Sample paper · GCSE Higher · grades 4–9
GCSE Higher 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 Higher qualification — number 15%, algebra 30%, ratio, proportion and rates of change 20%, geometry and measures 20%, probability 7.5%, statistics 7.5% — with a calculator allowed. AO1 / AO2 / AO3 at this tier: 40% / 30% / 30%.
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Answer key: GCSE Higher sample Paper 2 (calculator)
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- (c) 19 — Without any restriction there would be 5 × 4 = 20 different sandwiches. The restriction removes exactly one combination, cheese and mustard on gluten-free bread, so subtract 1: 20 − 1 = 19. 20 comes from ignoring the restriction completely. 15 comes from removing the gluten-free bread altogether, as if none of the fillings were available on it, 5 × 3 = 15. 16 comes from removing the cheese and mustard filling completely, as if it were not available on any bread, 4 × 4 = 16.
- (b) £115, and C = 40 + 25h is a formula — Substitute h = 3 into the rule, multiplying before adding. The hours cost 25 × 3 = 75, and adding the call-out fee gives 40 + 75 = 115, so the charge is £115. The rule itself links two different quantities, C and h, and lets one be worked out from the other, so it is a formula; an expression would have no equals sign in it. Adding the fee before multiplying gives 65 × 3 = 195, which charges the call-out fee three times over, and stopping at 40 + 25 leaves £65, the charge for a single hour.
- (c) 200 students — Method: find the percentage who have a brother or a sister, taking care that the students with both are not counted twice, then take that percentage from 100% and apply the result to the 400 students. Working: 25% + 40% = 65%, but the 15% with both has been counted in each of those figures, so 65% − 15% = 50% have a brother or a sister; that leaves 100% − 50% = 50%, and 50% of 400 = 200. Answer: 200 students. The distractors: 260 students is 65% of 400, the number with a brother or a sister when the 15% overlap is counted twice; 140 students comes from taking that same uncorrected 65% away from the 400; 300 students comes from subtracting only the 25% with a brother and ignoring the sisters altogether.
- (b) 10.5 m² — The area of a trapezium is half of the sum of the parallel sides, multiplied by the width. Add the parallel sides: 2.5 + 4.5 = 7. Multiply by the width: 7 × 3 = 21. Half of 21 is 10.5 m². 21 m² forgets to halve, using the full (2.5+4.5)×3. 3.5 m² averages the two parallel sides, (2.5+4.5)÷2 = 3.5, but forgets to multiply by the width. 6.75 m² treats the bed as a triangle, using only the longer parallel side: half of 4.5 × 3.
- (c) 1/56 — Method: for draws without replacement, multiply a chain of three fractions where both the numerator (reds remaining) and the denominator (counters remaining) fall by one after each draw. Working: P(all three red) = 3/8 × 2/7 × 1/6 = 6/336 = 1/56. Answer: 1/56. Watch out: using 3/8 for all three draws (27/512) treats the counters as if they were replaced each time. Reducing only the numerator each draw (3/8 × 2/8 × 1/8) forgets that the total number of counters left in the bag also falls. And reducing only the denominator while keeping the numerator at 3 each time (3/8 × 3/7 × 3/6) forgets that a red counter has actually left the bag.
- (a) The modal size, 9, bought by more customers than any other — Method: work out both averages from the frequencies, then choose the one the shop can act on. Working: for the mean, multiply each size by the number of pairs sold at it and add: 6 × 4 + 7 × 5 + 8 × 8 + 9 × 13 + 10 × 10 = 340, and 340 ÷ 40 = 8.5, so the mean size is 8.5. The largest frequency is 13, which belongs to size 9, so the modal size is 9. The mean 8.5 is a size no customer in the record asked for, so 40 pairs of it would sit unsold, while 13 of the 40 customers wanted size 9, more than wanted any other size. Answer: the modal size, 9, bought by more customers than any other. The distractors: the mean size 8.5 does take account of all 40 pairs, but a mean of sizes is a summary figure and not a size the month's customers were buying; the mean size 8 comes from averaging the five sizes on sale, 6 + 7 + 8 + 9 + 10 = 40 and 40 ÷ 5 = 8, which ignores how many pairs were sold at each size and so treats the 4 pairs of size 6 as equal in weight to the 13 pairs of size 9; the range 4 comes from 10 − 6 and measures spread, so it says how wide a set of sizes the shop must stock, not which size to stock most of.
- (d) 3√5 — Split 45 into a perfect square times a factor: 45 = 9 × 5. Take the square root of each part separately: √45 = √9 × √5 = 3√5, since √9 = 3. Writing the perfect-square factor itself (9) as the coefficient instead of its root would give 9√5 — that trap comes from forgetting the last step, rooting 9. Multiplying 3 and 5 together instead of keeping them as coefficient and radicand gives 15, which throws away the surd entirely. Doubling the correct coefficient by mistake gives 6√5.
- (d) x = −2, y = −5 — 3x² + 12x + 7 rewrites as 3(x² + 4x) + 7, then as 3[(x + 2)² − 4] + 7, which simplifies to 3(x + 2)² − 5, since −3 × 4 + 7 = −5. Substituting x = −2: 3 × (−2)² = 12, 12 × (−2) = −24, so 12 − 24 + 7 = −5, confirming the minimum value −5 at x = −2: turning point x = −2, y = −5. Halving b instead of halving b/a — using −4 as the shift instead of −2 — lands on turning point x = −4, y = −41, which is wrong because when a ≠ 1 the shift inside the bracket is b/(2a), not b/a alone. Reading the bracket's sign directly as the turning point's x-coordinate gives x = 2, y = −5 — wrong, because (x + 2)² is zero at x = −2, not x = 2. Computing 12 − 7 = 5 instead of 7 − 12 = −5 flips the sign of the constant, giving x = −2, y = 5 — wrong, since the completed square's constant must be evaluated as 7 minus 12, not 12 minus 7. Whenever a ≠ 1, factor a out of the x² and x terms first, and always check a turning point by substituting back into the original equation.
- (c) 7800 kg/m³ — Method: build the conversion factor from the two unit changes separately — one for the mass, one for the volume. Working: 1 kg = 1000 g, so the mass figure is divided by 1000; 1 m = 100 cm, so 1 m³ = 100 × 100 × 100 = 1000000 cm³ and the volume figure is multiplied by 1000000. The density figure is therefore multiplied by 1000000 ÷ 1000 = 1000, giving 7.8 × 1000 = 7800. So the density of the metal is 7800 kg/m³. Distractor 780 kg/m³ comes from multiplying by 100 instead of 1000. Distractor 78000 kg/m³ comes from multiplying by 10000, an extra zero. Distractor 7.8 kg/m³ comes from not converting the units at all.
- (c) 1 : 300 — First convert both lengths to the same unit: 15 m = 1500 cm. The scale compares 5 cm on the drawing to 1500 cm in real life, so dividing both parts by 5 gives a scale of 1 : 300. A candidate who compares 5 to 15 without converting units gets 1 : 3. A candidate who converts 15 m to 150 cm, using the wrong conversion factor, gets 1 : 30. A candidate who converts 15 m to 15000 cm, again using the wrong conversion factor, gets 1 : 3000. The scale of the drawing is 1 : 300.
- (b) £100 — Over 250 games, the expected total winnings are 250 × (1/5) × £12 = £600, since a player wins on 1 of the 5 equally likely sections. The total cost of playing is 250 × £2 = £500. The players' expected profit is the winnings minus the cost: £600 − £500 = £100. Writing £500 is wrong because that is only the total cost of playing, without any winnings included. Writing £600 is wrong because that is only the total expected winnings, without subtracting what was paid to play. Writing £2,500 is wrong because it assumes a win on every single game (250 × £12 = £3,000) instead of using the 1-in-5 probability, then subtracts the cost: £3,000 − £500 = £2,500. The players' expected profit over the 250 games is £100.
- (a) 10 — √49 = 7 and ∛27 = 3, so √49 + ∛27 = 7 + 3 = 10. Treating the cube root as dividing by 3 instead of finding the cube root gives 27 ÷ 3 = 9, then 7 + 9 = 16. Multiplying the two roots instead of adding them gives 7 × 3 = 21. Ignoring the cube root symbol and using 27 as it stands gives 7 + 27 = 34.
- (b) 12 — Method: substitute the total cost into the formula, then subtract the fixed cost and divide by the cost per person. Working: 92 = 6p + 20, so 6p = 92 − 20 = 72, p = 72 ÷ 6 = 12. Answer: 12 people. 15.33 comes from dividing the whole £92 by £6 without first subtracting the £20 fixed cost: 92 ÷ 6 ≈ 15.33. 4.3 comes from swapping the two amounts round, subtracting £6 and dividing by £20: (92 − 6) ÷ 20 = 4.3. 18.67 comes from adding the fixed cost instead of subtracting it: (92 + 20) ÷ 6 ≈ 18.67.
- (d) 12 — First find the y-values at the two ends of the chord. At x = 2, y = 2 × 2² + 5 = 2 × 4 + 5 = 8 + 5 = 13. At x = 4, y = 2 × 4² + 5 = 2 × 16 + 5 = 32 + 5 = 37. The gradient of the chord is the change in y divided by the change in x: 37 − 13 = 24, then 24 ÷ (4 − 2) = 24 ÷ 2 = 12. Stopping after finding the change in y and not dividing by the change in x gives 24, which is not a gradient at all. Inverting the fraction, change in x divided by change in y, gives 2 ÷ 24 ≈ 0.08. Averaging the two y-values instead of finding the change between them gives (13 + 37) ÷ 2 = 50 ÷ 2 = 25. A gradient is always change in y over change in x — never the other way round, and never a single y-value.
- (b) 13 — Method: the straight-line distance between two points on a grid is the hypotenuse of a right-angled triangle whose shorter sides are the horizontal gap and the vertical gap between them, so Pythagoras' theorem applies. Working: one point is the origin, so the horizontal gap is 5 and the vertical gap is 12. Then d² = 5² + 12² = 25 + 144 = 169, so d = √169 = 13. Answer: 13. The distractors: 17 comes from adding the two gaps, 5 + 12, instead of adding their squares and taking the root; 12 comes from reading off the vertical gap alone and offering that as the whole distance; 7 comes from subtracting the gaps, 12 − 5, as though a distance were a difference.
- (b) 3w − 5 ≥ 16 — "Three times w, minus 5" translates to 3w − 5, and "is at least 16" means it must be 16 or more, giving 3w − 5 ≥ 16. A candidate who reads "at least" as a strict inequality writes 3w − 5 > 16. A candidate who misreads the wording and applies the subtraction before the multiplication writes 3(w − 5) ≥ 16. A candidate who reverses the direction of the inequality writes 3w − 5 ≤ 16.
- (d) 3 : 8 — Multiply both parts by 4 to clear the decimal: 0.75 × 4 = 3 and 2 × 4 = 8, giving 3 : 8, which has no common factor other than 1. Giving 75 : 200 multiplies by 100 instead of 4, and has not then been simplified down to 3 : 8. Giving 0.75 : 2 has not been converted into whole numbers at all. Giving 3 : 2 converts the first part correctly but leaves the second part unscaled.
- (a) the lens-shaped region where the two circles overlap — The region within 30 km of the ship is a circle of radius 30 km around the ship; the region within 20 km of the lighthouse is a circle of radius 20 km around the lighthouse. Since 30 + 20 = 50 km is more than the 40 km between them, and 30 − 20 = 10 km is less than 40 km, the two circles genuinely overlap, in a lens-shaped region where both conditions hold at once. "the whole smaller circle, since it fits inside the larger one" would only be true if the 40 km distance between the centres were small enough for one circle to swallow the other, which it is not here. "no overlap — 30 km and 20 km don't add to 40 km" wrongly assumes the two radii must add up to exactly the distance between the centres for an overlap to exist; any total greater than that distance is enough. "a straight strip, exactly halfway between the two" confuses this with an equidistant locus, which is a different problem from the one asked here.
- (d) x = 1 and x = 2 — Method: where a line meets a curve the two expressions for y are equal, so set them equal and solve the quadratic that results. Working: x² − 1 = 3x − 3 collects to x² − 3x + 2 = 0; factorising gives (x − 1)(x − 2) = 0, so x = 1 or x = 2, and each value gives the same y on both graphs. Answer: x = 1 and x = 2. The distractors: x = −1 and x = −2 come from factorising as (x + 1)(x + 2) and so reversing the sign of both roots; x = −1 and x = 4 come from moving the −3 across the equals sign without changing its sign, which gives x² − 3x − 4 = 0; x = 1 and x = −1 come from setting each expression equal to zero separately instead of equal to each other.
- (d) 13 — Method: the numbers of matchsticks form a sequence with a term-to-term rule, so count the first square in full and then add the repeated amount once for every extra square. Working: one square uses 4 matchsticks; a row of 4 squares has 3 extra squares after the first, and each of those adds 3 matchsticks, giving 3 × 3 = 9 to add on to the 4. Answer: 13. The distractors: 16 comes from counting each square as a separate set of 4 matchsticks, 4 × 4, and ignoring the shared sides; 12 comes from using 3 matchsticks for all four squares, 3 × 4, and forgetting that the first square needs a fourth side; 10 comes from adding the 3 only twice, as though a row of four squares had two extra squares rather than three.
How the 20 questions are shared out
- Number — 3 questions (15% of the qualification)
- Algebra — 6 questions (30% of the qualification)
- Ratio, proportion and rates of change — 4 questions (20% of the qualification)
- Geometry and measures — 4 questions (20% 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.