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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- (c) 5/11 — Let x = 0.45 recurring, so x = 0.454545... . Since two digits repeat, multiply by 100: 100x = 45.454545... . Subtracting the original x removes the recurring part exactly, because it lines up digit for digit: 100x − x = 45.454545... − 0.454545... = 45, so 99x = 45, giving x = 45/99 = 5/11. Treating the decimal as if it terminated at two places gives 45/100 = 9/20, which is only 0.45 and drops the repeating part entirely. Subtracting 10x instead of x — using 100x − 10x = 90x = 45 — is the wrong power of ten for a two-digit repeating block, and gives x = 45/90 = 1/2. Making an arithmetic slip in the numerator, 45 − 1 = 44 instead of 45, gives 44/99 = 4/9.
- (a) −10 — Method: multiply each bracket out, treating the second bracket as being multiplied by −3 because it is subtracted, then collect like terms. Working: 4(2x − 1) = 8x − 4 and −3(x + 2) = −3x − 6, so the expression becomes 8x − 4 − 3x − 6 − 5x; the x terms give 8x − 3x − 5x = 0, so no term in x survives, and the numbers give −4 − 6 = −10. Answer: −10. The distractors: 2 comes from expanding −3(x + 2) as −3x + 6, leaving the numbers −4 + 6; 5x − 10 comes from forgetting the final −5x, so the x terms give 8x − 3x = 5x; −7 comes from multiplying the 4 over only the first term of its bracket, giving 8x − 1 and so the numbers −1 − 6.
- (c) 180 pages — The gradient is (9 − 24) ÷ (300 − 0) = −15 ÷ 300 = −0.05, so the cartridge uses 0.05 ml of ink per page. At p = 300 there are 9 ml left. The extra pages before the cartridge is empty is 9 ÷ 0.05 = 180 pages. Giving 9 as the answer confuses the millilitres of ink remaining with the number of pages remaining — they are different quantities with different units. Multiplying instead of dividing, 9 × 0.05 = 0.45, does not undo the rate correctly. Working out the total number of pages a full cartridge lasts, 24 ÷ 0.05 = 480 pages, answers how many pages the cartridge prints in total from full, not how many more pages it can print from the 300-page point.
- (d) 600 cm² — Enlarging by scale factor 2 makes the new dimensions 10 × 2 = 20 cm and 15 × 2 = 30 cm, so the poster's area = 20 × 30 = 600 cm². A pupil who scales the original area, 150 cm², by the scale factor itself instead of by its square gets 150 × 2 = 300 cm². A pupil who adds the scale factor to each dimension instead of multiplying gets (10 + 2) × (15 + 2) = 204 cm². A pupil who forgets to enlarge the postcard at all just uses the original area, 150 cm². The correct area of the poster is 600 cm².
- (b) 100 — The sample shows a proportion of 8/60 = 2/15 faulty. Apply that proportion to the new batch of 750: 750 × 2/15 = 100. Flipping the ratio, calculating 8/750 × 60 instead of 8/60 × 750, gives 0.64, which rounds to about 1. Assuming the same number of faulty items applies to the new batch, without scaling for its larger size, just repeats the sample's count of 8. Rounding the proportion 8/60 = 0.1333... down to 0.1 before multiplying gives 750 × 0.1 = 75.
- (b) 5.5 kg — Method: with an even number of values the median is the mean of the two middle values, taken once the data are in order of size. Working: the eight masses are already in order and 8 ÷ 2 = 4, so the middle pair are the 4th and 5th values, 5 kg and 6 kg; the median is (5 + 6) ÷ 2 = 5.5 kg. Answer: 5.5 kg. The distractors: 5 kg comes from reading the 4th value and stopping there instead of averaging the middle pair; 8 kg comes from working out the range, 11 − 3, which measures spread rather than centre; 4 kg comes from writing down the modal mass, the only value that occurs twice, instead of the median.
- (a) 2/3 — Total parts = 6 + 4 + 5 = 15. Sopranos and altos together are not tenors: 6 + 4 = 10 parts, so the fraction is 10/15, which simplifies to 2/3. 1/3 comes from finding the fraction of tenors instead of the fraction that is not tenors. 2/5 comes from counting only the sopranos as not tenors and leaving the altos out. 4/9 comes from leaving sopranos out of the total, 4 + 5 = 9, and then using only the altos as the fraction that is not tenors.
- (c) 6x(2x + 3) — The highest common factor of 12x² and 18x is 6x. Dividing each term by 6x gives 12x² ÷ 6x = 2x and 18x ÷ 6x = 3, so 12x² + 18x = 6x(2x + 3). A candidate who only takes out the number 6 (missing the x) gets 6(2x² + 3x), which is not fully factorised. A candidate who only takes out 2x (missing the extra factor of 3 in 6) gets 2x(6x + 9), also not fully factorised — the bracket still shares a common factor. A candidate who takes out 3x instead of the full 6x gets 3x(4x + 6), which again is not fully factorised since 4x + 6 shares a common factor of 2.
- (b) 1.00 litres — Total volume = 350 + 650 = 1000 cm³. Since 1000 cm³ = 1 litre, the smoothie is 1.00 litre. Using only the orange juice's 650 cm³ and converting that gives 0.65 litres, forgetting the mango juice entirely. Using only the mango juice's 350 cm³ gives 0.35 litres, forgetting the orange juice. Adding both volumes correctly to get 1000 cm³ but not converting to litres leaves the answer as 1000.00, which is the volume in the wrong unit.
- (a) 1/3 — Method: since OABC is a parallelogram, B = OA + OC = a + c. M = (1/2)c, since M is the midpoint of OC. Since AN is twice NC, N is 2/3 of the way along AC from A, so N = a + 2/3(c − a) = (1/3)a + (2/3)c. Working: MN = N − M = (1/3)a + (1/6)c, and MB = B − M = a + (1/2)c. Comparing term by term, 1/3 × (a + (1/2)c) = (1/3)a + (1/6)c, which matches MN exactly. Answer: k = 1/3, so M, N and B lie on a straight line. Giving 2/3 instead is the scalar linking N to B (NB = (2/3)MB), not M to N; giving 1/6 is just MN's c-coefficient read off on its own, without comparing it to MB's c-coefficient at all; and giving 3 is the scalar the wrong way up — it is MB that equals 3 × MN, not the other way round, since MN = k × MB was what was asked for. Always match the direction of the scalar to the vectors exactly as the question states them.
- (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) 8 hours 35 minutes — From 21:35 to midnight is 2 hours 25 minutes, and from midnight to 06:10 is a further 6 hours 10 minutes, giving a total of 8 hours 35 minutes. Misreading the departure time as 22:35 instead of 21:35 loses an hour from the calculation and gives 7 hours 35 minutes. Misreading the departure time as 20:35 instead of 21:35 gains an hour and gives 9 hours 35 minutes. Subtracting the times as if both fell on the same day, without crossing midnight, gives 15 hours 25 minutes.
- (c) 72 m — The first section is a triangle with base 4 s and height 8 m/s, area = 1/2 × 4 × 8 = 16 m. The middle section is a rectangle with base 6 s and height 8 m/s, area = 6 × 8 = 48 m. The final section is a triangle with base 2 s and height 8 m/s, area = 1/2 × 2 × 8 = 8 m. Total distance = 16 + 48 + 8 = 72 m. Treating both sloping sections as full rectangles instead of triangles — forgetting the factor of a half — gives 4 × 8 + 6 × 8 + 2 × 8 = 96 m, which is wrong because a sloping speed-time section is a triangle, not a rectangle. Leaving out the final, decreasing section entirely gives 16 + 48 = 64 m, which is wrong because the cyclist is still moving and covering distance while slowing down. Leaving out the middle, constant-speed section gives 16 + 8 = 24 m, which is wrong because the cyclist covers the most distance of all during the 6 seconds of constant speed.
- (b) £16 — Since cost is proportional to the square root of diameter, C = k√d. Using d = 9, C = 12: √9 = 3, so 12 = k × 3, giving k = 12 ÷ 3 = 4. The equation is C = 4√d. When d = 16: √16 = 4, so C = 4 × 4 = 16. Halving the new diameter instead of taking its square root gives 16 ÷ 2 = 8, and then C = 4 × 8 = 32 — halving a number is not the same as taking its square root, as √16 = 4, not 8. Multiplying k by the diameter itself instead of by its square root gives C = 4 × 16 = 64, skipping the square root altogether. Reporting √16 on its own, without multiplying by k, gives only 4, not the cost. The cost of manufacturing a lens of diameter 16 mm is £16.
- (c) minor segment — A chord splits a circle into two segments; the smaller of the two is called the minor segment and the larger one the major segment. "major segment" names the larger region, the opposite of what is asked for. "sector" is a different region altogether, enclosed by two radii and an arc, not by a chord. "arc" is a curved length along the circumference, not an enclosed region at all.
- (a) Litres — The area under a graph is found by multiplying a y-value by an x-value, so its units are the y-axis units multiplied by the x-axis units: litres per minute × minutes = litres, since the 'per minute' cancels with the 'minutes'. Answering 'litres per minute' keeps the y-axis units unchanged, as if multiplying by time did nothing to the units at all. Answering 'litres per minute squared' treats the x-axis as also being measured 'per minute', squaring a unit that should instead cancel. Answering 'minutes' keeps only the x-axis units and drops the rate altogether.
- (c) Yes — 700 ÷ 7 = 100 ml for the 1 part of concentrate — Method: add the ratio parts to find the total number of parts, divide the total volume by this, then use the ratio to find concentrate's share. Working: 1 + 6 = 7 parts. 700 ÷ 7 = 100 ml per part. Concentrate = 1 part = 100 ml, so Freya is correct. Wrong options: 'divide 700 by 6' uses only one of the ratio numbers instead of the total of 7 parts, giving about 117 ml; '600 ml is concentrate' swaps which ratio number belongs to the concentrate and which belongs to the water; 'half of 700 ml should be concentrate' ignores the ratio altogether and assumes an equal split.
- (b) (1/2)b − (1/2)a — Method: MN runs from M to N, so MN = ON − OM, with OM = (1/2)a and ON = (1/2)b. Working: MN = (1/2)b − (1/2)a. Answer: MN = (1/2)b − (1/2)a. Subtracting the other way round gives (1/2)a − (1/2)b, the reverse vector from N to M; subtracting the wrong way round AND forgetting to halve gives a − b, which is BA, not MN; and adding the two halved vectors instead of subtracting them gives (1/2)a + (1/2)b, which is the position vector of the midpoint of AB. Always subtract the START point's vector from the END point's vector, and halve OA and OB before you combine them, not after.
- (d) x = 2 — Method: expand the bracket by multiplying both terms inside it by 2, then undo the addition and the multiplication in turn. Working: expanding gives 2x + 6 = 10; subtracting 6 from both sides gives 2x = 4; dividing both sides by 2 gives x = 2. Answer: x = 2. The distractors: x = 5 comes from dividing both sides by 2 first, reaching x + 3 = 5 and writing 5 without taking the 3 away; x = 8 comes from adding 6 to both sides instead of subtracting it, giving 2x = 16; x = 3.5 comes from expanding 2(x + 3) as 2x + 3, multiplying only the x by the 2, which leads to 2x = 7.
- (b) £14 — Method: read the fixed charge (the cost at 0 miles) and the rate (the cost per extra mile) from the graph, then use them to work out the cost for a distance beyond the part that is plotted. Working: the graph shows a fixed charge of £2 at 0 miles, and the cost rises by £2 for every extra mile, so for 6 miles the cost is £2 + (£2 × 6) = £2 + £12 = £14. Answer: £14. Distractor refutation: £12 comes from multiplying the rate by the distance and leaving out the £2 fixed charge. £8 comes from misreading the rate as £1 per mile instead of £2 per mile. £24 comes from adding the fixed charge to the rate first and then multiplying the total by the distance, instead of multiplying the rate by the distance and then adding the fixed charge.
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.