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) 0.024 cm — Method: significant figures are counted from the first non-zero digit; the zeros in front of it only fix the place value and are not significant. Working: in 0.02384 the first significant figure is 2 and the second is 3, so the rounding is decided by the next digit, 8. As 8 is 5 or more, the second significant figure goes up from 3 to 4, in the same place value. Answer: 0.024 cm. The distractors: 0.023 cm comes from chopping the 84 off instead of rounding it; 0.02 cm comes from counting the leading zeros as significant figures, so the 2 is taken as the second figure and the rounding stops there; 0.0238 cm is 0.02384 correct to 3 significant figures, one figure too many.
- (d) (5x − 1)/(x² − x − 2) — The common denominator is (x + 1)(x − 2) = x² − x − 2. The numerator becomes 2(x − 2) + 3(x + 1) = (2x − 4) + (3x + 3) = 5x − 1, so the sum is (5x − 1)/(x² − x − 2). Choosing 5/(2x − 1) adds the numerators (2 + 3 = 5) and the denominators ((x + 1) + (x − 2) = 2x − 1) directly, which is not how fractions add. Choosing (5x + 1)/(x² − x − 2) comes from expanding 2(x − 2) as 2x − 2 instead of 2x − 4, losing part of the constant term. Choosing (5x − 1)/(x² + x − 2) has the correct numerator but the wrong sign on the x-term when (x + 1)(x − 2) is expanded.
- (d) £30.00 — The gradient is (130 − 70) ÷ (5 − 2) = 60 ÷ 3 = £20 per hour. Using the point (2, 70): the cost for 2 hours at £20 per hour is 20 × 2 = £40, so the call-out fee is 70 − 40 = £30. Taking the C-value of the first point as the fee without subtracting the hourly cost gives £70.00 — but that point already includes 2 hours of the hourly rate. Using the gradient itself as the fee, £20.00, confuses the rate per hour with the fixed charge. Subtracting 20 × 3 = 60 instead of 20 × 2 = 40 (using the wrong h-value) gives 70 − 60 = £10.00.
- (d) The angle between a tangent and a radius is 90° — OP is the radius drawn to the point of contact P, and a circle theorem states that a tangent always meets that radius at a right angle, so angle OPQ = 90°. A tangent is not parallel to the radius it touches — at the point of contact it is perpendicular to that radius, not parallel to it. A tangent does not pass through the centre — a straight line through the centre that also touches the circle at one point would have to be a diameter, which is a different line entirely. The angle-in-a-semicircle theorem needs a triangle drawn inside the circle with a diameter as its longest side; there is no such triangle here, just a tangent and a radius.
- (b) The relative frequency is settling near 0.5 — Method: turn each result into a relative frequency before comparing them, because it is the relative frequency, and not the difference between the two counts, that tends towards the theoretical probability. Working: after 10 flips the relative frequency of a head is 7 ÷ 10 = 0.7, which is a long way from 0.5. After 1000 flips it is 528 ÷ 1000 = 0.528, which is much closer to 0.5. Meanwhile the gap between the two counts has grown rather than shrunk: it was 7 − 3 = 4 after 10 flips and is 528 − 472 = 56 after 1000 flips. Answer: the relative frequency is settling near 0.5, which is what an unbiased experiment does as the sample grows. The distractors: saying the counts are levelling out is the usual form of this idea and the figures contradict it, since the gap went from 4 to 56; saying the coin is biased treats 28 extra heads in 1000 flips as proof, when 0.528 sits close to 0.5 and a fair coin gives results like this often; saying the next flip is more likely to be a tail is the gambler's fallacy, since each flip stays at 1/2 whatever came before.
- (c) 40 minutes — Method: for grouped data, estimate the mean using the midpoint of each class — multiply each midpoint by its frequency, add the results, then divide by the total frequency. Working: the midpoints are 10, 30, 50 and 70 minutes. 10 × 5 = 50. 30 × 10 = 300. 50 × 10 = 500. 70 × 5 = 350. Σfx = 50 + 300 + 500 + 350 = 1200. Σf = 5 + 10 + 10 + 5 = 30. Estimated mean = 1200 ÷ 30 = 40 minutes. Using the upper boundary of each class instead of the midpoint — 20 × 5 = 100, 40 × 10 = 400, 60 × 10 = 600, 80 × 5 = 400 — gives a total of 1500 and an estimate of 1500 ÷ 30 = 50 minutes, too high because a boundary is not the middle of the class. Averaging the frequencies themselves, 5, 10, 10 and 5, ignores the times altogether and gives 7.5. Stopping after Σfx = 1200 without dividing by the total frequency gives a number far too large to be a time in minutes. Always find the midpoint of each class before multiplying by the frequency, and always divide by Σf at the end.
- (a) 1/2, 3/5, 2/3, 5/6 — Convert all four fractions to a common denominator of 30: 2/3 is 20/30, 3/5 is 18/30, 5/6 is 25/30, and 1/2 is 15/30. Ordering by these numerators, smallest to largest, gives 15/30, 18/30, 20/30, 25/30, which is 1/2, 3/5, 2/3, 5/6. Ordering by the size of the numerator in the original fractions, 1, 2, 3, 5, rather than converting to a common denominator, gives the wrong order 1/2, 2/3, 3/5, 5/6, because it ignores that the denominators are different. Ordering largest to smallest instead of smallest to largest, as the question asks, gives 5/6, 2/3, 3/5, 1/2. Using the rule "the bigger the denominator, the smaller the fraction" to place the last two, so that 5/6 is put below 2/3 because 6 is bigger than 3, gives 1/2, 3/5, 5/6, 2/3 — that rule only holds when the numerators are the same, and here 20/30 really is smaller than 25/30. So the correct order, smallest to largest, is 1/2, 3/5, 2/3, 5/6.
- (c) 22 m — Rearranging P = 2l + 2w for l gives l = (P − 2w)/2. Substituting P = 60 and w = 8: l = (60 − 16)/2 = 44/2 = 22 m. Answering 30 m comes from ignoring the width altogether and working out l = P/2, which only holds when w = 0. Answering 38 m comes from adding 2w instead of subtracting it: (60 + 16)/2 = 38. Answering 11 m comes from dividing by 4 instead of 2, as if the formula were P = 4l + 4w. The pen's length is 22 m.
- (c) £450 — Method: use the equal ratios 4:5 = 200:x to find Grace's savings, then add the two amounts. Working: Noah's £200 is 4 parts, so one part is £200 ÷ 4 = £50; Grace has 5 parts, so 5 × £50 = £250; altogether £200 + £250 = £450. Answer: £450. The distractors: £250 is Grace's savings on their own, which is the middle step rather than the total the question asks for; £360 comes from reading £200 as the 5 parts instead of the 4, giving one part of £40 and a total of 9 × £40; £400 comes from doubling £200, which treats the two savings as equal and ignores the ratio altogether.
- (b) $\binom{-5}{5}$ — Method: add the two flights to get the single vector of the whole journey out, then reverse that vector to get the journey home. Working: across, 6 − 1 = 5; up, 4 − 9 = −5. So the drone finishes at (7, −8), which is 5 to the right of its start and 5 below it. The way home is therefore 5 to the left and 5 up. Answer: $\binom{-5}{5}$. Giving the combined journey itself, $\binom{5}{-5}$, describes the flight out rather than the flight home. Subtracting the second vector instead of adding it makes the journey out 7 across and 13 up, and reversing that gives $\binom{-7}{-13}$. Reversing the horizontal movement but leaving the vertical one alone gives $\binom{-5}{-5}$.
- (c) 1/2 — Method: restrict to the students who did NOT bring a packed lunch, then find what fraction of that group are girls. Girls without lunch = 42 − 24 = 18. Boys = 70 − 42 = 28, so boys without lunch = 28 − 10 = 18. Total without lunch = 18 + 18 = 36. Working: P(girl | no lunch) = 18 ÷ 36 = 1/2. Answer: 1/2. Watch out: dividing 18 by 42 (the total number of girls) instead of by 36 finds P(no lunch | girl), the reverse conditional. Dividing by 70 (the whole trip) ignores that you already know the student did not bring a lunch. And using the 'brought a lunch' numbers (24 out of 34) answers the question for the wrong group entirely — you were asked about the students who did NOT bring one.
- (b) 54 — Method: round each number to the nearest whole number, then subtract the rounded values. Working: 79.3 rounds to 79 (nearest whole number) and 24.6 rounds to 25 (nearest whole number). 79 − 25 = 54. Answer: 54. 54.7 is the exact value of 79.3 − 24.6, found without rounding first, so it is not an estimate. 55 comes from rounding 24.6 down to 24 instead of up to the nearest whole number, 25, giving 79 − 24. 59 comes from rounding 24.6 to the nearest 10, 20, instead of to the nearest whole number, 25, giving 79 − 20.
- (a) Subtracting 4x gives 3 = 10, which is never true. — Method: try to solve the equation as normal and see what happens. Working: subtract 4x from both sides: 4x + 3 − 4x = 4x + 10 − 4x, giving 3 = 10. This statement is false for every value of x, so the equation has no solution. Answer: subtracting 4x gives 3 = 10, which is never true. "x would have to be negative" invents a constraint on x that the equation never states. "It's true for every x" confuses this equation with an identity, where both sides would simplify to the same expression. "x = 7" misreads the false statement 3 = 10 as something to solve for x, rather than recognising it means no solution exists.
- (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).
- (c) base angles of an isosceles triangle are equal — Because AB = AC, triangle ABC is isosceles, and the base angles of an isosceles triangle — the two angles opposite the equal sides — are always equal, which is why angle C equals angle B, 70°. Angles in a triangle adding up to 180° is a true fact about the triangle as a whole, but it is not the reason two specific angles are equal to each other. Corresponding angles are equal is a fact about parallel lines cut by a transversal, which does not apply inside a single triangle like this. Vertically opposite angles are equal is a fact about two lines crossing, not about a triangle's base angles.
- (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.
- (d) 20 — Method: equivalent ratios are linked by a single multiplier, so find it from the part you know and apply it to the other part. Working: 15 ÷ 3 = 5, so the multiplier is 5, and 4 × 5 = 20. Answer: 20. The distractors: 16 comes from adding the difference between the ratio parts, 4 − 3 = 1, to 15, treating the ratio as a difference; 60 comes from multiplying 15 by 4 without first dividing by 3; 11.25 comes from using the ratio the wrong way round, working out 15 × 3 ÷ 4.
- (c) 9 cm — The scale factor from triangle ABC to triangle PQR is PQ ÷ AB = 12 ÷ 8 = 1.5. So PR = AC × 1.5 = 6 × 1.5 = 9 cm. (4 cm comes from using the scale factor upside down, AB ÷ PQ; 10 cm comes from adding the difference between AB and PQ, 4 cm, to AC instead of scaling; 16 cm comes from multiplying PQ by the ratio AB : AC instead of scaling AC by the correct scale factor.)
- (a) 3 — y = f(x + 2) is f(x) translated 2 units to the LEFT (inside the bracket, adding moves the graph in the negative x-direction). The root moves with the whole graph: 5 − 2 = 3. Moving right instead of left gives 7; assuming a bracket shift leaves the root unchanged gives 5; writing down the shift amount 2 itself skips the translation altogether.
- (a) k = 3 — Method: a solution of an equation makes both sides balance, so substitute it in and solve the equation in k that is left. Working: putting x = 3 gives 3² − (k + 1) × 3 + k = 0, that is 9 − 3k − 3 + k = 0, so 6 − 2k = 0 and k = 3; the equation is then x² − 4x + 3 = 0, whose solutions are 3 and 1. Answer: k = 3. The distractors: k = −3 comes from solving 6 − 2k = 0 as though it gave 2k = −6; k = 4 comes from expanding −3(k + 1) as −3k − 1, multiplying only the k by 3; k = 1.5 comes from working 3² as 3 × 2 = 6, which leaves 3 − 2k = 0.
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.