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) 1050 ml — Convert both amounts to millilitres: 1.8 litres = 1800 ml and 3/4 litre = 750 ml. Subtracting gives 1800 − 750 = 1050 ml. Confusing 3/4 with 75% and converting it as 75 ml instead of 750 ml gives 1800 − 75 = 1725 ml. Adding the amount poured out instead of subtracting it gives 1800 + 750 = 2550 ml. Misreading 1.8 litres as 0.8 litres, losing the whole litre, gives 800 − 750 = 50 ml.
- (a) An identity, because it is true for every value of x — Expanding the brackets multiplies both terms inside by 4, giving 4x + 12, which is exactly the right-hand side. The two sides are therefore equal whatever x is, and a statement true for every value of the letter is an identity. An equation is true only for particular values, and trying to solve this one leads to 0 = 0, which places no restriction on x at all. Being able to expand brackets is not what makes a statement an identity, since any equation with brackets can be expanded. Nor is it a formula: a formula links two different quantities, and only one letter appears here.
- (b) 120 minutes — Method: find the rate in bottles per minute, then divide the order size by the rate. Working: rate = 810 ÷ 45 = 18 bottles per minute. Time = 2,160 ÷ 18 = 120 minutes. Wrong options: 1,350 minutes comes from subtracting 810 from 2,160 instead of using the rate; 48 minutes comes from dividing the order size by the original time (2,160 ÷ 45) instead of the rate; 108 minutes comes from rounding the rate to 20 bottles per minute before dividing.
- (a) 7.5 cm — Method: two sides and the angle between them are given, so use the cosine rule a² = b² + c² − 2bc cos A, with BC as the side facing the 68° angle. Working: BC² = 7.4² + 5.9² − 2 × 7.4 × 5.9 × cos 68° = 54.76 + 34.81 − 87.32 × 0.3746 = 89.57 − 32.71 = 56.86, and the square root of 56.86 is 7.5405. Answer: BC = 7.5 cm to 1 decimal place. The distractors: 11.1 cm comes from adding the last term instead of subtracting it, 89.57 + 32.71 = 122.28, the commonest sign slip on the cosine rule; 56.9 cm is the value of BC² written down as though it were BC, stopping one step before the square root; 2.9 cm comes from pressing sin instead of cos, using 87.32 × sin 68° = 80.96 as the term to subtract.
- (b) 39 — Method: put the counts into a two-way table and fill each missing cell by subtracting along a row or down a column. Working: the number of female members is 150 − 80 = 70. The pool column holds 66 members and 35 of them are male, so the number of female pool users is 66 − 35 = 31. Subtracting along the female row, 70 − 31 = 39 female members use the gym. Answer: 39 female members use the gym. The distractors: 45 comes from subtracting along the male row instead, 80 − 35 = 45, which counts male gym users; 31 is the female pool cell, written down one step before the gym cell; 84 is 150 − 66 and counts every gym user, male and female together.
- (a) The means are equal, and Class B's marks are the more consistent because its range is smaller. — Method: comparing two distributions needs two things — a measure of average and a measure of spread — and each must be put into the context of the question. Working: both classes have a mean mark of 70, so on average the two classes scored the same; the range measures spread, and Class A's range of 30 marks is three times Class B's range of 10 marks, so Class B's marks sit closely around the mean while Class A's are far more spread out. Answer: the means are equal, and Class B's marks are the more consistent because its range is smaller. The distractors: the reply crediting Class A with more consistency reverses the meaning of the range, treating a larger range as tighter data when a larger range means more spread; the reply that Class A's mean mark is higher compares the wrong pair of figures, reading the range of 30 as an average; the reply that Class B's mean mark is higher reads the spread correctly but its claim about the means is false, since both means are 70.
- (d) £50.00 — The sale price is 85% of the original, so the original price = £42.50 ÷ 0.85 = £50.00. 15% of £42.50 is £6.375. A candidate who finds 15% of £42.50 and subtracts it from the sale price gets £42.50 − £6.375 = £36.125, which is £36.13 to the nearest penny. A candidate who adds 15% of £42.50 instead of reversing the decrease gets £42.50 + £6.375 = £48.875, which is £48.88 to the nearest penny. A candidate who divides by 0.15 instead of 0.85 gets £283.33.
- (d) 30 — Method: give each group a letter, turn each sentence of the stem into an equation in those letters, then eliminate one letter by adding the two equations. Working: let m be the number of men and w the number of women. The 50 adults give m + w = 50, and 10 more men than women gives m − w = 10. Adding the two equations term by term cancels w: (m + w) + (m − w) = 50 + 10, so 2m = 60 and m = 30. Substituting back into m + w = 50 gives w = 20, and 30 − 20 = 10 as required. Answer: 30. The distractors: 20 comes from subtracting the equations instead of adding them, which cancels m and gives 2w = 40, so w = 20 — the other unknown, not the one asked for; 40 comes from forming the second equation as w = 10, reading '10 more men than women' as '10 women', and then getting m = 50 − 10 from the first equation; 35 comes from using m + w = 50 alone with m = w, giving 25 each, and then adding the whole difference of 10 to the men instead of splitting it between the two groups, which leaves a difference of 20 rather than 10.
- (b) 5/7 — The enlargement multiplier is 1.4, which as a fraction is 7/5. To reverse an enlargement, use the reciprocal of the multiplier: flip 7/5 to get 5/7. 7/5 comes from using the enlargement multiplier again, instead of reversing it. 3/5 comes from treating the reverse as 'give back the extra amount', working out 1 − (1.4 − 1) = 0.6, instead of using the reciprocal. 5/2 comes from ignoring the whole number in 1.4 and inverting only the decimal part, 0.4, as if it were the whole multiplier.
- (b) £157 — Method: round the hours worked up to the next whole hour, multiply by the hourly rate, then add the call-out fee. Working: 3 hours 30 minutes rounds up to 4 hours; 28 × 4 = 112; 112 + 45 = 157. Answer: £157. A candidate who uses the unrounded time of 3.5 hours, working out 28 × 3.5 = 98 and adding 45, gets £143. A candidate who forgets the call-out fee, giving only 28 × 4, gets £112. A candidate who rounds down to 3 hours instead of up, working out 28 × 3 = 84 and adding 45, gets £129.
- (c) Noah — 11/32 = 0.34375, above 34%. — Converting 11/32 to a decimal gives 11 ÷ 32 = 0.34375, which is greater than 34% (0.34), so Noah has the better save rate: 'Noah — 11/32 = 0.34375, above 34%.' Comparing the raw numbers 34 and 11 directly, without converting the fraction to the same form, gives 'Elin — 34 is bigger than 11.' Treating a larger denominator as meaning a bigger value, rather than smaller equal shares, gives 'Noah — 32 is a bigger denominator.' Rounding 34.375% to 34% to the nearest whole percent hides the difference and gives 'Equal — both round to 34% to the nearest percent.'
- (b) 24√2 — The side length of the tile is √72. Since 72 = 36 × 2, √72 = √36 × √2 = 6√2 cm. A square has four equal sides, so the perimeter is 4 × 6√2 = 24√2 cm. Simplifying √72 by writing the perfect-square factor itself as the coefficient instead of its root, 36√2 instead of 6√2, and then multiplying by 4 lands on 144√2. Working out the correct side length, 6√2 cm, but then giving that as the final answer without multiplying by 4 for the perimeter gives 6√2. Doubling the side length instead of quadrupling it, as if the perimeter were 2 × 6√2 rather than 4 × 6√2, gives 12√2.
- (d) 2n² − n − 2 — First differences: 5, 9, 13, 17. Second differences: 4, 4, 4, so a = 4 ÷ 2 = 2. Subtracting 2n² (2, 8, 18, 32, 50) from the terms (−1, 4, 13, 26, 43) leaves −3, −4, −5, −6, −7, which is the linear expression −n − 2. So the nth term is 2n² − n − 2. Using the second difference itself as a, without halving it, gives 4n² − n − 2. Finding a = 2 correctly but dropping the linear remainder −n − 2 entirely leaves 2n². Treating the first first difference (5) as a common difference and building a + (n − 1)d = −1 + 5(n − 1) gives 5n − 6, which only matches the first two terms.
- (a) 3/8, 41%, 0.43 — Method: convert every value to a decimal so they can be compared on the same scale, then order them. Working: 3/8 = 0.375, 41% = 0.41, and 0.43 stays as 0.43, so from smallest to largest the decimals are 0.375, 0.41, 0.43, giving the order 3/8, 41%, 0.43. Answer: 3/8, 41%, 0.43. The order 3/8, 0.43, 41% comes from comparing 0.43 and 41% as raw digits (43 versus 41) without converting 41% into the decimal 0.41 first, wrongly placing 0.43 before 41%. The order 0.43, 41%, 3/8 comes from placing the values in completely reversed order, from largest to smallest instead of smallest to largest. The order 41%, 0.43, 3/8 comes from ordering the values by their TYPE (percentage, then decimal, then fraction) rather than by their actual size.
- (a) SSA - not sufficient to prove congruence — The angle given, angle A, is not the angle between sides AB and BC — it is not the included angle — so this data is SSA (side, side, angle), which is not sufficient to prove congruence on its own; two triangles can share this SSA information without being congruent. SAS is wrong because the given angle is not the one included between the two given sides. ASA is wrong because only one angle (A) is given, not two. AAS is wrong for the same reason — only one angle is given, not two.
- (c) 7n − 1 — Method: find the common difference, then find the constant that fits the first term. Working: 13 − 6 = 7, 20 − 13 = 7, 27 − 20 = 7, so the terms increase by 7 each time and the nth term has the form 7n + c. Substituting n = 1: 7(1) + c = 6, so c = −1. Answer: the nth term is 7n − 1. The value 7n comes from leaving out the constant altogether. The value 7n + 6 comes from using the first term as the constant directly, without subtracting the common difference first. The value 6n + 1 comes from using the first term, 6, as the coefficient of n instead of the common difference, and then attaching +1 with the sign of the constant flipped.
- (d) 218 — The rule x_{n+1} = 0.8x_n + 50 must be applied once for each step, using the result of the previous step every time — not the same starting value repeated. Starting from x_0 = 200: 0.8 × 200 = 160, so x_1 = 160 + 50 = 210. Then 0.8 × 210 = 168, so x_2 = 168 + 50 = 218. Stopping after one iteration leaves x_1 = 210, not x_2. Applying only the multiplier twice without adding 50 at each step uses 0.8² = 0.64, and 0.64 × 200 = 128, which drops the 50 completely. Adding 50 twice at the end instead of once per step, 128 + 100 = 228, still does not reproduce the actual recurrence, because the 50 added at the first step is itself multiplied by 0.8 at the second step. After two iterations, x_2 = 218.
- (b) 53.1° — The angle of elevation is opposite the height of the flagpole, 12 m, and adjacent to the distance from its base, 9 m, so tan θ = 12/9 = 1.333..., giving θ = tan⁻¹(1.333...) = 53.13...° ≈ 53.1°. "36.9°" finds the OTHER acute angle of the triangle, 90° − 53.1°, the angle at the top of the flagpole rather than the angle of elevation at Freya's position. "48.6°" comes from wrongly treating 9/12 as a sine ratio and finding sin⁻¹(0.75) = 48.6°, when neither side here is the hypotenuse. "41.4°" comes from wrongly treating 9/12 as a cosine ratio and finding cos⁻¹(0.75) = 41.4°, again without a hypotenuse in the ratio at all.
- (b) (n + 1)² − n² = 2n + 1 — (n + 1)² = n² + 2n + 1, so (n + 1)² − n² = n² + 2n + 1 − n² = 2n + 1, which is odd because it is one more than the even number 2n. Expanding (n + 1)² as n² + 1 uses the false rule (a + b)² = a² + b², and subtracting n² from that leaves just 1 — always expand (a + b)² as a² + 2ab + b². Writing n² + 2n + 1 expands correctly but never carries out the subtraction of n². Writing 2n forgets the constant term left after subtracting.
- (b) the first and the third — Method: the sign of a product depends only on the signs of the two numbers multiplied: like signs give a positive product and unlike signs a negative one, so look for the quadrants in which both coordinates carry the same sign. Working: in the first quadrant x and y are both positive, and positive times positive is positive; in the third quadrant both are negative, and negative times negative is positive as well; in the second quadrant x is negative while y is positive, and in the fourth x is positive while y is negative, so each of those gives a negative product. Answer: the first and the third. The distractors: 'the second and the fourth' comes from applying the rule for a NEGATIVE product, unlike signs, to a positive one; 'the first and the second' comes from testing only the y-coordinate and keeping the quadrants where the height is positive; 'the first and the fourth' comes from testing only the x-coordinate in the same way.
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.