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) 2/11 — Let x = 0.18 recurring, so x = 0.181818... . Since two digits repeat, multiply by 100: 100x = 18.181818... . Subtracting the original x removes the recurring part, because the digits line up exactly: 100x − x = 18.181818... − 0.181818... = 18, so 99x = 18, giving x = 18/99 = 2/11. Treating the decimal as if it terminated at two places gives 18/100 = 9/50, which is only 0.18 and drops the repeating part entirely. Subtracting 10x instead of x — using 100x − 10x = 90x = 18 — is the wrong power of ten for a two-digit repeating block, and gives x = 18/90 = 1/5. Making an arithmetic slip in the numerator, 18 − 1 = 17 instead of 18, gives 17/99.
- (a) 38 — Method: substitute the position number into the rule and follow the order of operations, so the squaring is carried out before the 2 is added. Working: n = 6 gives 6² + 2; 6² means 6 × 6 = 36, and then 2 is added to 36. Answer: 38. The distractors: 14 comes from multiplying the position by 2 instead of squaring it, 6 × 2 + 2; 36 comes from squaring correctly and then forgetting to add the 2; 64 comes from adding the 2 first and squaring afterwards, (6 + 2)².
- (a) Inverse proportion — A curve that decreases and never touches either axis is the standard shape for inverse proportion, y = k/x. Direct proportion graphs are straight lines through the origin, which this is not, so it must be inverse proportion rather than neither.
- (b) No — the angle given is not the included angle — Method: check whether the given angle sits between the two given sides, since SAS requires the included angle. Working: sides AB and BC meet at vertex B, so the angle between them is angle B — but the angle given is angle A, which is not between the two given sides, and the same mismatch happens in triangle DEF. Options: 'two sides and one angle match' restates SAS's ingredients without checking their positions, which is exactly Meera's mistake; 'SSS needs three equal sides' is a true fact about a different condition, but it is not the reason Meera is wrong here; 'SAS allows any equal angle' states a rule that is not how SAS works, since the angle must be the included one. Answer: no, the angle given is not the included angle.
- (a) 12/25 — P(red then blue) = 4/10 × 6/10 = 24/100. P(blue then red) is the same, 6/10 × 4/10 = 24/100. Since either order counts as different colours, add them: 24/100 + 24/100 = 48/100 = 12/25. Working out only one order, red-then-blue, and forgetting blue-then-red, gives 24/100 = 6/25. Working out the probability that the two counters are the SAME colour instead — 4/10 × 4/10 + 6/10 × 6/10 = 16/100 + 36/100 = 52/100 = 13/25 — answers a different question. Using 6/9 for the second draw, as if the first counter were not replaced, gives 2 × (4/10 × 6/9) = 48/90 = 8/15.
- (c) Branch A waits longer, and Branch A is more consistent — Method: compare the two branches using a measure of location (the median) for who waits longer, and a measure of spread (the interquartile range) for who is more consistent — a smaller interquartile range means more consistent. Working: Branch A's median, 12 minutes, is higher than Branch B's, 9 minutes, so Branch A's customers wait longer on average. Branch A's interquartile range, 5 minutes, is smaller than Branch B's, 11 minutes, so Branch A's waiting times vary less. Answer: Branch A waits longer, and Branch A is also the more consistent of the two. Watch that each half of the comparison uses the right statistic and reads it correctly: swapping both readings gives Branch B the longer wait and the greater consistency, when neither is true; keeping the median comparison right but reading a larger interquartile range as 'more consistent' has the direction of spread backwards; and swapping only the median comparison keeps the correct branch for consistency but gives the wrong branch the longer wait.
- (b) 30 — 70 is close to the perfect square 64, so √70 ≈ 8. 65 is close to the perfect cube 64, so ∛65 ≈ 4. Multiplying these estimates: 8 × 4 = 32, which rounds to 30 to 1 significant figure. Estimating ∛65 as 5 instead of 4, perhaps by confusing it with the nearby cube 125 = 5³ rather than the much closer 64 = 4³, and then multiplying by 8, gives 8 × 5 = 40. Adding the two estimates instead of multiplying them, 8 + 4 = 12, rounds to 10 to 1 significant figure. Rounding both estimates up to the next whole number using the wrong nearby power for each, taking √70 as 9 and ∛65 as 5, gives 9 × 5 = 45, which rounds to 50 to 1 significant figure.
- (a) −1 — Reflecting y = sin x in the x-axis gives y = −sin x, so g(x) = −sin x. Since sin 90° = 1, g(90) = −1. Reading sin 90° = 1 and forgetting to apply the reflection gives 1. Misreading the angle as 0° instead of 90° gives sin 0° = 0, so 0. Confusing sin 90° with sin 30° = 0.5, then reflecting it, gives −0.5.
- (d) The population is growing at 2500 people per year. — The gradient of a tangent to a graph gives the instantaneous rate of change of the quantity on the vertical axis with respect to the quantity on the horizontal axis, at that exact point — not the total change and not an average. Here the vertical axis is population in thousands and the horizontal axis is time in years, so the gradient is measured in thousands of people per year. A gradient of 2.5 means the population is growing at an instantaneous rate of 2.5 thousand people per year, and since P is measured in thousands, 2.5 × 1000 = 2500 people per year. This describes the rate of change at that instant, not the total increase over the 6 years and not an average population.
- (b) 60 units² — Method: the area of a triangle is half the base times the perpendicular height, so choose a side to act as the base and measure the perpendicular distance from the opposite vertex to it. Working: A(0, 0) and B(10, 0) both lie on the x-axis, so AB is horizontal and AB = 10 − 0 = 10. The perpendicular height is the distance of C from the x-axis, which is its y-coordinate, 12. Area = (10 × 12) ÷ 2 = 120 ÷ 2 = 60. Answer: 60 units². The distractors: 120 units² comes from multiplying base by height and forgetting to halve; 65 units² comes from using the slanting side AC, which is 13 long, as the height in place of the perpendicular distance 12; 30 units² comes from halving the base to 5 before multiplying and then halving the product as well, so the halving is done twice.
- (a) 2/3 — Method: 'in A or in B' means every score that belongs to at least one of the two sets; a score that belongs to both is still only one outcome, so it is listed once. Working: the even scores are 2, 4 and 6; the scores greater than 4 are 5 and 6. Listing the scores that appear in either set gives 2, 4, 5 and 6, with 6 written once. That is 4 of the 6 faces, or 4/6. Answer: the probability is 2/3. The distractors: 5/6 comes from adding the sizes of the two sets, 3 + 2, so that the score 6 is counted in both and appears twice; 1/2 comes from using the even scores alone; 1/6 comes from giving the probability that the score is in both sets, which is the single score 6, rather than in either of them.
- (c) 3:4 — The jacket costs 3/7 of £84, which is 3 × (84 ÷ 7) = 3 × 12 = £36. The bag then costs the rest of the money, £84 − £36 = £48. The ratio of the jacket to the bag is 36:48, which simplifies to 3:4. Writing the fraction spent on the jacket, 3/7, directly as the ratio, without working out that the bag's share is the remaining 4/7, gives 3:7. Giving the ratio the wrong way round, bag to jacket instead of jacket to bag, gives 4:3. Assuming the jacket and bag cost the same, ignoring the fraction given, gives 1:1.
- (d) 68, and it is a formula — Substituting C = 20 gives F = 1.8 × 20 + 32 = 36 + 32 = 68. The statement F = 1.8C + 32 is a formula because it expresses a general relationship between F and C that holds for any value of C, not just C = 20. A candidate who gets the correct value but calls it an equation is confusing "has an equals sign" with "is an equation" — an equation is only true for particular value(s) of the unknown, whereas this relationship is used to calculate rather than to be solved. A candidate who forgets to multiply C by 1.8 and just adds 32 to 20 gets 52. A candidate who multiplies correctly but forgets to add 32 gets 36.
- (b) 13/2 — Put the number of pages in the novel over the number of pages in the comic: 312/48. Divide both numbers by their highest common factor, 24: 312÷24 = 13, 48÷24 = 2, giving 13/2. (2/13 comes from writing the page counts the wrong way round. 11/2 comes from finding the difference in the page counts, 312 − 48 = 264, and writing it as a fraction of the comic's page count, 264/48. 13/15 comes from comparing the novel's page count to the total number of pages in both books, 312/360.)
- (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) x ≥ 5 — Expand the bracket: 2(3x − 1) = 6x − 2, so the inequality is 6x − 2 ≥ 4x + 8. Subtract 4x from both sides and add 2 to both sides: 2x ≥ 10. Divide both sides by 2: x ≥ 5. A candidate who only multiplies the 3x by 2 and forgets to multiply the −1 gets 6x − 1 ≥ 4x + 8, leading to x ≥ 4.5. A candidate who adds 4x instead of subtracting it gets 10x ≥ 10, leading to x ≥ 1. A candidate who multiplies by 2 instead of dividing gets x ≥ 20.
- (d) 48 — Find the rate first: 18 ÷ 3 = 6 bottles per minute. Then apply it to the new time: 6 × 8 = 48 bottles. Working out 18 + (8 − 3) = 23 adds the extra 5 minutes onto the number of bottles instead of scaling proportionally. Working out 18 × 8 = 144 multiplies the given number of bottles by the new number of minutes without finding the rate first. Writing 18 keeps the count the same, not realising it must change with the time. In 8 minutes the machine fills 48 bottles.
- (c) Equal sides and equal interior angles — Method: recall the full definition of 'regular' as applied to a polygon. Working: a regular polygon must have both equal side lengths and equal interior angles at the same time. Options: 'all sides equal' alone describes an equilateral but not necessarily equiangular shape, such as a rhombus, which is not regular; 'all angles equal' alone describes an equiangular but not necessarily equilateral shape, such as a rectangle, which is not regular; 'at least one line of symmetry' is a much weaker condition that many irregular shapes also satisfy. Answer: equal sides and equal interior angles.
- (c) y = 200 − 8x — Method: in a linear model the amount present at the start is the constant term and the steady rate of change is the gradient, which is negative because the amount is falling. Working: at x = 0 minutes there are 200 litres, so the constant term is 200; 8 litres are lost every minute, so after x minutes 8x litres have gone and the amount left is y = 200 − 8x. Answer: y = 200 − 8x. The distractors: y = 8x + 200 treats the draining as filling, so the butt would gain 8 litres a minute; y = 200x − 8 swaps the two numbers over, using the starting 200 litres as the rate per minute and the 8 litres per minute as the starting amount; y = −8x − 200 makes the starting amount negative as well as the rate, so the butt would begin 200 litres in deficit.
- (b) x² + y² = 625 — Since (7, 24) lies on the circle, x² + y² = 7² + 24² = 49 + 576 = 625, so the equation is x² + y² = 625. Choosing x² + y² = 31 adds the coordinates 7 and 24 directly instead of squaring them first. Choosing x² + y² = 576 uses only 24² and forgets to add 7². Choosing x² + y² = 49 uses only 7² and forgets to add 24².
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.