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.A window display has red baubles, gold baubles and green baubles in the ratio 4 : 5 : 6. What fraction of the baubles are not green?
- 2.Work out the coordinates of the midpoint of the line segment joining (−3, 5) and (7, 9).
- 3.300 g of a salt solution of concentration 10% is poured into 200 g of a salt solution of concentration 25%. Work out the concentration of salt in the mixture.
- 4.A packaging designer is told a box has 6 faces, 12 edges and 8 vertices, and every face is a rectangle. To fit exactly onto a shelf, the designer needs to know whether the box must be a cube. Based only on this information, which solid must the box be?
- 5.The probability that Priya scores with a netball shot is 3/4. She takes two shots, and the shots are independent. Work out the probability that she scores with her first shot but misses her second shot.
- 6.The daily high temperatures, in °C, in Leeds over five days were 14, 16, 18, 19 and 23. In York over the same five days they were 15, 17, 17, 18 and 18. Write a sentence comparing Leeds and York using both the median and the range.
- 7.A restaurant offers a lunch deal of one starter from 5 options, one main from 6 options and one dessert from 3 options. Work out how many different lunch deals are possible.
- 8.A market stall's cost of hiring n tables is modelled by two formulas: Formula A: C = 3(2n + 5); Formula B: C = 6n + 15, where C is in pounds. A stallholder says the two formulas always give the same cost. Work out the cost given by each formula when n = 4, and use your results to decide whether the stallholder is correct.
- 9.A tangent to a distance–time graph, with distance in kilometres and time in minutes, has gradient 0.78 kilometres per minute at a particular point. Work out the instantaneous rate of change of distance with time, in kilometres per hour.
- 10.A rectangle has vertices at (2, 1), (9, 1), (9, 5) and (2, 5). Work out the area of the rectangle.
- 11.At a sports club, the numbers of members who play tennis and badminton are: 50 members play tennis, 40 members play badminton, and 18 members play both tennis and badminton. A member who plays tennis is chosen at random. Work out the probability that this member also plays badminton.
- 12.During one night on a mountain, four hikers each recorded the temperature they felt: Ben −6 °C, Priya −2 °C, Sam −9 °C, Alex 1 °C. Work out the difference between the coldest and the warmest of these four temperatures.
- 13.A train's distance-time graph shows it travelling 100 km at a steady speed in the first 1.5 hours, waiting at a signal for 0.25 hours, then travelling a further 47 km in 0.75 hours. Work out the average speed for the whole journey, correct to 1 decimal place.
- 14.Water flows into a tank at a constant rate. After 8 minutes the tank holds 200 litres. Write the ratio of time in minutes to volume in litres, in its simplest form.
- 15.A roof truss has two horizontal parallel rafters, one above the other. A straight strut crosses both rafters. Where the strut crosses the lower rafter, the angle above the rafter and to the left of the strut is 65°. Work out the size of the angle above the upper rafter and to the right of the strut, where the strut crosses it.
- 16.The formula connecting speed, distance and time is v = s ÷ t. Work out v, in km/h, when s = 180 and t = 4.
- 17.Priya is paid £58.50 for working 7.5 hours on a Saturday. Work out her rate of pay, in £ per hour.
- 18.Work out the exact value of cos 0° − sin 30°.
- 19.Solve the inequality 6.3x + 1.8 ≤ 40. Hence write down the greatest number with exactly one decimal place that satisfies the inequality.
- 20.f(x) = 2x + 1. Work out the value of x for which f⁻¹(x) = 5.y = 2x + 1
Answer key
- (b) 3/5 — Method: add all three parts for the total, add together the parts that are not green, then write this over the total. Working: total parts = 4 + 5 + 6 = 15. Not green = 4 + 5 = 9. Fraction = 9/15 = 3/5. Answer: 3/5. 2/5 comes from finding the fraction that IS green (6/15 = 2/5) instead of not green. 4/15 comes from only counting the red baubles as 'not green' and forgetting the gold ones. 9/10 comes from adding only two of the three ratio parts to find the total (4+6=10), missing out the gold part, while still using 9 for the numerator.
- (a) (2, 7) — Midpoint = ((x1+x2)/2, (y1+y2)/2) = ((−3+7)/2, (5+9)/2) = (4/2, 14/2) = (2, 7). (4, 14) comes from adding the coordinates correctly but forgetting to divide by 2. (2, 9) comes from correctly averaging the x-coordinates but simply copying the y-coordinate of the second point instead of averaging the y-coordinates. (5, 2) comes from subtracting the coordinates instead of adding them before halving: ((7−(−3))/2, (9−5)/2) = (5, 2).
- (c) 16% — Method: scale each concentration to its own mass to find the salt it contains, add the two masses of salt, then write the ratio of salt to mixture per 100 g. Working: 10:100 = x:300 gives 30 g of salt, and 25:100 = y:200 gives 50 g of salt; the mixture holds 30 + 50 = 80 g of salt in 300 + 200 = 500 g of solution; 80:500 = 16:100. Answer: 16%. The distractors: 17.5% is the mean of 10% and 25%, which ignores that there is more of the weaker solution than of the stronger one; 19% comes from swapping the two concentrations over, working out (300 × 25% + 200 × 10%) ÷ 500; 26.7% comes from dividing the 80 g of salt by the 300 g of the first solution rather than by the 500 g of mixture.
- (b) Cuboid — not necessarily a cube — A solid with 6 faces, 12 edges and 8 vertices in which every face is a rectangle is a cuboid, but nothing here confirms that all the edges are the same length, so the box could be a cube or a non-cube cuboid; the most that can be concluded is that it is a cuboid, making 'Cuboid — not necessarily a cube' correct. 'Cube — only a cube fits this' is wrong because a cube is just one particular cuboid; a general cuboid with different length, width and height has exactly the same face, edge and vertex counts and rectangular faces. 'Triangular prism' is wrong because a triangular prism has 5 faces, 9 edges and 6 vertices, and two of its faces are triangles, so it matches neither the counts nor the face shape. 'Not enough information' is wrong because rectangular faces with these counts do pin the solid down to the cuboid family, even though they cannot pin down a cube specifically.
- (c) 3/16 — The probability that Priya misses a shot is 1 − 3/4 = 1/4. The two shots are independent, so multiply the probability of scoring the first shot by the probability of missing the second shot: 3/4 × 1/4 = 3/16. Choosing 9/16 comes from using the probability of scoring, 3/4, for both shots instead of switching to the miss probability for the second shot, 3/4 × 3/4 = 9/16, which is the probability that she scores with both shots. Choosing 1/2 comes from subtracting the two probabilities instead of multiplying them, 3/4 − 1/4 = 1/2. Choosing 3/4 comes from giving only the probability that she scores with the first shot, forgetting to combine it with what happens on the second shot.
- (a) Leeds has a higher median and a greater range than York. — In order, Leeds's temperatures are 14, 16, 18, 19 and 23, so the median is the middle value, 18, and the range is 23 − 14 = 9. York's temperatures in order are 15, 17, 17, 18 and 18, so the median is 17, and the range is 18 − 15 = 3. Since 18 is higher than 17, and 9 is greater than 3, Leeds has both the higher median and the greater range. Choosing 'Leeds has a higher median but a smaller range than York' gets the median comparison right but the range comparison backwards — Leeds's range of 9 is actually greater than York's range of 3. Choosing 'York has a higher median and a greater range than Leeds' reverses both comparisons. Choosing 'York has a higher median but a smaller range than Leeds' reverses the median comparison; York's median of 17 is lower than Leeds's 18, even though it is correct that York's range is the smaller one.
- (c) 90 — Multiply the number of choices for each course: 5 × 6 × 3 = 90. 14 comes from adding the three numbers instead of multiplying them. 30 comes from multiplying only the starters and mains, 5 × 6, and forgetting the dessert. 18 comes from multiplying only the mains and desserts, 6 × 3, and forgetting the starter.
- (d) Formula A gives £39 and Formula B gives £39, and since 3(2n + 5) expands to 6n + 15 for every value of n, the stallholder is correct. — Formula A: 3(2 × 4 + 5) = 3 × 13 = £39. Formula B: 6 × 4 + 15 = 24 + 15 = £39. Expanding Formula A algebraically gives 3(2n + 5) = 6n + 15, which is identical to Formula B for every value of n, not just n = 4, so the stallholder is correct — this is an identity, not a coincidence. The option giving £29 for Formula A comes from multiplying only the 2n by 3 and forgetting to also multiply the 5, then adding the unmultiplied 5: 3 × 2 × 4 = 24, + 5 = 29. The two options that reach the correct numbers but reject the stallholder's claim both use faulty reasoning — matching values at one value of n, or counting terms, does not decide whether two expressions are identical for every n; expanding the bracket does.
- (a) 46.8 km/h — Method: 1 hour = 60 minutes, so a rate given in kilometres per minute is converted to kilometres per hour by multiplying by 60. Working: 0.78 × 60 = 46.8, so the instantaneous rate of change is 46.8 kilometres per hour. Keeping the given value unchanged and only relabelling the unit gives 0.78 km/h, which ignores that the time unit has changed. Dividing by 60 instead of multiplying — as you would when converting to a larger length unit — gives 0.78 ÷ 60 = 0.013 km/h, the wrong direction for a rate measured against a larger time unit. Adding 60 to the given rate instead of multiplying by it gives 0.78 + 60 = 60.78 km/h. Converting a rate always means multiplying or dividing by the conversion factor between the units, never adding it, and the direction depends on whether the new time unit is bigger or smaller than the old one.
- (b) 28 — The width of the rectangle is the difference in x-coordinates, 9 − 2 = 7, and the height is the difference in y-coordinates, 5 − 1 = 4. The area is width × height = 7 × 4 = 28. 22 comes from using the perimeter formula, 2 × (7 + 4), instead of the area formula. 35 comes from multiplying 7 by 5 instead of 4, misreading one of the y-coordinates. 63 comes from multiplying 9 by 7, using an x-coordinate instead of the height.
- (a) 9/25 — Method: P(badminton | tennis) = n(tennis and badminton) ÷ n(tennis) — restrict to the tennis-players, then find what fraction of them also play badminton. Working: n(tennis and badminton) = 18, n(tennis) = 50, so P(badminton | tennis) = 18/50 = 9/25. Answer: 9/25. Watch out: dividing by 40 (the badminton total) finds P(tennis | badminton) instead of P(badminton | tennis) — the wrong direction. Dividing by 90 (all the members named in the question) ignores that you already know the member plays tennis. And dividing by 72 (50 + 40 − 18, the number who play at least one of the two sports) answers a question about the union, not the condition you were given.
- (a) 10 °C — Method: work out the coldest and warmest of the four temperatures, then subtract to find the difference. Working: the coldest temperature is Sam's, −9 °C, and the warmest is Alex's, 1 °C. The difference is 1 − (−9) = 1 + 9 = 10. Answer: 10 °C. 8 °C comes from working out 1 − 9 = −8 and reporting 8, dropping the negative sign on −9 instead of turning the subtraction into an addition. 3 °C comes from comparing the wrong pair, Sam's −9 °C and Ben's −6 °C, instead of the coldest and the warmest: −6 − (−9) = 3. 7 °C comes from comparing Ben's −6 °C with Alex's 1 °C, mistakenly treating Ben's reading as the coldest instead of Sam's.
- (b) 58.8 km/h — Total distance = 100 + 47 = 147 km. Total time = 1.5 + 0.25 + 0.75 = 2.5 hours. Average speed = 147 ÷ 2.5 = 58.8 km/h. Leaving out the 0.25 hours of waiting from the total time gives 147 ÷ 2.25 = 65.3 km/h (1 d.p.) — wrong, because the train is stationary but time is still passing on the whole journey. Using only the first leg gives 100 ÷ 1.5 = 66.7 km/h (1 d.p.) — wrong, because it ignores the second leg of the journey entirely. Working out each leg's own speed (100 ÷ 1.5 = 66.7 km/h and 47 ÷ 0.75 = 62.7 km/h) and then averaging those two speeds gives 64.7 km/h (1 d.p.) — wrong, because the average of two speeds over DIFFERENT times is not the same as total distance divided by total time.
- (b) 1:25 — Write the ratio time : volume using the numbers in the question: 8 : 200. Divide both parts by their highest common factor, 8, to give 1 : 25. (25:1 comes from writing the ratio the wrong way round, volume : time. 8:25 comes from dividing only the volume by 8 and leaving the time unchanged. 25:8 is that same mistake written the wrong way round.)
- (c) 115 — Method: use corresponding angles to carry the 65° angle from the lower rafter up to the upper rafter, then use angles on a straight line to move to the other side of the strut. Working: the angle above the upper rafter and to the left of the strut corresponds to the given angle, so it is 65°; the angle above the upper rafter and to the right of the strut lies on a straight line with it, so it is 180 − 65 = 115. Answer: 115°. A candidate who assumes the angle stays 65° without allowing for the move from the left of the strut to the right of it gives 65. A candidate who uses 90° instead of 180°, working out 90 − 65, gets 25. A candidate who adds instead of subtracting, working out 180 + 65, gets 245.
- (d) 45 — v = s ÷ t = 180 ÷ 4 = 45 km/h. 720 comes from multiplying s and t instead of dividing, 180 × 4. 184 comes from adding s and t instead of dividing, 180 + 4. 176 comes from subtracting t from s instead of dividing, 180 − 4.
- (a) £7.80 — Rate of pay = total pay ÷ number of hours. £58.50 ÷ 7.5 = £7.80 per hour. £438.75 comes from multiplying the pay by the hours instead of dividing (£58.50 × 7.5). £0.13 comes from dividing the hours by the pay instead of the pay by the hours (7.5 ÷ 58.50). £51.00 comes from subtracting the hours from the pay (£58.50 − 7.5) instead of dividing.
- (a) 1/2 — cos 0° = 1 and sin 30° = 1/2, so cos 0° − sin 30° = 1 − 1/2 = 1/2. 1 comes from writing down cos 0° alone and forgetting to subtract sin 30°. 3/2 comes from adding the two values instead of subtracting. −1/2 comes from working out sin 30° − cos 0°, the two terms the wrong way round.
- (d) 6.0 — Subtract 1.8 from both sides: 6.3x ≤ 38.2. Divide both sides by 6.3: x ≤ 6.0634… . The question asks for the greatest value to 1 decimal place that still satisfies the inequality. Testing 6.1: 6.3 × 6.1 + 1.8 = 40.23, which is more than 40, so 6.1 fails. Testing 6.0: 6.3 × 6.0 + 1.8 = 39.6, which is no more than 40, so 6.0 works and is the greatest such value. A candidate who simply rounds 6.0634… to 1 decimal place answers 6.1, without checking that it satisfies the inequality. A candidate who adds 1.8 instead of subtracting works out 41.8 ÷ 6.3 and answers 6.6. A candidate who forgets the 1.8 altogether divides 40 by 6.3 and answers 6.3.
- (d) 11 — f⁻¹(x) = 5 means x = f(5), since applying f to both sides undoes the inverse. f(5) = 2 × 5 + 1 = 11. Writing 2 comes from confusing f⁻¹(x) = 5 with f(x) = 5, and solving 2x + 1 = 5 instead: 2x = 4, x = 2. Writing 9 comes from finding f⁻¹(x) with a sign error, f⁻¹(x) = (x + 1)/2 instead of (x − 1)/2, then setting this equal to 5: x + 1 = 10, x = 9. Writing 6 comes from finding f⁻¹(x) without dividing by 2 at all, f⁻¹(x) = x − 1, then setting this equal to 5: x = 6.
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.