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) 1/5 — Work out the bracket first: 2 + 3 = 5. The reciprocal of 5 is 1/5. A candidate who forgot to take the reciprocal and just gave the value of the bracket wrote 5. A candidate who took the reciprocal but made a sign error wrote −1/5. A candidate who found the reciprocal of each number separately and added them, treating reciprocal as if it distributes over addition, worked out 1/2 + 1/3 = 5/6.
- (a) 8 — Method: rate = height gained ÷ time taken. Working: height gained = 52 − 20 = 32 m, time taken = 4 minutes, so rate = 32 ÷ 4 = 8 metres per minute. Answer: the rate is 8 metres per minute. 32 comes from working out the height gained but forgetting to divide by the time taken. 13 comes from dividing the final height, 52, by the time taken instead of the height gained. 5 comes from dividing the starting height, 20, by the time taken instead of the height gained.
- (b) 24 — Method: let one part of the ratio be worth x, write both groups in terms of x, and use the fact that the two groups end up equal. Working: the boys are 3x and the girls are 4x; after the 6 boys join, 3x + 6 = 4x, so x = 6; the girls are 4 parts, so 4 × 6 = 24. Answer: 24 girls. The distractors: 18 is the number of boys before the 6 join, which is 3 × 6; 30 comes from adding the 6 new members to the girls as well as to the boys; 42 is the total number of members in the choir before the 6 boys join, the 18 boys and the girls together.
- (c) (10.5, −2) — Method: the vector from the first tile to the third tile is the pattern's vector doubled, since two translations happen between them. Working: doubling (4.5, −2.5) gives (9, −5); adding this to the starting corner (1.5, 3) gives x-coordinate 1.5 + 9 = 10.5 and y-coordinate 3 − 5 = −2. Answer: (10.5, −2). A candidate who only applies the vector once, translating to the second tile instead of the third, gets (6, 0.5). A candidate who adds 2.5 instead of subtracting it in the y-coordinate gets (10.5, 8). A candidate who doubles the x-part of the vector correctly but forgets to change the y-coordinate at all gets (10.5, 3).
- (c) 7/8 — Method: 'at least one head' is the opposite of 'no heads at all', so work out the probability of three tails and take it away from 1. Working: a flip that is not a head has probability 1/2, and the flips are independent, so three tails in a row has probability 1/2 × 1/2 × 1/2 = 1/8. Taking this from 8/8 leaves 7/8. Answer: the probability is 7/8. The distractors: 1/8 is the probability of three tails, written down without the final subtraction; 3/8 is the probability of exactly one head, which comes from reading 'at least one' as 'exactly one'; 1/2 comes from giving the probability of a head on a single flip and ignoring that three flips are made.
- (c) 145 g — Method: find the position of the median from the total frequency, locate the class that contains it, then use linear interpolation inside that class, assuming the apples in it are spread evenly. Working: the median is the 100 ÷ 2 = 50th apple; the running totals are 10, then 10 + 30 = 40, then 40 + 40 = 80, so the 50th apple lies in the class 140 ≤ m < 160; it is the 50 − 40 = 10th of the 40 apples in that class, and the class is 20 g wide, so the median is 140 + (10 ÷ 40) × 20 = 140 + 5 = 145. Answer: an estimated median of 145 g. The distractors: 150 g comes from giving the midpoint of the class that contains the median instead of interpolating inside it; 140 g comes from stopping at the lower boundary of that class, which locates the class but not the value; 155 g comes from measuring the 5 g step down from the upper boundary, 160 − 5, instead of up from the lower boundary.
- (d) 60 mph — Average speed = distance ÷ time, with the time measured in hours. 45 minutes is 45/60 of an hour, which is 0.75 of an hour, so the journey takes 1.75 hours. Speed = 105 ÷ 1.75 = 60 mph. 52.5 mph rounds the time up to 2 hours, 72.4 mph writes 1 hour 45 minutes as 1.45 hours, and 183.75 mph multiplies the distance by the time instead of dividing.
- (d) 5 — Each term is found by multiplying the previous term by the common ratio, 0.5: 80, 40, 20, 10, and the next term is 10 × 0.5 = 5. A candidate who instead subtracts the same amount each time (repeating the last difference of 10) would reach 10 − 10 = 0. A candidate who divides by 4 instead of by 2 would reach 10 ÷ 4 = 2.5. A candidate who multiplies by 2 instead of dividing (reversing the direction of the sequence) would reach 10 × 2 = 20.
- (d) 7/3 — The ratio cranberry : apple is 3:7, so apple juice is 7 parts and cranberry juice is 3 parts. Write apple over cranberry: 7/3. (3/7 comes from writing the ratio the wrong way round, cranberry over apple. 7/10 comes from comparing the apple juice to the total amount of the mixture, 7 parts out of 10. 3/10 comes from comparing the cranberry juice to the total amount of the mixture, 3 parts out of 10.)
- (b) 68° — Method: two properties are needed. Angle A and angle D are co-interior angles between the parallel sides AB and DC, so they add up to 180°; and because the trapezium is isosceles, the two angles on the side AB are equal, so angle B = angle A. Working: angle A = 180° − 112° = 68°, and angle B = angle A = 68°. Answer: 68°. The distractors: 112° comes from assuming that angles B and D are equal, which is the property of a parallelogram, not of a trapezium; 90° comes from assuming that the angles on the other parallel side must be right angles; 248° comes from using the 360° angle sum of a quadrilateral and taking away only the one angle that is given.
- (b) 4 — With 150 rolls and probability 1/6 for each number, the expected count is 150 ÷ 6 = 25. Comparing each actual count with 25: 1 is 22 (3 below), 2 is 27 (2 above), 3 is 24 (1 below), 4 is 34 (9 above), 5 is 21 (4 below) and 6 is 22 (3 below). Number 4 is furthest above its expected count, so it is the most over-represented. Number 2 is also above its expected count, but by only 2, far less than 4's 9. Number 3's count of 24 is below the expected 25, so it is under-represented, not over. Number 6's count of 22 is also below the expected 25, so it too is under-represented.
- (b) 4.05 — Method: line up the decimal points (or place value columns) before adding. Working: 3.60 + 0.45 = 4.05. Answer: 4.05. 3.65 is Priya's answer, from adding the digits without lining up the place value columns, which effectively treats 0.45 as 0.05. 4.5 comes from rounding both numbers up first, 3.6 to 4 and 0.45 to 0.5, and adding those instead of adding the exact values. 0.81 comes from adding the digits 36 and 45 together to get 81, then placing the decimal point in the wrong position.
- (c) Statement (ii) — Statement (ii) opens with 'Since n(n + 1) is even', treating the very fact the proof is meant to establish as if it were already known — that is circular reasoning, assuming the conclusion to help derive itself. Statement (i) only names the two consecutive integers as n and n + 1; it makes no claim about whether their product is even, so it introduces nothing circular. Statement (iii) states the conclusion, and would be a valid final step if statement (ii) had reached 'one of n and n + 1 is even' by a genuine argument, such as considering the cases where n is even or odd separately. Saying the proof assumes nothing circular is wrong, because statement (ii)'s opening clause is exactly that assumption.
- (d) 25 minutes — Method: find the time for one kilometre, then multiply by the number of kilometres — the unitary method with a rate. Working: 10 ÷ 2 = 5 minutes per km, and 5 × 5 = 25. Answer: 25 minutes. The distractors: 20 minutes comes from multiplying the 10 minutes by 2, the distance in the given rate, instead of by the scale factor 2.5; 50 minutes comes from multiplying 10 by 5, treating the 10 minutes as the time for a single kilometre; 15 minutes comes from adding the 5 km on to the 10 minutes, adding quantities that are not the same kind.
- (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.
- (b) −0.5 — Method: gradient = change in height ÷ horizontal distance, and the height decreases so the change is negative. Working: change in height = 5 − 25 = −20, horizontal distance = 40, so gradient = −20 ÷ 40 = −0.5. Answer: the gradient is −0.5. 0.5 comes from dropping the negative sign, ignoring that the zip-line descends. 2 comes from inverting the gradient, dividing the horizontal distance by the drop in height instead of the other way round. −0.8 comes from dividing the drop by the platform height, 25, instead of by the horizontal distance, 40.
- (b) 5:8 — The mix has 5 parts sand and 3 parts cement, so 5 + 3 = 8 parts in total. Sand to total is 5 : 8, and since the highest common factor of 5 and 8 is 1, this is already in its simplest form. Giving 5 : 3 answers sand to cement, not sand to the total mix. Giving 3 : 8 is cement to total, the wrong part of the mix. Giving 8 : 5 has the total and the sand swapped round.
- (b) 28 — The width of the rectangle is the difference in x-coordinates, 9 − 2 = 7, and the height is the difference in y-coordinates, 5 − 1 = 4. The area is width × height = 7 × 4 = 28. 22 comes from using the perimeter formula, 2 × (7 + 4), instead of the area formula. 35 comes from multiplying 7 by 5 instead of 4, misreading one of the y-coordinates. 63 comes from multiplying 9 by 7, using an x-coordinate instead of the height.
- (a) 3x/2 — Factorise top and bottom first: 6x³ − 15x² = 3x²(2x − 5), and 4x² − 10x = 2x(2x − 5). The bracket (2x − 5) is common to both, so it cancels, leaving 3x²/(2x); dividing the power of x, 3x² ÷ x = 3x, gives 3x/2. Writing 3x²/2 cancels the (2x − 5) correctly and removes the denominator's x, but never reduces the power of x left in the numerator — 3x² ÷ x should give 3x, not stay as 3x². Writing 3/2 cancels the (2x − 5) correctly but then drops the x from the numerator altogether, treating 3x² over x as if it cancelled completely to 3 instead of reducing to 3x. Writing −3x/2 comes from factorising the denominator with the wrong sign, as 2x(5 − 2x) instead of 2x(2x − 5); cancelling (5 − 2x) against the numerator's (2x − 5) then needs an extra minus sign, which flips the answer to −3x/2.
- (a) t = 0 or t = 4 — The ball is at ground level exactly when h = 0. From the table, h = 0 at t = 0 and at t = 4, so those are the two times. Distractor origins: t = 1 or t = 3 picks the times with equal (but non-zero) height instead of ground level; t = 2 picks the time of maximum height instead of ground level; t = 0 finds only the starting time and misses the second one.
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.