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 an estimate for 2.9² + 3.1², by rounding each number to the nearest whole number.
- 2.Solve 4(x − 3) = 2x + 6.
- 3.An oven is preheating. Its temperature, T °C, is plotted against time, t minutes, on a straight-line graph. The line passes through the points (2, 60) and (6, 140). Work out the gradient of the line.
- 4.A shop sells a scale model of a van. The model and the real van are similar shapes, with lengths in the ratio 1 : 24. The model is 18.5 cm long. Work out the length of the real van, giving your answer in metres.
- 5.A spinner is divided into sectors of 144°, 90° and 126°, coloured purple, orange and grey in that order. The spinner is spun 300 times. Work out how many times you would expect it to land on purple.
- 6.In a histogram of the distances, d metres, thrown by some athletes, the bar covering 20 ≤ d < 60 has a constant frequency density of 1.8 per metre. Estimate the number of throws of at least 20 metres but less than 35 metres.
- 7.√700 lies between which two consecutive integers?
- 8.Simplify (5x² + 3x − 2) − (2x² − x + 5)
- 9.A tangent to a curve at the point where x = 3 passes through the points (1, 4) and (5, 20). Work out the gradient of the tangent at x = 3, and hence the instantaneous rate of change of y with respect to x at that point.
- 10.Two similar bottles have heights in the ratio 1 : 2. Write down the ratio of their volumes.
- 11.At a fun run, a raffle stall charges £1.50 per ticket. The probability that any one ticket wins a prize worth £8 is 0.12, and a losing ticket wins nothing. Nadia buys 25 tickets. Work out how much money Nadia should expect to lose in total.
- 12.Work out the value of .
- 13.The line y = 2x + 7 and the circle x² + y² = 4 are given. By finding the discriminant of the resulting quadratic, without solving it fully, work out how many points the line and the circle intersect at.y = 2x + 7
- 14.To make orange paint, red and yellow paint are mixed in the ratio 6:15. Write the amount of red paint as a fraction of the amount of yellow paint, in its simplest form.
- 15.Triangle ABC is extended so that side BC is extended beyond C to a point D, forming exterior angle ACD. Angle BAC = 48° and angle ABC = 67°. Which single angle fact gives angle ACD = 115° in one step, without first working out the interior angle at C?
- 16.Which expression is equivalent to 0.5(4x + 6) − x?
- 17.A tangent to a curve at the point where x = 3 passes through the points (1, 2) and (5, 14). A candidate says: "The gradient of the tangent at x = 3 is 3, because 14 − 2 = 12 and 12 ÷ 4 = 3." Which statement is correct?
- 18.A student is proving that opposite angles of a cyclic quadrilateral ABCD sum to 180°, letting a = angle DAB and c = angle BCD, and using the fact that the angle at the centre is twice the angle at the circumference. She has already written: 'By the angle at the centre theorem, the reflex angle BOD equals 2a and the non-reflex angle BOD equals 2c.' Which statement must come immediately after this one in a correct proof?
- 19.A straight line has equation y = 4 − 3x. Work out the gradient of the line.
- 20.The nth term of a sequence is 2n² + 1. Work out the 4th term of the sequence.
Answer key
- (a) 18 — Method: round each number to the nearest whole number, then square each rounded number and add the results. Working: 2.9 rounds to 3 and 3.1 rounds to 3, so the estimate is 3² + 3² = 9 + 9. Answer: 18. The distractors: 36 comes from adding before squaring, working out (3 + 3)² instead of 3² + 3²; 12 comes from doubling each rounded number instead of squaring it, adding 6 and 6; 6 comes from adding the two rounded numbers and forgetting to square them at all.
- (d) 9 — Method: expand the brackets fully first, then collect the x-terms and constants before dividing. Working: 4(x − 3) = 4x − 12, so 4x − 12 = 2x + 6; 4x − 2x = 6 + 12; 2x = 18; x = 9. Answer: x = 9. 4.5 comes from expanding only the x-term in the bracket and forgetting to multiply the 3, giving 4x − 3 = 2x + 6, then 2x = 9, x = 4.5. −3 comes from a sign error when expanding, giving 4x + 12 = 2x + 6, then 2x = −6, x = −3. 3 comes from a sign error collecting the x-terms, adding instead of subtracting: 4x + 2x = 6 + 12, so 6x = 18, x = 3.
- (d) 20 — Gradient = change in T ÷ change in t = (140 − 60) ÷ (6 − 2) = 80 ÷ 4 = 20. A student who subtracts in the wrong order gets −20. A student who divides 80 by 2 instead of 4 gets 40. A student who wrongly treats the line as passing through the origin and uses the point (2, 60) on its own gets 60 ÷ 2 = 30.
- (d) 4.44 m — The real van's length is 18.5 × 24 = 444 cm, which converts to 4.44 m. '0.77 m' comes from dividing by 24 instead of multiplying, using the ratio the wrong way round (18.5 ÷ 24 = 0.77 cm), and then writing that figure down as metres. '44.4 m' comes from converting 444 cm to metres with the decimal point in the wrong place. '444 m' comes from working out 444 cm correctly but forgetting to convert it into metres at all.
- (b) 120 — Purple's probability is its angle out of the full circle: 144° ÷ 360° = 2/5. Expected number on purple = 2/5 × 300 = 120. Assuming the three colours are equally likely, ignoring their different angles, gives 300 ÷ 3 = 100. Using orange's angle, 90°, instead of purple's 144° gives a probability of 1/4, so 300 × 1/4 = 75. Treating 144 as a percentage instead of finding its fraction of 360° gives 300 × 0.144 = 43.2, which rounds to 43.
- (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.
- (d) 26 and 27 — Find the two consecutive perfect squares either side of 700: 26² = 676 and 27² = 729. Since 676 < 700 < 729, √700 lies between 26 and 27. Answering 7 and 8 comes from stripping the two zeros off 700 and using the 7 itself as the size of the root, instead of comparing 700 with the perfect squares around it — dividing the number under the root by 100 divides the root by 10, so the digits do not simply carry across. Answering 25 and 26 comes from checking 25² = 625, seeing that it is less than 700, and stopping there without also checking the square directly above it. Answering 35 and 36 comes from halving 700 to 350 and then treating that halved value as if it were ten times the true root, drifting into the thirties instead of the twenties.
- (d) 3x² + 4x − 7 — Method: the minus sign in front of the second bracket changes the sign of every term inside it; then collect like terms. Working: removing the brackets gives 5x² + 3x − 2 − 2x² + x − 5; the squared terms give 5x² − 2x² = 3x², the x terms give 3x + x = 4x, and the number terms give −2 − 5 = −7. Answer: 3x² + 4x − 7. The distractors: 7x² + 2x + 3 comes from adding the two brackets instead of subtracting, giving 5x² + 2x², 3x − x and −2 + 5; 3x² + 2x + 3 comes from applying the minus sign to 2x² only, leaving −x and +5 unchanged so that 3x − x = 2x and −2 + 5 = 3; 3x² + 4x + 3 comes from changing the signs of the terms with letters but leaving +5 as it stood, so the number terms give −2 + 5 = 3.
- (c) 4 — The gradient of a tangent at a point equals the instantaneous rate of change of y with respect to x at that point. A straight line's gradient is found from the change in y divided by the change in x between two points on it. Here the tangent passes through (1, 4) and (5, 20), so the change in y is 20 − 4 = 16 and the change in x is 5 − 1 = 4. The gradient is 16 ÷ 4 = 4. Stopping after finding the change in y, 16, without dividing it by the change in x, is not a gradient at all — that rise happened across 4 units of x, not 1. Adding the two changes instead of dividing one by the other gives 16 + 4 = 20, which is not how a gradient is found. Subtracting in the wrong order, (4 − 20) ÷ (5 − 1), gives −16 ÷ 4 = −4, the wrong sign. The instantaneous rate of change of y with respect to x at x = 3 is 4.
- (a) 1 : 8 — Method: in similar solids the ratio of the volumes is the cube of the ratio of corresponding lengths. Working: the heights are in the ratio 1 : 2, so the volumes are in the ratio 1³ : 2³ = 1 : 8. Answer: 1 : 8. The distractors: 1 : 2 leaves the height ratio unchanged, as though a volume scaled in the same way as a length; 1 : 4 comes from squaring, 1² : 2², which is the rule for surface areas and not for volumes; 1 : 6 comes from multiplying each part by 3 instead of raising each part to the power 3.
- (b) £13.50 — The total cost of Nadia's 25 tickets is 25 × £1.50 = £37.50. The expected number of winning tickets is 25 × 0.12 = 3, so the expected prize money is 3 × £8 = £24.00. Nadia's expected loss is the cost minus the expected prize money: £37.50 − £24.00 = £13.50. A candidate who answers £24.00 has given the expected prize money and mistaken it for the loss. A candidate who answers £37.50 has given the total cost of the tickets, forgetting to subtract the expected prize money. A candidate who answers £34.50 has subtracted the expected number of wins, 3, from the cost instead of first converting it to prize money by multiplying by £8.
- (c) 0.001 — Method: a negative index means 'one over' the positive power, so $10^{-3}$ means one over $10^{3}$. Working: ten cubed is 1000, and one over 1000 is 0.001. 1000 comes from ignoring the negative sign and working out ten cubed instead of its reciprocal. −1000 comes from ignoring what the negative index does to the size, while still writing a negative sign on the large value. −0.001 comes from correctly finding the size, 0.001, but wrongly keeping a negative sign because the index was negative. Answer: 0.001.
- (a) 0 — the line does not intersect the circle — Substitute y = 2x + 7 into x² + y² = 4: x² + (2x + 7)² = 4, which expands to x² + 4x² + 28x + 49 = 4, giving 5x² + 28x + 45 = 0. The discriminant is b² − 4ac = 28² − 4 × 5 × 45. Since 28² = 784 and 4 × 5 × 45 = 900, the discriminant is 784 − 900 = −116. Since the discriminant is negative, the quadratic has no real solutions, so the line does not meet the circle at all: 0 intersection points. Distractor routes: "1 — the line is a tangent" confuses a negative discriminant with a zero one; a discriminant of exactly zero gives one point, a tangent, but −116 is not zero. "2 — the line crosses the circle at two points" assumes a positive discriminant without actually working it out. "It cannot be found without solving the quadratic" is the whole point the discriminant exists to avoid — its sign alone, without finding x, tells you the number of real solutions.
- (b) 2/5 — The ratio red : yellow is 6:15, so write red over yellow: 6/15. Divide both numbers by their highest common factor, 3: 6÷3 = 2, 15÷3 = 5, giving 2/5. (5/2 comes from writing the ratio the wrong way round, yellow over red, 15/6, which simplifies to 5/2. 2/7 comes from comparing the red paint to the total amount of paint, 6 parts out of 21. 5/7 comes from comparing the yellow paint to the total amount of paint, 15 parts out of 21.)
- (c) exterior angle = sum of the two interior opposite angles — Angle ACD is the exterior angle at C, and the exterior angle of a triangle is always equal to the sum of the two interior angles at the other two vertices — here, angle BAC and angle ABC — so angle ACD = 48° + 67° = 115°. "angles in a triangle add up to 180°" is a true fact, but on its own it only gives the INTERIOR angle at C (180 − 48 − 67 = 65°), not the exterior angle 115° — reaching 115° that way still needs a further step. "angles on a straight line add up to 180°" is the fact that links the interior and exterior angles at C to each other, not the fact that gives 115° directly from the two OTHER angles. "exterior angle is always double the smallest angle" is not a genuine angle fact: double the smallest angle here is 48° × 2 = 96°, not 115°.
- (c) x + 3 — Expand the bracket: 0.5(4x + 6) = 2x + 3. Then subtract the x: 2x + 3 − x = x + 3. The option 2x + 3 comes from expanding the bracket correctly but then forgetting to subtract the x at all. The option x + 6 comes from forgetting to multiply the 6 inside the bracket by 0.5 (treating it as 2x + 6), then subtracting x. The option 3x + 3 comes from adding the x instead of subtracting it: 2x + 3 + x = 3x + 3.
- (a) The gradient is 3; the candidate's method is right. — The gradient of a tangent, like any straight line, is the change in y divided by the change in x between two points on it. Here the tangent passes through (1, 2) and (5, 14), so the change in y is 14 − 2 = 12 and the change in x is 5 − 1 = 4. The gradient is 12 ÷ 4 = 3, so the candidate's calculation is correct. Subtracting in the wrong order, (2 − 14) ÷ (5 − 1), gives −12 ÷ 4 = −3, the wrong sign. Adding the two changes instead of dividing them, 12 + 4 = 16, does not find a gradient at all. Dividing the change in x by the change in y instead of the other way round, 4 ÷ 12 = 1/3, inverts the calculation completely. Before accepting or rejecting a claimed gradient, always redo the calculation yourself in the same order — change in y over change in x — rather than trusting the arithmetic as given.
- (a) Reflex + non-reflex angle BOD = 360°, so 2a + 2c = 360°. — A correct proof must build in a strictly logical order: the reflex and non-reflex angles at O, which together make a complete turn about the centre, must be added to 360° before 2a and 2c can be combined and divided. The statement 'Reflex + non-reflex angle BOD = 360°, so 2a + 2c = 360°' is the only one that follows directly from having 2a and 2c already established, and it is exactly what is needed before the final division step. Jumping straight to 'Divide by 2 throughout to get a + c = 180°' skips the step that justifies why 2a + 2c equals 360° in the first place — there is nothing yet to divide. Repeating 'Angle BOD is 2a and 2c, by the centre theorem' does not move the proof forward at all, since that fact has already been established in the sentence given. Introducing 'OB = OD are radii, so triangle OBD is isosceles' brings in an unrelated triangle and an unrelated method that plays no part in this particular proof.
- (d) −3 — Method: rewrite the equation in the form y = mx + c, then read off the gradient. Working: y = 4 − 3x can be written as y = −3x + 4, so comparing with y = mx + c gives m = −3. Answer: the gradient is −3. The value 3 comes from ignoring the negative sign on the x term. The value 4 comes from reading off the y-intercept instead of the gradient. The value −4 comes from a sign error, applying the negative sign to the intercept instead of the gradient.
- (a) 33 — Substitute n=4 into 2n²+1: 2×4²+1=2×16+1=33. A candidate who computes n² as 2×n instead of n×n would compute 2×(2×4)+1=2×8+1=17. A candidate who correctly finds 2×16 but forgets to add the constant 1 would stop at 32. A candidate who squares the whole term 2n, rather than squaring n before multiplying by 2, would compute (2×4)²+1=64+1=65.
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.