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.Simplify x⁵ × x³ ÷ x², giving your answer as a single power of x.
- 2.A student says 4(2x − 3) is equivalent to 8x − 3. Which statement gives the correct verdict and reason?
- 3.Write the ratio 0.75 : 2 as a ratio of whole numbers in its simplest form.
- 4.A cube has all of its edges the same length. How many planes of symmetry does it have?
- 5.A two-way table records how 130 pupils travel to school. 70 of the pupils are girls and the rest are boys. 42 of the girls walk to school and the rest of the girls cycle. 33 of the boys walk to school. Work out what fraction of the girls walk to school.
- 6.The daily high temperatures, in °C, in Leeds over five days were 14, 16, 18, 19 and 23. In York over the same five days they were 15, 17, 17, 18 and 18. Write a sentence comparing Leeds and York using both the median and the range.
- 7.Work out an estimate for 6.4 × 3.9, by rounding each number to the nearest whole number.
- 8.A car's journey is represented by a velocity-time graph. The total distance travelled, found from the area under the graph, is 350 metres, and the total time for the journey is 25 seconds. Work out the car's average speed for the whole journey, in m/s.
- 9.A journey takes 4 hours at an average speed of 60 km/h. The time taken is inversely proportional to the average speed. Work out how long the same journey takes at an average speed of 80 km/h.
- 10.The angle between north and a cycle path is 40°, but it is measured anticlockwise from north. What is the three-figure bearing of the cycle path?
- 11.A bag contains 4 red counters and 6 blue counters. A counter is taken at random, its colour noted, and it is put back in the bag before a second counter is taken at random. Work out the probability that the two counters are different colours. Give your answer as a fraction in its simplest form.
- 12.Work out (−2)³ + (−3)² − (−4)
- 13.Solve 2x + 7 = x + 13
- 14.A charity shop and a school share collection-box money in the ratio 5 : 8. The charity shop receives £47.50. Work out how much the school receives.
- 15.To construct the perpendicular from a point P to a line l, where P is a point above l, an arc centred at P is drawn to cross l at two points, X and Y. What is the correct next step?
- 16.A water butt holds 200 litres and is being drained at a steady 8 litres per minute. Write down the function for the amount of water y, in litres, left after x minutes.
- 17.A rectangular garden has its length and width in the ratio 5:3. Given that the perimeter of the garden is 64 m, work out the width of the garden.
- 18.Point P has coordinates (4, 2). P is rotated 90° clockwise about the point (1, 1), and the image is then reflected in the line x = 1. Work out the coordinates of the final image of P.
- 19.Which of these equations represents the graph of y = 2ˣ translated by 3 units in the positive y-direction?
- 20.The function f(x) = x² for all real values of x has no inverse function, but g(x) = x² for x ≥ 0 does have one. Which statement correctly explains this?y = x²
Answer key
- (a) x⁶ — Method: work through the powers in order — multiplying powers of the same base means adding indices, and dividing powers of the same base means subtracting indices. Working: first, x⁵ × x³ = x⁸ (adding 5 and 3); then x⁸ ÷ x² = x⁶ (subtracting 2 from 8). x⁴ comes from swapping the two rules — subtracting for the multiplication, 5 − 3 = 2, and then adding for the division, 2 + 2 = 4. x¹⁰ comes from adding all three indices, 5 + 3 + 2 = 10, treating the division the same as a multiplication. 6x comes from correctly reaching a total index of 6 but then writing it as a coefficient of x instead of as its power. Answer: x⁶.
- (a) False — 4(2x − 3) = 8x − 12, not 8x − 3. — Expand the bracket by multiplying both terms by 4: 4 × 2x = 8x and 4 × (−3) = −12, so 4(2x − 3) = 8x − 12, which is not 8x − 3 — the student is wrong. Saying 4(2x − 3) = 8x − 3 comes from multiplying only the 2x by 4 and copying the −3 across unchanged. Saying 4(2x − 3) = 2x − 12 comes from multiplying only the −3 by 4 and leaving 2x unmultiplied. Claiming it is true because both expressions are linear ignores that equivalence depends on the actual coefficients, not the type of expression.
- (d) 3 : 8 — Multiply both parts by 4 to clear the decimal: 0.75 × 4 = 3 and 2 × 4 = 8, giving 3 : 8, which has no common factor other than 1. Giving 75 : 200 multiplies by 100 instead of 4, and has not then been simplified down to 3 : 8. Giving 0.75 : 2 has not been converted into whole numbers at all. Giving 3 : 2 converts the first part correctly but leaves the second part unscaled.
- (a) 9 — A cube has 9 planes of symmetry in total: 3 that pass through the middles of pairs of opposite faces, and 6 more that pass through pairs of opposite edges diagonally. A candidate who counts only the 3 face-to-face planes — which is correct for a cuboid with three different edge lengths, but forgets that a cube's equal edges create 6 more diagonal planes — answers 3. A candidate who counts only the 6 diagonal planes and forgets the 3 face-to-face ones answers 6. A candidate who confuses the number of planes of symmetry with the number of edges on a cube answers 12. The correct total for a cube is 9.
- (b) 3/5 — The question asks about the girls only, so use the girls' total of 70 as the denominator: 42 out of 70 girls walk, giving 42/70 = 3/5. Choosing 21/65 comes from using the whole survey of 130 pupils as the denominator instead of just the 70 girls, 42/130 = 21/65. Choosing 33/70 comes from using the boys' walking count, 33, over the girls' total of 70, mixing up the two rows of the table. Choosing 2/5 comes from using the number of girls who CYCLE, 70 − 42 = 28, instead of the number who walk, giving 28/70 = 2/5.
- (a) Leeds has a higher median and a greater range than York. — In order, Leeds's temperatures are 14, 16, 18, 19 and 23, so the median is the middle value, 18, and the range is 23 − 14 = 9. York's temperatures in order are 15, 17, 17, 18 and 18, so the median is 17, and the range is 18 − 15 = 3. Since 18 is higher than 17, and 9 is greater than 3, Leeds has both the higher median and the greater range. Choosing 'Leeds has a higher median but a smaller range than York' gets the median comparison right but the range comparison backwards — Leeds's range of 9 is actually greater than York's range of 3. Choosing 'York has a higher median and a greater range than Leeds' reverses both comparisons. Choosing 'York has a higher median but a smaller range than Leeds' reverses the median comparison; York's median of 17 is lower than Leeds's 18, even though it is correct that York's range is the smaller one.
- (c) 24 — Method: round each number to the nearest whole number, then multiply the rounded values. Working: 6.4 rounds to 6 (nearest whole number) and 3.9 rounds to 4 (nearest whole number). 6 × 4 = 24. Answer: 24. 18 comes from rounding 3.9 down to 3 instead of up to the nearest whole number, 4, giving 6 × 3. 28 comes from rounding 6.4 up to 7 instead of down to the nearest whole number, 6, giving 7 × 4. 25 is the exact value of 6.4 × 3.9, which is 24.96, rounded to the nearest whole number after multiplying, rather than estimated by rounding first.
- (b) 14 m/s — Average speed for a whole journey is total distance divided by total time: 350 ÷ 25 = 14 m/s. Multiplying the distance and time instead of dividing gives 350 × 25 = 8750 m/s. Adding the distance and time instead of dividing gives 350 + 25 = 375 m/s. Inverting the division, working out time divided by distance, gives 25 ÷ 350, which rounds to 0.07 m/s.
- (c) 3 hours — Method: in inverse proportion the product of the two quantities is constant, and here that product is the distance. Working: 60 × 4 = 240 km, so at 80 km/h the time is 240 ÷ 80 = 3. Answer: 3 hours. The distractors: 5 hours 20 minutes comes from treating the relationship as direct, working out 4 × 80 ÷ 60; 2 hours 40 minutes comes from cutting the time by the fraction the speed rose by — the speed went up by one third, so the time was cut by one third — which is not how inverse proportion works; 4 hours comes from dividing the 240 km by the original speed of 60 km/h again instead of by the new speed.
- (b) 320° — A bearing is always measured clockwise from north. An angle measured anticlockwise must be converted by subtracting it from 360°: 360 − 40 = 320°, so the bearing is 320°. Choosing 040° treats the anticlockwise angle as if it were already a clockwise bearing, without converting it. Choosing 220° adds 180° to the angle, mixing this up with a back-bearing calculation (40 + 180 = 220). Choosing 400° adds the angle to 360° instead of subtracting it (40 + 360 = 400), giving a bearing greater than a full turn.
- (a) 12/25 — P(red then blue) = 4/10 × 6/10 = 24/100. P(blue then red) is the same, 6/10 × 4/10 = 24/100. Since either order counts as different colours, add them: 24/100 + 24/100 = 48/100 = 12/25. Working out only one order, red-then-blue, and forgetting blue-then-red, gives 24/100 = 6/25. Working out the probability that the two counters are the SAME colour instead — 4/10 × 4/10 + 6/10 × 6/10 = 16/100 + 36/100 = 52/100 = 13/25 — answers a different question. Using 6/9 for the second draw, as if the first counter were not replaced, gives 2 × (4/10 × 6/9) = 48/90 = 8/15.
- (a) 5 — Method: each index is worked out first, and subtracting a negative number is the same as adding the positive. Working: (−2)³ = (−2) × (−2) × (−2) = −8 and (−3)² = (−3) × (−3) = 9, while − (−4) becomes + 4, so the calculation becomes −8 + 9 + 4 = 5. Answer: 5. The distractors: −13 comes from taking (−3)² as −9, giving −8 − 9 + 4 = −13; −3 comes from reading − (−4) as − 4, giving −8 + 9 − 4 = −3; 21 comes from treating every power of a negative number as positive, so that (−2)³ is taken as 8 and the calculation becomes 8 + 9 + 4 = 21.
- (d) x = 6 — Method: with an unknown on both sides, first collect the x terms on one side by subtracting the smaller x term from both sides, then deal with the numbers. Working: subtracting x from both sides gives x + 7 = 13, and subtracting 7 from both sides gives x = 6. Answer: x = 6. The distractors: x = 20 comes from adding 7 to 13 instead of subtracting it once the x terms have been collected; x = 2 comes from collecting the x terms by adding them, giving 3x + 7 = 13 and then 3x = 6; x = −6 comes from subtracting 2x from both sides to get 7 = −x + 13, reaching −6 = −x and then copying the sign straight across instead of dividing by −1.
- (b) £76.00 — One part of the ratio is £47.50 ÷ 5 = £9.50. The school receives 8 parts, so its share is 9.50 × 8 = £76.00. Dividing £47.50 by 8 instead of 5, treating the charity's amount as if it were 8 parts, gives 47.50 ÷ 8 = 5.9375, then × 5 = £29.69. Adding the charity's amount to the school's amount instead of stopping at the school's own share gives the total collected, 9.50 × 13 = £123.50. Adding one part to the charity's amount instead of multiplying one part by 8 gives 47.50 + 9.50 = £57.00.
- (a) Draw equal arcs from X and Y, meeting below line l. — After the first arc marks two points X and Y on line l, compasses are opened to a new radius and arcs of equal radius are drawn centred at X and at Y, so that they meet on the opposite side of l from P; joining P to that meeting point gives the perpendicular. (Joining X and Y with a straight line only retraces part of line l itself, since X and Y both already lie on it; drawing an arc centred at P through only one of X or Y repeats part of the first step instead of moving on; drawing a circle through X, Y and P does not locate the new point needed to complete the perpendicular.)
- (c) y = 200 − 8x — Method: in a linear model the amount present at the start is the constant term and the steady rate of change is the gradient, which is negative because the amount is falling. Working: at x = 0 minutes there are 200 litres, so the constant term is 200; 8 litres are lost every minute, so after x minutes 8x litres have gone and the amount left is y = 200 − 8x. Answer: y = 200 − 8x. The distractors: y = 8x + 200 treats the draining as filling, so the butt would gain 8 litres a minute; y = 200x − 8 swaps the two numbers over, using the starting 200 litres as the rate per minute and the 8 litres per minute as the starting amount; y = −8x − 200 makes the starting amount negative as well as the rate, so the butt would begin 200 litres in deficit.
- (c) 12 m — Method: the perimeter of a rectangle is twice the sum of the length and the width, so half the perimeter is one length plus one width; split that half using the ratio. Working: half of 64 is 64 ÷ 2 = 32 m, the ratio 5:3 has 5 + 3 = 8 parts, so one part is 32 ÷ 8 = 4 m, and the width is 3 × 4 = 12 m. So the width is 12 m. Distractor 20 m is the length, 5 parts, not the width. Distractor 24 m comes from splitting the whole perimeter, 64 m, into 8 parts and multiplying by 3, forgetting to halve the perimeter first. Distractor 8 m comes from the same slip stopped one step earlier: splitting the whole perimeter into 8 parts, 64 ÷ 8 = 8, and giving that instead of the width.
- (d) (0, −2) — To rotate (4, 2) by 90° clockwise about (1, 1), first find its position relative to the centre: (4 − 1, 2 − 1) = (3, 1). A 90° clockwise rotation sends (a, b) to (b, −a), so (3, 1) becomes (1, −3); adding the centre back gives (1 + 1, 1 − 3) = (2, −2). Reflecting (2, −2) in the line x = 1 gives (2 × 1 − 2, −2) = (0, −2). Doing the two transformations in the opposite order, reflecting first and then rotating, gives a different result, (2, 4), which shows the order matters. Stopping after the rotation and forgetting the reflection gives (2, −2). Stopping after only reflecting P in x = 1 and forgetting the rotation entirely gives (−2, 2). Rotate first, then reflect, in that order, and the final image is (0, −2).
- (a) y = 2ˣ + 3 — A translation of 3 units in the positive y-direction shifts the whole graph up, which means adding to the output: y = f(x) + k with k = 3, so the image is y = 2ˣ + 3. Adding the 3 inside the power instead of outside it, which translates the graph horizontally instead of vertically, gives y = 2ˣ⁺³. Using a negative 3, which moves the graph down instead of up, gives y = 2ˣ − 3. Mistaking 2ˣ for the linear expression 2x and adding 3 inside brackets gives y = 2(x + 3).
- (d) g is one-to-one: f(3) = f(−3), removed by x ≥ 0 — A function has an inverse only if it is one-to-one: every output must come from exactly one input. f(3) = 9 and f(−3) = 9, so two different inputs give the same output, and there is no way to send 9 back to a single input — f is not one-to-one over all real x. Restricting the domain to x ≥ 0 removes one of the two inputs behind every such pair, so g is one-to-one and does have an inverse. 'g's outputs are positive; f's could be negative' is wrong because f(x) = x² also only gives outputs of 0 or more — the outputs of f and g are identical sets; it is the INPUTS that differ, not the outputs. 'Restricting any domain always creates an inverse' is wrong because a restriction only helps if it actually removes the repeated outputs: restricting f(x) = x² to x ≥ −3 still leaves f(1) = f(−1) = 1, so that restricted function is still not one-to-one and still has no inverse. 'Squares can never be reversed, under any conditions' is wrong because √9 = 3 does reverse 3² = 9 once you know the input was non-negative — a square root just cannot tell you WHICH of two inputs you started from unless the domain has already ruled one of them 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.