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%.
CalculatorGCSE Higher
GCSE Higher sample Paper 2 (calculator)
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- 1.A grain of fine sand has a mass of 0.001 grams. Write this mass in standard form.
- 2.The nth term of a sequence is n² + 2n − 4. Work out the 7th term.
- 3.A beaker holds 500 g of a salt solution of concentration 12%. Pure water is added until the concentration falls to 8%. Work out the mass of water added.
- 4.A bead starts at position (2, −1) on a grid, in centimetres. It is moved by the column vector u, with top number 3 and bottom number 5, and then moved by the column vector v, with top number −7 and bottom number 2. Work out the coordinates of the bead's final position.
- 5.A bag contains 3 yellow counters and 2 green counters. Sophie takes one counter at random and does not put it back. She then takes a second counter at random. Work out the probability that the first counter is yellow and the second is green.
- 6.The seven members of Team A took 20, 21, 22, 23, 24, 25 and 40 seconds to finish a task. The seven members of Team B took 20, 30, 32, 34, 36, 38 and 40 seconds. Tomás says that because the two teams have the same range, their times are spread out in the same way. Is Tomás right? Give a reason for your answer.
- 7.A plane flies 2340 km at an average speed of 780 km/h. It departs at 08:20. Work out the arrival time, using the 24-hour clock.
- 8.The graph of y = x² − 7x + 2 crosses the x-axis at two points. One root, read from the graph, is approximately x = 0.30. Using the fact that the sum of the two roots of x² − 7x + 2 = 0 is 7, estimate the other root, correct to 2 decimal places.y = x² − 7x + 2
- 9.Two numbers A and B are in the ratio 3:5. When 6 is added to B, the ratio of A to the new value of B is 1:2. Work out the value of A.
- 10.Two cylindrical storage tins are geometrically similar. The smaller tin has height 4 cm and the larger tin has height 6 cm. Work out the ratio of the surface area of the smaller tin to the surface area of the larger tin, giving your answer in the form a : b in its simplest form.
- 11.At a sports centre, 45% of the members are aged under 18. 60% of the members aged under 18 swim each week. 20% of the members aged 18 or over swim each week. Work out the percentage of all the members who swim each week.
- 12.Simplify x⁷ × x⁴, giving your answer as a single power of x.
- 13.The equation x² − 3x − 7 = 0 can be solved using the iterative formula xₙ₊₁ = √(3xₙ + 7). The starting value is x₀ = 4, so x₁ is the value after the formula has been used once. Work out x₃ correct to 3 decimal places.
- 14.The value of a rare coin increases by 12% each year. The coin is currently worth £270. Work out the value of the coin after 2 years, giving your answer to the nearest penny.
- 15.In triangle ABC, AB = x cm, AC = (x + 2) cm, BC = 15 cm and angle BAC = 100°. Work out the value of x. Give your answer to 1 decimal place.
- 16.Solve the simultaneous equations y = 3 − x and x² + y² = 9, giving both pairs of solutions.
- 17.Write the ratio 8 : 15 in the form 1 : n.
- 18.Lines GH and JK are parallel. A straight line crosses GH at point M and crosses JK at point N. Angle HMN = 71°. Angle HMN and angle MNK are alternate angles. Work out angle MNK.
- 19.A taxi company charges a £2.50 booking fee plus £1.80 per mile. A journey costs £15.10 in total. Work out the number of miles travelled.
- 20.Which expression is equivalent to 3(2x − 5) + 4x?
Answer key
- (d) 1 × 10⁻³ — Method: for a number smaller than 1 the index is negative, and it counts the places the decimal point moves to the right to leave a coefficient between 1 and 10. Working: the only significant digit is 1, so the coefficient is 1; the decimal point in 0.001 moves three places to the right to reach that 1, so the index is −3. Answer: 1 × 10⁻³. The distractors: 1 × 10³ comes from taking the index as positive, which describes one thousand grams rather than one thousandth of a gram; 0.1 × 10⁻² is the same mass written with a coefficient of 0.1, which is smaller than 1 and so is not standard form; 1 × 10⁻⁴ comes from counting the zero in front of the decimal point as well as the places after it.
- (c) 59 — Substitute n = 7: 7² + 2 × 7 − 4 = 49 + 14 − 4 = 59. A sign error on the +2n term, treating it as −2n, gives 49 − 14 − 4 = 31. Working out 7² + 2 × 7 but forgetting to subtract the final 4 gives 49 + 14 = 63. Using n = 6 instead of n = 7 gives 36 + 12 − 4 = 44.
- (b) 250 g — Method: adding water changes the total mass but not the mass of salt, so find the salt, hold it fixed, use the new ratio to find the new total mass and subtract the mass already in the beaker. Working: 12:100 = x:500 gives 12 ÷ 100 × 500 = 60 g of salt; that 60 g must be 8% of the new mixture, so 8:100 = 60:y gives y = 60 ÷ 8 × 100 = 750 g; the water added is 750 − 500 = 250 g. Answer: 250 g. The distractors: 750 g is the mass of the diluted solution, given without taking away the 500 g that was in the beaker to start with; 60 g is the mass of salt, the quantity that stays the same, given instead of the mass of water; 20 g comes from treating the fall from 12% to 8% as 4% of the original 500 g, which measures a change in concentration as though it were a mass of water.
- (d) (−2, 6) — Method: add the top numbers of both vectors to the starting x-coordinate, and the bottom numbers of both vectors to the starting y-coordinate. Working: x-coordinate 2 + 3 + (−7) = −2; y-coordinate −1 + 5 + 2 = 6. Answer: (−2, 6). A candidate who only applies vector u and forgets v gets (5, 4). A candidate who only applies vector v and forgets u gets (−5, 1). A candidate who works out the combined vector u + v but forgets to add it to the starting point gets (−4, 7).
- (b) 3/10 — Method: on a tree diagram, follow the path that matches the description and multiply the probabilities written along it; with nothing put back, the second set of branches is worked out from the counters that are left. Working: 3 of the 5 counters are yellow, so the first branch of the path is 3/5. A yellow counter has been kept out, so 4 counters remain and both green counters are still there, making the second branch 2/4. Multiplying along the path gives 6/20. Answer: the probability is 3/10. The distractors: 6/25 comes from using 2/5 on the second branch, which is the tree for a counter that is put back; 3/20 comes from taking one off the green count as well as off the total, using 1/4 on the second branch; 3/5 comes from reading the first branch only and never multiplying along the path.
- (d) No, the range uses only the fastest and slowest time — Method: check what the range is built from, then look at what it leaves out. Working: both teams have a fastest time of 20 seconds and a slowest of 40 seconds, so both ranges are 40 − 20 = 20 seconds and Tomás has that part right. But the range is calculated from those two values alone. Six of Team A's seven times lie between 20 and 25 seconds, with a single time far out at 40; Team B is the other way round, with six of its seven times at 30 seconds or more and a single time far out at 20. So Team A bunches at the fast end and Team B at the slow end. The two patterns are quite different, and the range cannot see the difference because the five middle times never enter the calculation. Answer: no, because the range uses only the fastest and slowest time. The distractors: comparing the means answers a different question, since a mean measures position rather than spread, and two sets with the same spread can have different means; saying that equal ranges mean equal spread is the very assumption that fails here; saying that seven times each forces the spreads to match confuses the size of a data set with how its values are arranged inside it.
- (a) 11:20 — Method: find the flight time using time = distance ÷ speed, then add this to the departure time. Working: 2340 ÷ 780 = 3 hours; 08:20 + 3 hours = 11:20. Answer: 11:20. 08:40 comes from dividing speed by distance instead of distance by speed, giving a flight time of 1/3 hour (20 minutes) rather than 3 hours. 11:00 comes from adding the 3-hour flight time to the hour of the departure time only, 8 + 3 = 11, and losing the 20 minutes. 03:00 comes from finding the flight time correctly but giving it as a clock time on its own, forgetting to add it to the departure time.
- (a) x ≈ 6.70 — Since the two roots sum to 7, the other root is 7 − 0.30 = 6.70. The option 7.30 comes from adding the given root to 7 instead of subtracting it. The option 6.30 comes from subtracting 0.70 (one minus the given root) rather than the given root itself. The option 0.70 confuses the required root with the amount by which the given root falls short of 1.
- (a) 18 — Method: write both numbers with the same multiplier, turn the second ratio into an equation by cross-multiplying, solve for the multiplier and then build A from it. Working: let A = 3k and B = 5k, so 3k : (5k + 6) = 1 : 2; cross-multiplying gives 2 × 3k = 5k + 6, so 6k = 5k + 6 and k = 6; A = 3 × 6 = 18. Answer: 18, and the check works, because B = 30, B + 6 = 36 and 18:36 = 1:2. The distractors: 9 comes from reading the 6 as the difference between the two numbers — 5 − 3 = 2 parts, so one part is 3 and A is 3 × 3 — but the 6 is added to B, it is not the gap between A and B; 6 comes from solving A : (A + 6) = 1 : 2, adding the 6 to A instead of to B; 30 is the value of B, found from the correct multiplier but given in place of A.
- (b) 4 : 9 — The heights are in the ratio 4 : 6, which simplifies to 2 : 3. For similar solids, area scales with the square of the length ratio, so the surface area ratio is 2² : 3² = 4 : 9. 2 : 3 is just the simplified length ratio, before squaring has been done. 8 : 27 comes from cubing the ratio (2³ : 3³) instead of squaring it — that's the rule for volumes, not areas. 9 : 4 has the correct squared values but in the wrong order, giving the larger tin's area first instead of the smaller.
- (b) 38% — Method: the two swimming percentages are quoted inside different age groups, so weight each one by the size of its group and add the two results. Working: the under 18s are 45% of the members and 60% of them swim, giving 0.45 × 60 = 27% of all the members. The members aged 18 or over are 55% of the members and 20% of them swim, giving 0.55 × 20 = 11% of all the members. Adding these gives 38%. Answer: 38% of the members swim each week. The distractors: 80% comes from adding 60% and 20% straight off, treating two rates quoted inside different groups as though they could be added; 40% is the mean of 60% and 20%, which would be right only if the two age groups were the same size, and they are not; 42% comes from pairing each swimming rate with the wrong age group, working out 0.45 × 20 added to 0.55 × 60.
- (a) x¹¹ — Method: when multiplying powers of the same base, add the indices. Working: 7 + 4 = 11, so x⁷ × x⁴ = x¹¹. x²⁸ comes from multiplying the indices, 7 × 4 = 28, instead of adding them. x³ comes from working out 7 − 4 = 3, which is the rule for dividing powers, not multiplying them. 11x comes from adding the indices to make 11 but then treating x as a coefficient instead of a power. Answer: x¹¹.
- (d) 4.521 — x₁ = √(3 × 4 + 7) = √19 = 4.358898944. x₂ = √(3 × 4.358898944 + 7) = √20.076696833 = 4.480702716. x₃ = √(3 × 4.480702716 + 7) = √20.442108148 = 4.521294964, which rounds to 4.521. Mislabelling the starting value x₀ as x₁, so that the working stops one iteration too early, reports the true x₂ = 4.480702716, which rounds to 4.481. Working out one iteration too many reports the true x₄ = √(3 × 4.521294964 + 7) = 4.534741987, which rounds to 4.535. Applying the formula in the wrong order, calculating √(3xₙ) + 7 at every step instead of √(3xₙ + 7), gives, from x₀ = 4: √12 + 7 = 10.464101615, then √(3 × 10.464101615) + 7 = 12.602883619, then √(3 × 12.602883619) + 7 = 13.148873950, which rounds to 13.149.
- (c) £338.69 — To increase by 12% each year, multiply by 1.12 twice. £270 × 1.12 × 1.12 = £338.688, which rounds to £338.69 (nearest penny, since the third decimal place is 8). £334.80 comes from treating the two 12% increases as a single flat 24% increase applied once instead of compounding: £270 × 1.24 = £334.80. £302.40 comes from applying the 12% increase only once, for 1 year instead of 2: £270 × 1.12 = £302.40. £338.68 comes from rounding £338.688 down to the nearest penny instead of up.
- (d) 8.8 — Method: substitute into the cosine rule BC² = AB² + AC² − 2 × AB × AC × cos(BAC) and solve the resulting quadratic in x, keeping only the positive root. Working: 15² = x² + (x + 2)² − 2x(x + 2) cos 100°, which expands to a quadratic with two roots, x = 8.8 and x = −10.8; since x is a length, x = 8.8. Expanding (x + 2)² as x² + 4 instead of x² + 4x + 4, missing the middle term, gives x = 9.6; using +2x(x + 2) cos 100° instead of −2x(x + 2) cos 100° (a sign error on the cosine term) gives x = 10.6; and reporting the size of the rejected negative root instead of discarding it gives x = 10.8. Always expand a squared bracket fully before collecting terms.
- (c) x = 0, y = 3 and x = 3, y = 0 — Substitute y = 3 − x into x² + y² = 9: x² + (3 − x)² = 9. Expanding (3 − x)² = 9 − 6x + x² gives x² + 9 − 6x + x² = 9, which simplifies to 2x² − 6x = 0, or 2x(x − 3) = 0, so x = 0 or x = 3. Using y = 3 − x: x = 0 gives y = 3; x = 3 gives y = 0. Distractor routes: x = 0, y = 3 alone stops after the factor 2x = 0 and never checks the second factor, x − 3 = 0. x = 0, y = 3 and x = −3, y = 6 comes from factorising 2x² − 6x as 2x(x + 3), a sign error that gives a second root of −3 instead of 3. x = 0, y = 3 and x = 6, y = −3 comes from expanding (3 − x)² as 9 − 6x, dropping the x² term, which changes the quadratic to x² − 6x = 0 and its second root to 6.
- (b) 1 : 1.875 — To write a ratio in the form 1 : n, divide both parts by the first part, 8: 8 ÷ 8 = 1 and 15 ÷ 8 = 1.875, giving 1 : 1.875. Giving 1 : 0.53 divides the wrong way round, computing 8 ÷ 15 instead of 15 ÷ 8. Giving 1.875 : 1 has the two parts of the answer swapped, which is the form n : 1, not 1 : n. Giving 8 : 1.875 divides only the second part by 8, so the first part is still 8, not 1.
- (d) 71° — Alternate angles between parallel lines are equal. Since angle HMN and angle MNK are alternate angles, angle MNK = angle HMN = 71°.
- (d) 7 — Method: set up the equation 2.50 + 1.80m = 15.10, then subtract the booking fee and divide by the cost per mile. Working: 1.80m = 15.10 − 2.50 = 12.60; m = 12.60 ÷ 1.80 = 7. Answer: 7 miles. 8.39 comes from dividing the whole £15.10 by £1.80 without first subtracting the booking fee: 15.10 ÷ 1.80 ≈ 8.39. 5.32 comes from swapping the two amounts round, subtracting £1.80 and dividing by £2.50: (15.10 − 1.80) ÷ 2.50 ≈ 5.32. 9.78 comes from adding the booking fee instead of subtracting it: (15.10 + 2.50) ÷ 1.80 ≈ 9.78.
- (c) 10x − 15 — Expand the bracket first: 3(2x − 5) = 6x − 15. Then add the 4x: 6x − 15 + 4x = 10x − 15. The option 10x − 5 comes from forgetting to multiply the 5 inside the bracket by 3 (treating it as 6x − 5), then adding 4x. The option 10x + 15 comes from a sign error when expanding, treating 3 × (−5) as +15 instead of −15, then adding 4x. The option 6x − 15 comes from expanding the bracket correctly but forgetting to add the 4x term at all.
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.