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) 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) 16 — Let a be the number of adult tickets and c the number of child tickets: a + c = 40 and a = c + 8. Substituting the second equation into the first: (c + 8) + c = 40, so 2c = 32, and c = 16. A candidate who reports the number of adult tickets instead of child tickets would get a = 24. A candidate who splits the 40 tickets evenly, ignoring the difference of 8, would get 20. A candidate who forgets to divide 2c = 32 by 2 would write c = 32.
- (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.
- (b) trapezium — A quadrilateral with exactly one pair of parallel sides is a trapezium; field ABCD has AB parallel to DC and no other pair of parallel sides, so trapezium is correct. Parallelogram requires BOTH pairs of opposite sides to be parallel, but the plan says only AB and DC are parallel. Rhombus requires all four sides to be equal in length, which is not stated here. A kite is defined by two pairs of adjacent equal sides, not by having a pair of parallel sides, so it does not match this description either.
- (c) 24 — Silver, bronze and copper cover every counter, so their probabilities sum to 1: P(copper) = 1 − 1/2 − 1/8 = 3/8. Number of copper counters = 3/8 × 64 = 24. Multiplying the silver probability by 64 gives 32, the number of silver counters, not copper. Multiplying the bronze probability by 64 gives 8, the number of bronze counters. Adding the silver and bronze probabilities (1/2 + 1/8 = 5/8) and multiplying by 64 gives 40, the combined number of silver and bronze counters, not the copper count.
- (c) 28 — Method: multiply the number of whole symbols by the value of one symbol, then add the value of any half symbol shown. Working: 3 whole symbols represent 3 × 8 = 24 cars. The half symbol represents 4 cars. Total cars sold in March = 24 + 4 = 28. Leaving out the half symbol, 3 × 8 = 24, undercounts by exactly the value of that half symbol. Treating the half symbol as if it were a full symbol, 4 × 8 = 32, overcounts because it doubles the value the half symbol is worth. Giving 3.5 reports the number of symbols shown, not the number of cars they represent — the key still needs to be applied. Always apply the key to every symbol shown, including a half symbol, rather than reading off the symbol count itself.
- (d) 1.2 × 10⁵ — 4 × 3 = 12, and 3 + 1 = 4, giving 12 × 10⁴ — but 12 is not between 1 and 10, so this must be rewritten as 1.2 × 10⁵. Stopping at 12 × 10⁴ without rewriting it leaves the coefficient out of range. Rewriting 12 as 1.2 but leaving the exponent at 4 instead of increasing it to 5 gives 1.2 × 10⁴, which is ten times too small. Adding the coefficients instead of multiplying them gives 4 + 3 = 7, so 7 × 10⁴.
- (a) P = 100I / (RT) — Method: undo the operations done to P in reverse order — multiply by 100, then divide by R and by T. Working: I = PRT/100, so 100I = PRT, so P = 100I / (RT). Answer: P = 100I / (RT). P = IRT/100 comes from leaving R and T in the numerator instead of moving them to the denominator. P = 100RT/I comes from swapping P and I when rearranging. P = 100I/R comes from dividing by R only and forgetting to also divide by T.
- (a) 2 : 3 — Simplify the volume ratio first: 64 : 216 divides by 8 to give 8 : 27. Volumes scale with the cube of the height ratio, so take the cube root of each part: the cube root of 8 is 2, and the cube root of 27 is 3, giving a height ratio of 2 : 3. Giving 3 : 2 has the ratio the right way round for larger to smaller, not smaller to larger. Giving 8 : 27 is the simplified volume ratio, without cube-rooting it. Giving 64 : 216 is the volume ratio before it has even been simplified.
- (a) 250° — The back bearing (the bearing of A from B) differs from the bearing of B from A by exactly 180°. Because the given bearing, 070°, is less than 180°, add 180°: 070 + 180 = 250°, so the bearing of A from B is 250°. Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 110° comes from subtracting 180° from 070° and dropping the negative sign (070 − 180 = −110) instead of adding 180°. Choosing 160° comes from adding only 90° instead of 180° (070 + 90 = 160).
- (a) unlikely — 3/8 = 3 ÷ 8 = 0.375, which is between 0 and 0.5, so the probability is best described as unlikely. Rounding 3/8 loosely to 'about a half' without converting it properly leads to evens, which is wrong since 0.375 is noticeably below 0.5. Reading 3/8 as the complement and using 1 − 3/8 = 5/8 = 0.625 instead gives likely, but the question asks about the probability of 3/8 itself, not its complement. Flipping the fraction to 8/3 ≈ 2.67, which is impossible for a probability, and rounding it towards the top of the scale gives certain.
- (d) 1 7/12 — Method: convert both mixed numbers to improper fractions with a common denominator, then subtract. Working: 3 1/4 = 39/12 and 1 2/3 = 20/12, so 39/12 − 20/12 = 19/12 = 1 7/12. Answer: 1 7/12. Chloe's method, subtracting whole numbers (3−1=2) and fraction parts (2/3−1/4=5/12) separately without exchanging, gives 2 5/12. 1 3/4 comes from converting 2/3 to twelfths incorrectly as 6/12 instead of 8/12, then subtracting. 8 comes from converting both mixed numbers to improper fractions correctly (13/4 and 5/3) but then subtracting numerators and denominators separately: (13−5)/(4−3) = 8/1.
- (d) (x − 8)(x + 8) — x² − 64 = x² − 8², a difference of two squares, which factorises as (x − 8)(x + 8). A candidate who treats it as a perfect square with a repeated negative factor gets (x − 8)(x − 8), which expands to x² − 16x + 64 — wrong on both the middle and constant terms. A candidate who uses a repeated positive factor gets (x + 8)(x + 8), which expands to x² + 16x + 64. A candidate who picks a different factor pair of 64, such as 4 and 16, without checking that the middle term cancels, gets (x − 4)(x + 16), which expands to x² + 12x − 64 — the wrong middle term.
- (d) 40 cm — Method: convert the real length to centimetres, then divide by the scale factor. Working: 10 m = 1000 cm. 1000 ÷ 25 = 40 cm. Wrong options: 0.4 cm comes from dividing 10 by 25 without converting metres to centimetres first; 25000 cm comes from multiplying instead of dividing (1000 × 25); 4 cm comes from converting 10 m to 100 cm (using ×10 instead of ×100) before dividing.
- (b) SAS — Method: check which condition matches two sides and the angle between them, since that is what has been measured for each bed. Working: the 60° angle is marked at the corner where the 5 m and 7 m edges meet, so it is the included angle — this is two Sides and the included Angle, SAS. Options: SSS would need a third side measured, but only two edges are known; ASA would need two angles and the side between them, but only one angle is measured; RHS needs a right angle, and 60° is not a right angle. Answer: SAS.
- (a) 12 — Method: for any two numbers, their highest common factor multiplied by their lowest common multiple equals the product of the two numbers. This holds because the HCF collects every prime factor the two numbers share, and the LCM collects every prime factor that appears in either number, so between them they use each prime factor of the two numbers exactly once — the same primes as the product. Working: 4 × 60 = 240, and 240 ÷ 20 = 12. 15 comes from working out 60 ÷ 4 = 15, dividing the wrong pair of numbers. 16 comes from working out 20 − 4 = 16, subtracting the highest common factor instead of using the product rule. 240 is 4 × 60, the product of the highest common factor and the lowest common multiple, left un-divided by 20. Answer: 12.
- (c) 8n + 7 — Method: find the weekly increase, then find the constant by adjusting the week 1 total. Working: the total goes up by £8 each week (23 − 15 = 8), so the coefficient of n is 8. The constant is the week 1 total minus the common difference: 15 − 8 = 7. Answer: the nth term is 8n + 7. 8n + 15 comes from using the week 1 total, 15, as the constant without subtracting the common difference. 8n − 1 comes from a slip in working out the constant, subtracting the common difference twice (15 − 8 − 8 = −1) instead of once. 7n + 8 comes from swapping the weekly increase and the constant.
- (a) 15.3 litres — Squash : water = 2 : 9, so water is 9 ÷ 2 = 4.5 times the amount of squash. Multiply: 3.4 × 4.5 = 15.3 litres. Using the multiplier upside down — treating squash as 9 ÷ 2 times water, when it is water that is 9 ÷ 2 times squash — and calculating 3.4 × (2 ÷ 9) gives about 0.8 litres (to 1 d.p.); that would be the squash needed for 3.4 litres of water, not the water needed for 3.4 litres of squash. Adding the difference between the ratio parts, 9 − 2 = 7, to the squash amount, 3.4 + 7 = 10.4, mistakes a ratio for a fixed extra amount. Using the total number of parts, 2 + 9 = 11, so the multiplier 11 ÷ 2 = 5.5, gives 3.4 × 5.5 = 18.7 litres — that finds the total mix from the squash amount, not the water alone.
- (d) 10 — Method: list the pairs systematically, taking each sweet in turn and pairing it only with the sweets that come after it, so that no pair is written down twice. Working: numbering the sweets 1 to 5, the first sweet pairs with 4 others, the second pairs with 3 sweets that come after it, the third with 2 and the fourth with 1, so the total is 4 + 3 + 2 + 1 = 10. Answer: 10. The distractors: 20 comes from working out 5 × 4 and never halving, which counts each pair twice, once in each order; 25 comes from working out 5 × 5, which allows the same sweet to be chosen twice; 9 comes from adding the 5 choices and the 4 remaining choices instead of combining them as a selection of two.
- (d) 3 : 7 — There are 9 apples and 21 oranges, giving the ratio 9 : 21. The highest common factor of 9 and 21 is 3: 9 ÷ 3 = 3 and 21 ÷ 3 = 7, so the simplest form is 3 : 7. Giving 9 : 21 has not been simplified. Giving 7 : 3 swaps the order, comparing oranges to apples instead of apples to oranges. Giving 9 : 12 compares the number of apples with the difference between the two amounts (21 − 9 = 12), not with the number of oranges.
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.