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.Rice is sold in a 400 g bag for £1.12 and in a 1.5 kg bag for £3.90. Work out how much less the rice in the larger bag costs per kilogram.
- 2.ABCD is a parallelogram. A has coordinates (−3, 1), B has coordinates (2, 1) and C has coordinates (4, 4). Work out the coordinates of D.
- 3.A dog walker's charge, C pounds, for walking a dog for m minutes is shown on a straight-line graph. The line passes through the points (20, 14) and (50, 26). Work out the charge for a 65-minute walk.
- 4.Every diameter of a circle is an example of a more general type of straight line joining two points on the circle. Which term names that more general type of line?
- 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 heights, h cm, of 80 plants are grouped like this: 0 ≤ h < 20, 14 plants; 20 ≤ h < 40, 22 plants; 40 ≤ h < 50, 16 plants; 50 ≤ h < 80, 28 plants. Write down the class interval that contains the lower quartile.
- 7.Work out (−3) × 4 + 2 × (−5)
- 8.To solve 5x − 4 = 2x + 8, Chloe's working is shown. Line 1: 5x − 4 = 2x + 8. Line 2: 3x − 4 = 8. Line 3: 3x = 4. Line 4: x = 4/3. Chloe has made a mistake in her working. Which line contains the mistake?
- 9.A map has a scale of 1:25000. A road on the map is 6 cm long. Work out the real-life length of the road, giving your answer in kilometres.
- 10.A tile manufacturer cuts a hexagonal tile so that all six sides measure exactly 4 cm, but a check shows only two pairs of the six interior angles are equal to each other, not all six angles. The customer's order specifies regular hexagonal tiles. Based on the check, is this tile a regular hexagon?
- 11.A drawer contains 9 black socks and 4 white socks. Three socks are taken out at random, one after another, without being replaced. Given that at least two of the three socks taken out are black, work out the probability that all three are black.
- 12.Write 60 as a product of its prime factors, using index notation.
- 13.A circle has centre O(0, 0) and equation x² + y² = 169. The point Q has coordinates (10, 11). Work out which of these gives the correct position of Q together with correct working.
- 14.A crate exerts a force of 84 N on the floor through a base with an area of 2.4 m². Work out the pressure on the floor, in N/m².
- 15.A ramp rises at 30° to the horizontal. Its sloping surface is 4.8 m long. Safety rules say the vertical rise of a ramp must be no more than 2.5 m. Work out how far below that limit the rise of this ramp is.
- 16.A triangle has vertices at (1, 1), (1, 5) and (6, 1). Work out the area of the triangle.
- 17.The rent on a flat increases by 10% one year and by a further 10% the following year. Work out the overall percentage increase over the two years.
- 18.Vector A is and vector B is . Write down how vector B compares to vector A.
- 19.Which expression is equivalent to 0.5(4x + 6) − x?
- 20.Rosie is asked whether 4(x + 3) = 4x + 12 is an equation or an identity. Which statement is correct? Give a reason for your answer.
Answer key
- (c) 20p — Turn each price into the same rate before comparing. The small bag is 400 g = 0.4 kg, so it costs £1.12 ÷ 0.4 = £2.80 per kg. The large bag costs £3.90 ÷ 1.5 = £2.60 per kg. The saving is £2.80 − £2.60 = £0.20, which is 20p per kg. 2p compares the prices per 100 g rather than per kilogram, £2.78 subtracts one bag price from the other without turning either into a rate, and £2.60 is the large bag's price per kilogram rather than the saving.
- (d) (−1, 4) — In parallelogram ABCD the side DC is parallel and equal to the side AB, so D = C − AB. The vector from A to B is (2 − (−3), 1 − 1) = (5, 0), so D = (4 − 5, 4 − 0) = (−1, 4). A candidate who adds this vector to C instead of subtracting it gets (4 + 5, 4 + 0) = (9, 4). A candidate who subtracts A's coordinates from C's rather than the vector AB, and drops the minus sign on −3 while doing so, works out (4 − 3, 4 − 1) and gets (1, 3). A candidate who makes only the y-part of that slip, working out 4 − 1 instead of 4 − 0, gets (−1, 3).
- (c) £32 — First find the gradient: (26 − 14) ÷ (50 − 20) = 12 ÷ 30 = £0.40 per minute. Using the point (20, 14), the charge for 65 minutes is 14 + 0.40 × (65 − 20) = 14 + 18 = £32. Choosing £26 comes from treating the charge as directly proportional to the time, multiplying the gradient by 65 minutes and ignoring the fixed part of the charge (0.40 × 65 = 26). Choosing £40 comes from treating £14 as if it were the charge at 0 minutes, then adding the gradient multiplied by the full 65 minutes (14 + 0.40 × 65 = 40), instead of multiplying by the extra time past 20 minutes. Choosing £33.80 comes from assuming the charge is directly proportional to the minutes already known, scaling up from the point (50, 26) in the ratio 65:50 (65 ÷ 50 × 26 = 33.80).
- (c) chord — A diameter is a straight line joining two points on the circle that happens to pass through the centre. The general term for any straight line joining two points on a circle is a chord — every diameter is a chord, but not every chord is a diameter. Calling it a radius is wrong because a radius runs from the centre to just ONE point on the circle, not between two points on the circle. A tangent touches the circle at only one point, so it cannot be a diameter at all. An arc is a curved part of the circle's edge, not a straight line.
- (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) 20 ≤ h < 40 — Method: with 80 values the lower quartile is the 80 ÷ 4 = 20th value in order, so build a running total until it first reaches 20. Working: the running totals are 14, then 14 + 22 = 36, then 52, then 80; the 20th plant is past 14 but not past 36, so it lies in the second class. Answer: the lower quartile lies in the class 20 ≤ h < 40. The distractors: 0 ≤ h < 20 comes from believing that the bottom quarter of the data must all sit in the first class, when that class holds only 14 of the 80 plants; 40 ≤ h < 50 comes from using the position 80 ÷ 2 = 40 and so locating the median rather than the lower quartile; 50 ≤ h < 80 comes from counting 20 plants down from the tallest instead of up from the shortest, which locates the upper quartile at the 60th plant.
- (a) −22 — Method: both multiplications are carried out before the addition, and a positive multiplied by a negative is negative. Working: (−3) × 4 = −12 and 2 × (−5) = −10, so the calculation becomes −12 + (−10) = −22. Answer: −22. The distractors: 22 comes from ignoring the minus signs and working out 3 × 4 + 2 × 5 = 22; 50 comes from working from left to right with no priority at all, giving −12 + 2 = −10 and then −10 × (−5) = 50; −2 comes from taking 2 × (−5) as +10, so that −12 + 10 = −2.
- (d) Line 3 — Method: check each line of Chloe's working against the correct algebraic step. Working: Line 1 to Line 2 is correct, subtracting 2x from both sides gives 3x − 4 = 8. But Line 2 to Line 3 should add 4 to both sides, giving 3x = 12, not 3x = 4 — the constant −4 has been dropped rather than removed correctly. Line 4 follows correctly from Chloe's own, incorrect, Line 3. Answer: Line 3. Line 1 is simply the original equation, copied out correctly. Line 2 correctly subtracts 2x from both sides of Line 1. Line 4 divides Chloe's own Line 3 by 3 correctly — the arithmetic there is fine, the mistake happened one line earlier.
- (b) 1.5 km — Multiply the map length by the scale factor: 6 × 25000 = 150000 cm. Convert to kilometres, using 100 cm = 1 m and 1000 m = 1 km, so 100000 cm = 1 km: 150000 ÷ 100000 = 1.5 km. (1500 km comes from converting only as far as metres, 150000 ÷ 100 = 1500 m, and then writing kilometres on the end. 15 km comes from dividing by 10000 instead of 100000. 0.15 km comes from dividing by 1000000 instead of 100000.)
- (d) No — angles must be equal too — A regular polygon must have both all sides equal and all angles equal. This tile has all six sides equal, but its interior angles are not all equal, so it fails the angle condition and is not regular — 'No — angles must be equal too' is correct. 'Yes — all sides are equal' is wrong because equal sides alone are not enough; a shape can have equal sides but unequal angles, as here. 'Yes — six equal sides means regular' is wrong for the same reason: equal sides do not automatically guarantee equal angles. 'No — hexagons can't be regular' is wrong because regular hexagons certainly exist (six equal sides and six equal 120° angles); it is this particular tile that fails to be regular, not hexagons in general.
- (c) 7/19 — Method: 'at least two black' covers two cases — all three black, and exactly two black. Work out the probability of each along a tree, add them, then use P(all three black | at least two black) = P(all three black) ÷ P(at least two black). Working: P(all three black) = 9/13 × 8/12 × 7/11 = 504/1716 = 42/143. For exactly two black, one order is black, black, white = 9/13 × 8/12 × 4/11 = 288/1716; the white sock could be drawn first, second or third, so there are 3 such orders, giving 3 × 288/1716 = 864/1716 = 72/143. P(at least two black) = 42/143 + 72/143 = 114/143. P(all three black | at least two black) = (42/143) ÷ (114/143) = 42/114 = 7/19. Answer: 7/19. Watch out: stopping at 42/143 gives the unconditioned probability that all three are black — it ignores that you already know at least two of them are. Dividing by the 'exactly two black' probability on its own gives 7/12, and forgets that the all-black outcomes are themselves part of the 'at least two black' group, so they must be inside the denominator, not left out of it. And 7/11 answers a different, easier question — the probability the THIRD sock is black given the FIRST TWO specifically are black — not 'at least two of the three, in any order, are black'.
- (b) 2² × 3 × 5 — Repeatedly divide 60 by prime numbers: 60 ÷ 2 = 30, 30 ÷ 2 = 15, 15 ÷ 3 = 5, and 5 is itself prime. So 60 is 2 × 2 × 3 × 5, which in index notation is 2² × 3 × 5. Stopping the factor tree after only three divisions and writing 2 × 3 × 5 misses that the 2 divides in twice, and gives only 30, not 60. Squaring the 3 as well as the 2 gives 2² × 3² × 5, which comes to 180, far too big. Squaring the 5 instead of the 2 gives 2 × 3 × 5², which comes to 150, also too big. So 60 = 2² × 3 × 5.
- (a) Outside: OQ² = 221 > r² = 169 — OQ² = 10² + 11² = 100 + 121 = 221. Comparing this with r² = 169: since 221 > 169, OQ > r, so Q lies outside the circle — this is the correct verdict AND the correct working. 'Outside: OQ = 21 (10 + 11) > r = 13' reaches the same Outside verdict, but by invalid working: it adds the coordinates instead of squaring them (10 + 11 = 21, rather than 10² + 11² = 221), so the stated 'OQ' of 21 is not a distance at all — the verdict happens to match, but the method is wrong. 'Inside: OQ ≈ 14.87 < r² = 169' correctly finds the distance OQ = √221 ≈ 14.87, but then compares that DISTANCE with r² = 169 instead of with r = 13 — comparing two different kinds of quantity gives a meaningless, and here wrong, verdict. 'Inside: OQ² = 221 < (2r)² = 676' confuses the radius with the diameter: it compares OQ² with the diameter squared, (2 × 13)² = 676, instead of with r² = 169.
- (b) 35 N/m² — Pressure = force ÷ area. 84 ÷ 2.4 = 35 N/m². 201.6 N/m² comes from multiplying the force by the area instead of dividing (84 × 2.4). 81.6 N/m² comes from subtracting the area from the force (84 − 2.4) instead of dividing. 0.03 N/m² comes from dividing the area by the force instead of the force by the area (2.4 ÷ 84).
- (a) 0.1 m — Method: the sloping surface is the hypotenuse and the vertical rise is the side opposite the 30° angle, so rise = 4.8 × sin 30°; then compare that rise with the limit. Working: the exact value of sin 30° is one half, so the rise = 4.8 × 1/2 = 2.4 m. The limit is 2.5 m, and 2.5 − 2.4 = 0.1. Answer: the ramp is 0.1 m below the limit. Working out the rise and stopping there gives 2.4 m, which answers a question that was not asked. Dividing by sin 30° instead of multiplying gives 4.8 ÷ 0.5 = 9.6 and then 9.6 − 2.5 = 7.1 m. Treating sine as proportional to the angle, so that sin 30° is a third of sin 90°, gives 4.8 ÷ 3 = 1.6 and then 2.5 − 1.6 = 0.9 m.
- (d) 10 — The right angle is at (1, 1). The vertical side has length 5 − 1 = 4 and the horizontal side has length 6 − 1 = 5, so the area is (4 × 5) ÷ 2 = 20 ÷ 2 = 10. A candidate who forgets to halve the product of the two sides gets 4 × 5 = 20. A candidate who forgets to subtract the shared vertex's coordinate and uses the raw coordinates 6 and 5 as the side lengths gets (6 × 5) ÷ 2 = 30 ÷ 2 = 15. A candidate who uses only one side length as the area gets 5.
- (c) 21% — Method: an increase of 10% is a multiplier of 1.1, and two successive increases are found by multiplying the multipliers. Working: 1.1 × 1.1 = 1.21, so the rent is 121% of the original, which is an increase of 21%. Answer: 21%. The distractors: 20% comes from adding the two percentages, which ignores that the second 10% is taken of a larger amount; 121% is the multiplier written as the change rather than the change itself; 11% comes from slipping in the multiplication and getting 1.11 instead of 1.21.
- (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.
- (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) An identity, because it is true for every value of x — Expanding the brackets multiplies both terms inside by 4, giving 4x + 12, which is exactly the right-hand side. The two sides are therefore equal whatever x is, and a statement true for every value of the letter is an identity. An equation is true only for particular values, and trying to solve this one leads to 0 = 0, which places no restriction on x at all. Being able to expand brackets is not what makes a statement an identity, since any equation with brackets can be expanded. Nor is it a formula: a formula links two different quantities, and only one letter appears here.
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.