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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- (d) 12 — Method: if the bracelets are identical and no beads are left over, the number of bracelets must divide exactly into both totals, so it is the highest common factor of 24 and 36. Working: 24 = 2³ × 3 and 36 = 2² × 3²; taking the lower index of each shared prime gives 2² × 3 = 4 × 3 = 12. Each bracelet then has 2 red beads and 3 blue beads. Answer: 12. The distractors: 6 comes from taking each shared prime once rather than at its lower index, giving 2 × 3, which is a common factor but not the highest; 72 is the lowest common multiple of 24 and 36, from taking the higher index of each prime instead of the lower; 60 comes from adding the two bead totals instead of looking for a common factor.
- (d) x² − 49 — Area = length × width = (x + 7)(x − 7). Expanding: x × x = x², x × (−7) = −7x, 7 × x = 7x, 7 × (−7) = −49. The two middle terms −7x and 7x cancel, leaving x² − 49. A candidate who misremembers the difference-of-two-squares result as a sum gets x² + 49. A candidate who makes a sign error and treats both middle terms as −7x instead of cancelling gets x² − 14x − 49. A candidate who confuses area with perimeter and simply adds the length and width gets 2x.
- (a) 150 g — Method: scale the recipe to find the total sugar needed, then subtract the sugar Sam already has. Working: 200 ÷ 8 × 20 = 500, so 500 g is needed in total; 500 − 350 = 150, so 150 g still to buy. Stopping after finding the total, 500, without subtracting what he has gives 500 g. Scaling the wrong way round, 200 × 8 ÷ 20 = 80, wrongly suggests he already has enough, giving 0 g. Adding the amount he has instead of subtracting it, 500 + 350 = 850, gives 850 g.
- (c) Rectangle — A rectangle has two pairs of parallel sides and four right angles, but does not require all sides to be equal — this matches exactly, so Rectangle is correct. A square also has four right angles and parallel sides, but additionally requires all four sides to be equal, which contradicts 'not all the same length', so it is wrong. A rhombus has two pairs of parallel sides and all four sides equal, but its angles are not generally 90° unless it is also a square, so it does not match the right-angle condition here. A kite has no pairs of parallel sides at all, so it does not match the first condition given.
- (a) 7/16 — Method: the adult picked is known to have a driving licence, so the sample space is everyone with a licence; divide the number of men with a licence by that total. Working: 45 women and 35 men have a licence, so 80 adults have one. The men with a licence give 35/80, and dividing the numerator and the denominator by 5 gives 7/16. Answer: the probability is 7/16. The distractors: 7/10 is 35/50, the probability that an adult has a licence given that he is a man, which is the condition and the event the wrong way round; 7/24 is 35/120, dividing by all 120 adults surveyed instead of by the 80 who have a licence; 5/12 is 50/120, the probability that an adult picked from the whole survey is a man, which uses none of the licence information the question supplies.
- (a) 22 — Method: the two subject totals overlap, because every pupil who passed both subjects has been counted once in the maths total and once again in the science total; adding the totals therefore counts those pupils twice, and the overlap has to be taken off once. Working: 18 + 12 = 30, and the 8 pupils who passed both have been counted twice in that 30, so the number who passed at least one subject is 30 − 8 = 22. Answer: 22 pupils, a count of pupils, and it is less than the 30 in the class, which leaves 8 pupils who passed neither. The distractors: 30 comes from adding the two subject totals and never removing the overlap, so it counts the 8 pupils twice; 14 comes from taking the 8 away twice, 18 + 12 − 8 − 8, removing an overlap that was only counted twice once too often; 18 comes from writing down the larger of the two subject totals on its own, which leaves out every pupil who passed science but not maths.
- (c) x⁽²⁄³⁾ — Method: a root can be written as a fractional index, with the root's index as the denominator and the power inside the root as the numerator. Working: the cube root gives a denominator of 3 and the square inside gives a numerator of 2, so ∛(x²) = x⁽²⁄³⁾. Answer: x⁽²⁄³⁾. The distractors: x⁽³⁄²⁾ comes from writing the fraction upside down, with the root's index on top; x⁽¹⁄⁶⁾ comes from treating the square as a second root and multiplying 1/3 by 1/2; x⁶ comes from multiplying the root's index by the power, 3 × 2, and keeping the result as a whole-number index.
- (a) 12 + 4√2 — a₁ = 4, a₂ = 4√2, and a₃ = a₁ × r² = 4 × (√2)² = 4 × 2 = 8. The sum of the first three terms is 4 + 4√2 + 8 = 12 + 4√2. 8 + 4√2 comes from leaving out a₁ and adding only a₂ + a₃ = 4√2 + 8. 36 + 4√2 comes from squaring the whole second term instead of applying the ratio to the first term: (4√2)² = 32 used as a₃, giving 4 + 4√2 + 32 = 36 + 4√2. 4 + 12√2 comes from using r³ instead of r² for the third term: 4 × (√2)³ = 4 × 2√2 = 8√2, giving 4 + 4√2 + 8√2 = 4 + 12√2.
- (b) t = 5 and t = 7 (closest, evenly spaced) — To estimate the instantaneous rate of change at t = 6, use the chord centred on t = 6 with the closest readings on either side, t = 5 and t = 7. The gradient of this chord is 15.4 − 17.5 = −2.1, then −2.1 ÷ 2 = −1.05 cm per minute. The interval t = 3 to t = 9 is also centred on t = 6 but is wider: 13.4 − 18.7 = −5.3, then −5.3 ÷ 6 ≈ −0.88 cm per minute — this brings in more of the curve's own change in steepness, so it is a worse estimate of the rate at the single instant t = 6. Using t = 6 and t = 7 only gives 15.4 − 16.6 = −1.2, then −1.2 ÷ 1 = −1.2 cm per minute, but this is not centred on t = 6 — it estimates the rate over (6, 7), not at t = 6 itself. Using t = 0 and t = 6 gives 16.6 − 20.0 = −3.4, then −3.4 ÷ 6 ≈ −0.57 cm per minute, the average rate for the whole first six minutes, not the rate at the instant t = 6. Always choose the chord that brackets the point as closely as possible.
- (d) No — angles must be equal too — A regular polygon must have both all sides equal and all angles equal. This tile has all six sides equal, but its interior angles are not all equal, so it fails the angle condition and is not regular — 'No — angles must be equal too' is correct. 'Yes — all sides are equal' is wrong because equal sides alone are not enough; a shape can have equal sides but unequal angles, as here. 'Yes — six equal sides means regular' is wrong for the same reason: equal sides do not automatically guarantee equal angles. 'No — hexagons can't be regular' is wrong because regular hexagons certainly exist (six equal sides and six equal 120° angles); it is this particular tile that fails to be regular, not hexagons in general.
- (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.
- (c) 31 — Method: work out each power separately, remembering that any non-zero base raised to the power 0 is 1 and a base raised to the power 1 is itself, then add the three values. Working: 5⁰ = 1, 5¹ = 5 and 5² = 25, so the total is 1 + 5 + 25 = 31. Answer: 31. The distractors: 30 comes from taking 5⁰ as 0 instead of 1; 35 comes from taking 5⁰ as 5, treating a zero index as leaving the base unchanged; 125 comes from adding the indices first, as though the three terms were being multiplied, and working out 5³.
- (d) 45 — v = s ÷ t = 180 ÷ 4 = 45 km/h. 720 comes from multiplying s and t instead of dividing, 180 × 4. 184 comes from adding s and t instead of dividing, 180 + 4. 176 comes from subtracting t from s instead of dividing, 180 − 4.
- (a) 1:8 — Convert 2 kg to grams: 2 kg = 2000 g. The ratio is 250 : 2000. The highest common factor of 250 and 2000 is 250. Divide both parts by 250: 250 ÷ 250 = 1 and 2000 ÷ 250 = 8, so the ratio is 1 : 8. Leaving the kilograms unconverted gives 250 : 2, which simplifies to 125 : 1 — the units on each side are different, so this does not compare like with like. Dividing by 50 instead of 250 gives 5 : 40, which still shares a common factor of 5, so it is not fully simplified. Swapping the order gives 8 : 1, grams to kilograms the wrong way round.
- (a) 24 cm — The line from the centre to the midpoint of a chord is perpendicular to the chord, so triangle OMA has a right angle at M. By Pythagoras' theorem, AM² = OA² − OM² = 169 − 25 = 144, so AM = 12 cm. AB is twice AM, since M is the midpoint: AB = 2 × 12 = 24 cm. Finding AM = 12 cm correctly but forgetting to double it for the full chord gives 12 cm. Working out 169 − 25 = 144 and forgetting to take the square root gives 144 cm. Subtracting first and then doubling the wrong way, (13 − 5) × 2, gives 16 cm. Doubling AM after finding it correctly is the step that's missing from all three — do it, and you get 24 cm.
- (c) t = 1 second and t = 5 seconds — The ball is at ground level when h = 0, which happens when either bracket is zero. t − 1 = 0 gives t = 1, and 5 − t = 0 gives t = 5. The option with t = −1 makes a sign error solving the first bracket. The option with t = −5 makes a sign error solving the second bracket, treating 5 − t = 0 as if it gave a negative solution. The option with t = 4 seconds is an arithmetic slip in solving 5 − t = 0.
- (a) 3/2 — Find each average speed: car = 180 ÷ 3 = 60 mph; lorry = 160 ÷ 4 = 40 mph. Put the car's speed over the lorry's speed: 60/40. Divide both numbers by their highest common factor, 20: 60÷20 = 3, 40÷20 = 2, giving 3/2. (2/3 comes from writing the speeds the wrong way round. 9/8 comes from comparing the distances travelled, 180/160, without working out the speeds. 3/4 comes from comparing the times taken, 3/4, instead of the speeds.)
- (c) They must also be equal — Once two triangles are proved congruent by any condition, including ASA, they are identical in every respect: every pair of corresponding sides and every pair of corresponding angles must be equal, not just the ones originally used to prove the congruence. So the two remaining pairs of corresponding sides must also be equal, making 'they must also be equal' correct. 'They might be equal or not' and 'not enough information to say' both wrongly suggest that congruence only guarantees the specific facts used to prove it, when congruence actually guarantees the triangles are identical overall. 'They must be different' is backwards: the triangles being identical is the entire point of proving congruence, not a reason for a side to differ.
- (c) The object was stationary (not moving) — On a distance-time graph, the gradient at any point represents the speed at that point. A gradient of 0 means distance is not changing over time, so the object is stationary. A straight, sloped line (not flat) shows constant nonzero speed; a curve bending one way shows acceleration or deceleration; a flat section is not a maximum speed — it is no speed at all. Read the shape of the graph, not just how steep it looks.
- (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.
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.