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.Maya buys 4 plants at £3.20 each and a bag of compost for £6.75. She pays with a £20 note. Work out her change.
- 2.Work out the value of the discriminant b² − 4ac for the equation 2x² + 3x − 2 = 0.
- 3.A graph shows y plotted against x. The graph is a curve that gets closer to both axes but never touches them, and y decreases as x increases. Write down whether this graph could show direct proportion, inverse proportion, or neither.
- 4.In a construction, a point P is marked at the foot of the perpendicular drawn from point X to a straight line. Which statement correctly compares the distance XP with the distance from X to any other point on the line?
- 5.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.
- 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.A photograph uses 4 × 10⁶ bytes of storage. A memory card holds 3.2 × 10¹⁰ bytes. Work out how many of these photographs the card can hold. Give your answer in standard form.
- 8.A square tile has an area of 144 cm². Work out the side length of the tile.
- 9.A furniture designer makes a scale model of a wardrobe using a scale of 1 : 8. The real wardrobe is 2.4 m tall and 1.2 m wide. Work out the perimeter of the front face of the model wardrobe, in centimetres.
- 10.OABC is a parallelogram, with OA = a and OC = c. X is the midpoint of the diagonal AC. Express the vector OX in terms of a and c, and use it to show that X also lies on the diagonal OB.
- 11.A grower knows that the probability that one of their seeds germinates is 0.6. The grower wants to expect 300 of the seeds to germinate. Work out how many seeds the grower should plant.
- 12.A plank has length a = 12 cm and a second plank has length b = 7 cm, each correct to the nearest centimetre. Work out the upper bound of a + b.
- 13.Make x the subject of the formula y = 2x − 9.y = 2x − 9
- 14.The price of a cycling helmet rises from £80 to £116. Work out the percentage increase.
- 15.Two bronze statues are geometrically similar and made from the same solid bronze. Their heights are 30 cm and 45 cm. The smaller statue has a mass of 12 kg. Work out the mass of the larger statue, in kg.
- 16.The graph of y = f(x) has a maximum turning point at (5, 8). Which of these correctly gives the corresponding turning point on the graph of y = −f(x + 1), and its type?
- 17.Write the ratio 0.75 : 2 as a ratio of whole numbers in its simplest form.
- 18.In triangle ABC, AB = x cm, AC = (x + 3) cm, BC = 7 cm and angle BAC = 60°. Work out the value of x.
- 19.Solve 3(2x − 1) = 27
- 20.A number machine multiplies its input by 2 and then subtracts 5. Work out the output when the input is 6.
Answer key
- (c) £0.45 — Method: find the total cost, then subtract from £20. Working: 4 × £3.20 = £12.80. £12.80 + £6.75 = £19.55. Change = £20.00 − £19.55 = £0.45. Answer: £0.45. (£7.20 comes from forgetting to include the compost and subtracting only the plants' cost from £20. £1.45 comes from dropping the carry when adding the pence: 80p + 75p = £1.55, but only the 55p is written down, giving £18.55 instead of £19.55. £10.05 comes from buying only one plant instead of four, using £3.20 + £6.75.)
- (c) 25 — Method: read off a, b and c with their signs and substitute them into b² − 4ac. Working: for 2x² + 3x − 2 = 0, a = 2, b = 3 and c = −2, so b² − 4ac = 3² − 4 × 2 × (−2) = 9 − (−16) = 9 + 16 = 25. Answer: 25. The distractors: −7 comes from taking c as +2, which gives 9 − 16; 22 comes from working b² as 2 × 3 = 6 and then 6 + 16; 13 comes from leaving the 4 out of 4ac and working 9 − 2 × (−2).
- (a) Inverse proportion — A curve that decreases and never touches either axis is the standard shape for inverse proportion, y = k/x. Direct proportion graphs are straight lines through the origin, which this is not, so it must be inverse proportion rather than neither.
- (d) XP is the shortest distance from X to the line — The perpendicular from a point to a line always gives the shortest possible distance to that line — joining X to any other point on the line forms the hypotenuse of a right-angled triangle with XP as one of the shorter sides, and a hypotenuse is always longer than either of the other two sides. So XP is shorter than the distance to every other point on the line. "XP is the longest distance from X to the line" reverses this relationship. "XP equals every other distance from X to the line" would only be true if X were equidistant from every point on the line, which is impossible for a point and a straight line. "XP cannot be compared without knowing the line's length" is false — the shortest-distance fact holds whatever the line's length, since only the local right angle matters.
- (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.
- (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) 8 × 10³ — Method: divide the capacity of the card by the size of one photograph, dividing the coefficients and subtracting the indices, then bring the coefficient back into the range 1 to 10. Working: 3.2 ÷ 4 = 0.8 and 10 − 6 = 4, which gives 0.8 × 10⁴; a coefficient of 0.8 is smaller than 1, so the decimal point moves one place to the right and the index falls by 1. Answer: 8 × 10³. The distractors: 8 × 10⁴ comes from correcting 0.8 to 8 without reducing the index, which makes the answer ten times too large; 1.28 × 10¹⁷ comes from multiplying the two numbers instead of dividing them, since 3.2 × 4 = 12.8 and 10 + 6 = 16; 8 × 10¹⁵ comes from dividing the coefficients but adding the indices instead of subtracting them.
- (c) 12 cm — For a square, area = side². So side² = 144, giving side = ±12. Since a length must be positive, the side length is 12 cm. A candidate who gives both square roots without rejecting the negative one, which cannot be a length, answers 12 cm or −12 cm. A candidate who halves 144 instead of taking its square root gets 72 cm. A candidate who divides 144 by 4, confusing the area formula with a perimeter calculation, gets 36 cm.
- (d) 90 cm — Method: scale each dimension by the scale factor, then find the perimeter. Working: model height = 240 ÷ 8 = 30 cm; model width = 120 ÷ 8 = 15 cm. Perimeter = 2 × (30 + 15) = 90 cm. Wrong options: 11.25 cm comes from squaring the scale factor as if finding an area (720 ÷ 64); 510 cm comes from scaling only one dimension and leaving the other at full size; 720 cm comes from finding the real perimeter (2 × (240 + 120)) but forgetting to scale it down at all.
- (d) (1/2)a + (1/2)c — Method: X is the midpoint of AC, so OX = OA + (1/2)AC, with AC = c − a. Working: OX = a + 1/2(c − a) = a − (1/2)a + (1/2)c = (1/2)a + (1/2)c. Answer: OX = (1/2)a + (1/2)c. Since OB = a + c, this is exactly half of OB, so OX = (1/2)OB, meaning X lies on OB at its midpoint too — the two diagonals bisect each other. Forgetting to halve AC at all gives a + c, which is OB itself, not its midpoint; halving only the c-term gives (1/2)a + c; and a sign error on the c-term gives (1/2)a − (1/2)c. Halve the whole of AC, both terms together, and add it to OA rather than to a alone.
- (c) 500 — Method: the expected number of successes is the number of trials multiplied by the probability, so to find the number of trials, divide the expected number by the probability. Working: let n be the number of seeds planted. Then n multiplied by 0.6 must come to 300, so n = 300 ÷ 0.6 = 500. Answer: the grower should plant 500 seeds. The distractors: 180 comes from multiplying instead of dividing, 300 × 0.6 = 180, which answers how many of 300 seeds would germinate; 750 comes from dividing by the probability of not germinating, 300 ÷ 0.4 = 750; 120 comes from multiplying by that same 0.4, 300 × 0.4 = 120.
- (b) 20 — Each length has its own error interval: 11.5 ≤ a < 12.5 and 6.5 ≤ b < 7.5. The upper bound of a sum is found by adding the upper bounds of both quantities: 12.5 + 7.5 = 20. Bounding only one of the two lengths and adding the other quantity's given value unbounded, 12.5 + 7 = 19.5, misses that both measurements carry their own uncertainty. Adding the lower bounds instead of the upper bounds, 11.5 + 6.5 = 18, gives the lower bound of the sum, not the upper one. Using a whole centimetre of error either side instead of half a centimetre, (12 + 1) + (7 + 1) = 21, comes from forgetting the error is half the rounding unit.
- (a) x = (y + 9) / 2 — Method: undo the subtraction of 9 first, then undo the multiplication by 2. Working: y = 2x − 9, so adding 9 to both sides gives y + 9 = 2x, then dividing both sides by 2 gives x = (y + 9) / 2. The value x = (y − 9) / 2 comes from a sign error, keeping the −9 instead of moving it to +9. The value x = 2(y + 9) comes from multiplying by 2 instead of dividing. The value x = y / 2 + 9 comes from dividing by 2 before adding 9, the wrong order of operations, instead of dividing the whole bracket.
- (a) 45% — Method: percentage increase = increase ÷ original amount × 100. Working: the increase is £116 − £80 = £36, and 36 ÷ 80 = 0.45, so 0.45 × 100 = 45%. Answer: 45%. The distractors: 36% comes from quoting the £36 increase as though pounds and per cent were the same thing; 31% comes from dividing the £36 increase by the new price £116 instead of by the original £80, which gives 31% to the nearest per cent; 145% is the new price written as a percentage of the original price, which is the whole of the new price rather than the increase.
- (c) 40.5 kg — Since the statues are made from the same material, mass is proportional to volume, which scales with the cube of the length scale factor. The length scale factor is 45 ÷ 30 = 1.5, so the volume (and mass) scale factor is 1.5³ = 3.375. The larger statue's mass is 12 × 3.375 = 40.5 kg. 18 kg comes from multiplying by the length factor 1.5 directly. 27 kg comes from using the area scale factor 1.5² = 2.25 instead of the volume scale factor. 54 kg comes from treating 'cubed' as 'multiplied by 3', giving 1.5 × 3 = 4.5 instead of 1.5³.
- (a) (4, −8), a minimum point — The transformation x → x + 1 inside f translates the graph 1 unit to the LEFT, so the x-coordinate becomes 5 − 1 = 4. The minus sign in front of f reflects the graph in the x-axis, so the y-coordinate becomes −8, and a reflection in the x-axis turns every maximum into a minimum, so (4, −8) is a minimum point. Writing '(4, −8), a maximum point' gets the coordinates right but forgets that a reflection in the x-axis swaps maximum and minimum points. Writing '(6, −8), a minimum point' comes from translating 1 unit to the RIGHT instead of the left — f(x + 1) always moves the graph in the negative x-direction. Writing '(4, 8), a minimum point' gets the x-coordinate and the type right, but forgets to actually negate the y-coordinate, even though it does correctly reclassify the point as a minimum.
- (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) 5 — Method: BC faces the 60° angle, so put the two algebraic sides into the cosine rule and solve the equation that results. By hand, cos 60° = 0.5. Working: 7² = x² + (x + 3)² − 2 × x × (x + 3) × 0.5, so 49 = x² + x² + 6x + 9 − x² − 3x = x² + 3x + 9. That rearranges to x² + 3x − 40 = 0, which factorises as (x + 8)(x − 5) = 0, giving x = −8 or x = 5. A length cannot be negative, so the negative root is rejected. Answer: x = 5. The distractors: 8 comes from expanding (x + 3)² as x² + 9, the error of assuming that squaring a bracket squares each term, which turns the equation into x² − 3x − 40 = 0; −8 is the negative root, kept by a candidate who solves the quadratic but never checks that x has to be a length; 37 comes from applying the cosine rule as though the 60° angle were at C facing AB, which gives x² = (x + 3)² + 49 − 7(x + 3) and a single linear solution.
- (b) 5 — Method: expand the bracket first, then solve the resulting linear equation. Working: 3(2x − 1) = 6x − 3, so 6x − 3 = 27. Add 3 to both sides: 6x = 30. Divide by 6: x = 5. Answer: 5. 4 2/3 comes from multiplying only the 2x inside the bracket by 3 and forgetting to multiply the −1 as well, giving 6x − 1 = 27. 4 comes from a sign error when expanding, writing 6x + 3 instead of 6x − 3. 4.5 comes from dividing 27 by 3 to get 2x − 1 = 9, then forgetting to add 1 back before dividing by 2, using 2x = 9 instead of 2x = 10.
- (a) 7 — Multiply the input by 2: 6 × 2 = 12. Then subtract 5: 12 − 5 = 7. A candidate who does the operations in the wrong order, subtracting 5 first and then multiplying by 2, gets (6 − 5) × 2 = 2. A candidate who only carries out the multiplication and forgets to subtract gets 12. A candidate who adds 5 instead of subtracting gets 6 × 2 + 5 = 17.
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.