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) £6 — First find 30% of £200, which is £60, then find 10% of that: £60 × 0.1 = £6. Adding the two percentages together instead of applying them one after the other, 10% + 30% = 40%, and finding 40% of £200 gives £80. Finding 30% of £200 = £60 correctly but stopping before applying the second percentage leaves £60 as the final answer. Finding only 10% of the original £200, ignoring the 30% entirely, gives £20.
- (b) One turning point. — Every quadratic graph, one with an x² term and no higher power of x, has exactly one turning point, since it is a single U-shaped or n-shaped curve. Saying two turning points describes a cubic graph, which can rise, turn, then turn again. Saying no turning points describes a straight line, which has none. Saying four turning points greatly overestimates how many times a simple quadratic curve changes direction — that would need a much higher power of x.
- (a) 12 — Speed × time is constant: k = 20 × 15 = 300. At 25 pages per minute, the time is 300 ÷ 25 = 12 minutes. Getting 18.75 comes from treating speed and time as directly proportional and working out 15 × 25 ÷ 20 instead of dividing k by the new speed. Getting 20 comes from adding the increase in speed (25 − 20 = 5) onto the time (15 + 5 = 20). Getting 10 comes from subtracting that same increase in speed from the time (15 − 5 = 10).
- (d) (6, 2) — Applying the first vector: (2, 1) + (5, −3) = (7, −2), which is the warehouse. Applying the second vector: (7, −2) + (−1, 4) = (6, 2), the delivery address. '(7, −2)' stops at the warehouse and forgets the second flight. '(8, −6)' comes from adding (1, −4) instead of (−1, 4) for the second vector, getting both signs wrong. '(11, −3)' comes from swapping the components of the second vector to (4, −1) before adding.
- (b) 3 pupils — Method: an expected frequency is the probability multiplied by the number of trials, so multiply the probability by the number of pupils. Working: 30 × 1/10 means finding one tenth of 30, and 30 ÷ 10 = 3. Answer: 3 pupils would be expected to have a nut allergy. The distractors: 27 pupils comes from working out how many are expected NOT to have the allergy, 30 − 3, instead of how many are; 10 pupils comes from reading the 10 in the fraction 1/10 as the number of pupils; 1 pupil comes from reading the numerator of the fraction as the expected number.
- (c) Route 2, because its interquartile range is smaller — Method: for a journey where turning up on time matters, what matters is not the typical (median) time but how predictable it is — a smaller interquartile range means the middle half of journeys cluster closer together. Working: Route 1's median, 34 minutes, is in fact lower than Route 2's, 41 minutes, so Route 1 is faster on average; but Route 1's interquartile range, 22 minutes, is far larger than Route 2's, 6 minutes, so Route 1's times are much less predictable. Answer: Priya should choose Route 2, because its interquartile range is smaller, even though it is slower on average. Watch which statistic answers the question actually asked: Route 1 does not have the smaller interquartile range, Route 2 does, so picking Route 1 for that reason misreads the table; Route 1's median genuinely is the lower one, but a lower median answers 'which is faster', not 'which is more reliable'; and Route 2's median is not the lower one, so that claim about Route 2 is simply false.
- (b) £1,000, so £900 is not enough — Method: round each number to 1 significant figure, multiply to estimate the total cost, then compare the estimate with the money available. Working: 187 rounds to 200 and £4.85 rounds to £5, so the estimate is 200 × 5 = 1,000, and £1,000 is more than the £900 the school has. Answer: £1,000, so £900 is not enough. The distractors: £800 comes from cutting £4.85 down to £4 instead of rounding it up to £5, giving 200 × 4 = 800, and that estimate wrongly suggests the money stretches; £935 comes from rounding the price only and keeping 187 lunches, giving 187 × 5 = 935; £950 comes from rounding 187 to the nearest 10 rather than to 1 significant figure, giving 190 × 5 = 950.
- (b) x ≤ 4 — Method: undo the number subtracted from 3x first, then divide by 3; the direction of the sign changes only for division by a negative number. Working: adding 1 to both sides of 3x − 1 ≤ 11 gives 3x ≤ 12; dividing both sides by 3, which is positive, gives x ≤ 4. Answer: x ≤ 4. The distractors: x ≥ 4 comes from turning the sign round while dividing by 3; x ≤ 10/3 comes from subtracting 1 from both sides instead of adding it, giving 3x ≤ 10; x ≤ 12 comes from stopping at 3x ≤ 12 and reading off the 12 without dividing by 3.
- (a) y is always 4 times x. — Check the multiplier for each pair: 12 ÷ 3 = 4, 20 ÷ 5 = 4, 32 ÷ 8 = 4 — the same multiplier every time, so y is always 4 times x. 'x is always 4 times y' comes from writing the multiplier the wrong way round. 'y is always x plus 9' only works for the first pair (3 + 9 = 12); it gives 14 for x = 5 and 17 for x = 8, not 20 and 32. 'y is always double x, plus 6' also only works for the first pair (3 × 2 + 6 = 12); it gives 16 for x = 5, not 20.
- (c) 1.5 — The area scale factor is the length scale factor squared, so if n is the length factor, n² = 2.25. Taking the positive square root gives n = 1.5 (check: 1.5² = 2.25). 2.25 is just the area factor restated, with no root taken. 1.125 comes from halving 2.25 instead of taking its square root. 5.0625 comes from squaring 2.25 instead of rooting it.
- (d) 1/20 — Method: add the counts on the faulty end branches, divide by the total number of items in the experiment, then cancel. Working: the faulty items number 15 + 5 = 20, and 400 items were checked, so the probability is 20/400. Dividing the top and the bottom by 20 gives 1/20. Answer: the probability is 1/20. The distractors: 3/50 is 15/250 and comes from dividing machine A's faults by machine A's output, which is that machine's own fault rate rather than the probability for the whole batch; 1/30 is 5/150 and does the same on machine B's branch; 19/20 is 380/400 and gives the probability that the item picked is not faulty.
- (c) 5 — Method: work out the volume of one box, divide the total volume by it, then round down since a partial box cannot fit. Working: volume of one box = 7³ = 343 cm³. 2000 ÷ 343 = 5.83 (2 d.p.). Since only whole boxes fit, the greatest number is 5. Answer: 5. (6 comes from rounding 5.83 up to the nearest whole number instead of rounding down to the number of boxes that actually fit. 343 comes from giving the volume of one box instead of the number of boxes. 5.8 comes from leaving the division as a decimal instead of rounding down to a whole number of boxes.)
- (c) 71 — Method: square x first, then multiply by 3, then subtract 4, following the order of operations. Working: x² = 5² = 25; 3 × 25 = 75; 75 − 4 = 71. Answer: 71. 221 comes from squaring (3x) as a whole first: (3 × 5)² = 225, then − 4 = 221, squaring the coefficient along with x. 75 comes from correctly working out 3x² but forgetting to subtract the 4. 3 comes from subtracting the 4 from x before squaring: (5 − 4)² × 3 = 3, doing the operations in the wrong order.
- (b) 62.5% — Total parts = 5 + 3 = 8. Apples make up 5 parts, so the percentage is 5/8 × 100 = 62.5%. A student who finds the oranges' share instead gets 3/8 × 100 = 37.5%. A student who assumes an even split gets 50%. A student who inverts the fraction gets 8/5 × 100 = 160%.
- (d) 77 — Method: angles that meet at a point add up to 360°. Working: 82 + 105 + 96 = 283; 360 − 283 = 77. Answer: 77°. A candidate who gives the sum of the three known angles and forgets to subtract it from 360° gets 283. A candidate who leaves out the 105° angle, working out 360 − 82 − 96, gets 182. A candidate who leaves out the 82° angle, working out 360 − 105 − 96, gets 159.
- (c) 4(n + 3) — 'Add 3 to n' must happen before 'multiply the result by 4', so the addition needs brackets to show it happens first: 4(n + 3). Writing 4n + 3 multiplies n by 4 immediately and only adds the 3 afterwards, which reverses the order the words describe. Writing n + 3 × 4 only multiplies the 3 by 4, and adds n as a separate, unmultiplied term — it treats 'the result' as just the 3, not the whole of n + 3. Writing 3(n + 4) keeps the correct structure but swaps which number is added and which is multiplied. The expression for 'add 3 to n, then multiply the result by 4' is 4(n + 3).
- (a) 24 N/m² — Pressure = force ÷ area. 68.4 ÷ 2.85 = 24 N/m². 194.94 N/m² comes from multiplying the force by the area instead of dividing (68.4 × 2.85). 65.55 N/m² comes from subtracting the area from the force (68.4 − 2.85) instead of dividing. 0.04 N/m² comes from dividing the area by the force instead of the force by the area (2.85 ÷ 68.4).
- (b) 9π cm² — Sector area is the fraction θ/360 of the full circle's area, πr². Substitute θ = 40 and r = 9: 40 out of 360 is one ninth, and the full circle's area is π × 9² = 81π. One ninth of 81π is 9π, so the sector area is 9π cm². Choosing 2π cm² is this sector's ARC LENGTH, not its area — a different formula and a different quantity. Choosing 81π cm² is the area of the WHOLE circle, forgetting to scale down by the sector's angle. Choosing 18π cm² is the CIRCUMFERENCE of the whole circle, 2π × 9, mixed up with an area.
- (c) x = 8 — The turning point lies exactly halfway between the two roots. If the other root is r, the midpoint of −2 and r must be 3, so (−2 + r) ÷ 2 = 3, giving r = 8. Choosing x = 5 comes from adding 2 and 3 rather than using the midpoint relationship correctly. Choosing x = 1 comes from subtracting 2 from 3 instead of reflecting −2 across the turning point. Choosing x = −8 finds the right distance but then reflects in the y-axis instead of in the line of symmetry x = 3, so the sign of the answer is flipped.
- (c) 2.26 — Method: apply the formula repeatedly, keeping the whole display each time, and stop when two values in a row round to the same 2 decimal places; that shared rounded value is the solution to that accuracy. Working: x₁ = ∛(2 × 2 + 7) = ∛11 = 2.22398…; x₂ = ∛(2 × 2.22398… + 7) = ∛11.44796… = 2.25377…; x₃ = ∛11.50754… = 2.25767…; x₄ = ∛11.51534… = 2.25818…. Now x₃ and x₄ both round to 2.26, so the sequence has settled. Answer: 2.26. The distractors: 2.22 is x₁ rounded, quoted by a candidate who stops after one use of the formula; 2.25 is x₂ rounded, quoted by a candidate who stops as soon as two values look close instead of waiting until two consecutive values round to the same figure; 1.91 is ∛7, which comes from ignoring the 2x term and solving x³ = 7 instead.
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.