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)
MathsUKwww.geekhero.co.uk
- (d) 4/3 — Method: the product of two negative numbers is positive, so work with 2/5 × 10/3 and then simplify. Multiply the numerators together and the denominators together. Working: 2 × 10 = 20 and 5 × 3 = 15, giving 20/15; both 20 and 15 divide by 5, so 20/15 = 4/3. Answer: 4/3. The distractors: −4/3 has the arithmetic right but keeps a minus sign, from treating negative × negative as negative; 3/25 comes from turning the second fraction upside down and multiplying, which divides instead of multiplying and gives 2/5 × 3/10 = 6/50; −56/15 comes from adding the two fractions instead of multiplying them, giving −6/15 − 50/15.
- (d) 3n + 2 — The common difference is 3 (8−5=3), so the expression starts 3n. To match the first term when n=1, 3×1+c=5, so c=2: the nth term is 3n+2. A candidate who uses the first term itself as the constant, instead of first term minus the common difference, would write 3n+5 (giving 8, 11, 14, 17 — one term too high throughout). A candidate who omits the constant term altogether would write just 3n (giving 3, 6, 9, 12, not matching the sequence at all). A candidate who adds the common difference to n instead of multiplying would write n+3 (giving 4, 5, 6, 7, far too small).
- (c) £2743.60 — Each year the balance is multiplied by 1.03. After the first year: 4000 × 1.03 = 4120. After the second year: 4120 × 1.03 = 4243.60, so that is what Leah takes out. She then spends £1500 of it, which leaves 4243.60 − 1500 = 2743.60. She has £2743.60 left.
- (b) 0 — At 0°, the opposite side of the right-angled triangle has shrunk to zero length, so sin 0° = 0. 1 is the value of sin 90° (mixing up the two angles). 1/2 is the value of sin 30°. 'Undefined' is what happens for tan 90°, not sin 0° — sin 0° has a perfectly good exact value.
- (b) 1/10 — Method: list every possible pair of subjects systematically, so that no pair is missed and no pair is counted twice, then compare the number of successful pairs with the size of the list. Working: pairing maths with each of the other four gives 4 pairs, physics with each subject after it gives 3, biology gives 2 and chemistry gives 1, so there are 4 + 3 + 2 + 1 = 10 pairs. Exactly one of them is maths with biology. Answer: the probability is 1/10. The distractors: 1/5 comes from counting maths-then-biology and biology-then-maths as two separate successes while still dividing by the 10 unordered pairs; 1/15 comes from a list that also pairs each subject with itself, giving 15 entries instead of 10; 1/4 comes from working out the second step alone, that 1 of the 4 subjects left after maths is biology, without allowing for the chance that maths is picked at all.
- (a) 1.24 — Method: for data given as a frequency table, the mean is Σfx ÷ Σf — multiply each value by its frequency, add the results, then divide by the total frequency. Working: 0 × 6 = 0. 1 × 10 = 10. 2 × 6 = 12. 3 × 3 = 9. So Σfx = 0 + 10 + 12 + 9 = 31. The total frequency is Σf = 6 + 10 + 6 + 3 = 25. Mean = 31 ÷ 25 = 1.24 siblings. Averaging the frequency column itself, (6 + 10 + 6 + 3) ÷ 4 = 6.25, mixes up the frequencies with the values they belong to. Writing down 1, the number of siblings with the highest frequency, gives the mode, not the mean. Writing down 31 stops after finding Σfx and forgets to divide by the total frequency, 25. Always divide Σfx by Σf — never stop at the top of the fraction.
- (c) 10 — Method: the multiplication is carried out before the subtraction. Working: 4 × 2 = 8, so the calculation becomes 18 − 8 = 10. Answer: 10. The distractors: 28 comes from working from left to right, giving (18 − 4) × 2 = 14 × 2 = 28; −10 comes from subtracting the 18 from the 8 rather than the 8 from the 18, giving 8 − 18 = −10; 12 comes from reading 4 × 2 as 4 + 2 = 6 and then working out 18 − 6 = 12.
- (c) 6 — Method: substitute the value, work out the top of the fraction first, then the division, and add the 3 last. Working: the top gives 10 − 4 = 6, dividing by 2 gives 6 ÷ 2 = 3, and adding 3 gives 3 + 3 = 6. Answer: 6. The distractors: 11 comes from dividing only the 4 by 2 instead of the whole of the top, giving 10 − 2 + 3; 4.5 comes from dividing the + 3 by 2 as well, giving (10 − 4 + 3) ÷ 2; 0 comes from subtracting the wrong way round on the top, giving (4 − 10) ÷ 2 = −3 and then −3 + 3.
- (a) £2.80 — Method: find the cost of one pencil first, then use it to find the cost of 7 pencils. Working: £1.20 ÷ 3 = £0.40 for one pencil, and £0.40 × 7 = £2.80. So 7 pencils cost £2.80. Distractor £8.40 comes from multiplying £1.20 by 7 without first dividing by 3 to find the cost of one pencil. Distractor £1.20 comes from assuming the price stays the same no matter how many pencils are bought. Distractor £0.40 is the cost of one pencil, found correctly but never multiplied by 7.
- (b) (5, −2) — Method: subtract column vectors by subtracting the bottom vector's top number from the top vector's top number, and doing the same for the bottom numbers. Working: top numbers 7 − 2 = 5; bottom numbers 3 − 5 = −2. Answer: u − v = (5, −2). A candidate who drops the negative sign gets (5, 2). A candidate who works out v − u instead of u − v gets (−5, 2). A candidate who adds instead of subtracts gets (9, 8).
- (c) 3/16 — The probability that Priya misses a shot is 1 − 3/4 = 1/4. The two shots are independent, so multiply the probability of scoring the first shot by the probability of missing the second shot: 3/4 × 1/4 = 3/16. Choosing 9/16 comes from using the probability of scoring, 3/4, for both shots instead of switching to the miss probability for the second shot, 3/4 × 3/4 = 9/16, which is the probability that she scores with both shots. Choosing 1/2 comes from subtracting the two probabilities instead of multiplying them, 3/4 − 1/4 = 1/2. Choosing 3/4 comes from giving only the probability that she scores with the first shot, forgetting to combine it with what happens on the second shot.
- (b) 1.45 ≤ m < 1.55 — The flour's mass is labelled 1.5 kg, correct to the nearest 0.1 kg, so half of 0.1 kg is added to and subtracted from 1.5 kg to find the interval: 1.5 − 0.05 = 1.45 and 1.5 + 0.05 = 1.55, giving 1.45 ≤ m < 1.55. '1.4 ≤ m < 1.6' comes from taking the whole 0.1 kg as the margin either side, instead of half of it. '1.45 < m ≤ 1.55' comes from writing the inequality signs the wrong way round — the lower bound should be included and the upper bound excluded, not the other way round. '1.45 ≤ m ≤ 1.55' comes from including the upper bound, when the convention is that the upper bound is never actually reached.
- (b) 4 — Adding the two equations: the y-terms, −2y and +2y, cancel, and the x-terms combine to 5x + 3x = 8x; the right-hand sides add to 16 + 16 = 32. This gives 8x = 32, so x = 4. A candidate who adds only one of the right-hand sides, instead of both, would get 8x = 16, so x = 2. A candidate who divides 32 by 4 instead of 8 would get x = 8. A candidate who subtracts the equations instead of adding them, getting 2x − 4y = 0, and then wrongly assumes y = 0, would get x = 0.
- (b) 20 — The gradient is the change in p divided by the change in t: (90 − 30) ÷ (5 − 2) = 60 ÷ 3 = 20. Choosing 0.05 comes from dividing the change in t by the change in p, the wrong way round (3 ÷ 60). Choosing −20 comes from subtracting the coordinates in the wrong order for one part of the calculation, for example (30 − 90) ÷ (5 − 2), giving a negative value. Choosing 18 comes from dividing the second p-coordinate by the second t-coordinate directly (90 ÷ 5) instead of using the change between the two points.
- (c) 20 cm² — The area of a triangle is half of base × height. First, base × height = 8 × 5 = 40. Half of 40 is 20 cm². 40 cm² forgets to halve and just gives base × height. 13 cm² adds the base and height together instead of multiplying them. 80 cm² doubles base × height instead of halving it.
- (c) 3⁻³ — Method: when dividing powers of the same base, subtract the index of the number you are dividing by from the index of the number being divided, keeping them in the order the question writes them. Working: 2 − 5 = −3, so 3² ÷ 3⁵ = 3⁻³. It is worth checking this against the numbers: 3² = 9 and 3⁵ = 243, and 9 ÷ 243 = 1/27, which is 3⁻³. 3³ comes from subtracting the other way round, 5 − 2 = 3, which reverses the sign of the index and gives 27 instead of 1/27. 3⁷ comes from working out 2 + 5 = 7, which is the rule for multiplying powers, not dividing them. 3¹⁰ comes from multiplying the indices, 2 × 5 = 10, instead of subtracting them. Answer: 3⁻³.
- (a) 17 — Substitute a = 3 into 5a + 2: 5 × 3 + 2 = 15 + 2 = 17. 10 comes from treating 5a as 5 + a instead of 5 × a, giving 5 + 3 + 2. 25 comes from adding the 2 to a before multiplying by 5, 5 × (3 + 2). 15 comes from working out 5 × 3 correctly but forgetting to add the 2.
- (d) 30 cm — Convert 300 m to centimetres: 300 m = 30 000 cm. The scale 1 : 1000 means the map length is the real length divided by 1000: 30 000 ÷ 1000 = 30, giving 30 cm. Dividing the unconverted 300 (metres, not centimetres) by 1000 gives 0.3 cm, far too small to be a real map measurement. Dividing by 10 000 instead of 1 000 gives 3 cm, ten times too small. Multiplying instead of dividing, 300 × 1000 = 300 000, gives 300 000 cm — using the scale the wrong way round, as if the map were bigger than real life.
- (c) 27 — Each of the 3 digits can be chosen independently for each of the 3 positions, so multiply: 3 × 3 × 3 = 27. 9 comes from multiplying only two of the three positions, 3 × 3, and forgetting the third. 6 comes from working out 3 × 2 × 1 = 6, which counts codes where digits do not repeat, but the question allows repeated digits. 3 comes from considering only one digit position.
- (c) £90 — Method: split the total amount into the number of parts shown by the ratio, then find Chloe's share. Working: the ratio 2:3 has 2 + 3 = 5 parts, so one part is £150 ÷ 5 = £30, and Chloe's share is 3 × £30 = £90. So Chloe receives £90. Distractor £60 is Ben's share, not Chloe's. Distractor £75 comes from splitting the money into two equal halves, ignoring the ratio. Distractor £30 is the value of one part, found correctly but never multiplied by 3.
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.