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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- (d) x⁴ — Method: dividing two powers of the same letter subtracts the index of the divisor from the index of the term being divided. Working: six factors of x on the top and two on the bottom cancel in pairs, leaving 6 − 2 = 4 factors of x. Answer: x⁴. The distractors: x³ comes from dividing the indices, 6 ÷ 2, instead of subtracting them; x⁸ comes from adding the indices, 6 + 2, as though the powers were being multiplied; x¹² comes from multiplying the indices, 6 × 2, as though a power were being raised to a power.
- (c) x + 2y = 6 — L has gradient 2, so the perpendicular gradient is −1/2. Substituting (4, 1) into y − 1 = −(1/2)(x − 4): y = −(1/2)x + 2 + 1 = −(1/2)x + 3, so 2y = −x + 6, giving x + 2y = 6. Distractor routes: x − 2y = 2 comes from using gradient +1/2 instead of −1/2, forgetting the negative sign the perpendicular rule needs. x + 2y = 9 comes from swapping the point's coordinates, substituting (1, 4) instead of (4, 1). x + 2y = −2 comes from a sign error distributing the negative gradient over the bracket, computing −(1/2)(x − 4) as −(1/2)x − 2 instead of −(1/2)x + 2.
- (d) 3:8 — Convert 2 hours to minutes: 2 hours = 120 minutes. The ratio is 45 : 120. The highest common factor of 45 and 120 is 15. Divide both parts by 15: 45 ÷ 15 = 3 and 120 ÷ 15 = 8, giving 3 : 8. Leaving the hours unconverted gives 45 : 2 — the units on each side are different, so this does not compare like with like. Dividing by 5 instead of 15 gives 9 : 24, which still shares a common factor of 3, so it is not fully simplified. Swapping the order gives 8 : 3, hours to minutes instead of minutes to hours.
- (c) 78° — The angles in any quadrilateral add up to 360°. Add the three given angles: 92° + 84° + 106° = 282°. Angle S = 360° − 282° = 78°. A pupil who only adds angle P and angle Q, forgetting angle R, gets 360° − (92° + 84°) = 184°. A pupil who only adds angle Q and angle R, forgetting angle P, gets 360° − (84° + 106°) = 170°. A pupil who makes a carrying slip adding the three angles, getting 292° instead of 282°, gets 360° − 292° = 68°. The correct answer is 78°.
- (d) 136 — 15% of 160 = 0.15 × 160 = 24 patients reported side effects, so 160 − 24 = 136 did not. Stopping after finding the number who reported side effects, 24, answers the wrong question — it is not the number who did NOT report them. Misreading '15%' as a raw count of 15 patients, rather than a percentage, gives 160 − 15 = 145. Subtracting 15% of 160 twice, 160 − 24 − 24 = 112, double-counts the side-effect group.
- (d) 1100 — Method: to combine two samples of different sizes, add the faulty counts together and add the sample sizes together before scaling up, rather than treating the two samples separately. Working: the combined sample found 34 + 21 = 55 scratched cases out of 100 + 50 = 150 cases checked, a proportion of 55 ÷ 150. Applying that proportion to the week's production of 3,000 gives an estimate of 55 ÷ 150 × 3000 = 1100 scratched cases. Averaging the two shifts' proportions instead of combining their totals, (34 ÷ 100 + 21 ÷ 50) ÷ 2 = 0.38, gives 0.38 × 3000 = 1140 — this treats the two samples as equally weighted even though Shift A checked twice as many cases as Shift B. Using only Shift A's sample, 34 ÷ 100 × 3000 = 1020, ignores Shift B's cases completely. Using only Shift B's sample, 21 ÷ 50 × 3000 = 1260, ignores Shift A's cases completely. When two samples are different sizes, combine their totals before finding the proportion — do not average the two proportions, and do not use only one shift's sample.
- (d) 3/5 — The box is 8 equal shares, of which 5 are milk, so the dark chocolates take 8 − 5 = 3 shares and dark : milk = 3 : 5. The comparison asked for is dark with milk, so the milk share count is the denominator and the fraction is 3/5. 5/3 compares milk with dark, 3/8 compares the dark chocolates with the whole box rather than with the milk ones, and 8/5 comes from reading 5/8 as the ratio milk : dark.
- (c) 26.6 — h² = 1.6² = 2.56, so B = 68 ÷ 2.56 = 26.5625, which rounds to 26.6 (1 d.p.). A candidate who forgets to square the height gets 68 ÷ 1.6 = 42.5. A candidate who squares the mass instead of the height gets 68² ÷ 1.6 = 2890.0. A candidate who truncates 26.5625 instead of rounding it gets 26.5.
- (d) 20 litres — The ratio of concentrate to water is 2 : 5, so water = concentrate × 5 ÷ 2. 8 × 5 ÷ 2 = 20, so Priya needs 20 litres of water. Giving 40 litres multiplies by 5 but forgets to divide by 2 (8 × 5 = 40). Giving 3.2 litres uses the ratio inverted, multiplying by 2 ÷ 5 instead of 5 ÷ 2 (8 × 2 ÷ 5 = 3.2). Giving 11 litres uses additive reasoning instead of multiplicative: it adds the difference between the ratio parts, 5 − 2 = 3, onto the amount of concentrate (8 + 3 = 11), but ratios scale by multiplying, not by adding a fixed amount.
- (b) 18.84 m² — The pond has radius 2.5 m (half of the 5 m diameter), so the outer edge of the path has radius 2.5 + 1 = 3.5 m. Path area = area of outer circle − area of pond = 3.14 × 3.5² − 3.14 × 2.5² = 38.465 − 19.625 = 18.84 m².
- (d) 12/17 — Method: the person picked is known to have passed, so the sample space is everyone who passed; divide the course takers who passed by that total. Working: 120 course takers and 50 others passed, so 170 people passed. The course takers who passed give 120/170, and dividing the numerator and the denominator by 10 gives 12/17. Answer: the probability is 12/17. The distractors: 4/5 is 120/150, the probability that someone passed given that they took the course, which reverses the condition and the event; 12/25 is 120/250, dividing by everyone who sat the test rather than by the 170 who passed; 17/25 is 170/250, the probability that a person picked from everyone sitting the test passed, which answers a different question altogether.
- (a) 14 — Multiply both numbers by 10 to clear the decimals: 8.4 becomes 84 and 0.6 becomes 6. Then divide: 84 ÷ 6 = 14, so 14 complete pieces can be cut. Scaling only the divisor by 10 and leaving the dividend as 8.4 gives 8.4 ÷ 6 = 1.4, which rounds down to 1 complete piece — the dividend was never converted. Scaling only the dividend by 10 and leaving the divisor as 0.6 gives 84 ÷ 0.6 = 140. Rounding the divisor from 0.6 to 0.7 before dividing, trading accuracy for a rounder number, gives 8.4 ÷ 0.7 = 12. So 14 complete pieces of ribbon can be cut.
- (b) x ≤ 5 — Method: divide out the bracket first, then undo the number term; the inequality sign turns round only if both sides are multiplied or divided by a negative number. Working: dividing both sides of 2(x − 1) ≤ 8 by 2 gives x − 1 ≤ 4, and 2 is positive so the ≤ is unchanged; adding 1 to both sides gives x ≤ 5. Answer: x ≤ 5. The distractors: x ≤ 3 comes from subtracting 1 from 4 instead of adding 1 to both sides; x ≤ 4 comes from stopping at 8 ÷ 2 = 4 and never undoing the −1 inside the bracket; x < 5 comes from reading ≤ as a strict inequality, which wrongly leaves the boundary value out of the solution set.
- (b) 7980 — After the first year: 8000 × 0.95 = 7600. After the second year: 7600 × 1.05 = 7980. 8000 comes from assuming a 5% decrease followed by a 5% increase returns exactly to the starting number — it does not, because the increase acts on the smaller, already-reduced number. 8400 comes from applying only the second year's 5% increase to the original number: 8000 × 1.05 = 8400. 7600 comes from applying only the first year's 5% decrease and stopping there, without applying the second year's increase.
- (c) 6 cm — For a parallelogram, area = base × height, so height = area ÷ base = 54 ÷ 9 = 6 cm. 12 cm comes from using the triangle's reverse formula, height = 2 × area ÷ base, which does not apply to a parallelogram. 45 cm comes from subtracting the base from the area, 54 − 9, instead of dividing. 486 cm comes from multiplying the area by the base, 54 × 9, instead of dividing.
- (b) n² + 2n + 3 — First differences: 5, 7, 9, 11. Second differences: 2, 2, 2, so a = 2 ÷ 2 = 1. Subtracting n² (1, 4, 9, 16, 25) from the terms (6, 11, 18, 27, 38) leaves 5, 7, 9, 11, 13, which is the linear expression 2n + 3. So the nth term is n² + 2n + 3. Using the second difference itself as a, without halving it, gives 2n² + 2n + 3. Finding a = 1 correctly but then dropping the linear part 2n, keeping only the constant, gives n² + 3. Finding a = 1 correctly but dropping the constant +3 gives n² + 2n.
- (b) 1.5 km — Multiply the map length by the scale factor: 6 × 25000 = 150000 cm. Convert to kilometres, using 100 cm = 1 m and 1000 m = 1 km, so 100000 cm = 1 km: 150000 ÷ 100000 = 1.5 km. (1500 km comes from converting only as far as metres, 150000 ÷ 100 = 1500 m, and then writing kilometres on the end. 15 km comes from dividing by 10000 instead of 100000. 0.15 km comes from dividing by 1000000 instead of 100000.)
- (c) 1/2 — sin 45° = √2/2 and cos 45° = √2/2, so sin 45° × cos 45° = √2/2 × √2/2 = 2/4 = 1/2. √2/2 comes from writing down only one of the two factors and forgetting to multiply by the other. √2 comes from adding the two exact values instead of multiplying them: √2/2 + √2/2 = √2. 1 comes from wrongly treating sin 45° × cos 45° as sin(45° + 45°) = sin 90° = 1 — multiplying two ratios is not the same as adding their angles.
- (a) x = 10 or x = −10 — Rearranging, x² = 100. Taking the square root of both sides gives x = ±10, i.e. x = 10 or x = −10. A candidate who forgets the negative root gives only x = 10. A candidate who halves 100 instead of taking its square root gets x = 50. A candidate who applies the ± sign to 100 itself instead of to its square root gets x = 100 or x = −100.
- (c) x = 0 or x = 3 — Method: take out the common factor 5x: 5x(x − 3) = 0, so 5x = 0 or x − 3 = 0, giving x = 0 or x = 3. Distractor origins: x = 0 or x = 15 comes from forgetting to divide the second term by the common factor 5x correctly, leaving 15 instead of 3; x = 3 loses the solution x = 0 by dividing both sides by x; x = 5 or x = 3 mistakes the coefficient 5 itself for one of the solutions.
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.