Sample paper · GCSE Foundation · grades 1–5
GCSE Foundation sample Paper 2 (calculator)
The real Paper 2 is 1.5 hour 30 minutes and 80 marks, 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 a calculator allowed. AO1 / AO2 / AO3 at this tier: 50% / 25% / 25%.
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Answer key: GCSE Foundation sample Paper 2 (calculator)
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- (d) 8 — Method: whatever is inside the brackets is worked out first, and the multiplication is carried out afterwards. Working: inside the brackets, −3 + 7 = 4, and then 2 × 4 = 8. Answer: 8. The distractors: 1 comes from ignoring the brackets and multiplying first, giving 2 × (−3) = −6 and then −6 + 7 = 1; −20 comes from reading −3 + 7 as −(3 + 7) = −10, so that 2 × (−10) = −20; 6 comes from working the brackets out correctly and then adding the 2 instead of multiplying by it, giving 2 + 4 = 6.
- (d) 8 — Method: write the nth term of the sequence, 2000 + 800(n − 1), and find the smallest whole n for which it is greater than 7000. Working: 2000 + 800(n − 1) > 7000, so 800(n − 1) > 5000, giving n − 1 > 6.25. Since n − 1 must be a whole number, the smallest value is 7, so n = 8. Check: year 8's total is 2000 + 800 × 7 = 7600, which is more than £7000, while year 7's total is 2000 + 800 × 6 = 6800, which is not. Answer: year 8. 7 comes from rounding 6.25 to the nearest whole number, 6, and then adding 1, instead of rounding up to the next whole number before adding 1. 6 comes from using 6.25 rounded down to 6 as the year number directly, without adding the 1 needed to convert from the number of increases to the year number. 9 comes from adding one extra year beyond the year that already satisfies the condition.
- (a) 3/4 — Write January's total over February's total: 360/480. Both numbers share a factor of 120, so dividing top and bottom by 120 gives 3/4. Choosing 4/3 comes from writing February's amount over January's amount, the wrong way round. Choosing 1/4 comes from finding the difference between the two months (480 − 360 = 120) and writing it over February's amount, instead of using January's amount. Choosing 3/7 comes from writing January's amount over the total received across both months (360 out of 840), instead of over February's amount alone.
- (d) 18.8 cm — Arc length is a fraction of the full circumference. Here 120 ÷ 360 = 1/3, and the full circumference is 2 × 3.14 × 9 = 56.52 cm, so the arc is 56.52 ÷ 3 = 18.84 cm, which rounds to 18.8 cm. Forgetting the fraction and giving the full circumference rounds to 56.5 cm. Using 9 cm as a diameter instead of a radius gives an arc of 9.4 cm. Using 120 ÷ 180 instead of 120 ÷ 360 for the fraction gives an arc of 37.7 cm.
- (b) 200 — Method: list the equally likely outcomes for the two coins before writing any probability, then multiply by the number of throws. Working: the equally likely outcomes are head then head, head then tail, tail then head, and tail then tail, so there are 4 of them. Two of those 4 give one head and one tail, so the probability is 2/4, which is 1/2. Over 400 throws the expected number is 400 × 1 ÷ 2 = 200. Answer: about 200 of the throws would be expected to give one head and one tail. The distractors: 133 comes from treating two heads, two tails and one of each as three equally likely results and working out 400 ÷ 3 = 133.3, then rounding; 100 comes from counting only head then tail as a success, giving 400 × 1 ÷ 4 = 100; 300 is the expected number of throws that do not give two heads, 400 × 3 ÷ 4 = 300.
- (c) No — 8 from one class is too small to represent the school. — Method: judge reliability by asking whether the sample is both large enough, and spread across the population, relative to what it is meant to represent. Working: 8 pupils is a tiny fraction of the school's 1,000 pupils, and all 8 come from a single class rather than a range of year groups, so the sample is both too small and too narrow to represent the whole school reliably. She is not right. Saying any sample size gives an equally reliable estimate ignores that reliability generally improves with a larger, more representative sample. Saying the method is unreliable because it was not done online is not a reason connected to sample size or representativeness at all. Saying 8 is reliable because it is more than half her class compares the sample to the wrong population — the school has 1,000 pupils, not one class. Always judge a sample's size against the population it is meant to represent, not against a smaller group within it.
- (b) £35 — Method: round each price to the nearest pound, then add the rounded prices. Working: £9.95 rounds to £10, £19.90 rounds to £20 and £4.99 rounds to £5, and £10 + £20 + £5 gives the estimate. Answer: £35. The distractors: £40 comes from rounding each price up to the nearest £10 rather than to the nearest pound, giving £10 + £20 + £10; £32 comes from cutting the pence off each price instead of rounding it, giving £9 + £19 + £4; £34.84 is the exact total, worked out in full when the question asks for an estimate.
- (b) y = 2x − 1 — Method: in a rule that multiplies and then adds, the multiplier is the step in the outputs for each step of 1 in the input, and the number added on is the output when the input is 0. Working: the inputs 0, 1, 2 rise in ones while the outputs −1, 1, 3 rise by 2 each time, so the input is multiplied by 2; an input of 0 gives 2 × 0 = 0 and the output must be −1, so 1 is subtracted. Answer: y = 2x − 1, checked against the last pair by 2 × 2 − 1 = 3. The distractors: y = 2x + 1 comes from finding the multiplier 2 correctly and then reading the output at an input of 0 as +1 instead of −1; y = x − 1 comes from taking the multiplier as 1 because the inputs go up in ones, instead of using the step in the outputs; y = 3x − 1 comes from reading the largest output, 3, as the multiplier.
- (b) 80% — Percentage = (40 ÷ 50) × 100 = 80%.
- (d) 0 — A vertex is a sharp corner point where edges meet. A cylinder has two curved, circular edges but no sharp corner points at all, so it has 0 vertices, making that the correct answer. '2' wrongly treats the two circular edges themselves as vertices, but an edge is not the same as a vertex. '4' overcounts by treating each circular edge as if it had two end-vertices, which does not apply to a continuous curved edge. '1' wrongly imagines the curved surface itself forming a single corner point, which it does not.
- (d) 140 — 90 pupils were in Year 8 on the coach branch. The minibus branch has 200 − 150 = 50 pupils, and all of them are Year 8 too, so the total number of Year 8 pupils is 90 + 50 = 140. Writing 90 alone is wrong because it only counts the coach's Year 8 pupils and misses the minibus ones. Writing 60 is wrong because that is the number of Year 9 pupils on the coach (150 − 90 = 60), not Year 8 at all. Writing 50 alone is wrong because it only counts the minibus pupils and misses the coach's Year 8 pupils. The total is 140.
- (d) the first mark after 0 — Method: when a unit length is split into equal parts, each gap is one part of the whole, so 1/4 is one gap along from 0. Working: four equal parts means each gap measures 1/4, so the marks after 0 stand for 1/4, 2/4, 3/4 and 4/4. One gap along from 0 is therefore the mark for 1/4. Answer: the first mark after 0. The distractors: the second mark after 0 comes from counting 0 itself as the first mark; the third mark after 0 comes from counting back from the 1 end instead of forward from 0; the fourth mark after 0 comes from reading the 4 in the denominator as the number of the mark rather than the number of parts.
- (b) 3 — Distance from the y-axis is given by the size of the x-coordinate, which is 3. 8 comes from using the size of the y-coordinate instead of the x-coordinate. −8 comes from using the y-coordinate and keeping its negative sign, but a distance is never negative. 11 comes from adding the sizes of both coordinates, 3 + 8, instead of using the x-coordinate alone.
- (d) 3/16 — Convert 4 kg to grams: 4 kg = 4000 g. Form the fraction 750/4000. Both numbers share a factor of 250, so 750 ÷ 250 = 3 and 4000 ÷ 250 = 16, giving 3/16. 16/3 comes from writing the fraction the wrong way round, as 4000/750. 15/8 comes from converting 4 kg using ×100 instead of ×1000, treating it as 400 g, then simplifying 750/400. 3/20 comes from dividing 750 by 250 correctly to get 3, but dividing 4000 by 200 instead of 250, giving 3/20.
- (d) RHS, using AM as common side — Triangle ABM and triangle ACM both have a right angle at M, since AM is perpendicular to BC. AB and AC are the hypotenuses of the two triangles and are equal, and AM is a side common to both triangles, giving a right angle, equal hypotenuses and one further equal side, exactly RHS, so 'RHS, using AM as common side' is correct. 'SAS, right angle as included angle' wrongly treats the right angle at M as included between AB and AM, but AB is the hypotenuse, not one of the two sides forming that right angle. 'SSS, using BM = CM as a fact' wrongly assumes BM equals CM as a given fact, when this is only true because of the RHS congruence, not before it, so it cannot be used to prove that congruence. 'ASA, AB as the included side' again wrongly labels a side as if it could sit between two angles when only one angle, the right angle, is actually known.
- (d) 600 — Method: the number of seats is the number of rows multiplied by the number of seats in each row, so round each number to 1 significant figure and then multiply the rounded values, which is quick because a product of two multiples of ten is found by multiplying the non-zero digits and attaching the zeros. Working: 21 rounds to 20 and 29 rounds to 30; 2 × 3 = 6, and 20 and 30 carry one zero each, so two zeros follow the 6. Answer: about 600 seats. The distractors: 50 comes from adding the two rounded numbers instead of multiplying them, 20 + 30; 60 comes from multiplying 20 by the 3 of 30 and forgetting the zero in 30; 6,000 comes from attaching three zeros to 2 × 3 when 20 and 30 provide only two between them.
- (c) x ≤ 4 — Subtract 9 from both sides: −2x ≥ 1 − 9 = −8. Divide both sides by −2, flipping the inequality: x ≤ 4. A candidate who divides by −2 but forgets to flip the inequality gets x ≥ 4. A candidate who divides −8 by −2 but keeps a negative sign gets x ≤ −4. A candidate who miscalculates 1 − 9 as −10 gets x ≤ 5.
- (d) 8 cm — Convert 4 m to centimetres: 4 m = 400 cm. The scale 1 : 50 means the real object is 50 times the plan, so the plan length is the real length divided by 50: 400 ÷ 50 = 8, giving 8 cm. Using the length in metres instead of centimetres, 4 ÷ 50 = 0.08, gives 0.08 cm, far too small to draw. Misplacing a digit in the division gives 80 cm, ten times too big. Multiplying instead of dividing, 400 × 50 = 20 000, gives 20 000 cm — using the scale in the wrong direction, as if going from plan to real life instead of real life to plan.
- (b) 0 > −1 — Method: test each statement by placing both of its numbers on a number line; the greater number is the one further to the right. Working: every negative number lies to the left of zero, so zero is greater than −1. Among the negatives, −3 lies to the right of −5, so −5 is not greater than −3. Answer: 0 > −1. The distractors: −5 > −3 comes from ordering the negatives by the size of their digits, so that 5 makes −5 look the larger; −3 > 0 comes from ignoring the minus sign and comparing 3 with 0; 0 < −1 comes from the belief that zero is the smallest number there is, so that even a negative number is above it.
- (d) 20 — Gradient = change in T ÷ change in t = (140 − 60) ÷ (6 − 2) = 80 ÷ 4 = 20. A student who subtracts in the wrong order gets −20. A student who divides 80 by 2 instead of 4 gets 40. A student who wrongly treats the line as passing through the origin and uses the point (2, 60) on its own gets 60 ÷ 2 = 30.
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.