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
Answer key: GCSE Foundation sample Paper 1 (non-calculator)
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- (d) 2 hours — Method: the time for a journey is the distance divided by the speed, so round the distance first and then divide by the speed. Working: 95 km rounds to 100 km, and 100 ÷ 50 = 2; the speed is in kilometres per hour, so the answer is a number of hours. Answer: 2 hours. The distractors: 1 hour comes from rounding the distance down to 50 km to match the speed, so that the journey looks like a single hour of driving; 30 minutes comes from dividing the speed by the distance, 50 ÷ 100, instead of the distance by the speed; 1 hour 54 minutes is the exact time, 95 ÷ 50 = 1.9 hours, worked out in full when the question asks for an estimate.
- (d) 8 — Method: write both present ages in terms of one letter, take four years off each of them, and form an equation from the comparison four years ago. Working: let Amelia be y, so her mother is 3y; four years ago they were y − 4 and 3y − 4, giving 3y − 4 = 5(y − 4); expanding gives 3y − 4 = 5y − 20; adding 20 and subtracting 3y from both sides gives 16 = 2y, and dividing by 2 gives y = 8. Checking: Amelia is 8 and her mother 24; four years ago they were 4 and 20, and 20 = 5 × 4. Answer: 8. The distractors: 4 comes from solving correctly but giving Amelia's age four years ago instead of her age now; 16 comes from stopping at 2y = 16 and giving 16 as her age; 2 comes from collecting the y terms by adding the 5y instead of subtracting it, giving 3y + 5y = 20 − 4 and so 8y = 16.
- (b) 1 : 2.25 — Method: to write a ratio in the form 1 : n, divide both parts by the first part. Working: 4 ÷ 4 = 1 and 9 ÷ 4 = 2.25, so 4 : 9 = 1 : 2.25. Working out 9 ÷ 4 = 2.25 correctly but then writing it as the first part gives 2.25 : 1, the two parts the wrong way round. Subtracting 9 − 4 = 5 gives 1 : 5, confusing the difference between the parts with the ratio. Multiplying 4 × 9 = 36 gives 1 : 36, confusing the product of the parts with the ratio.
- (d) 50 — The angles in any triangle add up to 180°, so x = 180 − 90 − 40 = 50. "90" comes from subtracting only the 90° angle from 180° and forgetting to also subtract the 40°. "130" comes from adding 90° and 40° together instead of subtracting them from 180°. "45" comes from halving the 90° angle instead of using the angle sum of the triangle at all.
- (b) 10 — The number who use the pool or the sauna (or both) is the total minus those who use neither: 70 − 12 = 58. Since pool + sauna double-counts the overlap, n(P ∩ S) = 38 + 30 − 58 = 10. Adding the pool and sauna counts without subtracting the overlap at all gives 38 + 30 = 68, more members than are in the whole gym. Subtracting the sauna count from the union, 58 − 30 = 28, actually finds the number who use ONLY the pool, not both. Reporting the 'neither' count, 12, confuses it with the 'both' region — they describe opposite corners of the diagram.
- (b) Tea — Method: the mode is the category with the highest frequency. Working: tea has a frequency of 8, which is higher than coffee's 7, hot chocolate's 4 and juice's 1. Mode = tea. Coffee has the second-highest frequency, 7, not the highest, so it is not the mode. Hot chocolate, with a frequency of 4, and juice, with a frequency of 1, both have lower frequencies still. Always compare every frequency in the table before naming the mode — the highest number wins, however close the next one is.
- (c) 8 — Each of the 4 spinner outcomes can be paired with each of the 2 coin outcomes, so multiply: 4 × 2 = 8. 6 comes from adding the two numbers instead of multiplying them. 4 comes from using only the spinner outcomes and ignoring the coin. 2 comes from using only the coin outcomes and ignoring the spinner.
- (a) 4(x + 3) = 20 and 4x + 3 = 11 when x = 2, so the two expressions are not equivalent, because the bracket means the 3 must be added before multiplying by 4. — Substituting x = 2: 4(x + 3) = 4 × 5 = 20, and 4x + 3 = 8 + 3 = 11. The two values are different, and expanding 4(x + 3) algebraically gives 4x + 12, which can never equal 4x + 3 (that would require 12 = 3) — so the two expressions are never equivalent, for any value of x. The option claiming they become equal for a larger x is wrong: 4x + 12 = 4x + 3 has no solution at all. The option claiming they are equivalent because they share the terms 4x and 3 ignores that the bracket changes the constant term. The option that calculates 4(x + 3) as 11 ignores the bracket completely, applying the 4 only to the x term.
- (b) 7.2 m — Multiply the model wingspan by the scale factor: 15 × 48 = 720. This is in centimetres, and 720 cm = 7.2 m, since 1 m = 100 cm. Giving 0.31 m divides by the scale factor instead of multiplying (15 ÷ 48 ≈ 0.31), scaling the model down rather than the real aircraft up. Giving 72 m converts centimetres to metres by dividing by 10 instead of 100. Giving 0.72 m converts by dividing by 1000 instead of 100.
- (d) (−5, −4) — Method: every vertex of a translated shape moves by the same vector, so find that vector from the one vertex whose image is given, then apply it to A. Working: C(4, 5) moves to (0, −1), so across 0 − 4 = −4 and up −1 − 5 = −6, giving the vector $\binom{-4}{-6}$. Applying it to A(−1, 2): −1 − 4 = −5 and 2 − 6 = −4. Answer: the image of A is (−5, −4). Working the vector out as object minus image gives 4 to the right and 6 up, which applied to A gives (3, 8). Getting the horizontal movement right but reversing the vertical one gives (−5, 8). Treating (0, −1) as the image of every vertex ignores that a translation carries each vertex to a different place.
- (a) 160 — Expected number = probability × number of trials = 0.08 × 2,000 = 160. Moving the decimal point one place too far, using 0.008 instead of 0.08, gives 2,000 × 0.008 = 16. Working out the expected number of customers who do NOT buy a bag, using the complement 1 − 0.08 = 0.92, gives 2,000 × 0.92 = 1,840. Rounding 0.08 up to 0.1 before multiplying gives 2,000 × 0.1 = 200.
- (c) 3 × 10⁶ — Daily revenue = (4 × 10³) × 2.5 = 1 × 10⁴ (£10,000). Multiplying by the number of days, (3 × 10²), gives annual revenue = (1 × 10⁴) × (3 × 10²) = 3 × 10⁶ (£3,000,000). A candidate who forgot to multiply by the price and just multiplied the number of items by the number of days worked out (4 × 10³) × (3 × 10²) = 1.2 × 10⁶. A candidate who added the number of days to the daily revenue instead of multiplying worked out 1 × 10⁴ + 3 × 10² = 1.03 × 10⁴. A candidate who misread £2.50 as £25 worked out a daily revenue of (4 × 10³) × 25 = 1 × 10⁵, giving an annual total of (1 × 10⁵) × (3 × 10²) = 3 × 10⁷.
- (b) x = 8 — Method: undo the addition first, then undo the division by 4. Working: subtracting 1 from both sides gives x/4 = 2, and multiplying both sides by 4 gives x = 8. Answer: x = 8. The distractors: x = 2 comes from stopping at x/4 = 2 and writing 2 as the value of x; x = 16 comes from adding 1 to both sides instead of subtracting it, giving x/4 = 4; x = 0.5 comes from dividing by 4 instead of multiplying by 4 at the last step.
- (b) The ratio C : m is not constant because the formula includes a fixed charge of £3 as well as the charge per mile. — For the ratio C : m to stay constant, C must be directly proportional to m, i.e. C = km with no constant term. Because of the +3 fixed charge, C is not directly proportional to m: for example m = 1 gives C = 5.5 (ratio 5.5 : 1), while m = 10 gives C = 28 (ratio 2.8 : 1) — the ratio has changed.
- (b) 2 — A rhombus has two lines of symmetry — along each of its two diagonals.
- (a) 23 — Division undoes multiplication, so the missing number is 391 ÷ 17 = 23. Writing down 17 repeats the number already given instead of solving for the missing one. Subtracting instead of dividing gives 391 − 17 = 374. Multiplying instead of dividing gives 391 × 17 = 6647.
- (c) x = 3 and x = −5 — To factorise x² + 2x − 15, find two numbers that multiply to −15 and add to 2: these are 5 and −3, since 5 × (−3) = −15 and 5 + (−3) = 2. So x² + 2x − 15 = (x + 5)(x − 3). Setting each factor to zero gives x = −5 and x = 3. Choosing x = −3 and x = 5 comes from swapping the signs of the correct roots. Choosing x = 5 and x = 3 uses the right pair of numbers, 5 and 3, but forgets that one of them must be negative for the product to equal −15. Choosing x = −15 and x = 1 mistakes the constant term, −15, for one of the roots, and pairs it oddly with x = 1.
- (a) 0.62 miles — The gradient of the line is the change in miles divided by the change in kilometres: 31 ÷ 50 = 0.62, so 1 kilometre converts to 0.62 miles. Dividing the wrong way round, 50 ÷ 31 = 1.612..., rounds to 1.61 miles — that finds how many kilometres are in 1 mile, not the reverse. Doubling the gradient, 1.24 miles, comes from using 62 ÷ 50 instead of 31 ÷ 50. Reading off the y-coordinate of the given point without dividing by the x-coordinate gives 31.00 miles, which is the number of miles for 50 kilometres, not for 1 kilometre.
- (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.
- (a) 4:3:1 — Convert every part to the same unit: 2 m = 200 cm, so the ratio is 200 : 150 : 50. Dividing all three parts by 50 gives 4 : 3 : 1. Writing 2 : 150 : 50 has not converted 2 m into centimetres, so the units do not match. Writing 3 : 4 : 1 has the first two parts the wrong way round. Writing 4 : 3 : 2 comes from an arithmetic slip on the last part: 50 ÷ 50 = 1, not 2.
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.