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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- (c) 0.5, 0.45, −0.09, −0.15, −0.6 — Method: compare the decimals by their position on a number line, remembering that with negative decimals the one closer to zero is larger. Working: 0.5 and 0.45 are positive, so they come first, with 0.5 the larger of the two. Among the negatives, −0.09 is closest to zero, then −0.15, then −0.6 is furthest from zero and so the smallest. Answer: 0.5, 0.45, −0.09, −0.15, −0.6. 0.5, 0.45, −0.15, −0.09, −0.6 swaps −0.09 and −0.15, treating the negative decimal with more digits after the point as closer to zero. −0.6, −0.15, −0.09, 0.45, 0.5 lists the numbers from smallest to largest instead of largest to smallest. 0.5, 0.45, −0.6, −0.15, −0.09 orders the negative decimals by the size of the digit (0.6 > 0.15 > 0.09) as if they were positive, instead of recognising that a bigger negative decimal is further from zero and so smaller.
- (c) x = (y + 24)/4 — Method: expand the bracket first, then undo the operations done to x in reverse order. Working: y = 4(x − 6) = 4x − 24, so y + 24 = 4x, so x = (y + 24)/4. Answer: x = (y + 24)/4. x = (y + 6)/4 comes from expanding the bracket incorrectly, treating 4(x − 6) as 4x − 6 instead of 4x − 24. x = 4y + 96 comes from multiplying by 4 instead of dividing by 4 to undo the multiplication, giving 4(y + 24) = 4y + 96. x = y/4 − 6 comes from dividing by 4 first without expanding the bracket, then subtracting 6 as if the bracket's operation still applied afterwards.
- (c) £450 — Method: use the equal ratios 4:5 = 200:x to find Grace's savings, then add the two amounts. Working: Noah's £200 is 4 parts, so one part is £200 ÷ 4 = £50; Grace has 5 parts, so 5 × £50 = £250; altogether £200 + £250 = £450. Answer: £450. The distractors: £250 is Grace's savings on their own, which is the middle step rather than the total the question asks for; £360 comes from reading £200 as the 5 parts instead of the 4, giving one part of £40 and a total of 9 × £40; £400 comes from doubling £200, which treats the two savings as equal and ignores the ratio altogether.
- (b) (3, −1) — Method: find the distance from the point to the mirror line, then place the image the same distance on the other side of the line. Working: B is 5 − 2 = 3 units above the line y = 2, so its image is 3 units below the line, at y = 2 − 3 = −1, giving (3, −1); the x-coordinate does not change, since the mirror line is horizontal. Options: (3, −5) comes from reflecting in the x-axis (y = 0) instead of the line y = 2; (−3, 5) comes from reflecting in the y-axis instead of a horizontal line; (3, −4) comes from doubling the distance from the line instead of reflecting it, giving 2 − 2×3 = −4. Answer: (3, −1).
- (b) 0.355 — Method: use the law of total probability across the two Monday branches: P(rain Tue) = P(rain Mon) × P(rain Tue | rain Mon) + P(no rain Mon) × P(rain Tue | no rain Mon). Working: P(no rain Mon) = 1 − 0.3 = 0.7. P(rain Tue) = (0.3 × 0.6) + (0.7 × 0.25) = 0.18 + 0.175 = 0.355. Answer: 0.355. Watch out: using only the rain-Monday branch (0.3 × 0.6) or only the no-rain-Monday branch (0.7 × 0.25) accounts for just one of the two ways Tuesday can turn out rainy — both branches must be added. And swapping which weekday-probability multiplies which branch (0.7 with the rain branch, 0.3 with the no-rain branch) uses the right numbers on the wrong branches.
- (c) 27 — Method: a frequency is the area of the part of the bar being asked about, so frequency = frequency density × the width of that part. Working: the part asked about runs from 20 to 35, so its width is 35 − 20 = 15 metres; the frequency density there is 1.8 per metre, so the estimate is 1.8 × 15 = 27. Answer: about 27 throws. The distractors: 72 comes from taking the whole bar, 1.8 × 40, and so counting every throw from 20 up to 60; 1.8 comes from reading the height of the bar as a frequency, when a height is a density and only an area is a count; 63 comes from using the upper value 35 as the width, 1.8 × 35, instead of the width 35 − 20.
- (a) 2.5 × 10⁷ — Method: place the decimal point so that the coefficient is at least 1 and less than 10, then count the places it has moved. Working: the digits give a coefficient of 2.5, and the decimal point travels from the end of 25,000,000 until it sits between the 2 and the 5, a move of 7 places. Answer: 2.5 × 10⁷. The distractors: 25 × 10⁶ is the same area but not in standard form, because the coefficient must be less than 10; 2.5 × 10⁸ comes from counting the eight digits of 25,000,000 instead of the seven places the decimal point moves; 2.5 × 10⁻⁷ comes from making the index negative because the decimal point was carried to the left.
- (a) y = −x³ + 5x — Reflecting a graph in the y-axis replaces every x in the equation with −x: y = (−x)³ − 5(−x) = −x³ + 5x. Writing y = −x³ − 5x comes from substituting −x into the x³ term only and leaving the −5x term as it was. Writing y = x³ + 5x comes from substituting −x into the −5x term only and leaving the x³ term as it was. Writing y = x³ − 5x is the original equation with no reflection applied at all — every term needs the substitution, not just one of them.
- (a) 2 km — Method: multiply the map distance by the scale to get the real distance in centimetres, then convert centimetres to kilometres using 100 cm = 1 m and 1000 m = 1 km. Working: 4 × 50 000 = 200 000 cm; 200 000 ÷ 100 = 2000 m; 2000 ÷ 1000 = 2. Answer: 2 km. The distractors: 200 km comes from dividing the 200 000 cm by 1000 in a single step, as if a kilometre were 1000 cm rather than the 100 000 cm it is; 20 km comes from converting to metres correctly, 200 000 ÷ 100 = 2000 m, and then dividing those metres by 100 instead of by 1000; 0.2 km comes from dividing by 1000 to reach metres, as if a metre were 1000 cm, and then dividing by 1000 again, so 200 000 is divided by 1 000 000 altogether.
- (b) 12 cm — Method: in similar shapes every length is multiplied by the same scale factor, and a ratio of 1 : 3 means that factor is 3 going from the smaller shape to the larger one. Working: 4 × 3 = 12. Answer: 12 cm. The distractors: 7 cm comes from adding 3 to the side instead of multiplying by it, which is what happens when a ratio is read as a difference; 36 cm comes from multiplying by 3² = 9, the factor that scales areas, and applying it to a length; 4 cm comes from treating the two shapes as congruent, so that corresponding sides stay equal — similar shapes have equal angles, but their sides are in proportion.
- (a) 5/8 — Winning, near-miss and losing are exhaustive, so the three probabilities sum to 1. Writing 1/4 as 2/8 so every fraction has the same denominator, 1 − 1/8 − 2/8 = 8/8 − 1/8 − 2/8 = 5/8. Subtracting only the winning probability and forgetting the near-miss probability gives 1 − 1/8 = 7/8. Subtracting only the near-miss probability and forgetting the winning probability gives 1 − 1/4 = 3/4. Adding the two given probabilities and stopping there gives 1/8 + 2/8 = 3/8, the probability that a ticket is winning or a near-miss, not the probability that it is losing.
- (a) 71 — 2.9⁴ = (2.9²)². First, 2.9² = 8.41. Then square that: 8.41² = 70.7281, since 841² = 707281 and the decimal point moves four places. To 2 significant figures this rounds to 71, because the figure after the first two significant figures (7 and 0) is a 7, which rounds the 0 up to 1. Rounding the working down instead of up — taking 70.7281 to 70 — ignores that the next figure is 5 or more. Rounding 2.9 up to 3 before doing any working at all, then computing 3⁴ = 81, uses a much cruder approximation and overshoots the true value. Rounding 8.41 all the way down to 8 before squaring, 8² = 64, rounds far too aggressively and loses the accuracy needed for 2 significant figures.
- (b) 4 — Adding the two equations: the y-terms, −2y and +2y, cancel, and the x-terms combine to 5x + 3x = 8x; the right-hand sides add to 16 + 16 = 32. This gives 8x = 32, so x = 4. A candidate who adds only one of the right-hand sides, instead of both, would get 8x = 16, so x = 2. A candidate who divides 32 by 4 instead of 8 would get x = 8. A candidate who subtracts the equations instead of adding them, getting 2x − 4y = 0, and then wrongly assumes y = 0, would get x = 0.
- (a) 5:3 — Divide both prices by their highest common factor, 3: 15 ÷ 3 = 5 and 9 ÷ 3 = 3, giving the ratio 5:3. Choosing 3:5 comes from writing the ratio the wrong way round, as child price to adult price. Choosing 2:3 comes from using the difference between the two prices (15 − 9 = 6) as the first part of the ratio instead of the adult price, then simplifying 6:9 by dividing by 3. Choosing 5:8 comes from comparing the adult price with the total cost of both tickets (£15 out of £24) instead of comparing it with the child price.
- (d) B is the reverse of A — Every component of vector B is the negative of the matching component of vector A (−2 is the negative of 2, and 5 is the negative of −5), so B undoes the translation that A performs — B is the reverse of A. 'B is the same as A' ignores that both signs have flipped. 'B is twice A' confuses a sign change with a scale-factor change; the sizes of the components have not changed, only their signs. 'B is unrelated to A in direction' misses that the two vectors are directly linked, just in opposite directions.
- (d) 48 — Each term is found by multiplying the previous term by the common ratio, 2: 3, 6, 12, 24, 48 — the 5th term is 48. A candidate who mistakes the common ratio for a common difference, and adds 2 four times, would reach 3+4×2=11. A candidate who works out the multiplier 2⁴=16 but forgets to multiply it by the first term would give 16. A candidate who multiplies one time too many (finding the 6th term instead of the 5th) would reach 3×2⁵=96.
- (c) 45% — Method: adding water changes the total volume but not the amount of fruit juice, so find the juice, find the new total volume, and write the first as a percentage of the second. Working: 3 × 0.6 = 1.8 litres of fruit juice; the new volume is 3 + 1 = 4 litres; 1.8 ÷ 4 = 0.45, which is 45%. Answer: 45%. The distractors: 60% is the strength before the water goes in, and assumes that adding water leaves the strength unchanged; 15% comes from dividing the 60% by the 4 litres of mixture instead of dividing the 1.8 litres of juice by the 4 litres; 75% is the fraction of the new mixture that came out of the original jug, 3 litres out of 4, which ignores that only 60% of that 3 litres was juice.
- (c) 8 — Method: the exterior angles of a polygon add up to 360°, so divide 360° by the size of one exterior angle. Working: 360 ÷ 45 = 8. Answer: 8 sides. A candidate who divides into a half turn instead of a full turn, working out 180 ÷ 45, gets 4. A candidate who reads off the given exterior angle as if it were the number of sides gets 45. A candidate who subtracts instead of dividing, working out 360 − 45, gets 315.
- (b) 40 — The differences between consecutive terms are 3, 5, 7, 9, increasing by 2 each time, so the next difference is 11, giving 29 + 11 = 40. A candidate who reuses the last difference of 9 without increasing it gets 29 + 9 = 38. A candidate who increases the difference by only 1 instead of 2 gets 29 + 10 = 39. A candidate who reuses the first difference of 3 instead of the pattern gets 29 + 3 = 32.
- (c) 55 — Continuing the pattern: 8+13=21 (5th term), 13+21=34 (6th term), 21+34=55 (7th term). A candidate who miscounts the position and stops one term early would give 34, the 6th term. A candidate who doubles the most recent term instead of adding the two before it would compute 34×2=68. A candidate who adds a non-adjacent pair — the 4th and 6th terms, skipping the 5th — would compute 13+34=47.
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.