³ and converting that gives 0.65 litres, forgetting the mango juice entirely. Using only the mango juice's 350 cm³ gives 0.35 litres, forgetting the orange juice. Adding both volumes correctly to get 1000 cm³ but not converting to litres leaves the answer as 1000.00, which is the volume in the wrong unit.
- (c) £7.25 — Find the total cost of the books first: 3 × 4.25 = 12.75, so the books cost £12.75 in total. Subtract this from the £20 note: 20.00 − 12.75 = 7.25, so the change is £7.25. Stopping after finding the cost and not subtracting it from £20 gives £12.75, which is the amount spent, not the change. Borrowing correctly in the pence column but forgetting to reduce the pounds column by 1 gives £8.25 instead of £7.25. Multiplying 3 × 4.25 as 12.25 instead of 12.75, a multiplication slip, makes the change come out £0.50 too high, at £7.75. So Jack receives £7.25 change.
- (d) 8 — The product of price and number of tickets is constant: k = 4 × 12 = 48. At £6 per ticket, the number of tickets is 48 ÷ 6 = 8. Getting 18 comes from treating price and tickets as directly proportional and working out 12 × 6 ÷ 4 instead of dividing k by the new price. Getting 12 assumes the number of tickets does not change when the price changes. Getting 6 comes from writing down the new price instead of working out the number of tickets.
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.