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
GCSE Higher sample Paper 1 (non-calculator)
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- 1.Work out 2 × 3 × 5 + 1 and decide whether the result is a prime number.
- 2.A plumber charges a call-out fee plus an hourly rate. The total cost, y in pounds, of a job lasting x hours is given by y = 45x + 60. Work out the total cost of a job that lasts 3 hours.y = 45x + 60
- 3.Write 350 ml : 1.4 l as a ratio in its simplest form.
- 4.Vector A is and vector B is . Write down how vector B compares to vector A.
- 5.A red fair dice and a blue fair dice are rolled together. Every outcome is listed as an ordered pair, red score then blue score. Work out the probability that the red score is less than the blue score.
- 6.Work out the median of these six numbers: 13, 21, 22, 36, 37, 47
- 7.Work out the value of 2⁻⁴
- 8.Isla says that the lines y = 2x + 1 and y = −2x + 3 are parallel. Write down the statement that gives the correct answer for the correct reason.y = 2x + 1y = -2x + 3
- 9.A mortar mix is made from sand and cement in the ratio 5 : 3. Write the ratio of sand to the total mix in its simplest form.
- 10.A tent has the cross-section of a right-angled triangle. The sloping side of the cross-section is 10 m long, and it makes an angle of 60° with the horizontal ground. Using the exact value of cos 60°, work out the horizontal distance covered by the sloping side.
- 11.Of the 70 students on the Year 10 geography trip, 42 are girls and the rest are boys. 24 of the girls and 10 of the boys brought a packed lunch. A student who did not bring a packed lunch is chosen at random. Work out the probability that this student is a girl.
- 12.Work out (−6) + (−4) × 3
- 13.f(x) = 2x − 1. Work out ff(x).y = 2x − 1
- 14.A metal alloy is made from copper and tin in the ratio 7:3. Work out the mass of tin in 250 g of the alloy.
- 15.Points A, B, C and D lie on a circle with centre O, in that order around the circle, and AC is a diameter. Which of the following circle facts does NOT apply to this diagram?
- 16.Which expression means 'y squared, multiplied by 3'?
- 17.A shop's profit, P pounds, is in direct proportion to the number of items sold, n. When 15 items are sold, the profit is £45. Write down the ratio n : P in its simplest form for this shop.
- 18.A tent's sloping side makes an angle of 45° with the ground. The sloping side is 3√2 m long. Using the exact value of sin 45°, work out the exact height of the tent.
- 19.Solve the simultaneous equations y = x + 1 and x² + y² = 25, giving both pairs of solutions.y = x + 1
- 20.Factorise x² − 64.
Answer key
- (d) 31, which is prime — Method: work out the value, remembering that multiplication comes before addition, then test it for primality by dividing by each prime up to its square root. Working: 2 × 3 × 5 = 30, so the value is 30 + 1 = 31. Since 6² = 36 is larger than 31, only 2, 3 and 5 need testing: 31 is odd, 31 ÷ 3 leaves a remainder of 1, and 31 does not end in 0 or 5. It therefore has exactly two factors, 1 and itself. Answer: 31, which is prime. The distractors: 30, which is not prime comes from working out 2 × 3 × 5 and forgetting to add the 1; the claim that 31 = 1 × 31 makes it non-prime comes from treating any factor pair as proof, forgetting that a prime is allowed the pair 1 and itself; the claim that 31 is a multiple of 3 comes from assuming that a number containing the digit 3 divides by 3, when in fact 31 ÷ 3 leaves a remainder.
- (d) £195 — Substituting x = 3 into y = 45x + 60 gives y = 45 × 3 + 60 = 135 + 60 = 195. A candidate who forgets to add the call-out fee would get only 45 × 3 = £135. A candidate who adds the hours to the fee and the rate instead of multiplying would get 45 + 60 + 3 = £108. A candidate who multiplies both the hourly rate and the call-out fee by the number of hours would get 45 × 3 + 60 × 3 = £315.
- (d) 1:4 — Convert 1.4 l to millilitres: 1.4 l = 1400 ml. The ratio is 350 : 1400. Divide both parts by 350: 350 ÷ 350 = 1 and 1400 ÷ 350 = 4, giving 1 : 4. Misreading 1.4 l as 14 (moving the decimal point) gives 350 : 14, which simplifies to 25 : 1 — a very different, implausible ratio. Dividing by 175 instead of 350 gives 2 : 8, which still shares a common factor of 2, so it is not fully simplified. Swapping the order gives 4 : 1, litres to millilitres the wrong way round.
- (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.
- (b) 5/12 — There are 36 equally likely ordered pairs. Listing the pairs where the red score is smaller than the blue score gives 15 outcomes, so the probability is 15/36 = 5/12. Choosing 1/2 comes from assuming the 36 outcomes split evenly between 'red smaller' and 'red bigger', which ignores the 6 outcomes where the two dice show the same score. Choosing 7/12 comes from including the outcomes where the two scores are equal, 15 + 6 = 21, giving 21/36 = 7/12, instead of using 'less than' strictly. Choosing 5/36 comes from counting only the outcomes where the two scores differ by exactly 1, such as (1, 2) and (2, 3), which is 5 outcomes, and missing the pairs that differ by more than 1.
- (d) 29 — Method: with an even number of values there is no single middle value, so the median is the mean of the two values either side of the middle. Working: the six numbers are already in order and 6 ÷ 2 = 3, so the middle pair are the third and fourth values, 22 and 36; their mean is (22 + 36) ÷ 2 = 58 ÷ 2 = 29. Answer: 29, which lies between the two middle values as a median of an even data set must. The distractors: 22 comes from taking the lower of the two middle values and stopping there instead of averaging the pair; 36 comes from taking the larger value of that pair because it sits just past the halfway point of the list; 34 comes from working out the range, 47 − 13, instead of a measure of centre.
- (d) 1/16 — Method: a negative index means the reciprocal of the power, so 2⁻⁴ is 1 divided by 2⁴. Working: 2⁴ = 2 × 2 × 2 × 2 = 16, so the value is 1/16. Answer: 1/16. The distractors: −16 comes from reading the negative index as a minus sign on the result; 16 comes from ignoring the minus sign and working out 2⁴; 1/8 comes from multiplying the base by the index, 2 × 4, and writing 1 over that product.
- (b) No, because their gradients are 2 and −2 — Method: two lines are parallel exactly when their gradients are equal as signed numbers, so m is read from each equation written in the form y = mx + c and the two are compared. Working: y = 2x + 1 has gradient 2 and y = −2x + 3 has gradient −2; those are not equal, so the lines are not parallel, and indeed one slopes upwards while the other slopes downwards. Answer: No, because their gradients are 2 and −2. The distractors: saying yes because both gradients have size 2 comes from comparing the sizes of the gradients and ignoring their signs; saying yes because the gradients add to 0 comes from using a sum of zero as the test for parallel lines instead of equality of gradients; saying no because the y-intercepts are 1 and 3 reaches the right verdict by the wrong route, since the intercepts decide where the lines sit rather than whether they are parallel.
- (b) 5:8 — The mix has 5 parts sand and 3 parts cement, so 5 + 3 = 8 parts in total. Sand to total is 5 : 8, and since the highest common factor of 5 and 8 is 1, this is already in its simplest form. Giving 5 : 3 answers sand to cement, not sand to the total mix. Giving 3 : 8 is cement to total, the wrong part of the mix. Giving 8 : 5 has the total and the sand swapped round.
- (c) 5 m — The horizontal distance is the side adjacent to the 60° angle, and the sloping side is the hypotenuse, so horizontal distance = hypotenuse × cos 60°. The exact value of cos 60° is 1/2, so horizontal distance = 10 × 1/2 = 5 m. Using the sloping side itself as the horizontal distance, without using any trigonometry at all, gives 10 m. Using sin 60° = √3/2 instead of cos 60° finds the vertical height of the tent rather than the horizontal distance: 10 × √3/2 = 5√3 = 8.7 m (1 d.p.). Dividing the sloping side by cos 60° instead of multiplying by it, 10 ÷ 0.5 = 20 m, treats the sloping side as though it were the adjacent side rather than the hypotenuse.
- (c) 1/2 — Method: restrict to the students who did NOT bring a packed lunch, then find what fraction of that group are girls. Girls without lunch = 42 − 24 = 18. Boys = 70 − 42 = 28, so boys without lunch = 28 − 10 = 18. Total without lunch = 18 + 18 = 36. Working: P(girl | no lunch) = 18 ÷ 36 = 1/2. Answer: 1/2. Watch out: dividing 18 by 42 (the total number of girls) instead of by 36 finds P(no lunch | girl), the reverse conditional. Dividing by 70 (the whole trip) ignores that you already know the student did not bring a lunch. And using the 'brought a lunch' numbers (24 out of 34) answers the question for the wrong group entirely — you were asked about the students who did NOT bring one.
- (b) −18 — Method: the multiplication is carried out before the addition, and a negative multiplied by a positive is negative. Working: (−4) × 3 = −12, so the calculation becomes (−6) + (−12) = −18. Answer: −18. The distractors: −30 comes from adding first and multiplying afterwards, giving (−6 + −4) × 3 = −10 × 3 = −30; 6 comes from treating (−4) × 3 as +12 on the grounds that a minus sign makes a product positive, giving −6 + 12 = 6; 18 comes from ignoring both minus signs and working out 6 + 4 × 3 = 18.
- (c) 4x − 3 — ff(x) means f(f(x)): substitute f(x) into f in place of x. f(f(x)) = 2 × f(x) − 1 = 2 × (2x − 1) − 1. Expanding the bracket: 2 × (2x − 1) = 4x − 2. Combining the constant terms: −2 − 1 = −3, so f(f(x)) = 4x − 3. Writing 4x − 2 comes from expanding 2(2x − 1) correctly to get 4x − 2, then forgetting to subtract the outer 1 at all. Writing 4x² − 4x + 1 comes from reading ff(x) as f(x) multiplied by itself, (2x − 1)(2x − 1) = 4x² − 4x + 1, instead of substituting f(x) into f. Writing 4x − 1 comes from doubling the coefficient of x in the original rule directly, without actually substituting f(x) into f at all.
- (b) 75 g — Method: split the total mass into the number of parts shown by the ratio, then find the mass of tin. Working: the ratio 7:3 has 7 + 3 = 10 parts, so one part is 250 ÷ 10 = 25 g, and the mass of tin is 3 × 25 = 75 g. So the alloy contains 75 g of tin. Distractor 175 g is the mass of copper, not tin. Distractor 125 g comes from splitting the alloy into two equal halves, ignoring the ratio. Distractor 25 g is the value of one part, found correctly but never multiplied by 3.
- (c) Tangent-chord angle = angle in the alternate segment. — This diagram has a diameter AC, a centre O, and a cyclic quadrilateral ABCD, but no tangent anywhere in it. 'Angle in a semicircle = 90°' applies directly, because AC is a diameter. 'Opposite angles of a cyclic quadrilateral sum to 180°' applies directly, because ABCD is a cyclic quadrilateral. 'Angle at the centre = twice angle at circumference' applies directly, because O is given as the centre of the circle. The tangent-chord fact relates the angle between a tangent and a chord to an angle elsewhere in the circle, and since this diagram has no tangent, there is nothing in it for that fact to describe — so it is the one theorem that does not apply here.
- (a) 3y² — 'y squared, multiplied by 3' means the square is applied to y only, and the result is then multiplied by 3, written as 3y². Writing y³ mistakes the multiplication by 3 for an extra factor of y, adding to the power instead of using a coefficient. Writing (3y)² squares the whole of 3y, including the 3, which gives 9y² rather than 3y² — the square should apply to y alone. Writing 3 + y² adds the 3 instead of multiplying by it. The expression for 'y squared, multiplied by 3' is 3y².
- (c) 1 : 3 — n : P = 15 : 45. Dividing both parts by their highest common factor, 15, gives 1 : 3. Inverting the ratio, 3 : 1, swaps profit and number of items. Dividing only the n-part by 15, getting 1, but leaving the P-part as 45 gives 1 : 45 — only one side has been simplified. Dividing only the P-part by 15, getting 3, but leaving the n-part as 15 gives 15 : 3, the opposite partial mistake.
- (a) 3 m — Height = sloping length × sin 45° = 3√2 × √2/2 = (3 × 2)/2 = 3 m, since √2 × √2 = 2. 3√2 m comes from forgetting to multiply by sin 45° at all. 3√2/2 m comes from using sin 30° = 1/2 instead of sin 45° = √2/2. 6 m comes from using √2 instead of √2/2 for sin 45°, dropping the denominator of the exact value: 3√2 × √2 = 6.
- (b) x = −4, y = −3 and x = 3, y = 4 — Substitute y = x + 1 into x² + y² = 25: x² + (x + 1)² = 25, which expands to x² + x² + 2x + 1 = 25, giving 2x² + 2x − 24 = 0, or x² + x − 12 = 0. Factorise: (x + 4)(x − 3) = 0, so x = −4 or x = 3. Using y = x + 1: when x = −4, y = −3; when x = 3, y = 4. Distractor routes: x = 4, y = 5 and x = −3, y = −2 comes from mis-factorising x² + x − 12 as (x − 4)(x + 3), reversing the sign of each root. x = −4, y = −3 alone stops after finding the first root of the quadratic and never goes back for the second pair. x = 12, y = 13 comes from misreading x² + y² = 25 as the linear equation x + y = 25 and solving that together with y = x + 1.
- (d) (x − 8)(x + 8) — x² − 64 = x² − 8², a difference of two squares, which factorises as (x − 8)(x + 8). A candidate who treats it as a perfect square with a repeated negative factor gets (x − 8)(x − 8), which expands to x² − 16x + 64 — wrong on both the middle and constant terms. A candidate who uses a repeated positive factor gets (x + 8)(x + 8), which expands to x² + 16x + 64. A candidate who picks a different factor pair of 64, such as 4 and 16, without checking that the middle term cancels, gets (x − 4)(x + 16), which expands to x² + 12x − 64 — the wrong middle term.
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.