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) 120 — There are 6 choices for the first digit. The second digit must be different from the first, leaving 5 choices, and the third digit must differ from both of the first two, leaving 4 choices. By the product rule, the number of codes is 6 × 5 × 4 = 120. Allowing every digit to repeat, ignoring the 'no digit twice' rule entirely, gives 6 × 6 × 6 = 216. Adding the number of choices at each position instead of multiplying them, 6 + 5 + 4, gives 15. Treating the three chosen digits as one unordered set, rather than as digits in a fixed order on the padlock, divides by the 3! = 6 ways of arranging them: 120 ÷ 6 = 20.
- (d) 20 km/h — Method: total distance = 9 + 21 = 30 km. Total time = 15 + 45 + 30 = 90 minutes = 1.5 hours. Average speed = total distance ÷ total time = 30 ÷ 1.5 = 20 km/h. Distractor origins: 40 km/h ignores the 45-minute stop, dividing the total distance by the travel time only (30 km ÷ 45 minutes = 40 km/h); 39 km/h averages the two section speeds, 36 km/h and 42 km/h, instead of using total distance over total time; 36 km/h uses only the first section's speed (9 km in 15 minutes), ignoring the second section and the stop.
- (b) 9 m/s — The gradient of line P is 21 ÷ 3 = 7, so P has a rate of 7 m/s. The gradient of line Q is 45 ÷ 5 = 9, so Q has a rate of 9 m/s. Because 9 is greater than 7, line Q is the steeper line, with gradient 9 m/s. Taking line P's gradient instead of Q's gives 7 m/s, the less steep line. Subtracting the two lines' coordinates directly, (45 − 21) ÷ (5 − 3) = 24 ÷ 2 = 12 m/s, mixes points from different lines rather than using one line's own two points. Adding the two gradients, 7 + 9 = 16 m/s, treats 'steeper' as a total rather than a comparison.
- (d) 6 m — The horizontal distance is adjacent to the 60° angle and the zip-wire is the hypotenuse, so horizontal distance = 12 × cos 60° = 12 × 1/2 = 6 m. 6√3 m comes from using sin 60° = √3/2 instead of cos 60°, which would give the vertical drop, not the horizontal distance. 4√3 m comes from treating 12 as the side adjacent to a tangent ratio and dividing by tan 60° = √3. 24 m comes from dividing 12 by cos 60° instead of multiplying by it.
- (d) 3/50 — Relative frequency is the number of faulty bolts divided by the total sample size: 15/250, which simplifies to 3/50 by dividing both the numerator and the denominator by 5. Using 235, the number of bolts that were NOT faulty, as the denominator instead of the total 250 gives 15/235, which simplifies to 3/47. Inverting the fraction, dividing the total by the number of faulty bolts instead of the other way round, gives 250/15, which simplifies to 50/3 — a value greater than 1, which cannot be a probability. Simplifying by dividing the numerator and the denominator by different numbers, 15 ÷ 15 = 1 and 250 ÷ 25 = 10, gives 1/10.
- (a) x = 0 gives y = −20: a negative number sold — The y-intercept is the value the line predicts when x = 0: y = 3 × 0 − 20 = −20. A kiosk cannot sell a negative number of ice creams, so this is not a sensible estimate. The 3 in the equation is the gradient, not the intercept, so an option claiming x = 0 gives y = 3 has swapped the two numbers around — substituting x = 0 makes the 3x term equal 0, leaving −20, not 3. The danger of extrapolating to very high temperatures is a real issue with this line, but it is a different issue from the y-intercept, so it does not answer this question. And whether x = 0 could occur on a trading day is beside the point: the model still makes that prediction, and it is the prediction itself, −20, that is impossible.
- (b) 8π cm² — A diameter of 8 cm gives a radius of 4 cm. The area of a full circle would be π × r² = π × 4² = 16π cm², and a semicircle is exactly half of this, giving 16π ÷ 2 = 8π cm². Forgetting to halve the area for the semicircle gives 16π cm², the area of the whole circle. Halving the diameter twice, using a radius of 2 instead of 4, gives π × 2² = 4π cm². Using the diameter itself as the radius, so π × 8² = 64π, and then halving that for the semicircle gives 32π cm².
- (a) Positive x-direction, 90°; image is y = sin x. — Writing cos(x − 90°) as cos(x − a) with a = 90 shows this is a horizontal translation, y = f(x − a), which moves the graph 90° in the positive x-direction; the identity cos(x − 90°) = sin x confirms the image is y = sin x. Choosing the negative x-direction reverses the sign inside the bracket — subtracting inside the bracket always translates in the positive x-direction, not the negative one, so that statement is wrong on direction. Getting the direction right but conflating the subtraction inside the bracket with an extra reflection of the output flips the sign of the resulting graph, wrongly giving y = −sin x. Treating the subtraction as if it changed the output directly, rather than the input, wrongly calls this a vertical translation even while still correctly recalling that the image simplifies to y = sin x.
- (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) Triangular-based pyramid (tetrahedron) — A solid with 4 triangular faces, 4 vertices and 6 edges is a triangular-based pyramid, also called a tetrahedron. A triangular prism also has triangular faces, but it has 2 triangular faces plus 3 rectangular faces, 6 vertices and 9 edges — the extra rectangular faces and edges rule it out here. A square-based pyramid has 5 faces (one square, four triangles), 5 vertices and 8 edges, which does not match. A cube has 6 faces, 8 vertices and 12 edges, all much higher than the numbers given. The solid described is a triangular-based pyramid.
- (c) 0.37 — Method: pool the two runs into one combined set of results, then find the relative frequency of red across all of the spins together. Working: total reds = 16 + 21 = 37. Total spins = 40 + 60 = 100. Relative frequency = 37 ÷ 100 = 0.37. Answer: 0.37. Watch out: writing down 0.40 uses only the first run, 16 ÷ 40, and throws away the extra evidence from the second 60 spins. Writing down 0.35 uses only the second run, 21 ÷ 60, and throws away the first run instead. And writing down 0.375 averages the two runs' separate rates, (0.40 + 0.35) ÷ 2, which treats a run of 40 spins and a run of 60 spins as equally weighted, when pooling the actual counts gives the larger run its fair share of influence.
- (a) 3/13 — Vegetables are 9 of the 9 + 4 = 13 parts, so vegetables are 9/13 of the plot. Potatoes are a third of the vegetable section, so potatoes are 1/3 of 9/13, which is 9/39, simplifying to 3/13, of the whole plot. 9/13 comes from stopping after finding the fraction of the plot that is vegetables, without taking the further third for potatoes. 1/3 gives the fraction of the vegetable section that is potatoes, not the fraction of the whole plot. 4/39 comes from taking a third of the flowers' fraction, 4/13, instead of the vegetables' fraction.
- (c) x = 0 or x = −7 — Factorising: x² + 7x = x(x + 7) = 0, so x = 0 or x + 7 = 0, giving x = 0 or x = −7. A candidate who divides both sides of the original equation by x, which loses the solution x = 0, gets only x = −7. A candidate who makes a sign error solving x + 7 = 0 gets x = 0 or x = 7. A candidate who misreads the coefficient and doubles it gets x = 0 or x = −14.
- (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) 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)(x + 2) — We need two numbers that multiply to −10 and add to −3: these are −5 and 2, since −5 × 2 = −10 and −5 + 2 = −3. So x² − 3x − 10 = (x − 5)(x + 2). A candidate who swaps the signs, using +5 and −2, gets (x + 5)(x − 2), which expands to x² + 3x − 10 — the wrong middle term. A candidate who picks the factor pair 1 and 10 instead of 2 and 5 gets (x − 10)(x + 1), which expands to x² − 9x − 10. A candidate who makes both factors negative gets (x − 5)(x − 2), which expands to x² − 7x + 10 — the wrong sign on the constant term.
- (d) 3/2 — Work out the number of altos: 40% of 30 = 12. The rest are sopranos, so there are 30 − 12 = 18 sopranos. Form the fraction 18/12; both numbers share a factor of 6, so 18 ÷ 6 = 3 and 12 ÷ 6 = 2, giving 3/2. 2/3 comes from writing the fraction the wrong way round, as altos over sopranos (12/18). 3/5 comes from comparing the sopranos with the whole choir (18/30), instead of with the altos. 7/3 comes from miscalculating 30 − 12 as 28 instead of 18, then forming 28/12.
- (b) 12 cm — Method: the area of a rectangle is one side multiplied by the other, so when the area and one side are known the other side is found by reversing that multiplication — divide the area by the side that is known. Working: 96 ÷ 8 = 12. Answer: 12 cm. The distractors: 88 cm comes from 96 − 8, subtracting the known side as though the area had been made by adding the two sides together; 768 cm comes from 96 × 8, running the area rule forwards on the two numbers given instead of reversing it; 40 cm comes from reading the 96 as a perimeter — halving it to 48 and taking the 8 cm side away — which reverses the perimeter rule rather than the area rule.
- (d) 30° and 150° — The first solution is x = 30°, since sin 30° = 0.5. The graph of y = sin x is symmetrical about x = 90° between 0° and 180°, so the second solution is 180° − 30° = 150°. Adding 180° to the first solution instead of subtracting it from 180° gives 30° and 210°, but sin 210° = −0.5, not 0.5. Reflecting the first solution about x = 90° by adding 30° to 90° instead of subtracting from 180° gives 30° and 120°, but sin 120° = √3/2, not 0.5. Misremembering the standard value and using sin 45° = 0.5 instead of sin 30° = 0.5 gives 45° and 135°, but sin 45° = √2/2, not 0.5.
- (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.
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.