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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- (a) 3.84 × 10⁵ — 384,000 = 3.84 × 100,000 = 3.84 × 10⁵, with the decimal point moved five places and the coefficient kept between 1 and 10. Moving the point six places instead of five gives 3.84 × 10⁶, ten times too large. Leaving the coefficient as 38.4 gives 38.4 × 10⁴, which is not between 1 and 10. Using a negative exponent instead of a positive one gives 3.84 × 10⁻⁵, a number far smaller than 1.
- (a) 12 km/h — Method: convert 15 minutes to hours: 15 ÷ 60 = 0.25 h. Speed = distance ÷ time = 3 ÷ 0.25 = 12 km/h. Distractor origins: 9 km/h is the speed for the second section (3 km in 20 minutes) instead of the first; 8 km/h is the average speed for the whole journey (6 km in 45 minutes) instead of just the first section; 0.2 km/h divides 3 km by 15 without converting the minutes into hours.
- (d) 90 cm — Method: scale each dimension by the scale factor, then find the perimeter. Working: model height = 240 ÷ 8 = 30 cm; model width = 120 ÷ 8 = 15 cm. Perimeter = 2 × (30 + 15) = 90 cm. Wrong options: 11.25 cm comes from squaring the scale factor as if finding an area (720 ÷ 64); 510 cm comes from scaling only one dimension and leaving the other at full size; 720 cm comes from finding the real perimeter (2 × (240 + 120)) but forgetting to scale it down at all.
- (d) 82.8° — Method: angle ABC is opposite the given side AC, so rearrange the cosine rule to cos B = (AB² + BC² − AC²) / (2 × AB × BC). Working: cos B = (3.5² + 2.8² − 4.2²) / (2 × 3.5 × 2.8), so angle ABC = 82.8°. Forgetting the negative sign in the rearrangement gives the supplementary angle 97.2°; leaving out the factor of 2 in the denominator gives 75.5°; and using AB and AC (the pair either side of angle A, not angle B) gives 41.4°, which is angle BAC, not angle ABC. Always check which angle sits opposite the side you left out of the pair you are dividing by.
- (c) 1/10 — On the adult branch, 210 − 189 = 21 appointments were missed. There are 300 − 210 = 90 child appointments, and 90 − 81 = 9 of those were missed. In total, 21 + 9 = 30 appointments were missed, out of 300: 30/300 = 1/10. Writing 7/100 is wrong because 21/300 simplifies to 7/100, and 21 only counts the adult branch, leaving out the 9 missed child appointments. Writing 3/100 is wrong because 9/300 simplifies to 3/100, and 9 only counts the child branch, leaving out the 21 missed adult appointments. Writing 1/9 is wrong because it divides the 30 missed appointments by the 270 that were attended (300 − 30) instead of by the whole 300 booked. The probability is 1/10.
- (d) Full-time workers, as most are at work at that time — Method: a sample is biased when the method of contact makes part of the population much less likely to be reached, so test each group against where its members actually are between 10 am and 2 pm on a weekday, and test each stated reason against the facts. Working: those hours are the middle of the working day, so people in full-time employment are at work and not beside a landline telephone, while people who are retired and people who are unemployed are far more likely to be at home and are reached at the usual rate; the method therefore collects far fewer replies from full-time workers than their share of the adult population the council is consulting. Answer: full-time workers, as most are at work at that time. The distractors: the reply naming retired people rests on the false claim that most retired people are at work in the daytime, when in fact a daytime call reaches them more easily than anyone; the reply naming unemployed people rests on the false claim that they are out during the day, when they too are among the easiest people to reach by a daytime call; the reply naming children rests on the false claim that children are at home at 11 am on a Tuesday, when they are at school and so are not reached by the call at all, and school-age children are in any case not the adults whose views the council is collecting.
- (c) 24 — Method: round each number to the nearest whole number, then multiply the rounded values. Working: 6.4 rounds to 6 (nearest whole number) and 3.9 rounds to 4 (nearest whole number). 6 × 4 = 24. Answer: 24. 18 comes from rounding 3.9 down to 3 instead of up to the nearest whole number, 4, giving 6 × 3. 28 comes from rounding 6.4 up to 7 instead of down to the nearest whole number, 6, giving 7 × 4. 25 is the exact value of 6.4 × 3.9, which is 24.96, rounded to the nearest whole number after multiplying, rather than estimated by rounding first.
- (d) The fixed call-out fee, charged before the hourly rate. — Method: compare the formula with C = (fixed charge) + (rate) × h, where the fixed charge is the part that does not depend on h. Working: in C = 40 + 25h, the term 25h depends on the number of hours, h, but 40 does not change however many hours the job takes. Answer: 40 is the fixed call-out fee. 'The hourly rate' confuses the fixed term with the coefficient of h, which is actually 25. 'The total cost for a job lasting 1 hour' comes from substituting h = 1 into the formula (40 + 25 = 65) rather than reading off the constant term. 'The number of hours worked before charging starts' wrongly treats the constant, which is in pounds, as if it were measured in hours.
- (b) 16 — Method: substitute x into each equation separately, then subtract the smaller value from the larger. Working: y = 3 × 4 = 12 and z = 7 × 4 = 28, so the difference is 28 − 12 = 16. Answer: 16. The distractors: 4 comes from subtracting the constants, 7 − 3, which is the difference between the two gradients rather than the difference between the values at x = 4; 40 comes from adding the two values, 12 + 28, instead of subtracting them; 28 is the value of z on its own, given instead of being compared with the value of y.
- (c) 5 cm — Volume of a cylinder = πr²h, so r² = V ÷ (πh) = 942 ÷ (3.14 × 12) = 942 ÷ 37.68 = 25, and r = √25 = 5 cm. A pupil who finds r² = 25 but forgets to take the square root gives 25 cm. A pupil who forgets to divide by π, using r² = 942 ÷ 12 = 78.5, gets r = √78.5 ≈ 8.9 cm. A pupil who forgets to divide by the height, using r² = 942 ÷ 3.14 = 300, gets r = √300 ≈ 17.3 cm. The correct radius is 5 cm.
- (c) 47 — Method: fill the first pair of branches of the frequency tree, then the failures on each branch, then add only the failing end branches. Working: 70% of 200 is 140, so 140 cars were more than 3 years old and 200 − 140 = 60 cars were 3 years old or less. One quarter of the older cars failed: 140 ÷ 4 = 35. The newer branch gives 12 failures. Adding the two failing branches gives 35 + 12 = 47. Answer: 47 of the cars failed the test. The distractors: 35 is the older branch on its own, with the 12 newer failures never added; 62 comes from taking 1 in 4 of all the cars, 200 ÷ 4 = 50, and then adding the 12; 153 is 200 − 47 and counts the cars that passed.
- (c) 3 × 10⁶ — Daily revenue = (4 × 10³) × 2.5 = 1 × 10⁴ (£10,000). Multiplying by the number of days, (3 × 10²), gives annual revenue = (1 × 10⁴) × (3 × 10²) = 3 × 10⁶ (£3,000,000). A candidate who forgot to multiply by the price and just multiplied the number of items by the number of days worked out (4 × 10³) × (3 × 10²) = 1.2 × 10⁶. A candidate who added the number of days to the daily revenue instead of multiplying worked out 1 × 10⁴ + 3 × 10² = 1.03 × 10⁴. A candidate who misread £2.50 as £25 worked out a daily revenue of (4 × 10³) × 25 = 1 × 10⁵, giving an annual total of (1 × 10⁵) × (3 × 10²) = 3 × 10⁷.
- (c) y = 3x + 7 — Method: the gradient is the change in y divided by the change in x with both taken in the same order, and a point whose x-coordinate is 0 gives the constant straight away because it lies on the y-axis. Working: m = (7 − 1) ÷ (0 − (−2)) = 6 ÷ 2 = 3; the point (0, 7) lies on the y-axis, so c = 7 and the line is y = 3x + 7. Answer: y = 3x + 7. The distractors: y = −3x + 7 comes from taking the y-difference as 1 − 7 while taking the x-difference as 0 − (−2), so the two subtractions run in opposite orders; y = 3x + 1 comes from using the y-coordinate of (−2, 1) as the constant instead of the point that actually lies on the y-axis; y = (1/3)x + 7 comes from writing the gradient upside down as the change in x over the change in y, 2 ÷ 6.
- (d) y = 3x/4 — y : x = 3 : 4 means y/x = 3/4. Rearranging to make y the subject gives y = (3/4)x = 3x/4. A student who mixes up which quantity goes on top gets y = 4x/3. A student who treats the ratio numbers as the coefficient and constant of a linear equation instead of a proportional relationship gets y = 3x + 4. A student who mistakes the relationship for inverse proportion gets y = 3/(4x).
- (a) £94.58 — Method: find the area with 1/2ab sin C, then multiply by the cost per square metre. Working: area = 1/2 × 1.8 × 1.3 × sin 72° = 1.11274 m², so cost = 1.11274 × £85 = £94.58. Leaving out the 1/2 gives an area of 2.22547 m² and a cost of £189.17; using cos 72° instead of sin 72° gives a cost of £30.73; and rounding the area to 1 decimal place (1.1 m²) before multiplying by the cost per square metre gives £93.50, which loses accuracy that the final answer needs. Even rounding to 1.113 m² is enough to shift the cost to £94.61 — keep the unrounded area in your calculator until the very last step.
- (a) 1.36 — Method: put the starting value into the right-hand side to get x₁, feed that value back in to get x₂, and round only once the second value has been found. Working: x₁ = 5 ÷ (1 + 2) = 5 ÷ 3 = 1.66666…; x₂ = 5 ÷ (1.66666… + 2) = 5 ÷ 3.66666… = 1.36363…. The digit in the third decimal place is 3, so x₂ = 1.36 correct to 2 decimal places. Answer: 1.36. The distractors: 1.67 is x₁, the value after a single use of the formula, given by a candidate who counts the starting value itself as x₁; 1.49 is x₃ = 1.48648…, one use of the formula too many; 1.37 comes from writing x₁ down as 1.66, truncating the display instead of keeping it in full, and then working out 5 ÷ 3.66 = 1.36612…, which rounds up to 1.37.
- (c) Firm B — £3 per mile against Firm A's £2 per mile — Firm A's gradient is (14 − 4) ÷ 5 = 2, so it charges £2 per mile. Firm B's gradient is (21 − 6) ÷ 5 = 3, so it charges £3 per mile. £3 is more than £2, so Firm B charges more per mile. Swapping the two firms' gradients gives the answer with Firm A at £3 and Firm B at £2, which has the labels the wrong way round. Dividing the change in miles by the change in cost, instead of the other way round, gives 5 ÷ 10 = £0.50 for Firm A and 5 ÷ 15 = £0.33 for Firm B and so names Firm A — that is the gradient upside down. And a positive fixed charge does not mean two firms charge the same rate: the rate is found from the gradient, not from whether the intercept is positive.
- (b) (3, −1) — Method: find the distance from the point to the mirror line, then place the image the same distance on the other side of the line. Working: B is 5 − 2 = 3 units above the line y = 2, so its image is 3 units below the line, at y = 2 − 3 = −1, giving (3, −1); the x-coordinate does not change, since the mirror line is horizontal. Options: (3, −5) comes from reflecting in the x-axis (y = 0) instead of the line y = 2; (−3, 5) comes from reflecting in the y-axis instead of a horizontal line; (3, −4) comes from doubling the distance from the line instead of reflecting it, giving 2 − 2×3 = −4. Answer: (3, −1).
- (d) The roots of x² − 6x + 5 = 0 are x = 1 and x = 5 (since it factorises to (x − 1)(x − 5)), so by symmetry the turning point has x-coordinate 3, not 6. — Factorising, x² − 6x + 5 = (x − 1)(x − 5), so the roots are x = 1 and x = 5. The turning point lies midway between the roots by symmetry: (1 + 5) ÷ 2 = 3. The coefficient of x has no direct role in locating the turning point this way. The option giving −6 makes an arbitrary sign change with no mathematical basis. The option giving 5 wrongly takes just one of the two roots instead of their midpoint.
- (a) (15, 0) — Method: the tangent is perpendicular to the radius at the point of contact, so find the gradient of the radius, take its negative reciprocal, write the equation of the tangent, then substitute y = 0 because every point on the x-axis has y-coordinate 0. Working: the radius from (0, 0) to (3, 6) has gradient 6 ÷ 3 = 2, so the tangent has gradient −1/2. Substituting into y − 6 = −1/2(x − 3) gives y = −0.5x + 7.5. Setting y = 0 gives 0.5x = 7.5, so x = 15 and P is (15, 0). Answer: (15, 0). The distractors: (0, 7.5) is where the same tangent crosses the y-axis, reached by setting x = 0 instead of y = 0; (0, 0) comes from using the gradient of the radius, 2, for the tangent, which gives the line y = 2x through the centre and so crosses the x-axis at the origin; (6, 0) comes from changing the sign of the radius gradient without turning it upside down, which gives y = −2x + 12.
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.