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 tap fills a 20 litre bucket in 2 minutes 30 seconds. Work out the rate of flow, in litres per minute.
- 2.A quadratic graph has equation y = (x − 4)². A student says this graph crosses the x-axis at two different points. Explain why the student is wrong.
- 3.The total cost, in pounds, of manufacturing a batch of components is plotted against the number of components produced. A tangent to this graph at a particular point has gradient 3.60. What does the value 3.60 represent?
- 4.A running track has a straight section and a semicircular bend. The bend is a semicircle with radius 32 m. Work out the length of the curved part of the bend, to 1 decimal place. (Use π = 3.14.)
- 5.A bag contains beads that are exactly one of red, white or black. The probability of taking a red bead is 3/10 and the probability of taking a white bead is 1/4. Work out the probability of taking a bead that is red or white.
- 6.The masses, m kg, of 60 parcels are grouped like this: 0 ≤ m < 5, 22 parcels; 5 ≤ m < 10, 20 parcels; 10 ≤ m < 20, 9 parcels; 20 ≤ m < 30, 5 parcels; 30 ≤ m < 50, 4 parcels. Write down the class interval that contains the median mass.
- 7.Expand and simplify (2 + √3)², giving your answer in the form a + b√3.
- 8.Which of these statements is an equation with exactly one solution?
- 9.Two numbers are in the ratio 3 : 7. Their sum is 60. Work out the positive difference between the two numbers.
- 10.Lines AB and CD are parallel. A straight line EF crosses AB at point P and crosses CD at point Q. At P, angle APE = 65°. Angle APE and angle BPQ are vertically opposite. Angle BPQ and angle PQD are co-interior (allied) angles. Work out angle PQD.
- 11.In a year group of 60 pupils, 25 play football, 18 play tennis and 8 play both football and tennis. A pupil who plays football or tennis or both is picked at random. Work out the probability that this pupil plays both sports.
- 12.For any two whole numbers, the product of the numbers is equal to the product of their highest common factor and their lowest common multiple. The highest common factor of 6 and 8 is 2, and 6 × 8 = 48. Work out the lowest common multiple of 6 and 8.
- 13.The equation x² = 5x − 3 is to be solved using iteration. Work out which of these iterative formulas comes from a correct rearrangement of the equation.
- 14.Two mathematically similar triangular flags have areas in the ratio 4 : 25. The height of the smaller flag is 6 cm. Work out the height of the larger flag.
- 15.Work out the exact value of cos 0° − sin 30°.
- 16.A student claims: 'For every positive integer n, n² + n + 1 is a prime number.' Which value of n shows that this claim is false?
- 17.y is directly proportional to x. When x = 4, y = 10. Work out the value of y when x = 6.
- 18.The diagram shows the plan view and the front elevation of a cuboid-shaped box, each drawn as a rectangle. Work out the area of the box's side elevation, in square centimetres.
- 19.A charity's fundraising total, T pounds, over d days follows T = (d − 3)(30 − d) for 3 ≤ d ≤ 30, where T = 0 marks the start and end of the campaign. Work out how many days the campaign runs for, from start to end.
- 20.A circle has centre (0, 0) and equation x² + y² = 100. A straight line is drawn from the origin through the point P(8, 9). Work out whether P lies inside, on, or outside the circle.
Answer key
- (a) 8 litres per minute — Method: write the time as a decimal number of minutes, then divide the volume by the time. Working: 30 seconds = 30/60 minute = 0.5 minute, so 2 minutes 30 seconds = 2.5 minutes. Rate = 20 ÷ 2.5 = 8 litres per minute. Answer: 8 litres per minute. (10 litres per minute comes from ignoring the extra 30 seconds and dividing by 2 minutes only. 8.7 litres per minute comes from misreading 2 minutes 30 seconds as 2.3 minutes instead of 2.5 minutes. 0.125 litres per minute comes from dividing the time by the volume instead of the volume by the time.)
- (a) It touches the x-axis once, only at x = 4. — (x − 4)² is a square, so it equals zero only when x − 4 = 0, that is at x = 4 — the curve just touches the x-axis there rather than crossing it, since a square cannot be negative on either side to cross through. Saying it crosses at x = 4 and x = −4 wrongly introduces a plus-or-minus, as if taking a square root of x, rather than recognising the bracket is already squared and only zero once. Saying it never touches the x-axis forgets that a squared term CAN equal zero, even though it can never be negative. Saying it crosses at x = 2 and x = −2 confuses (x − 4)² with the different expression x² − 4.
- (d) £3.60 per component — The gradient of a cost-against-components graph has units of pounds per component, since cost is measured in pounds and the horizontal axis counts components. So 3.60 means it costs an extra £3.60 to produce one more component at that point. Calling it '£3.60 total cost' confuses the gradient, a rate, with the y-value on the graph, which is the total cost itself. Giving it as 3.60 components per pound swaps which axis is on top, giving the units of the reciprocal gradient, not the gradient itself. Turning 3.60 into a percentage invents a unit that has no basis in the graph's axes — a gradient here is a number of pounds, not a percentage. Always build the gradient's units from the two axes' own units, in the order y-axis over x-axis.
- (c) 100.5 m — The curved part of a semicircular bend is half of a full circle's circumference. The full circumference would be 2 × 3.14 × 32 = 200.96 m, and half of that is 200.96 ÷ 2 = 100.48 m, which rounds to 100.5 m. Choosing 201.0 m uses the FULL circumference, forgetting to halve it for a semicircle. Choosing 50.2 m halves the radius as well as taking a semicircle, using 3.14 × 16 = 50.24 m instead of the correct radius of 32 m. Choosing 64.0 m uses the diameter, 2 × 32 = 64, as if it were the curved length, ignoring π and the semicircle shape entirely.
- (c) 11/20 — 'Red or white' combines two mutually exclusive events, so add their probabilities: 3/10 = 6/20 and 1/4 = 5/20, giving 6/20 + 5/20 = 11/20. Multiplying the two probabilities instead of adding them, 3/10 × 1/4, gives 3/40, which would be the probability of red and white together, not red or white — and a bead can't be both colours. Subtracting the sum from 1, 1 − 11/20 = 9/20, gives the probability of the bead being black instead of red or white. Converting 1/4 as 4/20 instead of 5/20 (dividing 20 by 4 but forgetting to scale the numerator) gives 6/20 + 4/20 = 1/2.
- (b) 5 ≤ m < 10 — Method: with 60 values the median is the 60 ÷ 2 = 30th value in order, so build a running total until it first reaches 30. Working: the running totals are 22 after the first class, 22 + 20 = 42 after the second, 51 after the third, 56 after the fourth and 60 after the fifth; the 30th parcel is past 22 but not past 42, so it lies in the second class. Answer: the median lies in the class 5 ≤ m < 10. The distractors: 0 ≤ m < 5 comes from giving the class with the greatest frequency, 22, which is the modal class and not the median class; 10 ≤ m < 20 comes from choosing the middle class in the list of five instead of counting to the middle value; 20 ≤ m < 30 comes from halving the range of the data, 50 ÷ 2 = 25, and giving the class that contains 25 kg rather than the class that contains the 30th parcel.
- (c) 7 + 4√3 — Expand the brackets fully: (2 + √3)² = 2² + 2 × 2 × √3 + (√3)² = 4 + 4√3 + 3. Adding the two whole-number terms, 4 + 3 = 7, gives 7 + 4√3. Using (a + b)² = a² + b² and skipping the middle cross term entirely gives just 4 + 3 = 7, with no surd term at all. Treating (√3)² as if it stayed √3 rather than becoming 3, then merging it with the existing surd term, gives 4 + 5√3. Squaring only the surd term correctly but carrying the whole-number term as 2 instead of squaring it to 4 gives 2 + 3 + 4√3 = 5 + 4√3.
- (a) 5x − 3 = 12 — 5x − 3 = 12 is an equation with exactly one solution: adding 3 and dividing by 5 gives x = 3, and no other value works. 5x − 3 = 5x − 3 is true for every value of x, since both sides are identical — it has infinitely many solutions, not one. 5x − 3 > 12 is an inequality: any value of x greater than 3 satisfies it, so it has a whole range of solutions, not a single one. 5x − 3 = 5x + 2 has no solution at all, since subtracting 5x from both sides leaves −3 = 2, which is never true. The equation with exactly one solution is 5x − 3 = 12.
- (d) 24 — Method: split 60 into 3 + 7 = 10 equal parts, find the value of one part, then use the difference in ratio parts. Working: 60 ÷ 10 = 6, so the numbers are 3 × 6 = 18 and 7 × 6 = 42, and their difference is 42 − 18 = 24. Answer: 24. 4 comes from finding the difference between the ratio numbers, 7 − 3, but forgetting to multiply by the value of one part. 60 comes from adding the two numbers back together instead of subtracting, which just repeats the given sum. 80 comes from dividing 60 by the first ratio number, 3, instead of by the total number of parts, 10, giving a part value of 20 and a difference of 7 × 20 − 3 × 20 = 80.
- (b) 115° — Angle APE and angle BPQ are vertically opposite, so angle BPQ = 65°, equal to angle APE. Angle BPQ and angle PQD are co-interior (allied) angles between the parallel lines, and co-interior angles always add up to 180°, so angle PQD = 180 − 65 = 115°. "65°" comes from treating co-interior angles as equal to each other, the way alternate angles are, instead of adding to 180°. "295°" comes from applying the angles-round-a-point fact (360° − 65°) directly to the original 65°, skipping the correct co-interior step. "25°" comes from misremembering co-interior angles as adding to 90° instead of 180° (90 − 65 = 25).
- (b) 8/35 — Method: the pupil picked is known to play at least one of the two sports, so first count how many pupils that is, then divide the number who play both by it. Working: 25 play football and 18 play tennis, but the 8 who play both have been counted in each figure, so the number who play at least one sport is 25 + 18 minus 8, which is 35. The pupils who play both give 8/35, which will not cancel. Answer: the probability is 8/35. The distractors: 2/15 is 8/60, dividing by the whole year group instead of by the 35 pupils who play at least one sport; 8/43 uses 25 + 18 as the denominator, forgetting that the 8 pupils who play both have been counted twice; 8/25 conditions on the footballers alone, answering the probability that a footballer also plays tennis rather than using every pupil who plays a sport.
- (a) 24 — Method: rearrange the relationship so that the lowest common multiple stands alone; it is the product of the two numbers divided by their highest common factor. Working: 48 = 2 × the lowest common multiple, so the lowest common multiple is 48 ÷ 2 = 24. Checking, 24 is in the 6 times table and in the 8 times table. Answer: 24. The distractors: 48 comes from giving the product of the two numbers and never dividing by the highest common factor; 96 comes from multiplying by the highest common factor instead of dividing by it; 12 comes from dividing by the highest common factor twice, once for each of the two numbers.
- (b) xₙ₊₁ = (xₙ² + 3) ÷ 5 — Starting from x² = 5x − 3, add 3 to both sides: x² + 3 = 5x. Divide both sides by 5: x = (x² + 3) ÷ 5. Writing this as an iteration gives xₙ₊₁ = (xₙ² + 3) ÷ 5. xₙ₊₁ = (xₙ² − 3) ÷ 5 comes from a sign error when moving the −3 across the equals sign — it should become +3, not stay as −3. xₙ₊₁ = 5(xₙ² + 3) comes from multiplying by 5 instead of dividing by 5 when isolating x. xₙ₊₁ = (xₙ + 3) ÷ 5 comes from dropping the index on x², using xₙ instead of xₙ².
- (a) 15 cm — Take the square root of each part of the area ratio to find the length ratio: the square root of 4 is 2 and the square root of 25 is 5, giving a length ratio of 2 : 5. Multiply the smaller flag's height by the scale factor 5 ÷ 2 = 2.5: 6 × 2.5 = 15, so the larger flag is 15 cm tall. Giving 37.5 cm uses the area ratio, 25 ÷ 4 = 6.25, directly as the scale factor without square-rooting it first (6 × 6.25 = 37.5). Giving 2.4 cm applies the length ratio the wrong way round, scaling the smaller flag down by 2 ÷ 5 instead of up by 5 ÷ 2 (6 × 0.4 = 2.4). Giving 27 cm adds the difference between the two area-ratio numbers, 25 − 4 = 21, onto the smaller height instead of using it as a scale factor (6 + 21 = 27).
- (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.
- (c) n = 4 — Testing n = 4: 4² = 16, and 16 + 4 + 1 = 21, and 21 = 3 × 7 is not a prime number, so this value disproves the claim. Testing n = 1: 1² = 1, and 1 + 1 + 1 = 3, which is prime, so it does not disprove the claim. Testing n = 2: 2² = 4, and 4 + 2 + 1 = 7, which is prime. Testing n = 3: 3² = 9, and 9 + 3 + 1 = 13, which is also prime — a counterexample has to give a result that isn't prime, and only n = 4 does that.
- (b) 15 — Find the multiplier connecting y to x: 10 ÷ 4 = 2.5. Then apply it to the new value of x: 2.5 × 6 = 15. Working out 10 + (6 − 4) = 12 adds the change in x straight onto y instead of scaling proportionally. Working out 10 × 6 = 60 multiplies the given y-value by the new x-value directly, without finding the multiplier first. Writing 10 keeps y the same as before, not realising it must change with x. When x = 6, y = 15.
- (d) 15 cm² — Method: identify the length that is common to both views, since it is the box's length; then read off the width from the plan view and the height from the front elevation, and multiply those two measurements to find the area of the side elevation. Working: both rectangles share a side of 8 cm, which is the box's length; the plan view's other side gives a width of 5 cm, and the front elevation's other side gives a height of 3 cm. The side elevation is bounded by the width and the height: 5 cm × 3 cm = 15 cm². Answer: 15 cm². The distractors: 24 cm² comes from giving the area of the front elevation shown (8 cm × 3 cm) instead of working out a new rectangle for the side elevation. 40 cm² comes from giving the area of the plan view shown (8 cm × 5 cm) instead of working out the side elevation. 120 cm² comes from multiplying all three measurements together (8 cm × 5 cm × 3 cm), finding the volume of the box instead of the area of one face.
- (c) 27 days — The campaign starts at d = 3 and ends at d = 30, so it runs for 30 − 3 = 27 days. Getting 33 days comes from adding the two values, 3 + 30 = 33, instead of subtracting them. Getting 30 days uses only the end day and ignores that the campaign did not start at day 0. Getting 24 days comes from subtracting the start day twice, 30 − 3 − 3 = 24, instead of once.
- (d) P lies outside the circle — The squared distance from the origin to P is 8² + 9² = 64 + 81 = 145, which is greater than 100, so P lies outside the circle. Choosing 'P lies inside the circle' comes from adding the coordinates directly (8 + 9 = 17) instead of squaring and adding them, and comparing 17 with 100. Choosing 'P lies on the circle' comes from squaring 9 incorrectly as 36 (as if it were 6²) instead of 81, giving 64 + 36 = 100, equal to the right-hand side. Choosing 'Cannot be determined' wrongly assumes the equation of the line itself is needed, when the radius and the distance from the origin to P are already enough.
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.