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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- (c) 24 — Method: round each number to the nearest whole number, then multiply the rounded values. Working: 6.4 rounds to 6 (nearest whole number) and 3.9 rounds to 4 (nearest whole number). 6 × 4 = 24. Answer: 24. 18 comes from rounding 3.9 down to 3 instead of up to the nearest whole number, 4, giving 6 × 3. 28 comes from rounding 6.4 up to 7 instead of down to the nearest whole number, 6, giving 7 × 4. 25 is the exact value of 6.4 × 3.9, which is 24.96, rounded to the nearest whole number after multiplying, rather than estimated by rounding first.
- (d) It has a solution between x = 2.2 and x = 2.3 — Since f(2.2) is negative and f(2.3) is positive, the graph of f crosses the x-axis somewhere between x = 2.2 and x = 2.3, so the equation has a solution in that interval. Choosing 'between x = −2.2 and x = −2.3' confuses the negative f-VALUE at 2.2 with a negative x-value. Choosing 'no solution' misreads a change of sign as meaning the opposite of what it shows. Choosing 'exactly two solutions' assumes a single change of sign must give two roots, which is not what the rule guarantees.
- (c) 8 m — Method: in the same sunlight every object has its height and its shadow in the same ratio, so write 2:3 = h:12, find the multiplier that takes 3 to 12 and apply it to the height. Working: 12 ÷ 3 = 4, so the tree's shadow is 4 times the post's shadow; the height must be scaled by the same 4, giving 4 × 2 = 8 m. Answer: 8 m. The distractors: 18 m comes from setting up the proportion upside down, 12 ÷ 2 × 3, which scales by shadow over height instead of height over shadow; 24 m comes from multiplying the 12 m shadow by the post's height of 2 m and never dividing by the post's shadow of 3 m; 4 m is the scale factor 12 ÷ 3, given as a length instead of being used to scale the 2 m post.
- (b) 58° — Method: the angle at the centre is TWICE the angle at the circumference when both stand on the same arc. Working: angle AOC is the angle at the centre standing on arc AC, and angle ABC is the angle at the circumference standing on the same arc AC, with B on the major arc, so angle AOC = 2 × angle ABC. Rearranging, angle ABC = 116 ÷ 2 = 58 degrees. Answer: 58°. Halve the centre angle, do not double it, and use the angle AOC exactly as given, 116°: you do not need its reflex angle here, because B sits on the major arc, which is precisely the arrangement the theorem is stated for.
- (c) 1/36 — Method: the two dice do not affect each other, so the probability that both results happen is the product of the two separate probabilities. Working: a 1 on the red dice has probability 1/6 and a 2 on the blue dice has probability 1/6. Multiplying gives 1 × 1 = 1 on the top and 6 × 6 = 36 on the bottom. Answer: the probability is 1/36. The distractors: 1/18 comes from allowing either order, counting a 2 on the red dice with a 1 on the blue dice as well, although the stem names which dice shows which score; 1/3 comes from adding 1/6 and 1/6 instead of multiplying; 1/6 comes from giving the probability for one dice only and treating the other as making no difference.
- (c) Class X has the higher median and the wider spread — Method: compare the two box plots statistic by statistic — median for location, and the interquartile range for spread — checking the true value of each rather than assuming a pattern. Working: Class X's median is 60 and Class Y's is 58, so Class X's median is the higher one. Class X's interquartile range is 70 − 45 = 25 and Class Y's is 65 − 50 = 15 (and the ranges follow the same order: 95 − 20 = 75 against 80 − 35 = 45), so Class X also has the wider spread. Answer: Class X has both the higher median and the wider spread. Watch that each half of a compound statement is checked separately: claiming Class Y has the higher median and the wider spread gets both comparisons backwards; claiming Class X has the higher median but the narrower spread keeps the median right while reading the spread the wrong way round; and claiming Class Y has the higher median but the narrower spread swaps the median comparison while getting the spread right.
- (a) 200 g — One quarter of 160 g is 40 g. Increasing the amount means adding this on: 160 + 40 = 200 g. Finding the increase, 1/4 of 160 = 40 g, but stopping there without adding it to the original amount leaves just 40 g. Using 4/5 instead of 5/4 as the scaling fraction, 160 × 4/5 = 128 g, actually decreases the amount rather than increasing it. Increasing by a half instead of a quarter, 160 + 80 = 240 g, uses the wrong fraction of 160.
- (b) (n + 1)² − n² = 2n + 1 — (n + 1)² = n² + 2n + 1, so (n + 1)² − n² = n² + 2n + 1 − n² = 2n + 1, which is odd because it is one more than the even number 2n. Expanding (n + 1)² as n² + 1 uses the false rule (a + b)² = a² + b², and subtracting n² from that leaves just 1 — always expand (a + b)² as a² + 2ab + b². Writing n² + 2n + 1 expands correctly but never carries out the subtraction of n². Writing 2n forgets the constant term left after subtracting.
- (c) 4 : 5 — Method: a formula of the form y = kx says that for every 1 of x there are k of y, so the ratio y : x is k : 1, which is then written with whole numbers and cancelled. Working: here k = 0.8, so y : x = 0.8 : 1; multiplying both parts by 10 gives 8 : 10, and dividing both parts by their highest common factor 2 gives 4 : 5. Answer: 4 : 5. The distractors: 5 : 4 comes from writing x : y, reversing the order asked for; 8 : 10 has the decimal cleared but the ratio left unsimplified, and the question asks for the simplest form; 8 : 1 comes from multiplying only the left-hand part by 10 and leaving the 1 unchanged, which changes the relationship instead of keeping it.
- (b) 166.3 cm² — Method: split the regular hexagon into 6 identical triangles meeting at the centre, each with two sides of 8 cm and a 60° angle between them, and use Area = (1/2)ab sin C on just one of them. Working: one triangle's area = 1/2 × 8 × 8 × sin 60° = 27.7 cm² (1 d.p.); the hexagon is 6 of these, so its area is 6 × 27.7 = 166.3 cm² (1 d.p.). Answer: 166.3 cm². Reporting just one triangle's area, without multiplying by 6, gives 27.7 cm²; treating the angle at the centre as a right angle instead of 60°, using 1/2 × 8 × 8 with no sine factor at all, gives 6 × 32 = 192.0 cm²; and multiplying by 5 instead of 6, miscounting the triangles in the hexagon, gives 5 × 27.7 = 138.6 cm². A regular hexagon always splits into exactly 6 triangles at its centre — count them before you multiply.
- (d) 3/50 — Relative frequency is the number of faulty bolts divided by the total sample size: 15/250, which simplifies to 3/50 by dividing both the numerator and the denominator by 5. Using 235, the number of bolts that were NOT faulty, as the denominator instead of the total 250 gives 15/235, which simplifies to 3/47. Inverting the fraction, dividing the total by the number of faulty bolts instead of the other way round, gives 250/15, which simplifies to 50/3 — a value greater than 1, which cannot be a probability. Simplifying by dividing the numerator and the denominator by different numbers, 15 ÷ 15 = 1 and 250 ÷ 25 = 10, gives 1/10.
- (c) 6 — The units digit must be even, so it can be 2 or 8, giving 2 choices. The tens digit can then be any of the remaining 3 digits, since one digit has been used for the units. Multiply: 2 × 3 = 6. 12 comes from working out how many two-digit numbers can be made in total, 4 × 3 = 12, ignoring the requirement that the number is even. 8 comes from choosing the units digit from 2 options and then wrongly allowing any of the 4 digits again for the tens digit, 2 × 4 = 8, which lets a digit repeat. 2 comes from counting only the choices for the units digit and forgetting the tens digit.
- (d) Line 3 — Method: check each line of Chloe's working against the correct algebraic step. Working: Line 1 to Line 2 is correct, subtracting 2x from both sides gives 3x − 4 = 8. But Line 2 to Line 3 should add 4 to both sides, giving 3x = 12, not 3x = 4 — the constant −4 has been dropped rather than removed correctly. Line 4 follows correctly from Chloe's own, incorrect, Line 3. Answer: Line 3. Line 1 is simply the original equation, copied out correctly. Line 2 correctly subtracts 2x from both sides of Line 1. Line 4 divides Chloe's own Line 3 by 3 correctly — the arithmetic there is fine, the mistake happened one line earlier.
- (c) 15 — Method: the number of cakes is in direct proportion to the mass of flour, so find the multiplier between the two masses and apply it to the number of cakes. Working: 6 ÷ 2 = 3, so there is three times as much flour, and 5 × 3 = 15. Answer: 15. The distractors: 10 comes from multiplying the 5 cakes by 2, the mass in the recipe, instead of by the multiplier 3; 20 comes from multiplying by the difference 6 − 2 = 4, treating a proportion problem as a difference problem; 12 comes from rounding 5 ÷ 2 down to 2 cakes per kilogram and working out 6 × 2.
- (b) Square — A square has all four sides equal, all four angles equal to 90°, and diagonals that are equal in length and bisect each other at right angles — every part of the description matches, so Square is correct. A rhombus has all four sides equal and diagonals bisecting at right angles, but its interior angles are not generally 90° (only a square, a special rhombus, has that), so it does not fully match. A rectangle has four 90° angles and equal diagonals, but its sides are not all equal in general, so it fails the equal-sides condition. A kite has two pairs of adjacent equal sides rather than all four sides equal, and its diagonals are not generally equal in length, so it fails both conditions.
- (d) 6 — Method: rearrange the formula to make d the subject, then substitute C = 133. Working: C = 25 + 18d, so subtracting 25 from both sides gives C − 25 = 18d, then dividing by 18 gives d = (C − 25) / 18. Substituting C = 133: d = (133 − 25) / 18 = 108 / 18 = 6. The value 7.4 comes from dividing 133 by 18 without subtracting the fixed £25 first (133 / 18 ≈ 7.4). The value 4.6 comes from pairing the numbers the wrong way round, working out (133 − 18) / 25 = 4.6. The value 90 comes from subtracting both 25 and 18 from 133 instead of dividing by 18.
- (a) 20 litres per minute — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, and its units are the vertical unit for each one of the horizontal unit. Working: from (2, 50) to (6, 130) the volume changes by 130 − 50 = 80 litres and the time changes by 6 − 2 = 4 minutes, so the gradient is 80 ÷ 4 = 20, measured in litres for each minute. Answer: 20 litres per minute. The distractors: 25 litres per minute comes from using one point on its own, 50 ÷ 2, which assumes the line starts at the origin when the tank already held 50 litres at 2 minutes; 0.05 litres per minute comes from dividing the change in time by the change in volume, 4 ÷ 80, which gives the time for each litre but is then labelled as litres for each minute; 20 minutes for each litre has the right value with the units the wrong way round, and a tank that needed 20 minutes to gain a single litre would be filling far more slowly than this one.
- (a) 46° — Method: use the angle at the centre, then the cyclic quadrilateral, then the angle sum of a triangle. Working: angle ABC = 152° ÷ 2 = 76° (angle at the centre is twice the angle at the circumference); angle ADC = 180° − 76° = 104° (opposite angles of a cyclic quadrilateral sum to 180°); angle DAC = 180° − 104° − 30° = 46° (angle sum of triangle ACD). Skipping the halving step and taking angle ABC = 152° gives angle ADC = 180° − 152° = 28° and then angle DAC = 122°; using angle ABC = 76° directly as angle ADC, skipping the cyclic quadrilateral step, gives 74°; and finding angle DAC as 180° − 104° without subtracting the given 30° gives 76°. Each of the three steps must be carried through in order — halve, then subtract from 180°, then subtract from 180° again with the extra angle.
- (d) 4 — Subtracting the second equation from the first: the y-terms cancel, and the x-terms combine as 3x − x = 2x; the right-hand sides give 14 − 6 = 8. This gives 2x = 8, so x = 4. A candidate who subtracts in the wrong order would get 2x = 6 − 14 = −8, so x = −4. A candidate who uses only the first equation, dividing 14 by 3 as if y were 0, would get x ≈ 4.67. A candidate who reports the value of y instead of x would get y = 6 − 4 = 2.
- (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.