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.£60 is shared between Amir, Bo and Chen in the ratio 1:2:3. Work out the fraction of the £60 that Amir and Bo receive together.
- 2.A taxi company charges a £2.50 booking fee plus £1.80 per mile. A journey costs £15.10 in total. Work out the number of miles travelled.
- 3.A map has a scale of 1 : 25 000. A footpath measures 6 cm on the map. Work out the real length of the footpath, in kilometres.
- 4.A circular tabletop has a radius of 15 cm. Work out the area of the tabletop. Use π = 3.14. Give your answer to 1 decimal place.
- 5.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.
- 6.The mean of three numbers is 50. A fourth number, 100, is added to the set. Work out the mean of the four numbers.
- 7.A semicircle has a diameter of 8 cm. Work out the exact area of the semicircle, in terms of π.
- 8.The perimeter of a rectangle is given by the formula P = 2(l + w), where l is the length and w is the width. A rectangular garden has a perimeter of 46 m and a length of 14 m. Rearrange the formula to make w the subject, then work out the width of the garden.
- 9.A scale drawing of a car park has a scale of 1 : 300. On the drawing, the car park has an area of 15 cm². Work out the real area of the car park, in square metres.
- 10.From an external point P, two tangents PA and PB touch a circle with centre O at A and B. Angle AOB = 112°. Work out the size of angle APB.
- 11.For two events A and B, P(A) = 0.6 and P(A and B) = 0.15. Work out P(B | A).
- 12.A plane flies 2340 km at an average speed of 780 km/h. It departs at 08:20. Work out the arrival time, using the 24-hour clock.
- 13.Work out the value of 3a − 2b when a = 5 and b = 4
- 14.The volume of water in a paddling pool, in litres, is plotted against the time since the tap was turned on, in minutes. What are the units of the gradient of a tangent to this graph?
- 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.The expression 6n + 15 is written as 3(2n + 5). Which statement is correct?
- 17.Amelia uses 2 kg of flour to bake 5 cakes. Using the same recipe, work out how many cakes she can bake with 6 kg of flour.
- 18.A construction for the perpendicular bisector of AB begins by opening a pair of compasses to more than half the length of AB, then drawing an arc centred at A. What is the next step?
- 19.A circle has centre (0, 0) and equation x² + y² = 8. Work out the radius of the circle, giving your answer as a surd in its simplest form.
- 20.The table shows some values of f(x): when x = 0, f(x) = 5; when x = 1, f(x) = 8; when x = 2, f(x) = 4; when x = 3, f(x) = 1. Work out the value of f(x − 1) when x = 2.
Answer key
- (a) 1/2 — The ratio 1:2:3 has 1 + 2 + 3 = 6 parts in total. Amir and Bo together receive 1 + 2 = 3 of those parts, so together they receive 3/6 of the £60, which simplifies to 1/2. Using only Amir's single part, 1/6, ignores Bo's share entirely. Adding Bo's and Chen's parts instead of Amir's and Bo's, 2 + 3 = 5, gives 5/6. Comparing Amir and Bo's 3 parts to Chen's 3 parts, rather than to the total of 6 parts, gives 3/3 = 1.
- (d) 7 — Method: set up the equation 2.50 + 1.80m = 15.10, then subtract the booking fee and divide by the cost per mile. Working: 1.80m = 15.10 − 2.50 = 12.60; m = 12.60 ÷ 1.80 = 7. Answer: 7 miles. 8.39 comes from dividing the whole £15.10 by £1.80 without first subtracting the booking fee: 15.10 ÷ 1.80 ≈ 8.39. 5.32 comes from swapping the two amounts round, subtracting £1.80 and dividing by £2.50: (15.10 − 1.80) ÷ 2.50 ≈ 5.32. 9.78 comes from adding the booking fee instead of subtracting it: (15.10 + 2.50) ÷ 1.80 ≈ 9.78.
- (a) 1.5 km — Multiply the map length by the scale: 6 × 25 000 = 150 000 cm. Convert to kilometres: 150 000 cm = 1.5 km. Dividing by only 1000 instead of the full conversion when changing units gives 150 km, a hundred times too large. Misreading the scale as 1 : 2500 instead of 1 : 25 000 gives 6 × 2500 = 15 000 cm = 0.15 km, a hundred times too small. Leaving the answer as 150 000 without converting units at all, and calling it 150 000 km, mistakes centimetres for kilometres completely.
- (d) 706.5 cm² — Area of a circle = πr². With r = 15 cm and π = 3.14, area = 3.14 × 15² = 3.14 × 225 = 706.5 cm². A student who uses the circumference formula 2πr instead of the area formula gets 2 × 3.14 × 15 = 94.2 cm². A student who uses πr instead, forgetting to double, gets 3.14 × 15 = 47.1 cm². A student who squares the diameter (30 cm) instead of the radius gets 3.14 × 900 = 2826.0 cm².
- (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.
- (a) 62.5 — Method: a mean cannot be averaged with a new value — rebuild the total, add the new value to it, then divide by the new count. Working: three numbers with a mean of 50 have a total of 50 × 3 = 150; adding 100 makes the total 150 + 100 = 250; there are now 4 numbers, so the new mean is 250 ÷ 4 = 62.5. Answer: 62.5. The distractors: 75 comes from averaging the old mean with the new value, (50 + 100) ÷ 2, which ignores that three numbers pull against one; 50 comes from assuming an extra value leaves the mean unchanged; 37.5 comes from dividing the old total of 150 by the new count of 4, adding the new value to the count but not to the total.
- (b) 8π cm² — A diameter of 8 cm gives a radius of 4 cm. The area of a full circle would be π × r² = π × 4² = 16π cm², and a semicircle is exactly half of this, giving 16π ÷ 2 = 8π cm². Forgetting to halve the area for the semicircle gives 16π cm², the area of the whole circle. Halving the diameter twice, using a radius of 2 instead of 4, gives π × 2² = 4π cm². Using the diameter itself as the radius, so π × 8² = 64π, and then halving that for the semicircle gives 32π cm².
- (a) 9 m — Undo the multiplication by the bracket first: dividing both sides by 2 gives P/2 = l + w. Subtracting the length from both sides gives w = P/2 − l. Substituting the measurements, 46 ÷ 2 = 23, and 23 − 14 = 9, so the width is 9 m. Taking the length off before halving gives (46 − 14) ÷ 2 = 16, which halves the length as well; expanding to P = 2l + 2w and then forgetting to divide by 2 gives 46 − 28 = 18; subtracting the length in the wrong direction gives 23 + 14 = 37.
- (d) 135 m² — Method: for area, the scale factor must be squared. Working: area scale = 300² = 90 000. 15 × 90 000 = 1,350,000 cm². Convert to m² by dividing by 10 000: 1,350,000 ÷ 10 000 = 135 m². Wrong options: 0.45 m² comes from using the linear scale factor (×300) instead of squaring it; 1,350,000 m² comes from forgetting to convert the answer from cm² to m²; 13,500 m² comes from dividing by 100 instead of 10 000 when converting units.
- (a) 68° — Method: a tangent meets a radius at 90°, so quadrilateral OAPB has two right angles at A and B; its four angles sum to 360°, so angle APB = 360° − 90° − 90° − angle AOB. Working: angle APB = 360° − 90° − 90° − 112° = 68°. Halving angle AOB, confusing this with the angle-at-centre theorem for a chord, gives 56°; assuming the angle between two tangents is always 90° (true only for the angles at A and B, not at P) gives 90°; and assuming angle APB simply equals angle AOB by symmetry gives 112°. The two right angles at A and B, from the tangent-radius property, are what link angle AOB and angle APB.
- (c) 0.25 — Method: P(B | A) = P(A and B) ÷ P(A). Working: P(B | A) = 0.15 ÷ 0.6 = 0.25. Answer: 0.25. Watch out: multiplying 0.6 by 0.15 instead of dividing gives 0.09, and subtracting 0.15 from 0.6 gives 0.45 — neither uses the conditional probability formula. Leaving the answer as 0.15 mistakes the probability of A and B happening together for the probability of B once you already know A has happened — those are different quantities.
- (a) 11:20 — Method: find the flight time using time = distance ÷ speed, then add this to the departure time. Working: 2340 ÷ 780 = 3 hours; 08:20 + 3 hours = 11:20. Answer: 11:20. 08:40 comes from dividing speed by distance instead of distance by speed, giving a flight time of 1/3 hour (20 minutes) rather than 3 hours. 11:00 comes from adding the 3-hour flight time to the hour of the departure time only, 8 + 3 = 11, and losing the 20 minutes. 03:00 comes from finding the flight time correctly but giving it as a clock time on its own, forgetting to add it to the departure time.
- (d) 7 — Method: substitute both values, work out the two multiplications first, and only then subtract. Working: 3a = 3 × 5 = 15 and 2b = 2 × 4 = 8, so the expression becomes 15 − 8 = 7. Answer: 7. The distractors: 23 comes from adding the two products instead of subtracting, giving 15 + 8; −7 comes from subtracting the wrong way round and working out 8 − 15; 52 comes from working from left to right instead of following the order of operations, giving 3 × 5 = 15, then 15 − 2 = 13, then 13 × 4.
- (b) litres per minute — The gradient of a tangent is the change in the quantity on the vertical axis divided by the change in the quantity on the horizontal axis, so its units come from both axes: litres on the vertical axis and minutes on the horizontal axis give litres per minute. Giving the units as minutes for each litre inverts the fraction, giving the units of the RECIPROCAL of the gradient, not the gradient itself. Writing just litres uses only the vertical axis's units and ignores that a gradient is a rate, not an amount. Writing just minutes uses only the horizontal axis's units. A gradient always combines both axes' units as one divided by the other.
- (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) 3 and 2n + 5 are both factors of 6n + 15 — Factors are the parts multiplied together to make an expression; terms are the parts added together. In 3(2n + 5) the 3 and the bracket 2n + 5 are multiplied, so both of them are factors of 6n + 15. The parts added together are 6n and 15, and those are its two terms, not its factors. A factor need not be a number: a bracket containing letters is a factor in exactly the same way.
- (c) 15 — Method: the number of cakes is in direct proportion to the mass of flour, so find the multiplier between the two masses and apply it to the number of cakes. Working: 6 ÷ 2 = 3, so there is three times as much flour, and 5 × 3 = 15. Answer: 15. The distractors: 10 comes from multiplying the 5 cakes by 2, the mass in the recipe, instead of by the multiplier 3; 20 comes from multiplying by the difference 6 − 2 = 4, treating a proportion problem as a difference problem; 12 comes from rounding 5 ÷ 2 down to 2 cakes per kilogram and working out 6 × 2.
- (b) an arc of the same radius, centred at B — The perpendicular bisector construction needs two arcs of equal radius, one centred at each end of the segment — after the arc from A, the next step is an arc of exactly the same radius from B, so the two arcs cross at two points; the line through those two crossing points is the perpendicular bisector. "an arc of the same radius, centred at the midpoint of AB" is not possible yet, since the midpoint is only found once both arcs are drawn — it is the RESULT of the construction, not a step in it. "a smaller arc, centred at A again" gives two different-sized arcs from the same point, which never cross to give the bisector. "a straight line joining the ends of the first arc" only connects points on one arc, and locates nothing.
- (d) 2√2 — The radius satisfies r² = 8, so r = √8 = √(4 × 2) = √4 × √2 = 2√2. Choosing 8 forgets to take the square root of r² at all. Choosing 4 comes from halving 8 instead of finding its square root. Choosing √2 splits off the factor of 4 correctly but forgets to multiply the 2 back in front of the root.
- (a) 8 — f(x − 1) means every input to f is reduced by 1 before it is looked up. At x = 2, the input to f becomes 2 − 1 = 1, so f(x − 1) at x = 2 is f(1) = 8. Reading f(2) = 4 directly from the table, without applying the shift, gives 4. Shifting in the wrong direction, using x + 1 = 3 instead of x − 1 = 1, gives f(3) = 1. Confusing f(x − 1) with f(x) − 1 — taking f(2) = 4 and then subtracting 1 — gives 3.
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.