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 −(−3)⁴ + (−3)³
- 2.A lift has a weight limit: it must carry no more than 630 kg. Let w be the total weight, in kg, carried by the lift. Which inequality describes this?
- 3.A shop's takings were £4500 in May. In June the takings fell by £900. Write the June takings as a fraction of the May takings, in its simplest form.
- 4.A shape is translated by the vector , and one vertex of the image is at the point (1, 6). The original shape is instead translated by the vector . Work out the coordinates of the image of that same vertex under this second translation.
- 5.At a sports club, the numbers of members who play tennis and badminton are: 50 members play tennis, 40 members play badminton, and 18 members play both tennis and badminton. A member who plays tennis is chosen at random. Work out the probability that this member also plays badminton.
- 6.The marks scored by 11 pupils in a test are given in order: 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42. Work out the lower quartile of these marks.
- 7.A shortbread recipe uses flour and butter in the ratio 5 : 2. Yuki changes the recipe by doubling the amount of butter but keeping the amount of flour the same. Work out the fraction of the new mixture that is butter.
- 8.The tangent to a curve at the point (6, 1) has gradient −3. Work out the y-coordinate of the point where this tangent crosses the y-axis.
- 9.Two mathematically similar hexagonal tiles have areas 18 cm² and 50 cm². Work out the ratio of the side length of the smaller tile to the side length of the larger tile, in simplest form.
- 10.A rectangular notice board measures 150 cm by 80 cm. Work out its area in m².
- 11.A charity raffle has three types of ticket: winning, near-miss and losing, and every ticket is exactly one of these. The probability that a ticket is winning is 1/8 and the probability that it is a near-miss is 1/4. Work out the probability that a ticket is losing.
- 12.A semicircle has a diameter of 8 cm. Work out the exact area of the semicircle, in terms of π.
- 13.A rectangular garden has length (x + 7) m and width (x − 7) m. Work out an expression for the area of the garden, giving your answer in its simplest form.
- 14.Line P passes through the origin and the point (3, 21). Line Q passes through the origin and the point (5, 45). Both lines represent the distance, in metres, run by an athlete against the time, in seconds. Work out the gradient of the steeper line.
- 15.Two similar shapes have lengths in the ratio 1 : 3. A side of the smaller shape is 4 cm long. Work out the length of the corresponding side of the larger shape.
- 16.By completing the square, find the turning point of the curve y = 3x² + 12x + 7.y = 3x² + 12x + 7
- 17.A river 120 km long is drawn as a line 6 cm long on a map. Work out the scale of the map.
- 18.Two metal window frames are checked before installation. Frame A has edges of 40 cm and 65 cm, with a marked angle of 35° at the far end of the 65 cm edge, away from where the two given edges meet. Frame B has matching edges of 40 cm and 65 cm, with a matching 35° angle in the same position. The fitter wants to know if these measurements alone guarantee the two frames are congruent. What should the fitter be told?
- 19.A taxi fare in pounds is given by F = 2.50 + 1.20m, where m is the number of miles travelled. Sam takes a taxi for a journey of 6 miles. Work out the fare.
- 20.Write down the expression that means the same as m² × m × 3.
Answer key
- (a) −108 — Method: a power is worked out before any minus sign written in front of it, while a minus sign inside the brackets is part of the base. Working: (−3)⁴ = 81, because four negative factors multiply to a positive result, so −(−3)⁴ = −81. (−3)³ = −27, because three negative factors multiply to a negative result. Adding gives −81 + (−27) = −108. Answer: −108. The distractors: 54 comes from attaching the leading minus sign to the base, working out (−(−3))⁴ = 81 and then adding −27; −54 comes from taking (−3)³ as +27, forgetting that an odd power keeps the negative sign; 108 comes from believing that any power of a negative number is positive and that the leading minus belongs to the base, giving 81 + 27.
- (d) w ≤ 630 — 'No more than 630 kg' means the weight can be exactly 630 kg or anything less, so the correct inequality is w ≤ 630, using 'less than or equal to' to include the limit itself. Writing w < 630 excludes 630 kg itself, as though the limit could not be reached exactly. Writing w ≥ 630 reverses the direction, describing a minimum weight rather than a maximum. Writing w > 630 both reverses the direction and excludes the boundary value. The inequality describing the lift's weight limit is w ≤ 630.
- (a) 4/5 — Method: find the June takings first, then write them over the May takings and cancel. Working: the takings fell by £900, so June is £4500 − £900 = £3600; the fraction is 3600/4500, and dividing the numerator and the denominator by 900 gives 4/5. Answer: 4/5 of the May takings. The distractors: 1/5 comes from writing the fall over the May takings, 900/4500, which answers how far the takings dropped rather than what June's takings are compared with May's; 5/4 comes from writing May over June, 4500/3600, reversing the order the question asks for; 4/9 comes from writing June over the two months added together, 3600/8100, a part-to-whole fraction when the comparison asked for is with May alone.
- (d) (11, −2) — First undo the original translation to find the vertex on the original shape: (1 − (−8), 6 − 3) = (9, 3). Then apply the second vector to that original vertex: (9 + 2, 3 + (−5)) = (11, −2). (3, 1) comes from applying the second vector to the image point (1, 6) instead of to the original vertex — (1 + 2, 6 + (−5)) = (3, 1). (7, 8) comes from subtracting the second vector from the original vertex (9, 3) instead of adding it — (9 − 2, 3 − (−5)) = (7, 8). (11, 3) comes from applying only the x-component of the second vector to the original vertex and leaving the y-coordinate unchanged.
- (a) 9/25 — Method: P(badminton | tennis) = n(tennis and badminton) ÷ n(tennis) — restrict to the tennis-players, then find what fraction of them also play badminton. Working: n(tennis and badminton) = 18, n(tennis) = 50, so P(badminton | tennis) = 18/50 = 9/25. Answer: 9/25. Watch out: dividing by 40 (the badminton total) finds P(tennis | badminton) instead of P(badminton | tennis) — the wrong direction. Dividing by 90 (all the members named in the question) ignores that you already know the member plays tennis. And dividing by 72 (50 + 40 − 18, the number who play at least one of the two sports) answers a question about the union, not the condition you were given.
- (c) 18 — Method: for n ordered values, GCSE convention places the lower quartile at position (n + 1) ÷ 4, counting from the smallest value. Working: there are 11 marks, so n + 1 = 11 + 1 = 12 and 12 ÷ 4 = 3, so the lower quartile is the 3rd value in the ordered list 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, which is 18. Answer: the lower quartile is 18 marks. Watch the position you count to: dividing 11 ÷ 4 = 2.75 without adding 1 first, then rounding down, lands on the 2nd value, 15, not the 3rd; reaching for the middle of the whole list instead gives the median, 27, a different statistic; and averaging the 3rd and 4th values, 18 + 21 = 39 and 39 ÷ 2 = 19.5, borrows a method for an even split where it is not needed here.
- (b) 4/9 — Method: double the butter part of the ratio, keeping flour the same, find the new total, then write butter's part over the new total. Working: the new ratio is flour : butter = 5 : 4, since butter doubles from 2 to 4. New total = 5 + 4 = 9. Fraction of butter = 4/9. Answer: 4/9. 2/9 comes from forgetting to double the butter part and using the original value 2 over the new total of 9. 4/7 comes from doubling the butter part correctly to 4 but keeping the old total of 7 instead of working out the new total. 2/5 comes from using the original ratio 5:2 directly as butter over flour without doubling anything.
- (b) 19 — A line through (x₁, y₁) with gradient m has equation y − y₁ = m(x − x₁). Substituting (6, 1) and m = −3: y − 1 = −3(x − 6), so y = −3x + 18 + 1, which simplifies to y = −3x + 19; at x = 0 this gives 19. Making a sign error when distributing, writing −3(x − 6) as −3x − 18 instead of −3x + 18, gives the wrong line y = −3x − 17, so −17 at x = 0. Forgetting to add the y₁ = 1 at the end, using y = −3(x − 6) alone, gives 18 at x = 0. Substituting the y-coordinate into the gradient term instead of using x, working out 1 + (−3 × 1), gives −2.
- (d) 3 : 5 — Simplify the area ratio: 18 : 50 divides by 2 to give 9 : 25. Areas scale with the square of the length ratio, so take the square root of each part: the square root of 9 is 3, and the square root of 25 is 5, giving a side length ratio of 3 : 5. Giving 5 : 3 has the ratio the right way round for larger to smaller, not smaller to larger. Giving 9 : 25 is the simplified area ratio, without square-rooting it. Giving 18 : 50 is the area ratio before it has even been simplified.
- (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) 5/8 — Winning, near-miss and losing are exhaustive, so the three probabilities sum to 1. Writing 1/4 as 2/8 so every fraction has the same denominator, 1 − 1/8 − 2/8 = 8/8 − 1/8 − 2/8 = 5/8. Subtracting only the winning probability and forgetting the near-miss probability gives 1 − 1/8 = 7/8. Subtracting only the near-miss probability and forgetting the winning probability gives 1 − 1/4 = 3/4. Adding the two given probabilities and stopping there gives 1/8 + 2/8 = 3/8, the probability that a ticket is winning or a near-miss, not the probability that it is losing.
- (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.
- (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.
- (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.
- (b) 12 cm — Method: in similar shapes every length is multiplied by the same scale factor, and a ratio of 1 : 3 means that factor is 3 going from the smaller shape to the larger one. Working: 4 × 3 = 12. Answer: 12 cm. The distractors: 7 cm comes from adding 3 to the side instead of multiplying by it, which is what happens when a ratio is read as a difference; 36 cm comes from multiplying by 3² = 9, the factor that scales areas, and applying it to a length; 4 cm comes from treating the two shapes as congruent, so that corresponding sides stay equal — similar shapes have equal angles, but their sides are in proportion.
- (d) x = −2, y = −5 — 3x² + 12x + 7 rewrites as 3(x² + 4x) + 7, then as 3[(x + 2)² − 4] + 7, which simplifies to 3(x + 2)² − 5, since −3 × 4 + 7 = −5. Substituting x = −2: 3 × (−2)² = 12, 12 × (−2) = −24, so 12 − 24 + 7 = −5, confirming the minimum value −5 at x = −2: turning point x = −2, y = −5. Halving b instead of halving b/a — using −4 as the shift instead of −2 — lands on turning point x = −4, y = −41, which is wrong because when a ≠ 1 the shift inside the bracket is b/(2a), not b/a alone. Reading the bracket's sign directly as the turning point's x-coordinate gives x = 2, y = −5 — wrong, because (x + 2)² is zero at x = −2, not x = 2. Computing 12 − 7 = 5 instead of 7 − 12 = −5 flips the sign of the constant, giving x = −2, y = 5 — wrong, since the completed square's constant must be evaluated as 7 minus 12, not 12 minus 7. Whenever a ≠ 1, factor a out of the x² and x terms first, and always check a turning point by substituting back into the original equation.
- (a) 1 : 2 000 000 — Method: a scale is a ratio between two lengths written in the same unit, reduced so that the map distance is 1. Working: 120 km = 120 × 1000 × 100 = 12 000 000 cm, so the ratio is 6 : 12 000 000, and dividing both parts by 6 gives 1 : 2 000 000. Answer: 1 : 2 000 000. The distractors: 1 : 12 000 000 comes from writing the real length in centimetres without dividing by the 6 cm on the map; 1 : 200 000 comes from taking 120 km as 1 200 000 cm, one conversion step short, before dividing by 6; 1 : 20 000 comes from converting 120 km to 120 000 m and treating those metres as centimetres.
- (a) No — the angle is not between the two sides — Method: check whether the given angle sits between the two given sides. Working: the 35° angle is marked away from the corner where the 40 cm and 65 cm edges meet, so it is not the included angle — this is SSA, which is not one of the four basic congruence conditions, so it does not guarantee congruence. Options: 'Yes — SAS' wrongly treats any two sides plus any angle as SAS, without checking the angle's position; 'Yes — SSS' wrongly counts the angle as if it were a third side; 'No — only two sides measured' is not the real reason, since SAS itself only needs two sides, so this reasoning is beside the point. Answer: no, the angle is not between the two sides.
- (c) £9.70 — F = 2.50 + 1.20 × 6 = 2.50 + 7.20 = £9.70. £7.20 comes from forgetting to add the £2.50 fixed charge. £10.20 comes from adding the 2.50 to the number of miles before multiplying by 1.20, 1.20 × (6 + 2.50). £22.20 comes from adding the fixed charge and the rate together first and then multiplying by the miles, (2.50 + 1.20) × 6.
- (d) 3m³ — Method: multiply the powers of m by adding their indices, then bring the number coefficient to the front. Working: m² × m has indices 2 and 1; add them to get 3, giving m³, then × 3 gives 3m³. Answer: 3m³. 3m² comes from multiplying the indices instead of adding them: 2 × 1 = 2, giving m², then × 3 = 3m². m³ comes from correctly combining the m's but dropping the coefficient 3. m⁶ comes from multiplying the index by the coefficient instead of writing the coefficient in front: taking the 2 in m² and the 3 to give m raised to the power 2 × 3, which is m⁶, with the lone m left out.
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.