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%.
Non-calculatorGCSE Higher
Answer key: GCSE Higher sample Paper 1 (non-calculator)
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- (a) 5 — Method: the number of glasses is the amount in the jug divided by the amount one glass holds. Write the mixed number as an improper fraction, then divide by multiplying by the reciprocal. Working: 3 1/3 = (3 × 3 + 1)/3 = 10/3, and 10/3 ÷ 2/3 = 10/3 × 3/2 = 30/6 = 5. Answer: 5. The distractors: 2 comes from writing 3 1/3 as 4/3, adding the whole number to the numerator instead of multiplying it by the denominator first, and then dividing 4/3 by 2/3; 20/9 comes from multiplying by 2/3 instead of dividing by it; 5/3 comes from dividing by 2 rather than by 2/3, as though each glass held 2 litres.
- (a) y = −2x − 1 — Method: the gradient is the change in y divided by the change in x with both differences taken in the same order, and the constant then comes from substituting either point into y = mx + c. Working: m = (−9 − 3) ÷ (4 − (−2)) = (−12) ÷ 6 = −2, so the line is y = −2x + c; substituting (−2, 3) gives 3 = −2 × (−2) + c = 4 + c, so c = 3 − 4 = −1. Answer: y = −2x − 1. The distractors: y = −2x + 1 comes from rearranging 3 = 4 + c the wrong way round and taking the constant as 4 − 3; y = 2x + 7 comes from losing the minus sign when −12 is divided by 6 and then substituting correctly, 3 = 2 × (−2) + c; y = −(1/2)x + 2 comes from writing the gradient upside down as the change in x over the change in y, 6 ÷ (−12).
- (c) 150% — Percentage = (180,000 ÷ 120,000) × 100 = 150%.
- (c) 155° — Turning clockwise adds to the bearing. Starting on a bearing of 065° and turning clockwise through 90° gives 065° + 90° = 155°. A candidate who instead subtracts, working out 90° − 65° = 25°, has performed the wrong operation, giving 025°. A candidate who turns anticlockwise instead of clockwise works out 065° − 90°, which gives a negative number, and adding 360° to fix this gives 335° — the bearing for turning the other way. A candidate who thinks turning does not change the bearing at all keeps the answer as 065°. The new bearing, turning clockwise, is 155°.
- (c) 100 — There are 3 even numbers on a fair dice (2, 4 and 6), so the probability of landing on an even number is 3/6 = 1/2, and 300 × 1/2 = 150. The probability of landing on a six is 1/6, so 300 × 1/6 = 50. The dice is expected to land on an even number 150 − 50 = 100 more times than on a six. Writing 50 is wrong because that is just the expected number of sixes on its own, without comparing it to the expected number of evens. Writing 150 is wrong because that is just the expected number of evens on its own, without subtracting the sixes. Writing 200 is wrong because it adds the two expected frequencies together (150 + 50 = 200) instead of finding the difference between them. The dice is expected to land on an even number 100 more times than on a six.
- (c) 15 — Method: the interquartile range is the upper quartile take away the lower quartile, IQR = Q3 − Q1. Working: n + 1 = 15 + 1 = 16, so the lower quartile sits at position 16 ÷ 4 = 4, the 4th value in the list, which is 12; the upper quartile sits at position 3 × 4 = 12, the 12th value, which is 27. So 27 − 12 = 15. Answer: the interquartile range is 15 minutes. Watch which values you use and which way round: taking the smallest time away from the largest, 34 − 5 = 29, finds the range, which uses every value between the extremes rather than just the middle half; taking the median away from the upper quartile instead of the lower quartile, 27 − 19 = 8, swaps the median in for the lower quartile; and reversing the subtraction, 12 − 27 = −15, finds the right two values but in the wrong order — an interquartile range is never negative.
- (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) 6th term — The terms are aₙ = 3 × (√3)ⁿ⁻¹: a₁ = 3, a₂ = 3√3 ≈ 5.196, a₃ = 9, a₄ = 9√3 ≈ 15.588, a₅ = 27, a₆ = 27√3 ≈ 46.765. Since a₅ = 27 is still below 30 and a₆ ≈ 46.8 is above 30, the 6th term is the first to exceed 30. 4th term comes from treating the common ratio as 3 instead of √3: 3, 9, 27, 81, … — under that wrong ratio, 81, the 4th term, is the first over 30. 5th term comes from misjudging a₅ = 27 as already greater than 30. 7th term comes from tracking only the whole-number terms (3, 9, 27, 81, …) and reporting the ORIGINAL position of 81 in the full sequence — 81 is indeed a₇, but a₆ ≈ 46.8 already exceeds 30 first.
- (c) Yes, correct, because 24 ÷ 8 × 3 = 9. — Method: divide the count you know by its ratio part to find the value of one part, then multiply by the part you want. Working: one part = 24 ÷ 8 = 3 counters, so green = 3 × 3 = 9 counters, and the student is correct. Dividing 24 by 3 instead of 8 gives 24 ÷ 3 × 8 = 64, the ratio parts the wrong way round. Stopping after 24 ÷ 8 = 3 gives 3, forgetting to multiply by the green ratio part. Adding 24 + 3 = 27 confuses adding a ratio part with scaling by it.
- (d) (−14, 5) — To undo a composition, reverse the order and invert each transformation: undo the rotation first, then undo the translation. The inverse of 'rotate 90° clockwise about the origin' is 'rotate 90° anticlockwise about the origin', which maps (x, y) to (−y, x); applied to (3, 9) this gives (−9, 3). Then undo the translation by subtracting the vector (5, −2), i.e. adding (−5, 2): (−9 − 5, 3 + 2) = (−14, 5). Undoing the two inverse steps in the same order as the original composition, rather than reversing it, gives (−11, −2). Rotating 90° clockwise again instead of inverting the rotation's direction gives (4, −1). Adding the translation vector again instead of subtracting it, after correctly inverting the rotation, gives (−4, 1).
- (d) 3/36 — Method: list the results as ordered pairs, decide which totals satisfy the condition, count the pairs that give those totals and divide by the number of pairs there are. Working: there are 6 × 6 = 36 equally likely ordered pairs. Greater than 10 means a total of 11 or a total of 12. A total of 11 comes from (5, 6) and (6, 5); a total of 12 comes from (6, 6) alone, because both dice must show a 6. That is 2 + 1 = 3 pairs out of the 36. Answer: the probability is 3/36. The distractors: 2/36 comes from counting the two ways of making 11 and forgetting that 12 is greater than 10 as well; 4/36 comes from writing (6, 6) down twice, applying the rule that a pair can be turned round to a double that can only happen one way; 33/36 comes from reading the condition the wrong way round and giving the probability that the total is 10 or less.
- (d) 11/12 — Convert both mixed numbers to improper fractions with a common denominator. 2 3/4 = 11/4, which is 33/12, and 1 5/6 = 11/6, which is 22/12. Subtracting, 33/12 − 22/12 gives 11/12, already in its simplest form. Forgetting to borrow, and instead subtracting the fraction parts the other way round to avoid a negative, 10/12 − 9/12 gives 1/12; adding that to the whole-number difference of 1 gives 13/12. Subtracting only the fraction parts, 9/12 − 10/12, and reporting just the size of that difference gives 1/12, which ignores the whole numbers altogether. Adding the two improper fractions instead of subtracting them, 33/12 + 22/12, gives 55/12. So 2 3/4 − 1 5/6 = 11/12.
- (a) x = 4 ± √13 — Method: halve the coefficient of x to form the bracket, subtract the square of that number to keep the expression equal to the original, then rearrange and take the square root of both sides. Working: half of −8 is −4, so x² − 8x + 3 = (x − 4)² − 16 + 3 = (x − 4)² − 13; setting this equal to zero gives (x − 4)² = 13, so x − 4 = ±√13 and x = 4 ± √13. Answer: x = 4 ± √13. The distractors: x = 8 ± √13 comes from putting the whole coefficient 8 inside the bracket instead of half of it; x = 4 ± √19 comes from adding 16 and 3 rather than subtracting 3 from 16; x = −4 ± √13 comes from writing the bracket as (x + 4)², which reverses the sign of the number that comes out of it.
- (c) 4/5 — Find 20% of £45: 10% is £4.50, so 20% is £9. The sale price is £45 − £9 = £36. Form the fraction 36/45; both numbers share a factor of 9, so 36 ÷ 9 = 4 and 45 ÷ 9 = 5, giving 4/5. 1/5 comes from writing the discount itself as a fraction of the normal price (9/45), instead of the sale price. 6/5 comes from adding the 20% instead of subtracting it, giving a sale price of £54, then 54/45 = 6/5. 5/9 comes from treating 'reduced by 20%' as 'reduced by £20', giving a sale price of £25, then 25/45 = 5/9.
- (d) (−2, 6) — Method: add the top numbers of both vectors to the starting x-coordinate, and the bottom numbers of both vectors to the starting y-coordinate. Working: x-coordinate 2 + 3 + (−7) = −2; y-coordinate −1 + 5 + 2 = 6. Answer: (−2, 6). A candidate who only applies vector u and forgets v gets (5, 4). A candidate who only applies vector v and forgets u gets (−5, 1). A candidate who works out the combined vector u + v but forgets to add it to the starting point gets (−4, 7).
- (b) (0, −2) — Method: a translation acts on the two coordinates separately: moving right or left changes the x-coordinate only, moving up or down changes the y-coordinate only, and right and up add while left and down subtract. Working: the point starts at (−4, 7); moving 4 units to the right gives an x-coordinate of −4 + 4 = 0; moving 9 units down gives a y-coordinate of 7 − 9 = −2. Answer: (0, −2). The distractors: (5, 3) comes from pairing each number with the wrong coordinate, adding 9 to −4 and taking 4 from 7; (0, 16) comes from treating 'down' as an addition, giving 7 + 9 = 16 for the second coordinate; (−8, −2) comes from treating 'to the right' as a subtraction, giving −4 − 4 = −8 for the first coordinate.
- (d) 7/5 — The number of unbroken crayons is 60 − 25 = 35. The comparison is with the broken crayons, so the broken crayons are the denominator: 35/25. Both parts divide by 5: 35 ÷ 5 = 7 and 25 ÷ 5 = 5. The fraction is 7/5, which is greater than 1 because there are more unbroken crayons than broken ones.
- (b) Yes — OA = OC (radii), so the base angles are equal. — Method: check the definition of a radius and the isosceles triangle property the step relies on. Working: every radius of a circle has the same length, so OA = OC regardless of where A and C sit on the circle — this alone makes triangle OAC isosceles, and the base angles opposite the two equal sides, angle OAC and angle OCA, must be equal. The verdict is Yes, for exactly that reason. Claiming OA and OC are only equal if they are drawn to 'the very same point' misunderstands what a radius is — A and C can be any two points on the circle and OA still equals OC. Claiming the triangle is isosceles because angle AOC is 90° reverses the logic: nothing in the step has fixed angle AOC at 90°, and even if it had, that alone would not explain why OA = OC. Claiming the triangle would need to be equilateral confuses isosceles (two equal sides) with equilateral (three equal sides) — only two sides, OA and OC, are being compared here.
- (b) 20 — The rate is the gradient: the change in litres divided by the change in time. From 2 to 7 minutes, the tank gains 150 − 50 = 100 litres over 7 − 2 = 5 minutes, so the rate is 100 ÷ 5 = 20 litres per minute. 100 comes from working out the change in litres but forgetting to divide by the change in time. 25 comes from using only the first reading, 50 ÷ 2, and ignoring the second reading entirely. 0.05 comes from dividing the change in time by the change in litres instead of the other way round, 5 ÷ 100.
- (d) 23 — Method: let the smallest integer be n, so the three consecutive integers are n, n + 1 and n + 2. Form the equation n + (n + 1) + (n + 2) = 72. Working: simplify the left side: 3n + 3 = 72, so 3n = 69, giving n = 23. Answer: 23. 24 comes from dividing 72 by 3 directly, 72 ÷ 3 = 24, which finds the middle integer rather than realising the three numbers differ. 25 comes from correctly finding n = 23 but reading off the largest integer, n + 2, instead of the smallest as asked. 21 comes from dividing first and subtracting after, (72 ÷ 3) − 3, instead of subtracting 3 before dividing 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.