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
Answer key: GCSE Higher sample Paper 1 (non-calculator)
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- (d) 28 — Method: round each number to the nearest whole number, then multiply the rounded numbers to get an estimate that can be compared with the assistant's answer. Working: 7.2 rounds to 7, and 3.9 rounds to 4, so the estimate is 7 × 4 = 28. Since 28 is much smaller than 56.16, the assistant's answer cannot be correct. 56 comes from rounding the assistant's answer to the nearest whole number, instead of rounding the two numbers being multiplied and then multiplying them. 35 comes from rounding both numbers correctly but then slipping in the seven times table, writing 7 × 5 = 35 in place of 7 × 4 = 28. 21 comes from rounding 3.9 down to 3 instead of 4, giving 7 × 3 = 21. Answer: 28.
- (d) 6th term — The terms are aₙ = 3 × (√3)ⁿ⁻¹: a₁ = 3, a₂ = 3√3 ≈ 5.196, a₃ = 9, a₄ = 9√3 ≈ 15.588, a₅ = 27, a₆ = 27√3 ≈ 46.765. Since a₅ = 27 is still below 30 and a₆ ≈ 46.8 is above 30, the 6th term is the first to exceed 30. 4th term comes from treating the common ratio as 3 instead of √3: 3, 9, 27, 81, … — under that wrong ratio, 81, the 4th term, is the first over 30. 5th term comes from misjudging a₅ = 27 as already greater than 30. 7th term comes from tracking only the whole-number terms (3, 9, 27, 81, …) and reporting the ORIGINAL position of 81 in the full sequence — 81 is indeed a₇, but a₆ ≈ 46.8 already exceeds 30 first.
- (a) 18/25 — First find the new number of rose bushes: 90 × 1.2 = 108 (a 20% increase multiplies by 1.2). Then write 108 over 150 and divide top and bottom by 6 to get 18/25. Choosing 3/5 comes from using the original 90 rose bushes without applying the 20% increase (90/150 = 3/5). Choosing 25/18 comes from writing the number of lavender bushes over the new number of rose bushes, the wrong way round. Choosing 3/25 comes from multiplying 90 by 0.2 instead of 1.2, finding only the increase (18) rather than the new total, then writing 18/150 = 3/25.
- (b) 2.8 m — Use Pythagoras' Theorem: the plank is the hypotenuse (3.5 m) of a right-angled triangle formed with the wall and the ground (2.1 m). height² = 3.5² − 2.1² = 12.25 − 4.41 = 7.84. height = √7.84 = 2.8 m. A student who subtracts the two given lengths directly instead of using Pythagoras gets 3.5 − 2.1 = 1.4 m. A student who doubles the distance from the wall by mistake gets 2.1 × 2 = 4.2 m.
- (c) 39.5% — Win, draw and lose are mutually exclusive and exhaustive, so their probabilities sum to 100%: 42% + 18.5% = 60.5% is the percentage that wins or draws. 100% − 60.5% = 39.5% is the percentage that neither wins nor draws. Adding 42% and 18.5% and stopping there, 60.5%, is the probability of winning or drawing, not of neither. Subtracting only the 42% from 100% gives 58.0%, ignoring the draw percentage. Subtracting only the 18.5% from 100% gives 81.5%, ignoring the win percentage.
- (c) 18 — Method: for n ordered values, GCSE convention places the lower quartile at position (n + 1) ÷ 4, counting from the smallest value. Working: there are 11 marks, so n + 1 = 11 + 1 = 12 and 12 ÷ 4 = 3, so the lower quartile is the 3rd value in the ordered list 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, which is 18. Answer: the lower quartile is 18 marks. Watch the position you count to: dividing 11 ÷ 4 = 2.75 without adding 1 first, then rounding down, lands on the 2nd value, 15, not the 3rd; reaching for the middle of the whole list instead gives the median, 27, a different statistic; and averaging the 3rd and 4th values, 18 + 21 = 39 and 39 ÷ 2 = 19.5, borrows a method for an even split where it is not needed here.
- (b) 3 — Method: solving an index equation like this means finding how many factors of the base multiply together to give the number on the right. Working: 4¹ = 4, 4² = 16 and 4³ = 64, so three factors of 4 are needed. Answer: 3. The distractors: 4 comes from listing 4, 16 and 64 and counting the base itself as a step, which gives one more than the index; 6 comes from solving the equation with 2 as the base instead of 4, since 2⁶ = 64; 16 comes from dividing 64 by 4, treating the index as an instruction to divide.
- (d) −2 — y = f(x) − 6 is f(x) shifted down by 6, so every y-value on the graph decreases by 6. At x = 0, f(0) = 4, so the new y-value is 4 − 6 = −2. Adding 6 instead of subtracting gives 10; writing down the shift itself, −6, or leaving the original value 4 unchanged both skip the translation altogether.
- (d) Gradient = acceleration; area = distance travelled. — Method: on a speed–time graph, the gradient of the graph at an instant is the rate of change of speed with time, which is acceleration; the area between the graph and the time-axis over an interval is the total distance covered in that interval, because it accumulates speed × time. Working: gradient = acceleration and area = distance travelled is the correct pairing. Swapping the two quantities completely, gradient = distance travelled and area = acceleration, is the reverse of what each actually measures. Keeping gradient = acceleration correct but then also claiming area = acceleration too is wrong because the area is a different physical quantity, distance, not a second way of finding the same rate. Claiming the gradient itself gives the speed confuses the RATE OF CHANGE of the plotted quantity with the plotted quantity itself — the gradient is how fast the speed is changing, not the speed. On any rate graph, the gradient of the graph is always the RATE at that instant, and the area under the graph is always the TOTAL AMOUNT accumulated — keep straight which of the two questions each one answers.
- (b) A semicircle of radius 4 m, away from the wall. — Every point the dog can reach is at most 4 m from the fixed ring, so without any wall the region would be a full circle of radius 4 m. The wall runs straight through the ring and blocks the dog from crossing it, and since the wall extends further than the lead in both directions, exactly half of that circle is cut off — leaving a semicircle of radius 4 m on the side of the wall the dog is tied on. (A full circle of radius 4 m ignores that the wall blocks half of the region; a quarter circle of radius 4 m would only be correct if the ring were fixed at a corner where two walls met, not along a single straight wall; a rectangle 4 m wide along the wall ignores that the lead lets the dog swing round in a curve, not stay a fixed distance out from the wall.)
- (b) The relative frequency is settling near 0.5 — Method: turn each result into a relative frequency before comparing them, because it is the relative frequency, and not the difference between the two counts, that tends towards the theoretical probability. Working: after 10 flips the relative frequency of a head is 7 ÷ 10 = 0.7, which is a long way from 0.5. After 1000 flips it is 528 ÷ 1000 = 0.528, which is much closer to 0.5. Meanwhile the gap between the two counts has grown rather than shrunk: it was 7 − 3 = 4 after 10 flips and is 528 − 472 = 56 after 1000 flips. Answer: the relative frequency is settling near 0.5, which is what an unbiased experiment does as the sample grows. The distractors: saying the counts are levelling out is the usual form of this idea and the figures contradict it, since the gap went from 4 to 56; saying the coin is biased treats 28 extra heads in 1000 flips as proof, when 0.528 sits close to 0.5 and a fair coin gives results like this often; saying the next flip is more likely to be a tail is the gambler's fallacy, since each flip stays at 1/2 whatever came before.
- (a) 7/8 — Method: write the decimal over the power of ten that matches the number of digits after the point, then divide the numerator and the denominator by their highest common factor. Working: 0.875 has three digits after the point, so it is 875 thousandths and can be written as 875/1000; the highest common factor of 875 and 1000 is 125, and 875 ÷ 125 = 7 with 1000 ÷ 125 = 8. Answer: 7/8. The distractors: 8/7 comes from cancelling correctly but writing the two parts the wrong way round; 9/10 comes from rounding 0.875 to one decimal place as 0.9 before converting; 7/80 comes from counting four decimal places instead of three and using a denominator of 10000, giving 875/10000.
- (d) y = (x + 3)(x − 5) — A root at x = −3 means the matching bracket must be zero when x = −3, so the bracket is (x − (−3)) = (x + 3). A root at x = 5 means the other bracket is (x − 5). So the equation is y = (x + 3)(x − 5); checking the y-intercept, (0 + 3)(0 − 5) = 3 × (−5) = −15, which matches the given value. The option (x − 3)(x + 5) swaps the signs of both roots. The option (x + 3)(x + 5) keeps the correct sign for the first root but gets the second wrong. The option (x − 3)(x − 5) gets the first root's sign wrong.
- (c) £450 — Method: use the equal ratios 4:5 = 200:x to find Grace's savings, then add the two amounts. Working: Noah's £200 is 4 parts, so one part is £200 ÷ 4 = £50; Grace has 5 parts, so 5 × £50 = £250; altogether £200 + £250 = £450. Answer: £450. The distractors: £250 is Grace's savings on their own, which is the middle step rather than the total the question asks for; £360 comes from reading £200 as the 5 parts instead of the 4, giving one part of £40 and a total of 9 × £40; £400 comes from doubling £200, which treats the two savings as equal and ignores the ratio altogether.
- (b) Rotate 180° about the origin, then translate by (6, 0). — Rotating 180° about the origin sends (x, y) to (−x, −y); applied to T's vertices (1, 1), (3, 1) and (1, 4) this gives (−1, −1), (−3, −1) and (−1, −4). Translating this image by the vector (6, 0) adds 6 to every x-coordinate, giving (5, −1), (3, −1) and (5, −4), which matches T′ exactly. Reflecting in the x-axis first changes the sign of the y-coordinate only, and translating that image by (6, 0) gives (7, −1), (9, −1) and (7, −4) — the wrong triangle. Using the correct rotation but translating by (4, 0) instead of (6, 0) gives (3, −1), (1, −1) and (3, −4), shifted 2 units too far left. Reflecting in the y-axis first changes the sign of the x-coordinate only, so translating that image by (6, 0) leaves every y-coordinate positive, giving (5, 1), (3, 1) and (5, 4) — the correct x-coordinates but the wrong sign throughout on y.
- (a) 18 — Method: substitute the value, apply the index before the multiplication, and remember that a negative number multiplied by itself gives a positive result. Working: x² = (−3) × (−3) = 9, and then 2 × 9 = 18. Answer: 18. The distractors: −18 comes from squaring only the 3 and leaving the minus sign outside the index, giving 2 × (−9); 36 comes from multiplying 2 by −3 first and squaring afterwards, giving (−6)²; −12 comes from reading x² as 2x, so that the calculation becomes 2 × 2 × (−3).
- (a) 3/4 — Write January's total over February's total: 360/480. Both numbers share a factor of 120, so dividing top and bottom by 120 gives 3/4. Choosing 4/3 comes from writing February's amount over January's amount, the wrong way round. Choosing 1/4 comes from finding the difference between the two months (480 − 360 = 120) and writing it over February's amount, instead of using January's amount. Choosing 3/7 comes from writing January's amount over the total received across both months (360 out of 840), instead of over February's amount alone.
- (c) 18 — Rearranging F + V − E = 2 gives E = F + V − 2. Substitute F = 8 and V = 12: 8 + 12 − 2 = 18 edges. Choosing 20 comes from adding the faces and vertices but forgetting to subtract the 2 (8 + 12 = 20). Choosing 22 comes from adding the 2 instead of subtracting it (8 + 12 + 2 = 22). Choosing 16 comes from subtracting 2 twice by mistake (8 + 12 − 2 − 2 = 16).
- (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) x = −2, y = −5 — 3x² + 12x + 7 rewrites as 3(x² + 4x) + 7, then as 3[(x + 2)² − 4] + 7, which simplifies to 3(x + 2)² − 5, since −3 × 4 + 7 = −5. Substituting x = −2: 3 × (−2)² = 12, 12 × (−2) = −24, so 12 − 24 + 7 = −5, confirming the minimum value −5 at x = −2: turning point x = −2, y = −5. Halving b instead of halving b/a — using −4 as the shift instead of −2 — lands on turning point x = −4, y = −41, which is wrong because when a ≠ 1 the shift inside the bracket is b/(2a), not b/a alone. Reading the bracket's sign directly as the turning point's x-coordinate gives x = 2, y = −5 — wrong, because (x + 2)² is zero at x = −2, not x = 2. Computing 12 − 7 = 5 instead of 7 − 12 = −5 flips the sign of the constant, giving x = −2, y = 5 — wrong, since the completed square's constant must be evaluated as 7 minus 12, not 12 minus 7. Whenever a ≠ 1, factor a out of the x² and x terms first, and always check a turning point by substituting back into the original equation.
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.