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.A charity shop buys a coat for £24 and sells it for a profit that is 3/8 of the buying price. Work out the selling price.
- 2.A market stall's cost of hiring n tables is modelled by two formulas: Formula A: C = 3(2n + 5); Formula B: C = 6n + 15, where C is in pounds. A stallholder says the two formulas always give the same cost. Work out the cost given by each formula when n = 4, and use your results to decide whether the stallholder is correct.
- 3.A recipe for 8 people needs 320 g of rice. Sanjay scales the recipe down to serve 5 people, then wants to scale his new amount back up to serve 8 people again. Work out the multiplier he should use to scale his amount for 5 people back up to the amount for 8 people.
- 4.A cuboid measures 5 cm long (left to right), 3 cm deep (front to back) and 2 cm tall. Its front elevation is 5 cm wide by 2 cm high. Its side elevation is 3 cm wide by 2 cm high. What are the dimensions of its plan view, looking down from above?
- 5.A garden centre recorded 240 plant sales using a frequency tree. The first stage splits the sales into three types: shrubs, bedding plants and trees. 96 sales were shrubs, 114 were bedding plants and the rest were trees. Work out the probability that a sale chosen at random from these 240 was a tree.
- 6.The weekly wages of the five people who work at a small garage in Norwich are £420, £440, £460, £480 and £1,500. Write down which average better describes a typical wage at this garage, and give a reason for your answer.
- 7.Work out the value of (√5)⁴
- 8.On Monday, a runner covers 15 km in 2.5 hours. On Tuesday, she covers 12 km in 1.5 hours. Using the formula speed = distance ÷ time, work out on which day she ran faster, and by how much.
- 9.The tangent to a curve at the point where x = 4 has equation y = 3x − 2. Work out the instantaneous rate of change of y with respect to x at x = 4.y = 3x − 2
- 10.A shop stacks identical storage crates in a display. The diagram shows the plan of the floor positions used, with one corner position marked. Every floor position is filled with crates stacked 3 high, except the marked corner position, which is left completely empty. Work out how many crates are used in total.
- 11.A fair six-sided dice is rolled 300 times. Work out how many more times you would expect it to land on an even number than on a six.
- 12.Simplify √8 + √18, giving your answer in the form k√2.
- 13.A candidate solves the inequality x² > 9 and writes: "x² > 9, so x > 3." Which of these is a correct comment on the candidate's work?
- 14.Write 250 g : 2 kg as a ratio in its simplest form.
- 15.A triangle has vertices A(0, 0), B(10, 0) and C(5, 12). Work out the area of the triangle.
- 16.A plumber charges a call-out fee plus an hourly rate. The total cost, y in pounds, of a job lasting x hours is given by y = 45x + 60. Work out the total cost of a job that lasts 3 hours.y = 45x + 60
- 17.The ratio of the number of pages Mia has read to the number of pages Leo has read is 7 : 4. Mia has read 84 pages. Work out how many pages they have read altogether.
- 18.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?
- 19.A rope of length L metres is cut into 6 equal pieces, and 4 metres is then removed from one piece. Write an expression, in metres, for the length of that piece after the cut.
- 20.A circle has centre (0, 0) and passes through the point (5, 12). Work out the equation of the circle.
Answer key
- (d) £33.00 — The profit is 3/8 of £24 = (£24 ÷ 8) × 3 = £3 × 3 = £9.00. Selling price = £24 + £9.00 = £33.00. A candidate who gives the profit instead of the selling price gets £9.00. A candidate who subtracts the profit instead of adding it gets £24 − £9 = £15.00. A candidate who works out one eighth of £24 and adds that on, forgetting to multiply by the numerator 3, gets £24 + £3 = £27.00.
- (d) Formula A gives £39 and Formula B gives £39, and since 3(2n + 5) expands to 6n + 15 for every value of n, the stallholder is correct. — Formula A: 3(2 × 4 + 5) = 3 × 13 = £39. Formula B: 6 × 4 + 15 = 24 + 15 = £39. Expanding Formula A algebraically gives 3(2n + 5) = 6n + 15, which is identical to Formula B for every value of n, not just n = 4, so the stallholder is correct — this is an identity, not a coincidence. The option giving £29 for Formula A comes from multiplying only the 2n by 3 and forgetting to also multiply the 5, then adding the unmultiplied 5: 3 × 2 × 4 = 24, + 5 = 29. The two options that reach the correct numbers but reject the stallholder's claim both use faulty reasoning — matching values at one value of n, or counting terms, does not decide whether two expressions are identical for every n; expanding the bracket does.
- (a) 8/5 — Scaling down from 8 people to 5 people uses the multiplier 5/8. To reverse this and scale back up from 5 people to 8 people, use the reciprocal of that multiplier: flip 5/8 to get 8/5. 5/8 comes from using the forward (scaling down) multiplier again, instead of reversing it. 3/5 comes from writing the difference in people (8 − 5 = 3) over 5, instead of using the reciprocal of 5/8. 25/64 comes from multiplying the forward multiplier by itself (5/8 × 5/8), instead of finding its reciprocal.
- (b) 5 cm by 3 cm — The plan view looks straight down on the cuboid's footprint, so it shows the length (5 cm, left to right) and the depth (3 cm, front to back) — the two dimensions that do not involve height. "5 cm by 2 cm" repeats the front elevation's dimensions, pairing the length with the height instead of the depth. "3 cm by 2 cm" repeats the side elevation's dimensions, again pairing the depth with the height rather than with the length. "5 cm by 5 cm" comes from mistakenly assuming the plan must be a square, pairing the length with itself instead of with the depth.
- (b) 1/8 — Shrubs, bedding plants and trees are the three branches at the first stage of the tree, so they must total 240: tree sales = 240 − 96 − 114 = 30. So P(tree) = 30/240 = 1/8. Using the shrub count instead, 96/240 = 2/5, is the probability of a shrub sale, not a tree sale. Using the bedding-plant count instead, 114/240 = 19/40, is the probability of a bedding-plant sale. Subtracting the bedding count from the shrub count (114 − 96 = 18) instead of subtracting both from 240 gives 18/240 = 3/40, which is not the number of tree sales at all.
- (a) The median, as the one very large wage does not move it — Method: an average describes a population well when it sits close to most of the values, so compare what each average does when one value lies far from the rest. Working: in order the wages are 420, 440, 460, 480 and 1,500, so the median is the third of the five, £460. The mean uses every wage: 420 + 440 + 460 + 480 + 1,500 = 3,300 and 3,300 ÷ 5 = 660, so the mean is £660. Four of the five people earn less than £660, and the nearest of those four wages is £180 below it, so £660 describes nobody at the garage; £460 sits inside the group of four similar wages. Answer: the median, as the one very large wage does not move it, while that same wage drags the mean £200 above the median. The distractors: saying the median is always larger than the mean is an invented rule, and here the median £460 is smaller than the mean £660; saying the mean is the only average that uses all five wages is true as far as it goes, but using a value and being dragged by it are the same thing when that value is £1,500; saying £660 lies between the smallest and largest wage is true of every mean ever calculated, so it proves nothing about whether this one is typical.
- (a) 25 — Method: squaring a square root removes the root, so a square root raised to the power 4 can be squared in two stages. Working: (√5)⁴ = ((√5)²)² = 5² = 5 × 5 = 25. Answer: 25. The distractors: 5 comes from squaring once and stopping, treating the fourth power as a square; 625 comes from raising 5 to the power 4 and ignoring the root sign altogether; 20 comes from multiplying 5 by the index 4 instead of raising 5 to that power.
- (a) Tuesday, by 2 km/h — Monday's speed is 15 ÷ 2.5 = 6 km/h and Tuesday's speed is 12 ÷ 1.5 = 8 km/h, so Tuesday was faster, by 8 − 6 = 2 km/h. A candidate who works out the correct speeds but mislabels which day is faster gets Monday, by 2 km/h. A candidate who divides 15 ÷ 2.5 incorrectly as 5 instead of 6 gets a difference of 8 − 5 = 3 km/h, still crediting Tuesday. A candidate who forgets to find Monday's speed and gives Tuesday's speed itself as the difference states Tuesday, by 8 km/h.
- (c) 3 — Method: for a tangent written in the form y = mx + c, the coefficient m is the gradient of the line, and the gradient of the tangent at its point of contact equals the curve's instantaneous rate of change there. Working: y = 3x − 2 has gradient 3, so the instantaneous rate of change of y with respect to x at x = 4 is 3. Reading the constant term as the rate instead of the coefficient of x gives −2, but −2 is only where the tangent crosses the y-axis, not a rate. Reading the x-coordinate of the point of contact as the rate gives 4, but 4 only tells you where on the curve the tangent touches, not how fast y is changing there. Substituting x = 4 into the tangent equation, 3 × 4 − 2 = 10, gives the y-coordinate of the point of contact, not the rate; a candidate who works out the height of the point instead of the gradient gives 10. Whenever a tangent is given as an equation, the rate of change is always the coefficient of x — do not let the constant term, the x-value or a substituted y-value stand in for it.
- (a) 33 — Method: total crates = (number of floor positions that actually have crates on them) × (the stack height). Working: there are 4 × 3 = 12 floor positions in the whole arrangement, but one corner position is left empty, leaving 11 filled positions; each filled position is stacked 3 crates high, so 11 × 3 = 33. Answer: 33. The distractors: 36 comes from forgetting to remove the empty corner and using all 12 positions (12 × 3). 35 comes from removing only one crate for the empty corner instead of the full stack of 3 (36 − 1). 11 comes from counting the filled floor positions and stopping there, forgetting that each one carries a stack 3 crates high.
- (c) 100 — There are 3 even numbers on a fair dice (2, 4 and 6), so the probability of landing on an even number is 3/6 = 1/2, and 300 × 1/2 = 150. The probability of landing on a six is 1/6, so 300 × 1/6 = 50. The dice is expected to land on an even number 150 − 50 = 100 more times than on a six. Writing 50 is wrong because that is just the expected number of sixes on its own, without comparing it to the expected number of evens. Writing 150 is wrong because that is just the expected number of evens on its own, without subtracting the sixes. Writing 200 is wrong because it adds the two expected frequencies together (150 + 50 = 200) instead of finding the difference between them. The dice is expected to land on an even number 100 more times than on a six.
- (a) 5√2 — Simplify each surd first: √8 = √4 × √2 = 2√2, and √18 = √9 × √2 = 3√2. Both terms are now multiples of √2, so they are like terms: 2√2 + 3√2 = 5√2. Adding the numbers under the two roots first, 8 + 18 = 26, and writing √26 treats unlike surds as if they combine under one root — they only combine once they share the same radicand, which is not how addition of surds works. Writing 9√2 for √18 instead of 3√2 (forgetting to root the 9) and then adding gives 2√2 + 9√2 = 11√2. Writing 4√2 for √8 instead of 2√2 (forgetting to root the 4) and adding gives 4√2 + 3√2 = 7√2.
- (d) Wrong — the correct solution is x < −3 or x > 3. — Squaring or unsquaring an inequality is not a safe one-step move: the solution must be split into two branches, since a number less than −3 also squares to more than 9. So x > 3 finds only half the solution set. The full answer is x < −3 or x > 3. Distractor routes: "Right — square rooting both sides gives x > 3" treats the square root of an inequality the same as the square root of an equation and misses the negative branch entirely. "Wrong — the correct solution is −3 < x < 3" applies the between-the-roots pattern that belongs to the opposite inequality, x² < 9. "Right, but x = −3 and x = 3 should also work" treats the inequality as if it also allowed equality, when x² > 9 is strict and excludes both x = 3 and x = −3.
- (a) 1:8 — Convert 2 kg to grams: 2 kg = 2000 g. The ratio is 250 : 2000. The highest common factor of 250 and 2000 is 250. Divide both parts by 250: 250 ÷ 250 = 1 and 2000 ÷ 250 = 8, so the ratio is 1 : 8. Leaving the kilograms unconverted gives 250 : 2, which simplifies to 125 : 1 — the units on each side are different, so this does not compare like with like. Dividing by 50 instead of 250 gives 5 : 40, which still shares a common factor of 5, so it is not fully simplified. Swapping the order gives 8 : 1, grams to kilograms the wrong way round.
- (b) 60 units² — Method: the area of a triangle is half the base times the perpendicular height, so choose a side to act as the base and measure the perpendicular distance from the opposite vertex to it. Working: A(0, 0) and B(10, 0) both lie on the x-axis, so AB is horizontal and AB = 10 − 0 = 10. The perpendicular height is the distance of C from the x-axis, which is its y-coordinate, 12. Area = (10 × 12) ÷ 2 = 120 ÷ 2 = 60. Answer: 60 units². The distractors: 120 units² comes from multiplying base by height and forgetting to halve; 65 units² comes from using the slanting side AC, which is 13 long, as the height in place of the perpendicular distance 12; 30 units² comes from halving the base to 5 before multiplying and then halving the product as well, so the halving is done twice.
- (d) £195 — Substituting x = 3 into y = 45x + 60 gives y = 45 × 3 + 60 = 135 + 60 = 195. A candidate who forgets to add the call-out fee would get only 45 × 3 = £135. A candidate who adds the hours to the fee and the rate instead of multiplying would get 45 + 60 + 3 = £108. A candidate who multiplies both the hourly rate and the call-out fee by the number of hours would get 45 × 3 + 60 × 3 = £315.
- (d) 132 — Method: find the value of one part of the ratio, use it to find Leo's pages, then add both amounts together. Working: 84 ÷ 7 = 12 (value of one part). Leo's pages = 12 × 4 = 48. Total = 84 + 48 = 132. Wrong options: 48 gives only Leo's pages and forgets to add Mia's; 147 comes from reversing the ratio parts (84 ÷ 4 × 7 = 147) and stopping there; 231 comes from reversing the ratio parts and then adding Mia's pages (84 + 147).
- (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) L/6 − 4 — Method: find the length of one equal piece first (divide by 6), then apply the later change (subtract 4) to that piece. Working: one piece is L/6 metres; removing 4 metres from it gives L/6 − 4. Answer: L/6 − 4. L/6 + 4 comes from adding the 4 metres instead of removing it. (L − 4)/6 comes from removing the 4 metres from the whole rope before cutting it into pieces, the wrong order. 4 − L/6 comes from subtracting the piece length from 4 instead of the other way round.
- (b) x² + y² = 169 — Method: a circle centred on the origin has equation x² + y² = r², and every point on it satisfies that equation, so substituting the coordinates of a point that lies on the circle gives r² directly. Working: substituting x = 5 and y = 12 gives 5² + 12² = 25 + 144 = 169, so r² = 169 and the circle is x² + y² = 169. Answer: x² + y² = 169. The distractors: x² + y² = 13 uses the radius, √169 = 13, where r² belongs, which is the confusion between r and r² made in the other direction; x² + y² = 17 adds the two coordinates, 5 + 12, instead of adding their squares; x² + y² = 119 subtracts the squares, 144 − 25, treating 12 as the hypotenuse of the right-angled triangle rather than as one of the shorter sides.
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.