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)
MathsUKwww.geekhero.co.uk
- 1.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.
- 2.Solve the inequality 3x − 1 ≤ 11.
- 3.A recipe for 8 muffins needs 200 g of sugar. Sam wants to make 20 muffins for a bake sale, and he already has 350 g of sugar. Work out how many more grams of sugar he needs to buy.
- 4.A scale model of a bridge is built so that 2 cm on the model represents 5 m of the real bridge. The real bridge is 45 m long. Work out the length of the model bridge, in centimetres.
- 5.A market research company found that, out of 120 shoppers surveyed, 42 preferred paper bags to plastic bags. A supermarket expects 900 shoppers this week. Using this relative frequency, work out how many of the 900 shoppers would be expected to prefer paper bags.
- 6.In a spelling test the 20 pupils in Group A had a mean mark of 80, and the 30 pupils in Group B had a mean mark of 70. Work out the mean mark of all 50 pupils.
- 7.The number 72 can be written as 2³ × 3², and the number 108 can be written as 2² × 3³. Work out the highest common factor of 72 and 108.
- 8.Yusuf has £40 saved and adds £6 each week. His sister Aisha has £10 saved and adds £11 each week. After how many weeks will they have saved the same amount?
- 9.The speed of a car is given by v = d ÷ t. A car travels 180 km in 3 hours. Work out the average speed of the car in km/h.
- 10.Work out the exact value of cos 0° − sin 30°.
- 11.A basketball player has a free-throw success probability of 0.75. She wants to expect to score 60 successful free throws. Work out how many free throws she needs to attempt.
- 12.A digital timer truncates every time to 1 decimal place. It shows a swimmer's time for one length as 12.3 seconds. Using t for the swimmer's actual time in seconds, write down the error interval for t.
- 13.The graph of y = f(x) has a root (an x-intercept) at x = 5. Work out the x-coordinate of the corresponding root on the graph of y = f(x + 2).
- 14.Two mathematically similar cubes have edge lengths 2 cm and 6 cm. Write the ratio of the volume of the smaller cube to the volume of the larger cube in its simplest form.
- 15.The diagram shows the plan of the base layer of a solid built from identical cubes. A second layer of cubes is added on top, filling the entire base layer to a height of 2 cubes everywhere. Work out how many squares are visible in the plan view of this solid.
- 16.A student is asked whether 3(x − 4) = 3x − 4 is an identity. Which statement gives the correct verdict and reason?
- 17.In a bag the ratio of red counters to blue counters is 7 : 4. Write the number of red counters as a fraction of the number of blue counters.
- 18.A packaging designer is told a box has 6 faces, 12 edges and 8 vertices, and every face is a rectangle. To fit exactly onto a shelf, the designer needs to know whether the box must be a cube. Based only on this information, which solid must the box be?
- 19.The point (9, 12) lies on the circle x² + y² = 225, which has centre (0, 0). The tangent to the circle at (9, 12) crosses the x-axis at the point P. Work out the x-coordinate of P.
- 20.The point (3, 6) lies on the circle x² + y² = 45. The tangent to the circle at (3, 6) crosses the x-axis at the point P. Work out the coordinates of P.
Answer key
- (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.
- (b) x ≤ 4 — Method: undo the number subtracted from 3x first, then divide by 3; the direction of the sign changes only for division by a negative number. Working: adding 1 to both sides of 3x − 1 ≤ 11 gives 3x ≤ 12; dividing both sides by 3, which is positive, gives x ≤ 4. Answer: x ≤ 4. The distractors: x ≥ 4 comes from turning the sign round while dividing by 3; x ≤ 10/3 comes from subtracting 1 from both sides instead of adding it, giving 3x ≤ 10; x ≤ 12 comes from stopping at 3x ≤ 12 and reading off the 12 without dividing by 3.
- (a) 150 g — Method: scale the recipe to find the total sugar needed, then subtract the sugar Sam already has. Working: 200 ÷ 8 × 20 = 500, so 500 g is needed in total; 500 − 350 = 150, so 150 g still to buy. Stopping after finding the total, 500, without subtracting what he has gives 500 g. Scaling the wrong way round, 200 × 8 ÷ 20 = 80, wrongly suggests he already has enough, giving 0 g. Adding the amount he has instead of subtracting it, 500 + 350 = 850, gives 850 g.
- (c) 18 — Method: divide the real length by 5 to find how many 'units' of 5 m it contains, then multiply by 2 cm for each unit. Working: 45 ÷ 5 = 9, so the real bridge is 9 lots of 5 m; each lot is represented by 2 cm on the model, so the model length is 9 × 2 = 18 cm. Options: 9 comes from stopping after the division, without multiplying by the 2 cm per unit; 90 comes from multiplying the real length by 2 directly, without dividing by 5 first; 4.5 comes from dividing by 5 and then dividing by 2 again, instead of multiplying by 2. Answer: 18.
- (d) 315 — The relative frequency from the survey is 42 ÷ 120 = 0.35, so the expected number who prefer paper bags among 900 shoppers is 0.35 × 900 = 315. Giving 42 as the answer reuses the original survey count without scaling it up to 900 shoppers at all. Finding the expected number who prefer PLASTIC bags instead of paper, using the relative frequency 78 ÷ 120 = 0.65, gives 0.65 × 900 = 585. Using 1000 shoppers instead of the 900 actually stated gives 0.35 × 1000 = 350.
- (b) 74 marks — Method: the two groups are different sizes, so their means cannot simply be averaged — rebuild each group's total mark, add the totals and divide by all 50 pupils. Working: Group A scored 20 × 80 = 1600 marks and Group B scored 30 × 70 = 2100 marks, giving 1600 + 2100 = 3700 marks altogether, so the overall mean is 3700 ÷ 50 = 74 marks. Answer: 74 marks, which sits nearer to 70 than to 80 because the larger group scored 70. The distractors: 75 marks comes from averaging the two group means, (80 + 70) ÷ 2, as though the groups were the same size; 76 marks comes from attaching each mean to the other group's size, (20 × 70 + 30 × 80) ÷ 50; 150 marks comes from adding the two means together and never dividing at all.
- (a) 36 — Method: for the highest common factor, take the LOWER power of each prime that appears in both numbers. Working: for 2, the lower power is 2² (from 108); for 3, the lower power is 3² (from 72), so the highest common factor is 2² × 3² = 4 × 9 = 36. 216 comes from taking the higher power of each prime instead, 2³ × 3³ = 8 × 27 = 216, which gives the lowest common multiple, not the highest common factor. 108 is simply one of the two numbers, not their highest common factor. 6 comes from multiplying the primes without any powers at all, 2 × 3 = 6. Answer: 36.
- (d) 6 — Method: write an expression for each person's savings after w weeks, and set them equal. Working: 40 + 6w = 10 + 11w. Subtract 6w from both sides: 40 = 10 + 5w. Subtract 10: 30 = 5w, so w = 6. Answer: 6 weeks. 0 comes from setting only the weekly amounts equal, 6w = 11w, and ignoring the different starting amounts entirely. 10 comes from adding the two starting amounts and dividing by the difference in weekly amounts, (40 + 10) ÷ (11 − 6), instead of forming and solving the correct equation. 1.76 comes from adding the two weekly amounts instead of subtracting them when rearranging, (40 − 10) ÷ (11 + 6).
- (d) 60 km/h — Method: use the formula v = d ÷ t with the distance and time given. Working: 180 ÷ 3 = 60 km/h. So the average speed is 60 km/h. Distractor 540 km/h comes from multiplying the distance and time instead of dividing. Distractor 90 km/h comes from dividing by 2 instead of 3. Distractor 18 km/h comes from dividing by 10 instead of 3, a decimal-point slip.
- (a) 1/2 — cos 0° = 1 and sin 30° = 1/2, so cos 0° − sin 30° = 1 − 1/2 = 1/2. 1 comes from writing down cos 0° alone and forgetting to subtract sin 30°. 3/2 comes from adding the two values instead of subtracting. −1/2 comes from working out sin 30° − cos 0°, the two terms the wrong way round.
- (b) 80 — To find the number of attempts needed, divide the target number of successes by the probability of success: 60 ÷ 0.75 = 80. Writing 45 is wrong because 60 × 0.75 = 45 multiplies instead of dividing — that is the number of successes expected from 60 attempts, not the number of attempts needed for 60 successes. Writing 240 is wrong because 60 ÷ 0.25 = 240 uses 0.25, the probability of MISSING, instead of 0.75, the probability of scoring. Writing 90 is wrong because it comes from misremembering 0.75 as 2/3 and dividing by that instead: 60 ÷ (2/3) = 90. She needs to attempt 80 free throws.
- (c) 12.3 ≤ t < 12.4 — Method: truncating cuts the later digits off instead of rounding them, so nothing is ever pushed upwards. The displayed value is therefore the smallest the time can be, and the time can run up to, but not reach, the next value the display can show. Working: the display reads 12.3, so the actual time is at least 12.3 seconds; as soon as the time reaches 12.3 + 0.1 = 12.4 seconds the display would read 12.4, so 12.4 is not included. Answer: 12.3 ≤ t < 12.4. The distractors: 12.25 ≤ t < 12.35 is the interval for a time rounded to 1 decimal place, and this display does not round; 12.3 < t ≤ 12.4 excludes the one value the display certainly allows and includes the one it rules out; 12.3 ≤ t ≤ 12.4 treats 12.4 seconds as possible, but at 12.4 seconds the display would no longer read 12.3.
- (a) 3 — y = f(x + 2) is f(x) translated 2 units to the LEFT (inside the bracket, adding moves the graph in the negative x-direction). The root moves with the whole graph: 5 − 2 = 3. Moving right instead of left gives 7; assuming a bracket shift leaves the root unchanged gives 5; writing down the shift amount 2 itself skips the translation altogether.
- (c) 1 : 27 — The edge lengths are in the ratio 2 : 6, which simplifies to 1 : 3. Volumes scale with the cube of the length ratio, so the volume ratio is 1³ : 3³ = 1 : 27. Giving 1 : 3 uses the length ratio without cubing it. Giving 1 : 9 squares the length ratio, which is the rule for areas, instead of cubing it, which is the rule for volumes. Giving 27 : 1 has the ratio the right way round for larger to smaller, not smaller to larger as the question asks.
- (c) 6 squares — Method: the plan view shows the footprint of the solid; a second layer stacked on top of floor positions that are already covered does not create any new squares in the plan. Working: the row of 4 cubes and the row of 2 cubes attached at the end do not overlap, so the footprint has 4 + 2 = 6 distinct squares. Answer: 6 squares. The distractors: 12 squares comes from counting the total number of cubes used, including the second layer (6 floor positions × 2 layers = 12), instead of the footprint. 4 squares comes from counting only the row of four and forgetting the attached row of two. 5 squares comes from wrongly treating the corner square as shared between the two rows (4 + 1 instead of 4 + 2).
- (b) It is not even an ordinary equation with a solution: expanding the left-hand side gives 3x − 12, and 3x − 12 = 3x − 4 would require −12 = −4, which is never true. — Expanding the left-hand side, 3(x − 4) = 3x − 12. Setting this equal to the right-hand side, 3x − 12 = 3x − 4, gives −12 = −4 once the 3x terms are removed from both sides — a statement that is never true, so no value of x satisfies the equation at all, and it is certainly not an identity. The option about substituting a specific value misunderstands algebraic expansion, which holds for every x, not one chosen value. The option matching the first term wrongly assumes that is enough to prove equivalence. The option about multiplying the 4 by 3 on both sides is nonsensical, since there is only one bracket to expand, on the left-hand side.
- (d) 7/4 — A part-to-part ratio a : b gives the fraction a/b when the first quantity is written as a fraction of the second, so 7 : 4 gives 7/4. Writing 4/7 puts the parts the wrong way round — blue as a fraction of red, not red as a fraction of blue. Writing 7/11 uses the total number of counters, 7 + 4 = 11, as the denominator instead of the number of blue counters — that is red as a fraction of the whole bag, not red as a fraction of blue. Writing 11/7 has both the wrong denominator and the parts inverted.
- (b) Cuboid — not necessarily a cube — A solid with 6 faces, 12 edges and 8 vertices in which every face is a rectangle is a cuboid, but nothing here confirms that all the edges are the same length, so the box could be a cube or a non-cube cuboid; the most that can be concluded is that it is a cuboid, making 'Cuboid — not necessarily a cube' correct. 'Cube — only a cube fits this' is wrong because a cube is just one particular cuboid; a general cuboid with different length, width and height has exactly the same face, edge and vertex counts and rectangular faces. 'Triangular prism' is wrong because a triangular prism has 5 faces, 9 edges and 6 vertices, and two of its faces are triangles, so it matches neither the counts nor the face shape. 'Not enough information' is wrong because rectangular faces with these counts do pin the solid down to the cuboid family, even though they cannot pin down a cube specifically.
- (b) 25 — The tangent at (9, 12) is 9x + 12y = 225. Setting y = 0 (the x-axis): 9x = 225, so x = 25. Choosing 18.75 comes from swapping the coefficients in the tangent equation (using 12x + 9y = 225) before setting y = 0. Choosing 15 is where the circle itself meets the x-axis (from x² = 225), not where the tangent does. Choosing 9 is just the x-coordinate of the original point (9, 12), not the point P.
- (a) (15, 0) — Method: the tangent is perpendicular to the radius at the point of contact, so find the gradient of the radius, take its negative reciprocal, write the equation of the tangent, then substitute y = 0 because every point on the x-axis has y-coordinate 0. Working: the radius from (0, 0) to (3, 6) has gradient 6 ÷ 3 = 2, so the tangent has gradient −1/2. Substituting into y − 6 = −1/2(x − 3) gives y = −0.5x + 7.5. Setting y = 0 gives 0.5x = 7.5, so x = 15 and P is (15, 0). Answer: (15, 0). The distractors: (0, 7.5) is where the same tangent crosses the y-axis, reached by setting x = 0 instead of y = 0; (0, 0) comes from using the gradient of the radius, 2, for the tangent, which gives the line y = 2x through the centre and so crosses the x-axis at the origin; (6, 0) comes from changing the sign of the radius gradient without turning it upside down, which gives y = −2x + 12.
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.