Sample paper · GCSE Higher · grades 4–9

GCSE Higher sample Paper 1 (non-calculator)

The real Paper 1 is 1.5 hour 30 minutes and 80 marks, non-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 no calculator-only items. AO1 / AO2 / AO3 at this tier: 40% / 30% / 30%.

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GCSE Higher sample Paper 1 (non-calculator)

Maths · GCSE Higher · Paper 1 · Non-calculator · 20 questions

MathsUKwww.geekhero.co.uk

1.A recipe uses 0.625 kg of flour. Write this mass as a fraction of a kilogram, in its simplest form.
(a)31/50
(b)25/4
(c)5/8
(d)8/5
2.A graph shows the total monthly cost of a mobile phone tariff against the number of minutes of calls. The line starts at £12 and stays level until 200 minutes, then rises by 5p for each further minute of calls. Use the graph to work out the total cost of a month in which 250 minutes of calls are made.
(a)£14.50
(b)£13.50
(c)£12.50
(d)£24.50
3.A necklace is made using gold beads and silver beads in the ratio 7:3. There are 40 more gold beads than silver beads. Work out the total number of beads in the necklace.
(a)100
(b)40
(c)70
(d)30
4.A shape is reflected in the line y = 1, and the image is then reflected in the line y = 4. Which single transformation is equivalent to this combination, for every point?
(a)Translation by the vector (0, 6)
(b)Reflection in the line y = 2.5
(c)Translation by the vector (0, −6)
(d)Translation by the vector (0, 3)
5.Of the 70 students on the Year 10 geography trip, 42 are girls and the rest are boys. 24 of the girls and 10 of the boys brought a packed lunch. A student who did not bring a packed lunch is chosen at random. Work out the probability that this student is a girl.
(a)9/35
(b)3/7
(c)1/2
(d)12/17
6.The masses, m kg, of 150 boxes are summarised by these cumulative frequencies: m < 5, 18 boxes; m < 10, 52 boxes; m < 20, 96 boxes; m < 35, 130 boxes; m < 60, 150 boxes. Work out the number of boxes with a mass in the class 10 ≤ m < 20.
(a)96
(b)54
(c)34
(d)44
7.Simplify √45.
(a)9√5
(b)6√5
(c)15
(d)3√5
8.The graph of y = f(x) has x-intercepts at x = −1 and x = 4. Which statement correctly describes the x-intercepts of y = f(2 − x)?
(a)Translate +2 in x only (no reflection); roots 1, 6
(b)Solved 2 − x = k as x = k − 2; roots −3, 2
(c)No overall change; roots −1, 4
(d)Reflect in the y-axis, +2 in x; roots 3, −2
9.Write 400 g : 1.5 kg as a ratio in its simplest form.
(a)4 : 15
(b)40 : 150
(c)4 : 1.5
(d)15 : 4
10.OABC is a parallelogram, with OA = a and OC = c. X is the midpoint of the diagonal AC. Express the vector OX in terms of a and c, and use it to show that X also lies on the diagonal OB.
(a)a + c
(b)(1/2)a + c
(c)(1/2)a − (1/2)c
(d)(1/2)a + (1/2)c
11.A bag contains 3 red counters and 5 blue counters. Three counters are taken out at random, one after another, without being replaced. Work out the probability that all three counters taken out are red.
(a)3/256
(b)9/112
(c)1/56
(d)27/512
12.A garden path is measured by Jon as 12 m, correct to the nearest metre, and by Mia as 12.6 m, correct to the nearest 0.1 m. Which statement about the two measurements is correct?
(a)They cannot both be describing the same path
(b)Jon's measurement must be wrong
(c)Mia's measurement must be wrong
(d)They must both be describing the same path
13.The first five terms of a quadratic sequence are 4, 7, 12, 19, 28. Work out an expression, in terms of n, for the nth term.
(a)3n + 1
(b)2n² + 3
(c)n²
(d)n² + 3
14.A lorry is carrying a load of mass m tonnes. Given that 1 tonne = 1000 kg, write down an expression for the mass of the load in kilograms.
(a)1000m
(b)m − 1000
(c)m/1000
(d)1000 + m
15.A market stall sells apples at £2.40 per kilogram. Which of these expressions gives the cost, in pounds, of buying m kilograms of apples?
(a)2.40m
(b)m/2.40
(c)m − 2.40
(d)2.40 + m
16.A delivery van's distance-time graph shows the following: it travels 9 km in the first 15 minutes at a steady speed, stops for 45 minutes to unload, then travels a further 21 km in the next 30 minutes, also at a steady speed. Work out the average speed, in km/h, for the whole journey.
(a)36 km/h
(b)39 km/h
(c)40 km/h
(d)20 km/h
17.Two quantities y and z are each in direct proportion to x, and are given by y = 3x and z = 7x. Work out the difference between the value of y and the value of z when x = 4.
y = 3x
(a)40
(b)16
(c)28
(d)4
18.A roof truss is shaped like an isosceles triangle resting on a horizontal ceiling joist. The truss's sloping edges are equal in length, and the angle at its apex is 40°. The ceiling joist continues in a straight line beyond the foot of the truss. Work out the angle between the truss's sloping edge and the extended joist, on the outside of the truss.
(a)110°
(b)70°
(c)140°
(d)20°
19.Subtract, giving your answer as a single fraction in its simplest form: 4/(x − 1) − 2/(x + 3)
(a)(2x + 11)/((x − 1)(x + 3))
(b)(2x + 14)/((x − 1)(x + 3))
(c)2/((x − 1)(x + 3))
(d)(2x − 10)/((x − 1)(x + 3))
20.Solve 6x − 5 = 3x + 13.
(a)18
(b)2
(c)2.67
(d)6

Answer key

(c) 5/8 — Method: write the decimal over 1000 using its three decimal places, then simplify. Working: 0.625 = 625/1000 = 5/8 (dividing both numerator and denominator by 125). Answer: 5/8. 25/4 comes from writing the decimal over 100 instead of 1000, as if there were only two decimal places. 31/50 comes from rounding 0.625 to 0.62 before converting. 8/5 comes from simplifying correctly to 5/8 and then writing the fraction upside down.
(a) £14.50 — Method: the first 200 minutes are already covered by the £12, so only 250 − 200 = 50 minutes are charged extra. Extra charge = 50 × £0.05 = £2.50. Total cost = £12 + £2.50 = £14.50. Distractor origins: £24.50 charges 5p for all 250 minutes instead of only the 50 minutes over 200 (250 × £0.05 = £12.50, plus £12 = £24.50); £12.50 makes that same slip of charging all 250 minutes but then forgets to add the £12 monthly fee; £13.50 works out 250 − 200 wrongly as 30 extra minutes instead of 50 (30 × £0.05 = £1.50, plus £12 = £13.50).
(a) 100 — The difference between the parts of the ratio is 7 − 3 = 4 parts, and this is worth 40 beads. Divide to find one part: 40 ÷ 4 = 10. The total number of parts is 7 + 3 = 10, so the total number of beads is 10 × 10 = 100. (40 is just the given difference between gold and silver, not the total. 70 is the number of gold beads only, using 7 parts. 30 is the number of silver beads only, using 3 parts.)
(a) Translation by the vector (0, 6) — Method: reflecting twice in two parallel horizontal lines is always equivalent to a single translation, at right angles to the lines, of twice the distance between them. Working: the two lines are 4 − 1 = 3 units apart, so the translation is 2 × 3 = 6 units in the positive y-direction. Answer: translation by the vector (0, 6). Using just the gap itself, without doubling it, gives (0, 3); translating in the negative y-direction, from the second line back towards the first, gives (0, −6); and describing the combination as a single reflection in the line halfway between them, y = 2.5, confuses this combination with the effect of a single reflection — two reflections in parallel lines are always equivalent to a translation, never to another reflection. Always double the gap between the lines, and translate in the direction from the first line towards the second.
(c) 1/2 — Method: restrict to the students who did NOT bring a packed lunch, then find what fraction of that group are girls. Girls without lunch = 42 − 24 = 18. Boys = 70 − 42 = 28, so boys without lunch = 28 − 10 = 18. Total without lunch = 18 + 18 = 36. Working: P(girl | no lunch) = 18 ÷ 36 = 1/2. Answer: 1/2. Watch out: dividing 18 by 42 (the total number of girls) instead of by 36 finds P(no lunch | girl), the reverse conditional. Dividing by 70 (the whole trip) ignores that you already know the student did not bring a lunch. And using the 'brought a lunch' numbers (24 out of 34) answers the question for the wrong group entirely — you were asked about the students who did NOT bring one.
(d) 44 — Method: a cumulative frequency counts everything below a value, so the frequency of a class is the running total at the top of the class minus the running total at the bottom of it. Working: the running total below 20 kg is 96 and the running total below 10 kg is 52, so the number of boxes in the class 10 ≤ m < 20 is 96 − 52 = 44. Answer: 44 boxes. The distractors: 96 comes from quoting the running total at 20 kg itself, which counts every box below 20 kg rather than only those in this class; 34 comes from subtracting the wrong pair, 52 − 18, which gives the class 5 ≤ m < 10 instead; 54 comes from subtracting from the grand total, 150 − 96, which gives the boxes of 20 kg or more.
(d) 3√5 — Split 45 into a perfect square times a factor: 45 = 9 × 5. Take the square root of each part separately: √45 = √9 × √5 = 3√5, since √9 = 3. Writing the perfect-square factor itself (9) as the coefficient instead of its root would give 9√5 — that trap comes from forgetting the last step, rooting 9. Multiplying 3 and 5 together instead of keeping them as coefficient and radicand gives 15, which throws away the surd entirely. Doubling the correct coefficient by mistake gives 6√5.
(d) Reflect in the y-axis, +2 in x; roots 3, −2 — f(2 − x) is zero exactly when 2 − x equals one of f's roots: 2 − x = −1 or 2 − x = 4. Solving each correctly (x = 2 − (−1) = 3, and x = 2 − 4 = −2) gives the new roots x = 3 and x = −2. This is the same as reflecting y = f(x) in the y-axis to get f(−x), then translating 2 units in the positive x-direction to get f(−(x − 2)) = f(2 − x). Solving 2 − x = k as x = k − 2 instead of x = 2 − k is a sign slip when rearranging, and gives x = −3 and x = 2. Translating +2 in x with no reflection at all uses f(x − 2), whose roots are the original roots plus 2: x = 1 and x = 6 — this misses the reflection completely. Assuming 'no overall change' wrongly treats a reflection-and-translation pair as always cancelling out, when here the roots genuinely move, from x = −1 and x = 4 to x = 3 and x = −2.
(a) 4 : 15 — Convert to the same unit first: 1.5 kg = 1500 g, since 1 kg = 1000 g. This gives the ratio 400 : 1500. Divide both parts by their highest common factor, 100, to get 4 : 15. Giving 40 : 150 divides by 10 only, which is not the highest common factor, so it is not fully simplified. Giving 15 : 4 swaps the order. Giving 4 : 1.5 has not converted 1.5 kg into grams, so the two parts are not measured in the same unit.
(d) (1/2)a + (1/2)c — Method: X is the midpoint of AC, so OX = OA + (1/2)AC, with AC = c − a. Working: OX = a + 1/2(c − a) = a − (1/2)a + (1/2)c = (1/2)a + (1/2)c. Answer: OX = (1/2)a + (1/2)c. Since OB = a + c, this is exactly half of OB, so OX = (1/2)OB, meaning X lies on OB at its midpoint too — the two diagonals bisect each other. Forgetting to halve AC at all gives a + c, which is OB itself, not its midpoint; halving only the c-term gives (1/2)a + c; and a sign error on the c-term gives (1/2)a − (1/2)c. Halve the whole of AC, both terms together, and add it to OA rather than to a alone.
(c) 1/56 — Method: for draws without replacement, multiply a chain of three fractions where both the numerator (reds remaining) and the denominator (counters remaining) fall by one after each draw. Working: P(all three red) = 3/8 × 2/7 × 1/6 = 6/336 = 1/56. Answer: 1/56. Watch out: using 3/8 for all three draws (27/512) treats the counters as if they were replaced each time. Reducing only the numerator each draw (3/8 × 2/8 × 1/8) forgets that the total number of counters left in the bag also falls. And reducing only the denominator while keeping the numerator at 3 each time (3/8 × 3/7 × 3/6) forgets that a red counter has actually left the bag.
(a) They cannot both be describing the same path — Jon's measurement means the true length, l, satisfies 11.5 m ≤ l < 12.5 m. Mia's measurement means the true length satisfies 12.55 m ≤ l < 12.65 m. These two ranges do not overlap, so the two measurements cannot both be describing the same path. 'They must both be describing the same path' ignores that the two ranges do not overlap at all. 'Jon's measurement must be wrong' wrongly assumes Jon is the one at fault, when the mismatch does not show which measurement, if either, is wrong. 'Mia's measurement must be wrong' makes the same unjustified assumption in the other direction.
(d) n² + 3 — First differences: 3, 5, 7, 9. Second differences: 2, 2, 2, so the sequence is quadratic and the coefficient of n² is half the second difference: a = 2 ÷ 2 = 1. Subtracting n² (1, 4, 9, 16, 25) from the terms (4, 7, 12, 19, 28) leaves 3, 3, 3, 3, 3, a constant, so the nth term is n² + 3. Using the second difference itself as a, without halving it, gives 2n² + 3. Finding a = 1 correctly but then dropping the constant remainder gives n². Treating the first first difference (3) as a common difference and building a linear formula a + (n − 1)d = 4 + 3(n − 1) gives 3n + 1, which fits only the first term.
(a) 1000m — Method: kilograms are a smaller unit than tonnes, so change tonnes into kilograms by multiplying by 1000. Working: m tonnes = m × 1000 kg = 1000m kg. So the expression is 1000m. Distractor m/1000 comes from dividing by 1000 instead of multiplying, which would make the number of kilograms smaller than the number of tonnes, the wrong way round. Distractor 1000 + m comes from adding the conversion factor instead of multiplying by it. Distractor m − 1000 comes from subtracting the conversion factor instead of multiplying by it.
(a) 2.40m — Method: a cost found from a rate is the rate multiplied by the amount bought. Working: the rate is £2.40 per kilogram and the amount is m kilograms, so the cost is 2.40 × m. Answer: 2.40m. A candidate who divides the amount by the rate instead of multiplying writes m/2.40. A candidate who adds the rate to the amount instead of multiplying writes 2.40 + m. A candidate who subtracts the rate from the amount instead of multiplying writes m − 2.40.
(d) 20 km/h — Method: total distance = 9 + 21 = 30 km. Total time = 15 + 45 + 30 = 90 minutes = 1.5 hours. Average speed = total distance ÷ total time = 30 ÷ 1.5 = 20 km/h. Distractor origins: 40 km/h ignores the 45-minute stop, dividing the total distance by the travel time only (30 km ÷ 45 minutes = 40 km/h); 39 km/h averages the two section speeds, 36 km/h and 42 km/h, instead of using total distance over total time; 36 km/h uses only the first section's speed (9 km in 15 minutes), ignoring the second section and the stop.
(b) 16 — Method: substitute x into each equation separately, then subtract the smaller value from the larger. Working: y = 3 × 4 = 12 and z = 7 × 4 = 28, so the difference is 28 − 12 = 16. Answer: 16. The distractors: 4 comes from subtracting the constants, 7 − 3, which is the difference between the two gradients rather than the difference between the values at x = 4; 40 comes from adding the two values, 12 + 28, instead of subtracting them; 28 is the value of z on its own, given instead of being compared with the value of y.
(a) 110° — The truss is isosceles, so its two base angles are equal; since the three angles of the triangle add up to 180° and the apex is 40°, each base angle is (180 − 40) ÷ 2 = 70°. The angle between the sloping edge and the joist, extended beyond the base of the truss, sits on a straight line with that 70° base angle, and angles on a straight line add up to 180°, so the required angle is 180 − 70 = 110°. "70°" comes from stopping after finding the base angle and giving it directly, without also using the straight-line fact to find the angle on the OUTSIDE of the truss. "140°" comes from doubling the base angle instead of using the straight-line fact. "20°" comes from taking half of the apex angle (40° ÷ 2), mistaking it for the required angle instead of properly using the triangle's angle sum.
(b) (2x + 14)/((x − 1)(x + 3)) — To subtract these fractions, write them over the common denominator (x − 1)(x + 3): the numerator becomes 4(x + 3) − 2(x − 1). Expanding gives 4x + 12 − 2x + 2, which simplifies to 2x + 14, so the answer is (2x + 14)/((x − 1)(x + 3)). Writing (2x + 11)/((x − 1)(x + 3)) comes from not distributing the minus sign fully across the second bracket, treating −2(x − 1) as −2x − 1 instead of −2x + 2. Writing 2/((x − 1)(x + 3)) comes from subtracting the two original numerators directly, 4 − 2 = 2, the same mistake as subtracting fractions without a common denominator. Writing (2x − 10)/((x − 1)(x + 3)) comes from swapping which numerator multiplies which bracket, forming 4(x − 1) − 2(x + 3) instead of 4(x + 3) − 2(x − 1).
(d) 6 — Method: collect the x-terms on one side and the constants on the other, then divide by the remaining coefficient of x. Working: 6x − 3x = 13 + 5, so 3x = 18, x = 18 ÷ 3 = 6. Answer: x = 6. 2.67 comes from a sign error when moving the 5, subtracting instead of adding: 3x = 13 − 5 = 8, x = 8 ÷ 3 ≈ 2.67. 2 comes from a sign error when moving the x-term, adding instead of subtracting: 9x = 18, x = 2. 18 comes from correctly finding 3x = 18 but forgetting to divide by 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.