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) 360 — The inverse of ÷ 15 is × 15, so the missing number is 24 × 15 = 360. Subtracting instead of multiplying gives 24 − 15 = 9. Dividing by 15 again instead of multiplying gives 24 ÷ 15 = 1.6. Adding instead of multiplying gives 24 + 15 = 39.
- (a) y = −3x − 5 — Gradient = (−8 − 4) ÷ (1 − (−3)) = −12 ÷ 4 = −3. Using the point (1, −8): −8 = −3(1) + c, so c = −5, giving y = −3x − 5. A candidate who drops the negative sign on the gradient, using m = 3 instead, would then solve −8 = 3(1) + c to get c = −11, writing y = 3x − 11. A candidate who makes a sign error isolating c, writing c = 5 instead of −5, would write y = −3x + 5. A candidate who mixes up both mistakes — keeping the correct gradient but the wrong, positive value of c from the flipped-gradient calculation — would write y = −3x + 11.
- (c) 3 hours — Method: inverse proportion means speed × time is constant for the journey, so find that constant and divide it by the new speed. Working: 60 × 2 = 120, which is the distance in kilometres; at 40 km/h the time is 120 ÷ 40 = 3 hours. Answer: 3 hours. The distractors: 1.5 hours is the ratio of the speeds, 60 ÷ 40, given as a time instead of being used to scale the original 2 hours; 1 hour 20 minutes comes from treating time as directly proportional to speed, 2 × 40 ÷ 60, which has the slower train arriving sooner; 2 hours comes from finding the constant 120 and then dividing it by the original 60 km/h again, so the time never changes.
- (d) 1.2 m² — Method: an area in square metres needs lengths in metres, so convert first and then multiply. Working: 100 cm = 1 m, so 150 cm = 1.5 m and 80 cm = 0.8 m, and the area = 1.5 × 0.8 = 1.2 m². Answer: 1.2 m². The same result comes from working in centimetres: 150 × 80 = 12 000 cm², and a square metre is a square of side 100 cm, so 100 × 100 = 10 000 cm² make one square metre and 12 000 ÷ 10 000 = 1.2. Dividing the 12 000 cm² by 100 instead, as though a square metre held only 100 square centimetres, gives 120 m²; dividing by 1000 gives 12 m². Working out the perimeter rather than the area gives 1.5 + 0.8 + 1.5 + 0.8 = 4.6, which is a length and not an area.
- (a) 3/8 — There are 4 × 6 = 24 equally likely outcomes. The outcomes with no 3 at all have Spinner E showing 1, 2 or 4 and Spinner F showing 1, 2, 4, 5 or 6, giving 3 × 5 = 15 outcomes. So the outcomes with at least one 3 are 24 − 15 = 9, and the probability is 9/24 = 3/8. Choosing 5/12 comes from adding the two individual probabilities of a 3, 1/4 + 1/6, which counts the outcome where both spinners show 3 twice over. Choosing 1/4 comes from only counting the case where Spinner E shows 3, and forgetting the outcomes where Spinner F shows 3 instead. Choosing 5/8 comes from working out the probability of getting no 3 at all, 15/24 = 5/8, and giving that as the final answer instead of subtracting it from 1.
- (c) 28 — Method: multiply the number of whole symbols by the value of one symbol, then add the value of any half symbol shown. Working: 3 whole symbols represent 3 × 8 = 24 cars. The half symbol represents 4 cars. Total cars sold in March = 24 + 4 = 28. Leaving out the half symbol, 3 × 8 = 24, undercounts by exactly the value of that half symbol. Treating the half symbol as if it were a full symbol, 4 × 8 = 32, overcounts because it doubles the value the half symbol is worth. Giving 3.5 reports the number of symbols shown, not the number of cars they represent — the key still needs to be applied. Always apply the key to every symbol shown, including a half symbol, rather than reading off the symbol count itself.
- (d) 4/3 — Method: the product of two negative numbers is positive, so work with 2/5 × 10/3 and then simplify. Multiply the numerators together and the denominators together. Working: 2 × 10 = 20 and 5 × 3 = 15, giving 20/15; both 20 and 15 divide by 5, so 20/15 = 4/3. Answer: 4/3. The distractors: −4/3 has the arithmetic right but keeps a minus sign, from treating negative × negative as negative; 3/25 comes from turning the second fraction upside down and multiplying, which divides instead of multiplying and gives 2/5 × 3/10 = 6/50; −56/15 comes from adding the two fractions instead of multiplying them, giving −6/15 − 50/15.
- (b) y = 2x − 1 — Method: in a rule that multiplies and then adds, the multiplier is the step in the outputs for each step of 1 in the input, and the number added on is the output when the input is 0. Working: the inputs 0, 1, 2 rise in ones while the outputs −1, 1, 3 rise by 2 each time, so the input is multiplied by 2; an input of 0 gives 2 × 0 = 0 and the output must be −1, so 1 is subtracted. Answer: y = 2x − 1, checked against the last pair by 2 × 2 − 1 = 3. The distractors: y = 2x + 1 comes from finding the multiplier 2 correctly and then reading the output at an input of 0 as +1 instead of −1; y = x − 1 comes from taking the multiplier as 1 because the inputs go up in ones, instead of using the step in the outputs; y = 3x − 1 comes from reading the largest output, 3, as the multiplier.
- (a) 4:3:1 — Convert every part to the same unit: 2 m = 200 cm, so the ratio is 200 : 150 : 50. Dividing all three parts by 50 gives 4 : 3 : 1. Writing 2 : 150 : 50 has not converted 2 m into centimetres, so the units do not match. Writing 3 : 4 : 1 has the first two parts the wrong way round. Writing 4 : 3 : 2 comes from an arithmetic slip on the last part: 50 ÷ 50 = 1, not 2.
- (c) a + (1/2)c — Method: OABC is a parallelogram, so OB = OA + AB, and since AB is equal and parallel to OC, AB = c; this gives OB = a + c. M is the midpoint of AB, so AM = (1/2)AB = (1/2)c. Working: OM = OA + AM = a + (1/2)c. Answer: OM = a + (1/2)c. Adding the whole of AB instead of half of it gives a + c, which is OB, not OM; halving the whole diagonal OB instead of just AB gives (1/2)a + (1/2)c, the midpoint of OB rather than of AB; and flipping the sign on the c-term gives a − (1/2)c, which points back the wrong way along AB. Halve only the side you are told to halve, and check the sign before you commit to an answer.
- (d) 1/15 — The probability that the first bead is red is 3/10. Since the first bead is not put back, there are now only 2 red beads left out of 9 beads in total, so the probability that the second bead is also red is 2/9. Multiplying these, 3/10 × 2/9 = 6/90 = 1/15. A candidate who answers 9/100 has treated the beads as replaced, using 3/10 twice. A candidate who answers 3/50 has correctly reduced the red count to 2 for the second pick but forgotten that the total also falls to 9, using 2/10 instead. A candidate who answers 5/19 has added the numerators and added the denominators, (3+2)/(10+9), instead of multiplying.
- (b) 3²⁰ is larger — Method: two powers with different bases and different indices can be compared once they are rewritten with a common index, which is possible whenever the indices share a factor. Working: 30 and 20 have a highest common factor of 10, so 2³⁰ = (2³)¹⁰ = 8¹⁰ and 3²⁰ = (3²)¹⁰ = 9¹⁰. Both are now tenth powers, and since 9 is larger than 8, 9¹⁰ is larger than 8¹⁰. Answer: 3²⁰ is larger. The distractors: 2³⁰ is larger comes from comparing only the indices and choosing the power with the bigger index; They are equal comes from multiplying base by index, 2 × 30 and 3 × 20, and finding 60 each time; They cannot be compared without a calculator comes from assuming that powers this large can only be ranked by evaluating them in full.
- (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.
- (c) 3 : 4 — Method: a fraction compares a part with the whole, while this ratio compares one part with the other part, so find the fraction that is black before writing the ratio. Working: if 3/7 are blue then the black pens make up 7/7 − 3/7 = 4/7 of the box, so out of every 7 pens 3 are blue and 4 are black, and blue : black = 3 : 4. Answer: 3 : 4. The distractors: 3 : 7 comes from reading the numerator and the denominator of 3/7 straight off as the two parts, which compares the blue pens with the whole box rather than with the black pens; 4 : 3 comes from writing the black pens before the blue pens, reversing the order asked for; 4 : 7 is the same numerator-and-denominator reading applied to the black fraction 4/7, again comparing a part with the whole box.
- (b) 360 cm² — Method: the total surface area is the square base plus the four triangular faces, and the height used for a triangular face is the slant height of 13 cm, not the vertical height of 12 cm. Working: the base is 10 × 10 = 100 cm²; one triangular face is (10 × 13) ÷ 2 = 65 cm², so four faces give 4 × 65 = 260 cm²; the total is 100 + 260 = 360. Answer: 360 cm². The distractors: 260 cm² comes from adding the four triangular faces and leaving out the base; 340 cm² comes from using the vertical height of 12 cm as the height of each triangle, 100 + 4 × 60; 620 cm² comes from working out each face as 10 × 13 without halving, 100 + 4 × 130.
- (b) y = −(1/3)x + 2 — Method: two lines are perpendicular when the product of their gradients is −1, so the gradient of a line perpendicular to one of gradient m is the negative reciprocal, −1 divided by m. Working: reading m from y = 3x + 1 gives a gradient of 3, so the perpendicular gradient is −1 ÷ 3 = −1/3, and the line carrying that gradient is y = −(1/3)x + 2; the check 3 × (−1/3) = −1 confirms it. Answer: y = −(1/3)x + 2. The distractors: y = 3x − 4 comes from using an equal gradient, which is the test for parallel lines rather than perpendicular ones; y = (1/3)x + 2 comes from turning the gradient upside down but leaving out the change of sign, giving a product of 1 instead of −1; y = −3x + 2 comes from changing the sign of the gradient without turning it upside down, giving a product of −9.
- (b) £60 — £48 represents 100% − 20% = 80% of the original price. 1% = £48 ÷ 80 = £0.60, so 100% = £0.60 × 100 = £60.
- (d) 77 — Method: angles that meet at a point add up to 360°. Working: 82 + 105 + 96 = 283; 360 − 283 = 77. Answer: 77°. A candidate who gives the sum of the three known angles and forgets to subtract it from 360° gets 283. A candidate who leaves out the 105° angle, working out 360 − 82 − 96, gets 182. A candidate who leaves out the 82° angle, working out 360 − 105 − 96, gets 159.
- (d) An inequality, because ≤ compares the two sides — The symbol ≤ means 'is less than or equal to', so the statement compares the sizes of the two sides instead of saying they are equal: that makes it an inequality. Solving it gives n ≤ 5, a whole range of values rather than the single value an equation would give. An identity has to be true for every value of the letter, and this fails at n = 6, so it is not one. A formula works one quantity out from another, and there is only one letter here.
- (a) x = 4, y = −3 — Method: both equations contain +y with the same coefficient, so subtracting one equation from the other removes y. Working: (2x + y) − (x + y) = 5 − 1 gives x = 4; substituting x = 4 into x + y = 1 gives 4 + y = 1, so y = −3. Answer: x = 4, y = −3, which also satisfies 8 − 3 = 5. The distractors: x = 4, y = 5 comes from rearranging x + y = 1 as y = 1 + x; x = −2, y = 3 comes from eliminating x by doubling the first equation and then reading −y = 3 as y = 3; x = 6, y = −5 comes from adding the constants instead of subtracting them while eliminating y, taking x as 5 + 1.
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.