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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GCSE Higher sample Paper 2 (calculator)
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- 1.An allotment is divided into two plots in the ratio 2:3. The larger plot has an area of 18 m². Work out the fraction of the total area taken up by the smaller plot.
- 2.The first term of an arithmetic sequence is 5, and each term after that increases by 6. Work out the 8th term of the sequence.
- 3.A cyclist rides 19.3 km in 47 minutes. Work out her average speed, in km/h, to 1 decimal place.
- 4.A map has a scale of 1 : 20 000. The distance between two villages is 3.4 km in real life. How long is this distance on the map, in centimetres?
- 5.A garage tested 200 cars in one week. 70% of the cars were more than 3 years old and the rest were 3 years old or less. Of the cars more than 3 years old, 1 in 4 failed the test. 12 of the cars that were 3 years old or less failed the test. Work out how many of the 200 cars failed the test altogether.
- 6.The midday temperature in Leeds was recorded on each day of one week: 22 °C, 24 °C, 23 °C, 25 °C, 26 °C, 21 °C, 24 °C. Work out the mean midday temperature, giving your answer to 2 decimal places.
- 7.Work out 3/4 − 5/12 exactly, giving your answer in its simplest form.
- 8.Solve 2x + 7 = x + 13
- 9.A train travels at a constant speed of 90 km/h. Work out the speed in m/s.
- 10.Two similar model cars have a length scale factor of 2 from the small model to the large model. Priya works out that the area scale factor is 2² = 4, and then says the volume scale factor must also be 4, since that is the number she just calculated. Is Priya correct?
- 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.Petrol costs £1.48 per litre. Sam buys 35 litres and pays with three £20 notes. Work out his change.
- 13.The discriminant of a quadratic equation is greater than zero. Which statement about the solutions of that equation is correct?
- 14.Two printers' running costs are shown on separate straight-line graphs of total cost against number of pages printed. Printer A's graph has gradient £0.04 per page. Printer B's graph has gradient £0.05 per page. Work out how much cheaper it is to print 200 pages on Printer A than on Printer B, based on these running costs alone.
- 15.A fan-shaped rug is in the shape of a sector of a circle with radius 4 m and angle 180°. It costs £4 per square metre to make. Using π = 3.14, work out the total cost of the rug.
- 16.A circle has centre (0, 0) and equation x² + y² = 50. Work out the length of the diameter of the circle, correct to 1 decimal place.
- 17.A company's turnover this year is £180,000. Last year's turnover was £120,000. Write down this year's turnover as a percentage of last year's turnover.
- 18.A student is asked to rearrange the cosine rule a² = b² + c² − 2bc cos A to make cos A the subject. They write: cos A = (a² − b² − c²) / (2bc). Is the student's rearrangement correct?
- 19.x = 5. Work out the value of 3x² − 4.
- 20.A company's profit was £2000 in its first year. Each following year, the profit increases by £800. Work out the first year in which the profit is more than £7000.
Answer key
- (a) 2/5 — The ratio 2:3 has 2 + 3 = 5 parts in total, and the larger plot is 3 of those parts. Since the larger plot is 18 m², each part is 18 ÷ 3 = 6 m², so the total area is 5 × 6 = 30 m² and the smaller plot is 2 × 6 = 12 m². The fraction of the total area taken up by the smaller plot is 12/30, which simplifies to 2/5. Giving the fraction for the larger plot instead of the smaller one gives 3/5. Comparing the smaller plot to the larger plot instead of to the total area gives 2/3. Assuming the two plots split the area evenly, ignoring the given ratio altogether, gives 1/2.
- (d) 47 — To reach the 8th term from the 1st term, the difference of 6 is added 7 times (8 − 1 = 7): 5 + 7 × 6 = 47. A candidate who multiplies by the term number itself, rather than one less, would compute 5 + 8 × 6 = 53. A candidate who uses one step too few (6 instead of 7) would reach 5 + 6 × 6 = 41. A candidate who forgets to include the first term at all would compute just 8 × 6 = 48.
- (d) 24.6 km/h — Average speed = distance ÷ time, with time in hours. 47 minutes = 47 ÷ 60 hours. 19.3 ÷ (47 ÷ 60) = 19.3 ÷ 47 × 60 = 24.638…, which rounds to 24.6 km/h (1 d.p.). 0.4 km/h comes from dividing the distance by 47 and treating the result as km/h directly, without converting the minutes to hours at all. 41.1 km/h comes from converting minutes to hours by dividing by 100 instead of 60 (19.3 ÷ 47 × 100). 0.3 km/h comes from dividing the distance by 60 instead of converting the 47 minutes to hours first.
- (a) 17 cm — Convert the real distance to centimetres first: 3.4 km = 340 000 cm. Then divide by the scale factor, 20 000, since the map is 20 000 times smaller than real life: 340 000 ÷ 20 000 = 17, so the map distance is 17 cm. Choosing 1700 cm divides by 200 instead of 20 000, losing two zeros from the scale factor (340 000 ÷ 200 = 1700). Choosing 170 cm divides by 2 000 instead of 20 000, losing one zero from the scale factor (340 000 ÷ 2 000 = 170). Choosing 0.17 cm divides by 2 000 000 instead of 20 000, adding two extra zeros to the scale factor (340 000 ÷ 2 000 000 = 0.17).
- (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.
- (d) 23.57 °C — Method: add all seven temperatures, divide by the number of readings and round only at the end. Working: 22 + 24 + 23 + 25 + 26 + 21 + 24 = 165, and 165 ÷ 7 = 23.5714…, which rounds to 23.57 to 2 decimal places. Answer: 23.57 °C. The distractors: 24 °C comes from writing down the mode, the only temperature recorded twice, instead of the mean; 27.5 °C comes from dividing the total by 6 instead of by the 7 days recorded; 5 °C comes from working out the range, 26 − 21, which is a measure of spread and not an average.
- (a) 1/3 — To subtract these fractions, first write 3/4 with a denominator of 12: 3/4 = 9/12. Then 9/12 − 5/12 = 4/12, which simplifies to 1/3. Subtracting the numerators and the denominators separately, (3 − 5)/(4 − 12), gives −2/−8, which simplifies to 1/4. Changing 3/4 to twelfths by only changing the denominator, without scaling the numerator to match, gives 3/12 − 5/12 = −2/12, which simplifies to −1/6. Adding the fractions instead of subtracting them, 9/12 + 5/12, gives 14/12, which simplifies to 7/6.
- (d) x = 6 — Method: with an unknown on both sides, first collect the x terms on one side by subtracting the smaller x term from both sides, then deal with the numbers. Working: subtracting x from both sides gives x + 7 = 13, and subtracting 7 from both sides gives x = 6. Answer: x = 6. The distractors: x = 20 comes from adding 7 to 13 instead of subtracting it once the x terms have been collected; x = 2 comes from collecting the x terms by adding them, giving 3x + 7 = 13 and then 3x = 6; x = −6 comes from subtracting 2x from both sides to get 7 = −x + 13, reaching −6 = −x and then copying the sign straight across instead of dividing by −1.
- (b) 25 m/s — Convert km/h to m/s by multiplying by 1000 (km to m) and dividing by 3600 (hours to seconds): 90 × 1000 ÷ 3600 = 25 m/s. Working out 90 ÷ 60 = 1.5 converts using 60, as if going from hours to minutes rather than to seconds. Working out 90 × 3.6 = 324 multiplies by 3.6 instead of dividing by it, going the wrong way between the units. Working out 90 × 1000 = 90000 converts kilometres to metres but forgets to convert hours to seconds at all. The train's speed is 25 m/s.
- (d) No — the volume factor is 2³ = 8, not the area factor 4. — For similar solids, area scales with the square of the length scale factor and volume scales with its cube — different powers of the same number, so they are not usually equal. The length scale factor here is 2, so the area scale factor is 2² = 4 (Priya's figure) but the volume scale factor is 2³ = 8. Priya is wrong: the correct volume scale factor is 8, not 4. The option claiming area and volume always scale by the same factor treats the two as interchangeable, which is only true for the length factor itself. The option giving 6 comes from multiplying the length factor by the number of dimensions (2 × 3) rather than cubing it. The option repeating '4' for volume simply reuses Priya's own area calculation.
- (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.
- (c) £8.20 — Method: find the total cost, then subtract from the amount paid. Working: 35 × £1.48 = £51.80. Amount paid = 3 × £20 = £60.00. Change = £60.00 − £51.80 = £8.20. Answer: £8.20. (£58.52 comes from forgetting to multiply the price by the 35 litres and subtracting only £1.48 from £60. £7.50 comes from rounding £1.48 up to £1.50 before multiplying, giving a total of £52.50 instead of £51.80. £9.20 comes from miscarrying in the pence column when subtracting £51.80 from £60.00.)
- (c) It has two different real solutions — Method: the discriminant is the quantity under the square root sign in the quadratic formula, so its sign decides how many real values the formula produces. Working: when the discriminant is positive its square root is a real number that is not zero, and the formula adds that root and subtracts it in turn, so the two results are different; a positive discriminant that is not a perfect square still gives two solutions, but they are not whole numbers. Answer: it has two different real solutions. The distractors: one repeated real solution is what a discriminant of zero gives, because adding and subtracting zero changes nothing; no real solutions is what a negative discriminant gives, because a negative number has no real square root; two whole-number solutions happens only when the discriminant is a perfect square and the division works out exactly, which a positive discriminant does not guarantee.
- (d) £2.00 — Cost for Printer A = 200 × £0.04 = £8. Cost for Printer B = 200 × £0.05 = £10. Difference = £10 − £8 = £2.00.
- (d) £100.48 — Area of the rug = (angle ÷ 360) × π × r² = (180 ÷ 360) × 3.14 × 16 = 0.5 × 50.24 = 25.12 m². Cost = 25.12 × £4 = £100.48. (£200.96 comes from finding the area of a full circle, 3.14 × 16 = 50.24 m², and forgetting the angle fraction before costing it; £25.12 comes from forgetting to square the radius, using 0.5 × 3.14 × 4 = 6.28 m², and then costing that; £50.24 comes from using the arc length formula, 0.5 × 2 × 3.14 × 4 = 12.56, in place of the area, and costing that.)
- (c) 14.1 — Method: in x² + y² = r² the right-hand side is the square of the radius, so take its square root to find the radius, then double the radius because the diameter is twice the radius. Working: r² = 50, so r = √50 = 7.07106…, and the diameter is 2 × 7.07106… = 14.14213…, which is 14.1 correct to 1 decimal place. Answer: 14.1. The distractors: 7.1 is the radius, worked out correctly but never doubled, so it answers a question about the radius rather than the diameter; 100.0 comes from doubling the 50 on the right-hand side, treating r² as though it were already the radius; 25.0 comes from halving the 50, treating r² as though it were already the diameter.
- (c) 150% — Percentage = (180,000 ÷ 120,000) × 100 = 150%.
- (b) No — should divide by −2bc, not +2bc; sign is wrong. — Method: rearrange a² = b² + c² − 2bc cos A step by step and compare with the student's version. Working: subtracting b² + c² from both sides gives a² − b² − c² = −2bc cos A, then dividing both sides by −2bc gives cos A = (a² − b² − c²) / (−2bc), which is the same as cos A = (b² + c² − a²) / (2bc). The student divided by +2bc instead of −2bc, so their expression is the negative of the correct one — the verdict is No. Saying the algebra is fine because 'either sign order gives a valid result' ignores that only one of the two signed expressions matches the original equation. Saying the denominator should be bc rather than 2bc is a different, unrelated error — the coefficient 2bc in the original formula is correct and must stay.
- (c) 71 — Method: square x first, then multiply by 3, then subtract 4, following the order of operations. Working: x² = 5² = 25; 3 × 25 = 75; 75 − 4 = 71. Answer: 71. 221 comes from squaring (3x) as a whole first: (3 × 5)² = 225, then − 4 = 221, squaring the coefficient along with x. 75 comes from correctly working out 3x² but forgetting to subtract the 4. 3 comes from subtracting the 4 from x before squaring: (5 − 4)² × 3 = 3, doing the operations in the wrong order.
- (d) 8 — Method: write the nth term of the sequence, 2000 + 800(n − 1), and find the smallest whole n for which it is greater than 7000. Working: 2000 + 800(n − 1) > 7000, so 800(n − 1) > 5000, giving n − 1 > 6.25. Since n − 1 must be a whole number, the smallest value is 7, so n = 8. Check: year 8's total is 2000 + 800 × 7 = 7600, which is more than £7000, while year 7's total is 2000 + 800 × 6 = 6800, which is not. Answer: year 8. 7 comes from rounding 6.25 to the nearest whole number, 6, and then adding 1, instead of rounding up to the next whole number before adding 1. 6 comes from using 6.25 rounded down to 6 as the year number directly, without adding the 1 needed to convert from the number of increases to the year number. 9 comes from adding one extra year beyond the year that already satisfies the condition.
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.