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 the exact value of √(2² + 3²)
- 2.A gym charges a joining fee plus a monthly fee. Anna paid £100 in total after 3 months of membership. Ben paid £160 in total after 6 months of membership (same joining fee and monthly fee as Anna). Work out the monthly fee.
- 3.Two mathematically similar cylinders have heights in the ratio 3 : 4. Write the ratio of their volumes in its simplest form.
- 4.Two triangular offcuts of wood, PQR and STU, are cut for a construction project. PQ = 8 cm, QR = 6 cm and angle PQR = 90°. ST = 8 cm, TU = 6 cm and angle STU = 90°. A carpenter wants to check the two pieces are identical in shape and size before using them as a matching pair. Using only the measurements given, and without working out any further lengths, which condition proves that triangle PQR is congruent to triangle STU?
- 5.An ordinary fair dice is rolled once. Set A is the set of even scores and set B is the set of scores greater than 4. Work out the probability that the score rolled is in set A or in set B or in both.
- 6.A scatter graph shows the number of days, x, that each of 16 tomato plants was watered and its height, y cm. The line of best fit has equation y = 1.5x + 4. Write down what the 1.5 in this equation tells you about the plants.y = 1.5x + 4
- 7.Work out 3/7 × 14/9. Give your answer as a fraction in its simplest form.
- 8.Simplify (2x² + 7x + 3) ÷ (x² − 9), giving your answer as a single fraction.
- 9.Which of these ratios is equivalent to 6 : 10 : 14?
- 10.A tap fills a tank at a rate of 18 litres per minute. The tank holds 0.45 m³. Work out how long the tap takes to fill the tank.
- 11.A bag contains 5 yellow marbles and 3 blue marbles. Noah takes a marble at random, notes its colour and puts it back in the bag. He then takes a second marble at random. Work out the probability that the first marble is yellow and the second marble is blue.
- 12.The two shorter sides of a right-angled triangle are √12 cm and √24 cm. Work out the exact length of the hypotenuse.
- 13.The graph of y = f(x) has x-intercepts at x = −2 and x = 6 and crosses the y-axis at (0, −12). Work out the x-intercepts and the y-intercept of y = −f(x).
- 14.A cleaning solution is made by mixing disinfectant and water in the ratio 1 : 9. Priti wants to make 6 litres of the solution in total. Work out how much disinfectant she needs.
- 15.Each of these has an exact value. Write down the one whose value is the greatest.
- 16.A circular pond has equation x² + y² = 20, with lengths in metres from the centre of the garden. A straight path runs along the line y = 2x, entering the pond and leaving it again. Work out the coordinates of the two points where the path meets the edge of the pond.y = 2x
- 17.Write the ratio 2/3 : 1/2 as a ratio of whole numbers, in its simplest form.
- 18.A tangent to a circle with centre O touches the circle at point P, where OP = 5 cm. Point Q lies on the tangent so that PQ = 12 cm. Using the fact that a tangent is perpendicular to the radius at the point of contact, work out the length OQ.
- 19.Points A(8, 15) and B(−8, 15) both lie on the circle x² + y² = 289, which has centre (0, 0). Work out the length of the chord AB.
- 20.A table of values is being drawn for the graph of y = x³. Work out the value of y when x = −2.y = x
Answer key
- (d) √13 — Method: everything under a root sign is worked out first, because a square root cannot be taken term by term across an addition. Working: 2² = 4 and 3² = 9, so the expression under the root is 4 + 9 = 13. As 13 is not a square number, the exact value is left in root form as √13. Answer: √13. The distractors: 5 comes from rooting each square separately and adding, 2 + 3, which treats the root of a sum of squares as the sum of the numbers; 13 comes from working out the sum under the root correctly and then forgetting to take the root; √5 comes from subtracting the two squares, 9 − 4, instead of adding them.
- (d) £20 — Let f be the joining fee and m the monthly fee: f + 3m = 100 and f + 6m = 160. Subtracting the first equation from the second eliminates f: 3m = 60, so m = 20. A candidate who finds the joining fee instead of the monthly fee would get f = 100 − 3(20) = £40. A candidate who divides Ben's total by his number of months, ignoring that part of the cost is a fixed joining fee, would get 160 ÷ 6 ≈ £26.67. A candidate who divides the difference in cost by the total number of months instead of the difference in months would get (160 − 100) ÷ 9 ≈ £6.67.
- (c) 27 : 64 — For similar solids, the ratio of volumes is the ratio of lengths cubed: 3³ : 4³ = 27 : 64. 3 : 4 comes from using the height ratio itself as the volume ratio, without cubing it at all. 9 : 16 comes from squaring each part instead of cubing (3² : 4²) — squaring is the rule for area, not volume. 27 : 4 comes from cubing only the first part of the ratio (3³ = 27), and leaving the second part uncubed.
- (b) SAS (two sides and the included angle) — PQ = ST and QR = TU are two pairs of matching sides, and the angle between them (angle Q and angle T) is 90° in both triangles — two sides and the angle between them are equal, so the triangles are congruent by SAS, using only the measurements given. RHS needs the hypotenuses to be known equal; here only the two legs are given, so you would first have to work out each hypotenuse by Pythagoras' theorem — extra working the question rules out. SSS has the same problem: the third side of each triangle is not given, only calculable. ASA needs two angles and the side between them, but here it is two sides and the angle between them that are given, not two angles.
- (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.
- (b) On average a plant grew 1.5 cm taller for each extra day — Method: in the equation of a line, the number multiplying x is the gradient, and a gradient states the change in y produced by an increase of 1 in x, read in the units of the two axes. Working: here x is measured in days and y in centimetres, so the gradient 1.5 carries the units centimetres per day. Testing it on the line, 5 days gives 1.5 × 5 + 4 = 11.5 cm and 6 days gives 1.5 × 6 + 4 = 13 cm, a rise of 1.5 cm for the one extra day. Answer: on average a plant grew 1.5 cm taller for each extra day of watering. The distractors: 1.5 cm as the height before any watering is the value of y when x is 0, which is the other number in the equation, 4 cm, so this swaps the gradient and the intercept; 1.5 cm as the gap between the tallest and the shortest plant reads the gradient as a range, when a range is a difference between two of the 16 plants and a gradient is a rate; 1.5 days for each extra centimetre inverts the rate, dividing days by centimetres instead of centimetres by days, and the line gives 1 cm of growth in two thirds of a day.
- (c) 2/3 — Method: multiply the numerators together and the denominators together, then divide both parts of the result by their highest common factor. Working: 3 × 14 = 42 and 7 × 9 = 63, giving 42/63; the highest common factor of 42 and 63 is 21, and 42 ÷ 21 = 2 with 63 ÷ 21 = 3. Answer: 2/3. The distractors: 17/16 comes from adding the numerators and adding the denominators, giving (3 + 14)/(7 + 9); 27/98 comes from turning the second fraction upside down and multiplying, which divides instead of multiplying and gives 3/7 × 9/14; 2/21 comes from cancelling the 7 into the 14 in the numerator but leaving the 7 in the denominator, giving 6/63.
- (b) (2x + 1)/(x − 3) — Factorise both the numerator and the denominator before you cancel anything. The numerator 2x² + 7x + 3 factorises to (2x + 1)(x + 3), and the denominator x² − 9 is a difference of two squares, factorising to (x − 3)(x + 3). The (x + 3) factor is common to both, so it cancels, leaving (2x + 1)/(x − 3). Writing (2x + 1)/(x + 3) comes from factorising x² − 9 as (x + 3)² instead of (x − 3)(x + 3) — a difference of two squares always has one plus and one minus bracket. Writing (2x + 3)/(x − 3) comes from mis-factorising the numerator as (2x + 3)(x + 1) and then wrongly cancelling the (x + 1) against the denominator's (x + 3) as though they were the same bracket. Writing 2x + 1 comes from cancelling the (x + 3) factor correctly but then dropping the remaining (x − 3) on the denominator altogether.
- (c) 9:15:21 — 6 : 10 : 14 simplifies to 3 : 5 : 7 (divide every part by 2). Multiplying every part of 3 : 5 : 7 by 3 gives 9 : 15 : 21, so 9 : 15 : 21 is equivalent to 6 : 10 : 14. Adding 2 to every part of 6 : 10 : 14 gives 8 : 12 : 16, which is not equivalent — ratios are equivalent when every part is multiplied by the same number, not when the same number is added to every part. Doubling only the first two parts, 6 × 2 = 12 and 10 × 2 = 20, but leaving the third part unchanged at 14, gives 12 : 20 : 14 — a scaling applied to two parts and not the third. Cancelling the first two parts correctly, 6 ÷ 2 = 3 and 10 ÷ 2 = 5, then treating the three numbers as a sequence and making the third part the sum of the first two, 3 + 5 = 8, gives 3 : 5 : 8 — the third part was never divided by 2 at all.
- (b) 25 minutes — Method: a rate in litres per minute can only be used on a volume measured in litres, so convert the tank first and then divide. Working: 1 m³ = 1000 litres, so the tank holds 0.45 × 1000 = 450 litres, and the time is 450 ÷ 18 = 25. Answer: 25 minutes. Using 1 m³ = 100 litres gives 45 ÷ 18 = 2.5 minutes. Using 1 m³ = 1 000 000 litres, which is the factor that turns cubic metres into cubic centimetres, gives 450 000 ÷ 18 = 25 000 minutes. Multiplying by the rate instead of dividing by it gives 450 × 18 = 8100.
- (d) 15/64 — Method: because the first marble is put back, the second draw is made from the same 8 marbles, so the two draws are independent and their probabilities are multiplied. Working: the bag holds 8 marbles both times, so yellow first has probability 5/8 and blue second has probability 3/8. Multiplying fractions gives 5 × 3 = 15 on the top and 8 × 8 = 64 on the bottom. Answer: the probability is 15/64. The distractors: 9/64 comes from working out the probability that BOTH marbles are blue, 3/8 × 3/8, instead of one of each colour in the stated order; 15/16 comes from multiplying the numerators but adding the denominators; 5/8 comes from writing down the first draw only and never combining it with the second.
- (d) 6 — By Pythagoras' theorem, the square of the hypotenuse equals the sum of the squares of the other two sides: (√12)² + (√24)² = 12 + 24 = 36. The hypotenuse is √36 = 6 cm. Adding the two side lengths directly instead of squaring them first, treating the theorem as if it were a straight sum of the sides, gives √12 + √24 = 2√3 + 2√6. Multiplying the two squared values, 12 × 24 = 288, instead of adding them, then taking the root, gives √288 = 12√2. Adding the squares correctly to get 36 but forgetting to take the square root at the end leaves 36 as the answer instead of the hypotenuse itself.
- (d) x = −2, x = 6; y-intercept (0, 12) — Reflecting y = f(x) in the x-axis, to get y = −f(x), negates every y-value but leaves every x-value fixed. The x-intercepts happen where y = 0, and −0 = 0, so they are unaffected: y = −f(x) still crosses the x-axis at x = −2 and x = 6. The y-intercept is the value at x = 0: f(0) = −12, so −f(0) = 12, giving the point (0, 12) — the sign flips because the y-intercept is a nonzero y-value, unlike the roots. Writing 'x = 2, x = −6; y-intercept (0, −12)' comes from confusing −f(x) with f(−x) — reflecting in the y-axis instead of the x-axis, which negates the x-values of the intercepts (turning −2 into 2 and 6 into −6) but leaves f(0) unchanged, since f(−0) = f(0) = −12. Writing 'x = −2, x = 6; y-intercept (0, −12)' comes from forgetting that −f(x) is a reflection at all, and assumes both intercepts stay exactly as they were. Writing 'x = 2, x = −6; y-intercept (0, 12)' correctly negates the y-intercept but wrongly negates the x-intercepts too, as if a reflection in the x-axis also flipped the sign of every x-value.
- (a) 0.6 litres — The ratio 1 : 9 means the solution has 1 + 9 = 10 equal parts in total. Each part is 6 ÷ 10 = 0.6 litres, and disinfectant is 1 part, so Priti needs 0.6 litres of disinfectant. Giving 0.667 litres divides by 9, the number of parts of water, instead of the total number of parts, 10 (6 ÷ 9 ≈ 0.667). Giving 6 litres is the total amount of solution, not just the disinfectant's share of it. Giving 5.4 litres works out the water's share (6 × 9 ÷ 10 = 5.4), not the disinfectant's.
- (c) tan 45° — Method: replace each ratio by its exact value, then compare. Working: a right-angled triangle with a 45° angle is isosceles, so its opposite and adjacent sides are equal and the tangent of 45° is exactly 1. The others are cos 30° = √3/2, about 0.87; sin 45° = √2/2, about 0.71; and cos 60° = 1/2. Answer: tan 45°, the only one of the four that reaches 1. Reading √3/2 as though it were √3, about 1.73, makes cos 30° look the largest, but the division by 2 is part of the value. Ranking by the size of the angle also fails here, because the cosine of an angle falls as the angle grows.
- (b) (2, 4) and (−2, −4) — Substitute y = 2x into x² + y² = 20: x² + (2x)² = 20, which gives x² + 4x² = 20, so 5x² = 20, x² = 4, and x = 2 or x = −2. Using y = 2x for each x-value: x = 2 gives y = 4; x = −2 gives y = −4. The path meets the pond's edge at (2, 4) and (−2, −4). Distractor routes: (2, −4) and (−2, 4) swaps the sign pairing, matching each x-value with the wrong sign of y instead of keeping each x with its own correctly-signed y. (2, 1) and (−2, −1) comes from using y = x/2 instead of y = 2x when finding the y-coordinates. (2, 4) alone stops after the positive square root of x² = 4 and never finds the second point from x = −2.
- (a) 4:3 — Multiply both parts by the lowest common denominator, 6: (2/3) × 6 = 4 and (1/2) × 6 = 3, giving the ratio 4 : 3, which is already in simplest form.
- (d) 13 — Method: OP and PQ meet at a right angle because of the tangent–radius fact, so triangle OPQ is right-angled at P; use Pythagoras' theorem. Working: OQ² = OP² + PQ² = 5² + 12² = 25 + 144 = 169; OQ = √169 = 13. A student who answers 17 has simply added the two given lengths (5 + 12) instead of using Pythagoras' theorem. A student who answers 7 has subtracted the two given lengths (12 − 5) instead of using Pythagoras' theorem. A student who answers 144 has correctly squared 12 but stopped there, forgetting to add 5² and take the square root. Answer: 13 cm.
- (a) 16 — A and B share the same y-coordinate, so the distance between them is the horizontal difference between their x-coordinates: 8 − (−8) = 16, giving a chord of length 16. Choosing 8 gives only the x-coordinate of one point, not the full distance between the two points. Choosing 34 is the diameter of the circle (2 × 17), not the length of this particular chord. Choosing 17 is the radius — the distance from the centre to A or to B, not from A to B.
- (c) −8 — (−2)³ = (−2) × (−2) × (−2) = −8, since multiplying three negative numbers gives a negative result. A candidate who forgets the sign of a negative number when cubing it might treat (−2)³ as if it were 2³ = 8. A candidate who multiplies −2 by 3 instead of cubing it might get −2 × 3 = −6. A candidate who combines both mistakes — multiplying by 3 and dropping the sign — might get 2 × 3 = 6.
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.