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 car is hired from 10:00 on Monday until 16:00 on Thursday. Work out the total length of the hire, in hours.
- 2.On a map, where each grid unit represents 1 km, Ravi's house is at (−3, 4) and the library is at (6, 4). Work out the distance between Ravi's house and the library.
- 3.A force of 126 N acts on an area of 3.5 m². Work out the pressure on the area, in N/m².
- 4.Line 1 passes through (0, 1) and (2, 5). Line 2 passes through (3, 2) and (5, 6). Work out the gradient of each line. Then write down whether the two lines are parallel.
- 5.The probability that Kofi passes his driving test on any attempt is 0.6, and each attempt is independent of the others. Work out the probability that he fails both his first two attempts.
- 6.Two classes sat the same maths test, both marked out of 100. Class A had a mean mark of 70 and a range of 30 marks. Class B had a mean mark of 70 and a range of 10 marks. Compare the marks of the two classes.
- 7.A digital timer truncates every time to 1 decimal place. It shows a swimmer's time for one length as 12.3 seconds. Using t for the swimmer's actual time in seconds, write down the error interval for t.
- 8.Which of these is a factor of 15x + 20?
- 9.Two mathematically similar rectangles have lengths in the ratio 3 : 5. The area of the smaller rectangle is 27 cm². Work out the area of the larger rectangle.
- 10.OABC is a parallelogram, with OA = a and OC = c. M is the midpoint of AB. Express the vector MC in terms of a and c.
- 11.Two ordinary fair dice are rolled and the two scores are multiplied together. Work out the probability that the product is odd.
- 12.A concrete mix is made from cement and sand in the ratio 3 : 5 by mass. A builder mixes 96 kg of concrete. Work out the mass of cement in the mix.
- 13.A ball is dropped and bounces. The height of each bounce after the first is 8 cm less than the bounce before it. The first bounce reaches 60 cm. Work out the height of the 5th bounce.
- 14.The exchange rate is £1 = 1.28 US dollars. Convert £350 into US dollars.
- 15.A circle has centre O and radius 13 cm. AB is a chord of the circle, and M is the midpoint of AB. OM is drawn, and OM = 5 cm. Work out the length of AB.
- 16.A tram sets off from a stop. Its velocity-time graph rises in a straight line from 0 m/s to 12 m/s over the first 8 seconds, then stays constant at 12 m/s for a further 10 seconds. Work out the total distance the tram travels in these 18 seconds, using the area under the graph.
- 17.A map has a scale of 1 : 100 000. Two railway stations are 8 km apart in real life. Work out the distance between the stations on the map, in centimetres.
- 18.In triangle ABC, AB = 15.6 cm, AC = 8.9 cm and angle BAC = 112°, the angle between them. Work out the area of triangle ABC. Give your answer to 1 decimal place.
- 19.A straight line passes through the points (2, 7) and (5, 16). Work out the value of y when x = 0.
- 20.A line segment has one endpoint at (−6, 2) and its midpoint at (−1, 5). Work out the coordinates of the other endpoint.
Answer key
- (b) 78 hours — From 10:00 on Monday to 10:00 on Thursday is exactly 3 complete days, which is 3 × 24 = 72 hours. From 10:00 to 16:00 on the Thursday is a further 6 hours, giving a total of 72 + 6 = 78 hours. Counting Monday to Thursday as 4 full calendar days instead of 3 complete 24-hour periods gives 4 × 24 = 96 hours. Undercounting the number of complete days as 2 instead of 3 gives 2 × 24 + 6 = 54 hours. Subtracting the extra 6 hours instead of adding them to the 3 complete days gives 72 − 6 = 66 hours.
- (a) 9 km — Ravi's house and the library have the same y-coordinate, so the distance between them is the difference between their x-coordinates: 6 − (−3) = 9 km. 3 km comes from dropping the negative sign on Ravi's x-coordinate and working out 6 − 3 instead. 0 km comes from using the y-coordinates, which are equal, instead of the x-coordinates. −9 km comes from working out −3 − 6 = −9 and not converting it to a positive distance.
- (a) 36 — Pressure = force ÷ area, so 126 ÷ 3.5 = 36 N/m². (0.03 comes from dividing the area by the force instead of the force by the area, the wrong way round. 129.5 comes from adding 126 and 3.5 instead of dividing. 441 comes from multiplying 126 by 3.5 instead of dividing.)
- (a) Parallel, since both gradients are 2 — Gradient of line 1 = (5 − 1) ÷ (2 − 0) = 4 ÷ 2 = 2. Gradient of line 2 = (6 − 2) ÷ (5 − 3) = 4 ÷ 2 = 2. The two gradients are equal, so the lines are parallel. "Not parallel, since the gradients are 2 and 4" uses 4 for line 2, which comes from dividing the change in y, 4, by 1 rather than by the change in x, 5 − 3 = 2. "Not parallel, since the two lines cross the y-axis at different points" confuses parallel lines with the same line repeated — parallel lines are distinct lines, and where a line crosses the y-axis has no effect on its gradient. "Parallel, since both lines pass through positive coordinates" reaches the right conclusion for the wrong reason: lines are parallel because their gradients are equal, not because the coordinates given happen to be positive.
- (a) 0.16 — The probability that Kofi fails a single attempt is 1 − 0.6 = 0.4. Since the attempts are independent, the probability he fails both is 0.4 × 0.4 = 0.16. Choosing 0.36 comes from squaring the probability of PASSING instead, 0.6 × 0.6 = 0.36, which is the probability of passing both attempts, not failing both. Choosing 0.4 comes from giving the probability of failing just one attempt, forgetting to combine two attempts. Choosing 0.24 comes from multiplying the fail probability by the pass probability, 0.4 × 0.6 = 0.24, mixing up passing and failing between the two attempts.
- (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.
- (c) 12.3 ≤ t < 12.4 — Method: truncating cuts the later digits off instead of rounding them, so nothing is ever pushed upwards. The displayed value is therefore the smallest the time can be, and the time can run up to, but not reach, the next value the display can show. Working: the display reads 12.3, so the actual time is at least 12.3 seconds; as soon as the time reaches 12.3 + 0.1 = 12.4 seconds the display would read 12.4, so 12.4 is not included. Answer: 12.3 ≤ t < 12.4. The distractors: 12.25 ≤ t < 12.35 is the interval for a time rounded to 1 decimal place, and this display does not round; 12.3 < t ≤ 12.4 excludes the one value the display certainly allows and includes the one it rules out; 12.3 ≤ t ≤ 12.4 treats 12.4 seconds as possible, but at 12.4 seconds the display would no longer read 12.3.
- (a) 5 — A factor of the whole expression must divide every term exactly. 15x + 20 = 5(3x + 4), so 5 is a factor. Distractor origins: 3 divides 15x exactly but does not divide 20 exactly; 4 divides 20 exactly but does not divide 15x exactly; 10 also divides 20 exactly but does not divide 15x exactly, so it is a factor of only one term, not of the whole expression.
- (b) 75 cm² — Areas of similar shapes are in the ratio of the squares of their lengths. Squaring both parts of 3 : 5 gives an area ratio of 9 : 25, so the larger area is 25/9 of the smaller one. Working with the smaller area: 27 ÷ 9 = 3, and 3 × 25 = 75. The area of the larger rectangle is 75 cm².
- (c) (1/2)c − a — Method: in parallelogram OABC, AB is equal and parallel to OC, so AB = c; M is the midpoint of AB, so AM = (1/2)c and OM = OA + AM = a + (1/2)c. MC runs from M to C, so MC = OC − OM. Working: MC = c − (a + (1/2)c) = (1/2)c − a. Answer: MC = (1/2)c − a. Subtracting in the wrong order gives a − (1/2)c, the same vector pointing the opposite way, from C to M rather than M to C; forgetting to halve the c-term gives c − a, which is AC, not MC; and adding instead of subtracting gives (1/2)c + a, which is OM itself. Always subtract the vector for the START of the journey, OM, from the vector for its END point, OC — and keep the fraction from the halving step.
- (d) 1/4 — Method: a product is odd only when BOTH factors are odd, so list the ordered pairs where both scores are odd and divide by 36. Working: the odd scores on a dice are 1, 3 and 5, so there are 3 × 3 = 9 ordered pairs where both scores are odd, out of the 36 equally likely pairs, cancelling down to 1/4. Answer: 1/4. Watch out: writing down 3/4 finds the probability that AT LEAST ONE score is odd, 1 minus the probability both are even, which is a different, easier condition to meet than both being odd. Considering only the first dice's score and ignoring the second gives 1/2, since 3 of the first dice's 6 scores are odd — but the product also depends on what the second dice shows. And writing down 1/12 comes from counting only the pairs where the SAME odd number appears twice, (1, 1), (3, 3) and (5, 5), missing pairs like (1, 3) and (5, 1) where the two odd scores differ.
- (a) 36 kg — The mix is 3 + 5 = 8 equal shares, so the cement is 3/8 of the mass. One share is 96 ÷ 8 = 12 kg, and the cement is 3 shares: 3 × 12 = 36 kg. Working out 3/5 of 96 gives 57.6 kg, which uses the sand as the denominator instead of the whole mix; 60 kg is the mass of the sand; 12 kg is one share only.
- (d) 28 — The height decreases by 8 cm at each bounce after the first, so the nth bounce reaches 60−(n−1)×8 cm. For the 5th bounce: 60−4×8=60−32=28. A candidate who subtracts 8 one time too many, five times instead of four, would compute 60−5×8=20. A candidate who adds the decrease instead of subtracting it, a sign error, would compute 60+4×8=92. A candidate who works out only the total decrease and forgets to include the starting height of 60 cm would compute just 5×8=40.
- (c) 448.00 US dollars — Method: multiply the amount in pounds by the exchange rate. Working: £350 × 1.28 = 448.00 US dollars. Wrong options: 273.44 US dollars comes from dividing by the rate instead of multiplying (350 ÷ 1.28); 351.28 US dollars comes from adding the rate to the amount instead of multiplying; 4,480.00 US dollars comes from a decimal-point slip, using 12.8 instead of 1.28.
- (a) 24 cm — The line from the centre to the midpoint of a chord is perpendicular to the chord, so triangle OMA has a right angle at M. By Pythagoras' theorem, AM² = OA² − OM² = 169 − 25 = 144, so AM = 12 cm. AB is twice AM, since M is the midpoint: AB = 2 × 12 = 24 cm. Finding AM = 12 cm correctly but forgetting to double it for the full chord gives 12 cm. Working out 169 − 25 = 144 and forgetting to take the square root gives 144 cm. Subtracting first and then doubling the wrong way, (13 − 5) × 2, gives 16 cm. Doubling AM after finding it correctly is the step that's missing from all three — do it, and you get 24 cm.
- (d) 168 m — Split the area under the graph into two parts. The rising section (0 to 8 s) is a triangle: area = 0.5 × 8 × 12 = 48. The constant section, from 8 s to 18 s (10 s), is a rectangle: area = 12 × 10 = 120. Total distance = 48 + 120 = 168 m. Leaving out the 0.5 doubles the triangle, giving 8 × 12 + 120 = 216 m; ignoring the triangle section altogether gives only 120 m; averaging the start and final speeds over the whole 18 seconds, (0 + 12) ÷ 2 = 6, then 6 × 18 = 108 m, wrongly treats the tram as accelerating the whole time, when the graph is flat for the last 10 seconds.
- (a) 8 cm — Method: a scale of 1 : n means the real distance is n times the distance on the map, so to go from the real distance back to the map distance, put both lengths in the same unit and then divide by the scale. Working: 1 km = 100 000 cm, so 8 km = 8 × 100 000 = 800 000 cm; 800 000 ÷ 100 000 = 8. Answer: 8 cm. The distractors: 800000 cm comes from converting the 8 km into centimetres and stopping there, so the division by the scale — the inverse operation the question asks for — is never done; 80 cm comes from taking a metre to be 1000 cm, which turns 8 km into 8 × 1000 × 1000 = 8 000 000 cm and gives 8 000 000 ÷ 100 000 = 80; 0.08 cm comes from taking a kilometre to be 1000 cm, which turns 8 km into 8000 cm and gives 8000 ÷ 100 000 = 0.08.
- (c) 64.4 cm² — Method: with two sides and the angle between them, use Area = (1/2)ab sin C. Working: Area = 1/2 × 15.6 × 8.9 × sin 112° = 64.4 cm² (1 d.p.). Answer: 64.4 cm². Leaving out the 1/2 gives 128.7 cm²; using cos 112° instead of sin 112° gives a negative value, which a candidate who drops the minus sign reads as 26.0 cm²; and squaring one side instead of multiplying the two different given sides together gives 112.8 cm². Sin C is never negative for an angle between 0° and 180°, so a negative area is always a sign that cos was used by mistake — check you used sin before you trust your answer.
- (a) 1 — Method: the value of y when x = 0 is where the line meets the y-axis, which is the constant c in y = mx + c, so the gradient is worked out from the two given points first and the constant follows by substituting one of them. Working: m = (16 − 7) ÷ (5 − 2) = 9 ÷ 3 = 3, so the line is y = 3x + c; substituting x = 2 and y = 7 gives 7 = 3 × 2 + c, so c = 7 − 6 = 1, and the value of y when x = 0 is that constant. Answer: 1. The distractors: 3 comes from stopping at the gradient and offering it as the intercept; 4 comes from stepping back from x = 2 to x = 0 by one unit of x instead of two, 7 − 3 = 4; −1 comes from working the constant out as mx − y, 3 × 2 − 7 = −1, instead of y − mx.
- (d) (4, 8) — The other endpoint is found from 2 × midpoint − known endpoint: x = 2 × (−1) − (−6) = −2 + 6 = 4, y = 2 × 5 − 2 = 10 − 2 = 8, giving (4, 8). (−3.5, 3.5) comes from averaging the given endpoint and the midpoint as if they were the two endpoints of a segment, instead of working backwards from the midpoint. (5, 3) comes from working out (−1 − (−6), 5 − 2) instead of doubling the midpoint before subtracting. (4, 5) comes from correctly finding the x-coordinate but copying the midpoint's y-coordinate of 5 instead of doubling it.
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.