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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- (c) 18 — 45 ÷ 5 = 9, and 2 × 9 = 18. A candidate who stops after finding one fifth gets 9. A candidate who uses 3/5 instead of 2/5 gets 27. A candidate who uses 4/5 instead of 2/5 gets 36.
- (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.
- (d) 600 g — Find the flour used for one person: 240 ÷ 6 = 40 g per person. Then scale up to 15 people: 40 × 15 = 600 g. Dividing 240 by 15 and then multiplying by 6 gives 96 g — that uses the ratio the wrong way round, scaling down instead of up. Stopping at 40 g only gives the amount for one person, not for 15. Adding 15 − 6 = 9 to 240 gives 249 g, which treats the number of people and the grams of flour as if they were the same kind of quantity — proportion means scaling, not adding. The recipe needs 600 g of flour.
- (d) 1.5 m — By Pythagoras' theorem, diagonal² = 1.2² + 0.9² = 1.44 + 0.81 = 2.25, so diagonal = √2.25 = 1.5 m. 2.1 m comes from simply adding the two sides (1.2 + 0.9) instead of using Pythagoras' theorem. 0.3 m comes from subtracting the two sides (1.2 − 0.9) instead. 2.25 m comes from correctly finding 1.2² + 0.9² = 2.25 but forgetting to take the square root at the end.
- (c) 26 — The number who play football or basketball is 27 + 21 − 9 = 39, subtracting the 9 who play both so they are not counted twice. The number who play neither is then 65 − 39 = 26. Giving 39 stops after finding the number who play football or basketball, without taking the complement within the 65 members. Subtracting all three given numbers from 65, 65 − 27 − 21 − 9 = 8, treats the 9 who play both as a separate group rather than a correction for double-counting. Adding 27 and 21 without removing the double-counted 9, then subtracting that total from 65, gives 65 − (27 + 21) = 17.
- (b) 5.5 kg — Method: with an even number of values the median is the mean of the two middle values, taken once the data are in order of size. Working: the eight masses are already in order and 8 ÷ 2 = 4, so the middle pair are the 4th and 5th values, 5 kg and 6 kg; the median is (5 + 6) ÷ 2 = 5.5 kg. Answer: 5.5 kg. The distractors: 5 kg comes from reading the 4th value and stopping there instead of averaging the middle pair; 8 kg comes from working out the range, 11 − 3, which measures spread rather than centre; 4 kg comes from writing down the modal mass, the only value that occurs twice, instead of the median.
- (d) 2⁷ — When multiplying powers of the same base, the indices add: 3 + 4 = 7, so 2³ × 2⁴ = 2⁷. Multiplying the indices instead of adding them gives 3 × 4 = 12, so 2¹². Subtracting the indices instead of adding them gives 4 − 3 = 1, so 2¹. Multiplying the bases together as well as adding the indices gives 2 × 2 = 4, so 4⁷.
- (b) y = 7 — Method: one of the equations already gives the value of x, so that value is substituted into the other equation. Working: replacing x with 2 in x + y = 9 gives 2 + y = 9, so y = 9 − 2 = 7. Answer: y = 7, and the pair x = 2 with y = 7 satisfies both equations. The distractors: y = 11 comes from rearranging x + y = 9 as y = 9 + x and adding instead of subtracting; y = 9 comes from ignoring the x term and reading the total as the value of y; y = 4.5 comes from halving 9, as though x and y had to be equal.
- (c) 9:15:21 — 6 : 10 : 14 simplifies to 3 : 5 : 7 (divide every part by 2). Multiplying every part of 3 : 5 : 7 by 3 gives 9 : 15 : 21, so 9 : 15 : 21 is equivalent to 6 : 10 : 14. Adding 2 to every part of 6 : 10 : 14 gives 8 : 12 : 16, which is not equivalent — ratios are equivalent when every part is multiplied by the same number, not when the same number is added to every part. Doubling only the first two parts, 6 × 2 = 12 and 10 × 2 = 20, but leaving the third part unchanged at 14, gives 12 : 20 : 14 — a scaling applied to two parts and not the third. Cancelling the first two parts correctly, 6 ÷ 2 = 3 and 10 ÷ 2 = 5, then treating the three numbers as a sequence and making the third part the sum of the first two, 3 + 5 = 8, gives 3 : 5 : 8 — the third part was never divided by 2 at all.
- (b) $\binom{8}{3}$ — The column vector is (end − start) in each coordinate: (9 − 1, 4 − 1) = (8, 3), written as $\binom{8}{3}$. $\binom{3}{8}$ swaps the horizontal and vertical components. $\binom{10}{5}$ comes from adding the coordinates instead of subtracting them. $\binom{−8}{−3}$ comes from working out start minus end instead of end minus start.
- (c) 325 — Method: first find the relative frequency of NOT landing on red from the 200 spins, then scale that up to 500 spins. Working: non-red results = 50 + 80 = 130, out of 200 spins, so P(not red) = 130 ÷ 200 = 0.65. Expected non-red results in 500 spins = 500 × 0.65 = 325. Answer: 325. Watch out: writing down 175 finds the expected number of RED results instead, 70 ÷ 200 × 500 = 175, answering the opposite of what was asked. Writing down 250 assumes landing red or not landing red must be a fair 50-50 split, but the spinner is biased and the actual results do not split evenly. And writing down 130 stops after finding how many of the 200 spins were non-red and forgets to scale that figure up to the 500 spins asked for.
- (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.
- (a) They always charge the same, since 3(2n + 4) = 6n + 12. — Expand Advert A's formula by multiplying both terms inside the bracket by 3: 3 × 2n = 6n, and 3 × 4 = 12, giving 3(2n + 4) = 6n + 12, which is identical to Advert B's formula — so the two adverts always charge the same amount, whatever n is. Getting 6n + 4 comes from multiplying the 2n by 3 but leaving the 4 unmultiplied. Getting 2n + 7 comes from adding 3 to the bracket instead of multiplying by it. Saying it depends on n avoids expanding the bracket at all — once expanded, both formulas are identical for every value of n, so the cost can be compared directly.
- (b) 19.0 g/cm³ — Density = mass ÷ volume. 342.6 ÷ 18 = 19.0333…, which rounds to 19.0 g/cm³ (1 d.p.). 6166.8 g/cm³ comes from multiplying the mass by the volume instead of dividing (342.6 × 18). 324.6 g/cm³ comes from subtracting the volume from the mass (342.6 − 18) instead of dividing. 0.1 g/cm³ comes from dividing the volume by the mass instead of the mass by the volume (18 ÷ 342.6 = 0.0525…, rounded to 1 d.p.).
- (b) No — the angle given is not the included angle — Method: check whether the given angle sits between the two given sides, since SAS requires the included angle. Working: sides AB and BC meet at vertex B, so the angle between them is angle B — but the angle given is angle A, which is not between the two given sides, and the same mismatch happens in triangle DEF. Options: 'two sides and one angle match' restates SAS's ingredients without checking their positions, which is exactly Meera's mistake; 'SSS needs three equal sides' is a true fact about a different condition, but it is not the reason Meera is wrong here; 'SAS allows any equal angle' states a rule that is not how SAS works, since the angle must be the included one. Answer: no, the angle given is not the included angle.
- (a) 8.15 ≤ y < 8.25 — Rounding to 1 decimal place means the error interval spans half of 0.1, so 0.05, either side of 8.2: 8.2 − 0.05 = 8.15 and 8.2 + 0.05 = 8.25. The lower bound uses ≤ because 8.15 itself rounds to 8.2, but the upper bound uses < because 8.25 would round up to 8.3. So the error interval is 8.15 ≤ y < 8.25. A candidate who used the wrong rounding band gave 8.1 ≤ y < 8.2. A candidate who used a strict inequality at both ends wrote 8.15 < y < 8.25, wrongly excluding 8.15 itself. A candidate who added the full 0.1 instead of half of it wrote 8.2 ≤ y < 8.3.
- (d) 5x² — 5x² is a single term: a product of a number and a power of a letter, with no + or − sign attached to anything else. A candidate who picks 5x² + 3x is choosing the expression that contains that term, rather than the term on its own. A candidate who picks 5x² = 20 is drawn in by the equals sign and misses that this makes it an equation. A candidate who picks A = 5x² thinks any statement with a letter and a number combined must be a formula.
- (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.
- (a) 3 — Method: the fourth root of a number is the positive value that gives that number when it is multiplied by itself four times. Working: 2 × 2 × 2 × 2 = 16, which is too small, and 3 × 3 × 3 × 3 = 9 × 9 = 81. Answer: 3. The distractors: 9 comes from taking the square root of 81 instead of its fourth root; 4.5 comes from taking the square root and then halving it, as though a fourth root were half a square root; 20.25 comes from dividing 81 by 4, treating the root's index as a divisor.
- (a) 1260 — To increase by 5%, multiply by 1.05 (100% + 5%). 1200 × 1.05 = 1260. 60 comes from working out only the increase (1200 × 0.05) and forgetting to add it to the original population. 1205 comes from adding 5 directly to 1200 instead of 5% of 1200. 1800 comes from multiplying by 1.5, using 50% instead of 5%.
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.