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.Nadia has £84. She spends 3/7 of it on a jacket and the rest on a bag. Work out the ratio of the amount spent on the jacket to the amount spent on the bag, in its simplest form.
- 2.The graph of y = 2x² + 6x − 1 is translated by the vector (−2, 0). Work out the equation of the image, giving your answer in the form y = 2x² + bx + c.y = 2x² + 6x − 1y = 2x²
- 3.Two mathematically similar hexagonal tiles have areas 18 cm² and 50 cm². Work out the ratio of the side length of the smaller tile to the side length of the larger tile, in simplest form.
- 4.A cyclist's journey from her house to the shop is represented by the column vector with top number 2 and bottom number 5, where each unit is 1 km east and 1 km north. She then cycles from the shop to the park, represented by the column vector with top number 4 and bottom number −3. What single column vector represents her journey from her house directly to the park?
- 5.A spinner can land on red, blue, green or yellow. The probability that it lands on red is 0.24 and the probability that it lands on yellow is 0.16. The probabilities of landing on blue and on green are equal, and each is called x. Work out the value of x.
- 6.A school has 1,500 pupils. The head teacher takes a random sample of 150 of them from the school register and asks how long they spend on homework. Rory says the sample is too small for the result to mean anything. Is Rory right? Give a reason for your answer.
- 7.Work out √3 × √12, giving your answer as an integer.
- 8.The line y = 3x + 6 meets the x-axis at one point. Work out the coordinates of that point.y = 3x + 6
- 9.A sequence is defined by the iterative rule x_{n+1} = 0.5x_n + 20, with x_0 = 0. By working out x_1, x_2 and x_3, find the value that the sequence is approaching.
- 10.The bearing of a harbour B from a ferry's position A is 260°. What is the bearing of A from B?
- 11.At a fun run, a raffle stall charges £1.50 per ticket. The probability that any one ticket wins a prize worth £8 is 0.12, and a losing ticket wins nothing. Nadia buys 25 tickets. Work out how much money Nadia should expect to lose in total.
- 12.Simplify (y³)⁴, giving your answer as a single power of y.
- 13.The equation x³ = 6x + 20 can be solved using the iterative formula xₙ₊₁ = ∛(6xₙ + 20). Taking x₀ = 3, x₁ = 3.3620 correct to 4 decimal places. Using the full unrounded value of x₁, work out x₂ correct to 3 decimal places.
- 14.y is inversely proportional to x, so y = k ÷ x. When x = 5, the value of y is 8. Work out the value of k.
- 15.A room measures 4.2 m by 3.6 m. A flooring company needs the area in cm² to order the right amount of vinyl flooring. Work out the area of the room in cm².
- 16.The graph of y = sin x is reflected in the x-axis to give the graph of y = g(x). Work out the value of g(90), where x is measured in degrees.y = sin(x)
- 17.A car travels at a constant speed. It travels 150 miles in 3 hours. At the same speed, work out how far it travels in 5 hours.
- 18.A square-based pyramid has a square base and four identical triangular sloping faces, with its apex directly above the centre of the base. How many planes of symmetry does it have?
- 19.A rectangular vegetable plot has an area of 30 m² and its length is 4 m greater than its width, x metres. This gives x² + 4x − 30 = 0, which can be solved using the iterative formula xₙ₊₁ = √(30 − 4xₙ). The starting value is x₀ = 3, so x₁ is the value after the formula has been used once. Work out x₃, and use it to find an estimate for the length of the plot, giving your answer correct to 1 decimal place.
- 20.The first four terms of a sequence are 4, 9, 14, 19. Work out an expression, in terms of n, for the nth term.
Answer key
- (c) 3:4 — The jacket costs 3/7 of £84, which is 3 × (84 ÷ 7) = 3 × 12 = £36. The bag then costs the rest of the money, £84 − £36 = £48. The ratio of the jacket to the bag is 36:48, which simplifies to 3:4. Writing the fraction spent on the jacket, 3/7, directly as the ratio, without working out that the bag's share is the remaining 4/7, gives 3:7. Giving the ratio the wrong way round, bag to jacket instead of jacket to bag, gives 4:3. Assuming the jacket and bag cost the same, ignoring the fraction given, gives 1:1.
- (b) y = 2x² + 14x + 19 — A translation by the vector (−2, 0) moves the graph 2 units in the negative x-direction, which means replacing x with (x + 2): 2(x + 2)² + 6(x + 2) − 1. Expanding 2(x + 2)² gives 2x² + 8x + 8, and 6(x + 2) gives 6x + 12; collecting terms 2x² + 8x + 8 + 6x + 12 − 1 simplifies to 2x² + 14x + 19. Substituting (x − 2) instead of (x + 2) — translating in the wrong direction — gives 2(x−2)² + 6(x−2) − 1, which simplifies to 2x² − 2x − 5. Distributing the outer 2 only onto the x² term of the expansion, instead of onto every term of (x + 2)², gives 2x² + 10x + 15. Dropping the −1 when collecting the constant terms, treating 8 + 12 as 20 rather than 8 + 12 − 1 as 19, gives 2x² + 14x + 20.
- (d) 3 : 5 — Simplify the area ratio: 18 : 50 divides by 2 to give 9 : 25. Areas scale with the square of the length ratio, so take the square root of each part: the square root of 9 is 3, and the square root of 25 is 5, giving a side length ratio of 3 : 5. Giving 5 : 3 has the ratio the right way round for larger to smaller, not smaller to larger. Giving 9 : 25 is the simplified area ratio, without square-rooting it. Giving 18 : 50 is the area ratio before it has even been simplified.
- (b) (6, 2) — The overall journey from house to park is the sum of the two vectors: top = 2 + 4 = 6, bottom = 5 + (−3) = 2, giving (6, 2). A candidate who subtracts the second vector from the first instead of adding gets (2 − 4, 5 − (−3)) = (−2, 8). A candidate who subtracts the other way round gets (4 − 2, −3 − 5) = (2, −8). A candidate who forgets the negative sign on the second vector's bottom number and adds 3 instead of −3 gets (6, 8). Because the journeys join end to end, the correct resultant vector is (6, 2).
- (a) 0.30 — Red, blue, green and yellow are exhaustive, so all four probabilities sum to 1: 0.24 + 0.16 + x + x = 1, so 2x + 0.40 = 1, giving 2x = 0.60 and x = 0.30. Stopping at 2x = 0.60 without dividing by 2 leaves 0.60, the combined probability of both blue and green together, not the value of x on its own. Sharing the 0.60 across all four colours instead of just the two unknown ones gives 0.60 ÷ 4 = 0.15. Leaving out the 0.16 for yellow gives 2x + 0.24 = 1, so 2x = 0.76 and x = 0.38.
- (a) No, 150 pupils are a tenth of the school, chosen at random — Method: judge a sample on two things, whether every member of the population had the same chance of being chosen, and whether the sample is large enough to carry a pattern. Working: the 150 pupils were drawn from the register of every pupil in the school, so no year group or set is shut out and no pupil chooses to take part; and 150 ÷ 1,500 = 0.1, so one pupil in ten has been asked. A random sample of that share is ample for an estimate of how long the school's pupils spend on homework. Answer: no, because 150 pupils are a tenth of the school and were chosen at random. The distractors: saying a random sample always gives the exact school figure reaches the same verdict for a reason that is false, since a second random sample of 150 would give a slightly different mean; saying 150 pupils cannot be picked at random from 1,500 treats randomness as something only a whole population can have, when drawing names from the register is exactly how a random sample is taken; saying that only asking all 1,500 could show anything rejects sampling altogether, which would leave no way to study any population too large to count.
- (a) 6 — Use √a × √b = √(ab): √3 × √12 = √(3 × 12) = √36 = 6. Adding the numbers under the roots instead of multiplying them, 3 + 12 = 15, gives √15 — that comes from applying the rule for adding surds to a multiplication question. Multiplying the two numbers under the roots but then forgetting to take the square root at the end leaves 36. Simplifying only √12 to 2√3 and then dropping the other √3 factor entirely gives 2√3.
- (b) (−2, 0) — Method: a graph meets the x-axis where the y-value is 0, so setting y = 0 turns the equation into a linear equation in x. Working: 0 = 3x + 6 gives 3x = −6, so x = (−6) ÷ 3 = −2 and the meeting point is (−2, 0). Answer: (−2, 0). The distractors: (2, 0) comes from solving 3x = −6 and then dropping the minus sign from the result; (0, 6) is the y-axis crossing, found by substituting x = 0 instead of y = 0; (6, 0) comes from reading the constant 6 straight off as the x-coordinate, without dividing by 3 and without changing its sign.
- (c) 40 — Working out successive terms shows where the sequence is heading, but the terms themselves keep changing — the limit is the value where the sequence stops changing, so x_{n+1} = x_n = L there. Substituting into the rule: L = 0.5L + 20. Subtracting 0.5L from both sides: L − 0.5L = 20, so 0.5L = 20, and L = 20 ÷ 0.5 = 40. The individual terms are x_1 = 0.5 × 0 + 20 = 20, x_2 = 0.5 × 20 + 20 = 30 and x_3 = 0.5 × 30 + 20 = 35, getting closer to this value but not equal to it — 35 is only the third term, not the limit. Multiplying by 0.5 instead of dividing at the final step, 20 × 0.5 = 10, undoes the rearrangement rather than completing it, and gives a value smaller than terms the sequence has already passed. Writing the fixed-point equation with the wrong sign, L = 0.5L − 20, gives 0.5L = −20 and L = −40, which cannot be right since every term in the sequence is positive and increasing. The value the sequence is approaching is 40.
- (a) 080° — The back bearing differs from the given bearing by 180°. Because 260° is greater than 180°, subtract 180°: 260 − 180 = 80°, so the bearing of A from B is 080°. Choosing 440° adds 180° instead of subtracting it, even though the result would be more than a full turn (260 + 180 = 440). Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 100° comes from measuring the reflex angle the other way round the circle (360 − 260 = 100) instead of applying the 180° back-bearing rule.
- (b) £13.50 — The total cost of Nadia's 25 tickets is 25 × £1.50 = £37.50. The expected number of winning tickets is 25 × 0.12 = 3, so the expected prize money is 3 × £8 = £24.00. Nadia's expected loss is the cost minus the expected prize money: £37.50 − £24.00 = £13.50. A candidate who answers £24.00 has given the expected prize money and mistaken it for the loss. A candidate who answers £37.50 has given the total cost of the tickets, forgetting to subtract the expected prize money. A candidate who answers £34.50 has subtracted the expected number of wins, 3, from the cost instead of first converting it to prize money by multiplying by £8.
- (a) y¹² — Method: when a power is raised to another power, multiply the two indices. Working: (y³)⁴ means y³ × y³ × y³ × y³, which is four lots of three y's multiplied together, so the index is 3 × 4 = 12 and (y³)⁴ = y¹². y⁷ comes from adding the indices, 3 + 4 = 7, which is the rule for multiplying two separate powers, not for raising a power to a power. y⁸¹ comes from working out 3⁴ = 81 and using that as the index, raising the inner index to the outer power instead of multiplying the two indices. 12y comes from multiplying the indices to make 12 but then treating y as a coefficient instead of a power. Answer: y¹².
- (d) 3.425 — x₁ = ∛(6 × 3 + 20) = ∛38 = 3.3620 (unrounded, 3.36198...). x₂ = ∛(6 × 3.3620 + 20) = ∛40.172 = 3.425 (3 d.p.). Choosing 3.362 stops at x₁ instead of continuing to x₂. Choosing 2.722 leaves out the '+ 20' inside the root, working out ∛(6 × 3.3620) = ∛20.172 = 2.722. Choosing 0.556 subtracts 20 instead of adding it, working out ∛(6 × 3.3620 − 20) = ∛0.172 = 0.556.
- (d) 40 — Substitute x = 5 and y = 8 into y = k ÷ x to get 8 = k ÷ 5, so k = 8 × 5 = 40. Getting 13 comes from adding the two numbers (5 + 8) instead of multiplying. Getting 1.6 comes from dividing 8 by 5 instead of multiplying. Getting 3 comes from subtracting the two numbers (8 − 5) instead of multiplying.
- (c) 151,200 cm² — First find the area in m²: 4.2 × 3.6 = 15.12 m². To convert m² to cm², multiply by 10,000, since 1 m² = 100 cm × 100 cm = 10,000 cm²: 15.12 × 10,000 = 151,200 cm². 15.12 cm² gives the area in m² without converting the units at all. 1,512 cm² multiplies by 100 instead of 10,000, treating the conversion as if it were for a single length rather than an area. 15,120 cm² multiplies by 1,000 instead of 10,000.
- (a) −1 — Reflecting y = sin x in the x-axis gives y = −sin x, so g(x) = −sin x. Since sin 90° = 1, g(90) = −1. Reading sin 90° = 1 and forgetting to apply the reflection gives 1. Misreading the angle as 0° instead of 90° gives sin 0° = 0, so 0. Confusing sin 90° with sin 30° = 0.5, then reflecting it, gives −0.5.
- (c) 250 miles — Find the distance travelled in 1 hour: 150 ÷ 3 = 50 miles. Multiply by 5 hours: 50 × 5 = 250 miles. Giving 300 miles doubles the original distance (150 × 2 = 300) using a scale factor of 2 instead of the correct 5 ÷ 3. Giving 200 miles adds only one extra hour's distance, 50, instead of the two extra hours actually needed (150 + 50 = 200, rather than 150 + 100). Giving 90 miles divides by the scale factor instead of multiplying (150 × 3 ÷ 5 = 90).
- (d) 4 — Method: a plane of symmetry must pass through the apex and cut the base along one of the base's own lines of symmetry. Working: a square has 4 lines of symmetry (2 through opposite edge midpoints, 2 through opposite corners), and each of these, combined with the apex, gives one plane of symmetry of the pyramid. A student who answers 2 has only found the planes through the edge midpoints, or only the ones through the corners, and missed the other pair. A student who answers 8 has doubled the correct count, perhaps confusing it with a different solid. A student who answers 1 has only spotted the one obvious front-to-back plane. Answer: 4.
- (a) 7.9 m — Method: the iteration converges on the width of the plot, so run the formula three times from the starting value and then add 4 m, because the length is 4 m greater than the width. Working: x₁ = √(30 − 4 × 3) = √18 = 4.24264…; x₂ = √(30 − 4 × 4.24264…) = √13.02943… = 3.60963…; x₃ = √(30 − 4 × 3.60963…) = √15.56147… = 3.94480…. The estimate for the length is 3.94480… + 4 = 7.94480…, which is 7.9 m correct to 1 decimal place. Answer: 7.9 m. The iteration is still oscillating at x₃, so this is the estimate that three uses of the formula give, not a settled value. The distractors: 3.9 m is x₃ itself, the width, given by a candidate who runs the iteration correctly and then stops before the step the question actually asks for; 7.6 m uses x₂ in place of x₃, one use of the formula short, and then adds the 4 m correctly; 15.8 m multiplies the width by 4 instead of adding 4 m to it, reading greater than as a multiplier.
- (d) 5n − 1 — The common difference is 5 (9−4=5), so the expression starts 5n. To match the first term when n=1, 5×1+c=4, so c=−1: the nth term is 5n−1. A candidate who uses the first term itself as the constant, instead of first term minus the common difference, would write 5n+4 (giving 9, 14, 19, 24 — one term too high throughout). A candidate who omits the constant term altogether would write just 5n (giving 5, 10, 15, 20, not matching the sequence). A candidate who adds the common difference to n instead of multiplying would write n+5 (giving 6, 7, 8, 9, far too small).
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.