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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- (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.
- (b) (2, 4) and (−2, −4) — Substitute y = 2x into x² + y² = 20: x² + (2x)² = 20, which gives x² + 4x² = 20, so 5x² = 20, x² = 4, and x = 2 or x = −2. Using y = 2x for each x-value: x = 2 gives y = 4; x = −2 gives y = −4. The path meets the pond's edge at (2, 4) and (−2, −4). Distractor routes: (2, −4) and (−2, 4) swaps the sign pairing, matching each x-value with the wrong sign of y instead of keeping each x with its own correctly-signed y. (2, 1) and (−2, −1) comes from using y = x/2 instead of y = 2x when finding the y-coordinates. (2, 4) alone stops after the positive square root of x² = 4 and never finds the second point from x = −2.
- (d) 7/3 — Put the kettle's energy over the toaster's energy: 2.1/0.9. Multiply both numbers by 10 to clear the decimals: 21/9. Divide both by their highest common factor, 3: 21÷3 = 7, 9÷3 = 3, giving 7/3. (3/7 comes from writing the energy values the wrong way round. 4/3 comes from finding the difference, 2.1 − 0.9 = 1.2 kWh, and writing it as a fraction of the toaster's energy, 1.2/0.9. 7/10 comes from comparing the kettle's energy to the total energy used by both appliances, 2.1/3.0.)
- (c) I is the same distance from all three sides. — The angle bisector from A is the locus of points equidistant from sides AB and AC, and the angle bisector from B is the locus of points equidistant from sides AB and BC. Point I lies on both bisectors, so I is equidistant from AB and AC, and also equidistant from AB and BC — meaning I is the same distance from all three sides. (Being the same distance from all three vertices instead describes the circumcentre, found from the perpendicular bisectors of the sides, not the angle bisectors; I being the midpoint of AB confuses the angle bisector construction with the perpendicular bisector of a side; I lying on side AC is wrong because the angle bisectors meet inside the triangle, not on one of its sides.)
- (a) unlikely — 3/8 = 3 ÷ 8 = 0.375, which is between 0 and 0.5, so the probability is best described as unlikely. Rounding 3/8 loosely to 'about a half' without converting it properly leads to evens, which is wrong since 0.375 is noticeably below 0.5. Reading 3/8 as the complement and using 1 − 3/8 = 5/8 = 0.625 instead gives likely, but the question asks about the probability of 3/8 itself, not its complement. Flipping the fraction to 8/3 ≈ 2.67, which is impossible for a probability, and rounding it towards the top of the scale gives certain.
- (a) 1,400 and 1,240, so combine the samples for one estimate — Method: scale each sample up to the whole stock, then use the fact that a larger sample gives a more reliable estimate than a smaller one. Working: the first sample gives 35 ÷ 50 = 0.7 and 0.7 × 2,000 = 1,400 paperbacks; the second gives 31 ÷ 50 = 0.62 and 0.62 × 2,000 = 1,240 paperbacks. Two random samples of the same size are expected to differ a little, so neither estimate is wrong. Putting the two together gives 35 + 31 = 66 paperbacks in 100 books, and 66 ÷ 100 = 0.66 with 0.66 × 2,000 = 1,320, an estimate resting on twice as many books as either volunteer checked. Answer: 1,400 and 1,240, so combine the samples for one estimate. The distractors: keeping 1,400 because it is larger picks an estimate by its size, when both samples held 50 books and neither has a stronger claim; saying a volunteer must have miscounted assumes two random samples ought to agree exactly, which is precisely what random sampling does not promise; 1,750 and 1,550 come from 35 × 50 = 1,750 and 31 × 50 = 1,550, multiplying each count by the size of the sample instead of scaling by 2,000 ÷ 50.
- (d) 0.3 — Converting the fractions to decimals, 1/4 = 0.25 and 2/5 = 0.4, so any decimal between 0.25 and 0.4 is a valid answer, and 0.3 fits. Confusing 1/4 with 1/5 and converting it as 0.2 instead of 0.25 gives a value below the true lower bound. Confusing 2/5 with 1/2 and converting it as 0.5 instead of 0.4 gives a value above the true upper bound. Converting the fractions correctly but choosing a decimal above the true upper bound of 0.4 instead of between the two values gives 0.45.
- (d) 47 — To reach the 8th term from the 1st term, the difference of 6 is added 7 times (8 − 1 = 7): 5 + 7 × 6 = 47. A candidate who multiplies by the term number itself, rather than one less, would compute 5 + 8 × 6 = 53. A candidate who uses one step too few (6 instead of 7) would reach 5 + 6 × 6 = 41. A candidate who forgets to include the first term at all would compute just 8 × 6 = 48.
- (a) 70 km/h — Method: average speed for a whole journey is the total distance divided by the total time, not the mean of the separate speeds. Working: 3 × 80 = 240 km and 1 × 40 = 40 km, giving 280 km in 3 + 1 = 4 hours, so 280 ÷ 4 = 70. Answer: 70 km/h. The distractors: 60 km/h comes from taking the mean of 80 and 40, which would only be right if equal times were spent at each speed; 280 km/h is the total distance written with a speed unit, from forgetting to divide by the total time; 80 km/h comes from quoting the speed of the longer leg as the average for the whole journey.
- (a) −2 — Let the centre be C = (c, 0). The vector from the centre to the image equals the scale factor k times the vector from the centre to the object: (7 − c, −4) = k(1 − c, 2). The y-coordinate gives −4 = 2k, so k = −2 — this doesn't depend on knowing c. (Checking: substituting k = −2 into the x-equation gives c = 3, consistent with a centre on the x-axis.) '2' comes from taking the magnitude of the ratio without noticing the image is on the opposite side of the centre from the object, so the sign should be negative. '−1/2' comes from inverting the scale factor, dividing the object's coordinate by the image's instead of the other way round. '3' is the x-coordinate of the centre, mistaken for the scale factor.
- (d) No, because 3/8 + 5/12 + 1/6 = 23/24 — Using a common denominator of 24: 3/8 = 9/24, 5/12 = 10/24 and 1/6 = 4/24. Adding these numerators gives 9 + 10 + 4 = 23, so the three probabilities sum to 23/24, which is less than 1 — Zara is not correct. Adding the original numerators (3 + 5 + 1 = 9) over a denominator of 12 instead of converting each fraction properly gives 9/12 = 3/4, still less than 1 but the wrong fraction. Converting 1/6 to 5/24 instead of 4/24 (using the wrong scaling) makes the total 9/24 + 10/24 + 5/24 = 24/24 = 1, wrongly suggesting the probabilities are valid. Judging validity from the fact that each individual fraction lies between 0 and 1 ignores that an exhaustive set must sum to exactly 1, not merely contain valid individual values.
- (d) √13 — Method: everything under a root sign is worked out first, because a square root cannot be taken term by term across an addition. Working: 2² = 4 and 3² = 9, so the expression under the root is 4 + 9 = 13. As 13 is not a square number, the exact value is left in root form as √13. Answer: √13. The distractors: 5 comes from rooting each square separately and adding, 2 + 3, which treats the root of a sum of squares as the sum of the numbers; 13 comes from working out the sum under the root correctly and then forgetting to take the root; √5 comes from subtracting the two squares, 9 − 4, instead of adding them.
- (a) x = 10 or x = −10 — Rearranging, x² = 100. Taking the square root of both sides gives x = ±10, i.e. x = 10 or x = −10. A candidate who forgets the negative root gives only x = 10. A candidate who halves 100 instead of taking its square root gets x = 50. A candidate who applies the ± sign to 100 itself instead of to its square root gets x = 100 or x = −100.
- (a) 7 — The average rate of change of y with respect to x over an interval is the change in y divided by the change in x between its two endpoints — the gradient of the chord joining them, not a rate at a single point. At x = 1: 2 × 1 = 2, so y = 1 + 2 = 3. At x = 4: 2 × 4 = 8, so y = 16 + 8 = 24. The change in y is 24 − 3 = 21 and the change in x is 4 − 1 = 3, so the average rate of change is 21 ÷ 3 = 7. Reporting the change in y, 21, on its own is not a rate of change, because it has not been divided by the 3 units of x over which it happened. Reporting 3 is not a rate either — 3 is the width of the interval, and also the value of y at x = 1, and neither of those measures how fast y is changing. Subtracting in the wrong order gives −21 ÷ 3 = −7, the wrong sign. The average rate of change of y with respect to x over the interval is 7.
- (d) 15 cm — Method: any two sides of a triangle must together be longer than the third, so the third side must be longer than the difference of the two given sides and shorter than their sum. Working: the difference is 20 − 8 = 12 cm and the sum is 20 + 8 = 28 cm, so the third side must be between 12 cm and 28 cm, and 15 cm lies inside that range. Answer: 15 cm. The distractors: 12 cm is exactly the difference, so the three lengths would lie flat along a straight line and never close into a triangle; 5 cm is shorter than the difference — 5 + 8 = 13 cm cannot reach across the 20 cm side — and is chosen by candidates who check no lower limit at all; 30 cm is longer than the sum of the other two, so those two sides could never meet, and it is chosen by candidates who check no upper limit.
- (c) 6 — Method: substitute the value, work out the top of the fraction first, then the division, and add the 3 last. Working: the top gives 10 − 4 = 6, dividing by 2 gives 6 ÷ 2 = 3, and adding 3 gives 3 + 3 = 6. Answer: 6. The distractors: 11 comes from dividing only the 4 by 2 instead of the whole of the top, giving 10 − 2 + 3; 4.5 comes from dividing the + 3 by 2 as well, giving (10 − 4 + 3) ÷ 2; 0 comes from subtracting the wrong way round on the top, giving (4 − 10) ÷ 2 = −3 and then −3 + 3.
- (b) 6 litres — Method: find the volume of the cuboid in cm³, then change cm³ into litres using 1 litre = 1000 cm³. Working: 30 × 20 × 10 = 6000 cm³, and 6000 ÷ 1000 = 6. Answer: 6 litres. The distractors: 60 litres comes from using 1 litre = 100 cm³; 600 litres comes from using 1 litre = 10 cm³; 0.6 litres comes from using 1 litre = 10 000 cm³.
- (d) (11, 0.75) — Add the moves to the starting point one component at a time. x: 12.5 + (−3.75) + 2.25 = 11; y: −4.25 + 6.5 + (−1.5) = 0.75, giving (11, 0.75). (8.75, 2.25) stops after the first move only and never applies the second vector. (6.5, 3.75) comes from subtracting the second vector instead of adding it. (0.75, 11) comes from swapping the final x-coordinate and y-coordinate.
- (b) 2n + 3 — Method: multiply the cost per mile by the number of miles to get an expression, then add the fixed fee as a separate term. Working: n miles at £2 each is 2n; add the £3 fixed fee: 2n + 3. Answer: 2n + 3. 3n + 2 comes from swapping the fee and the rate round, treating £3 as the rate per mile and £2 as the fixed fee. 5n comes from adding the fee and the rate together first (3 + 2 = 5) and multiplying the result by n, instead of keeping the fixed fee as its own term. 2n − 3 comes from subtracting the fixed fee instead of adding it.
- (d) 6 — Method: collect the x-terms on one side and the constants on the other, then divide by the remaining coefficient of x. Working: 6x − 3x = 13 + 5, so 3x = 18, x = 18 ÷ 3 = 6. Answer: x = 6. 2.67 comes from a sign error when moving the 5, subtracting instead of adding: 3x = 13 − 5 = 8, x = 8 ÷ 3 ≈ 2.67. 2 comes from a sign error when moving the x-term, adding instead of subtracting: 9x = 18, x = 2. 18 comes from correctly finding 3x = 18 but forgetting to divide by 3.
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.