Sample paper · GCSE Higher · grades 4–9
GCSE Higher sample Paper 2 (calculator)
The real Paper 2 is 1.5 hour 30 minutes and 80 marks, 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 a calculator allowed. AO1 / AO2 / AO3 at this tier: 40% / 30% / 30%.
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Answer key: GCSE Higher sample Paper 2 (calculator)
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- (d) 40 miles — Method: round each number to 1 significant figure first, then divide to estimate the daily distance. Working: 830 rounds to 800, and 19 rounds to 20, and 800 ÷ 20 = 40, so the estimate is 40 miles per day. 41.5 miles comes from rounding only the number of days and working out 830 ÷ 20 = 41.5, without rounding the distance too. 830 miles is the total distance for the whole trek, given as the answer without dividing by the number of days at all. 4 miles comes from working out 80 ÷ 20 = 4, misplacing a digit in the rounded distance. Answer: 40 miles.
- (c) Only x = 5 satisfies 3x + 5 = 20, not every value of x. — 3x + 5 = 20 is only true when x = 5, since 3 × 5 + 5 = 20; for any other value of x the two sides are not equal, so it is an equation, not an identity. Saying it cannot be simplified confuses simplifying with the equation/identity distinction, which is about how many values of x make it true. Saying it has an = sign is not a valid test, since identities are also written with an = or ≡ sign. A number on the right-hand side does not decide it either — what matters is whether both sides match for every value of x, not the form of the right-hand side.
- (a) 3 hours — Method: for a fixed pool the rate of flow multiplied by the time taken is constant, so multiplying the rate by a factor divides the time by that same factor. Working: tap B's rate is 2 times tap A's rate, so tap B's time is 6 ÷ 2 = 3 hours. Answer: 3 hours. The distractors: 12 hours comes from multiplying the time by 2 as well, which treats the time as directly proportional to the rate and has the faster tap taking longer; 4 hours comes from reading ‘twice as fast’ additively, as two hours quicker, and working out 6 − 2 instead of scaling the time by a factor of 2; 1.5 hours comes from applying the factor of 2 twice, halving 6 to 3 and then halving again.
- (a) Draw equal arcs from X and Y, meeting below line l. — After the first arc marks two points X and Y on line l, compasses are opened to a new radius and arcs of equal radius are drawn centred at X and at Y, so that they meet on the opposite side of l from P; joining P to that meeting point gives the perpendicular. (Joining X and Y with a straight line only retraces part of line l itself, since X and Y both already lie on it; drawing an arc centred at P through only one of X or Y repeats part of the first step instead of moving on; drawing a circle through X, Y and P does not locate the new point needed to complete the perpendicular.)
- (b) 200 — Method: list the equally likely outcomes for the two coins before writing any probability, then multiply by the number of throws. Working: the equally likely outcomes are head then head, head then tail, tail then head, and tail then tail, so there are 4 of them. Two of those 4 give one head and one tail, so the probability is 2/4, which is 1/2. Over 400 throws the expected number is 400 × 1 ÷ 2 = 200. Answer: about 200 of the throws would be expected to give one head and one tail. The distractors: 133 comes from treating two heads, two tails and one of each as three equally likely results and working out 400 ÷ 3 = 133.3, then rounding; 100 comes from counting only head then tail as a success, giving 400 × 1 ÷ 4 = 100; 300 is the expected number of throws that do not give two heads, 400 × 3 ÷ 4 = 300.
- (a) The modal size, 9, bought by more customers than any other — Method: work out both averages from the frequencies, then choose the one the shop can act on. Working: for the mean, multiply each size by the number of pairs sold at it and add: 6 × 4 + 7 × 5 + 8 × 8 + 9 × 13 + 10 × 10 = 340, and 340 ÷ 40 = 8.5, so the mean size is 8.5. The largest frequency is 13, which belongs to size 9, so the modal size is 9. The mean 8.5 is a size no customer in the record asked for, so 40 pairs of it would sit unsold, while 13 of the 40 customers wanted size 9, more than wanted any other size. Answer: the modal size, 9, bought by more customers than any other. The distractors: the mean size 8.5 does take account of all 40 pairs, but a mean of sizes is a summary figure and not a size the month's customers were buying; the mean size 8 comes from averaging the five sizes on sale, 6 + 7 + 8 + 9 + 10 = 40 and 40 ÷ 5 = 8, which ignores how many pairs were sold at each size and so treats the 4 pairs of size 6 as equal in weight to the 13 pairs of size 9; the range 4 comes from 10 − 6 and measures spread, so it says how wide a set of sizes the shop must stock, not which size to stock most of.
- (c) 7 + 4√3 — Expand the brackets fully: (2 + √3)² = 2² + 2 × 2 × √3 + (√3)² = 4 + 4√3 + 3. Adding the two whole-number terms, 4 + 3 = 7, gives 7 + 4√3. Using (a + b)² = a² + b² and skipping the middle cross term entirely gives just 4 + 3 = 7, with no surd term at all. Treating (√3)² as if it stayed √3 rather than becoming 3, then merging it with the existing surd term, gives 4 + 5√3. Squaring only the surd term correctly but carrying the whole-number term as 2 instead of squaring it to 4 gives 2 + 3 + 4√3 = 5 + 4√3.
- (c) £2.80 — 3n + 1.50 = 9.90, so 3n = 8.40 and n = 2.80, so each notebook costs £2.80. A candidate who forgets to subtract the cost of the pen and divides the total by 3 gets n = 9.90 ÷ 3 = £3.30. A candidate who adds the cost of the pen instead of subtracting it gets 3n = 11.40 and n = £3.80. A candidate who treats the pen as a fourth notebook and divides the total by 4 gets n = 9.90 ÷ 4 = £2.48 (2 d.p.).
- (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.
- (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.
- (c) £3.10 — Aisha's tickets cost 8 × 50p = £4.00. Her expected winnings are (8/400) × £45 = £0.90, since she holds 8 of the 400 tickets. Her expected loss is the cost minus the expected winnings: £4.00 − £0.90 = £3.10. Writing £4.00 is wrong because it is only the cost of her tickets, with no account taken of the expected winnings she might get back. Writing £0.90 is wrong because that is her expected WINNINGS, not her loss — the cost has not been subtracted. Writing £3.89 is wrong because it uses 1 ticket instead of her actual 8 tickets when working out the expected winnings: (1/400) × £45 = £0.1125, giving £4.00 − £0.11 = £3.89. Aisha should expect to lose £3.10.
- (a) 7.5 × 10⁴ — Standard form is A × 10ⁿ, where A is between 1 and 10 (1 ≤ A < 10) and n is an integer — 7.5 × 10⁴ satisfies all of this. 12 × 10³ fails because 12 is not below 10. 0.5 × 10⁴ fails because 0.5 is not at least 1. 6.2 × 4⁵ fails because standard form always uses a power of 10, not a power of 4.
- (c) x = 2, y = 3 — Method: no letter cancels straight away, so make one letter the subject of one equation and substitute that expression into the other. Working: 2x + y = 7 gives y = 7 − 2x, so x + 2y = 8 becomes x + 2(7 − 2x) = 8, that is x + 14 − 4x = 8, so −3x = −6 and x = 2; substituting back gives y = 7 − 4 = 3. Answer: x = 2, y = 3, which checks in 2 + 6 = 8. The distractors: x = 3, y = 2 comes from adding the equations to x + y = 5 and then subtracting them the wrong way round, writing x − y = 1 instead of x − y = −1; x = 2, y = 5 comes from substituting x = 2 into 2x + y = 7 as 2 + y = 7, dropping the coefficient; x = −2, y = 11 comes from dividing −3x = −6 as though two minus signs gave a negative answer.
- (d) 218 — The rule x_{n+1} = 0.8x_n + 50 must be applied once for each step, using the result of the previous step every time — not the same starting value repeated. Starting from x_0 = 200: 0.8 × 200 = 160, so x_1 = 160 + 50 = 210. Then 0.8 × 210 = 168, so x_2 = 168 + 50 = 218. Stopping after one iteration leaves x_1 = 210, not x_2. Applying only the multiplier twice without adding 50 at each step uses 0.8² = 0.64, and 0.64 × 200 = 128, which drops the 50 completely. Adding 50 twice at the end instead of once per step, 128 + 100 = 228, still does not reproduce the actual recurrence, because the 50 added at the first step is itself multiplied by 0.8 at the second step. After two iterations, x_2 = 218.
- (c) 155° — Turning clockwise adds to the bearing. Starting on a bearing of 065° and turning clockwise through 90° gives 065° + 90° = 155°. A candidate who instead subtracts, working out 90° − 65° = 25°, has performed the wrong operation, giving 025°. A candidate who turns anticlockwise instead of clockwise works out 065° − 90°, which gives a negative number, and adding 360° to fix this gives 335° — the bearing for turning the other way. A candidate who thinks turning does not change the bearing at all keeps the answer as 065°. The new bearing, turning clockwise, is 155°.
- (a) (2, 7) — Midpoint = ((x1+x2)/2, (y1+y2)/2) = ((−3+7)/2, (5+9)/2) = (4/2, 14/2) = (2, 7). (4, 14) comes from adding the coordinates correctly but forgetting to divide by 2. (2, 9) comes from correctly averaging the x-coordinates but simply copying the y-coordinate of the second point instead of averaging the y-coordinates. (5, 2) comes from subtracting the coordinates instead of adding them before halving: ((7−(−3))/2, (9−5)/2) = (5, 2).
- (b) 250 g — Method: adding water changes the total mass but not the mass of salt, so find the salt, hold it fixed, use the new ratio to find the new total mass and subtract the mass already in the beaker. Working: 12:100 = x:500 gives 12 ÷ 100 × 500 = 60 g of salt; that 60 g must be 8% of the new mixture, so 8:100 = 60:y gives y = 60 ÷ 8 × 100 = 750 g; the water added is 750 − 500 = 250 g. Answer: 250 g. The distractors: 750 g is the mass of the diluted solution, given without taking away the 500 g that was in the beaker to start with; 60 g is the mass of salt, the quantity that stays the same, given instead of the mass of water; 20 g comes from treating the fall from 12% to 8% as 4% of the original 500 g, which measures a change in concentration as though it were a mass of water.
- (b) angle B = angle E — AB and BC meet at vertex B, so the angle INCLUDED between them is angle B; making angle B = angle E completes SAS. Angle A sits between AB and AC, not between AB and BC, so it is not the included angle needed for SAS here. Angle C sits between BC and CA, not between AB and BC, so it is not included either. AC = DF would give a third pair of equal sides, which proves congruence by SSS instead of SAS.
- (b) −3/4 — The tangent to a circle at a point is always perpendicular to the radius drawn to that point. The radius from (0, 0) to (3, 4) has gradient 4/3. The gradient of a line perpendicular to a line with gradient m is the negative reciprocal, −1/m, so the tangent's gradient here is −3/4. Using the radius's own gradient, forgetting that the tangent is perpendicular to it, gives 4/3. Negating the radius's gradient without also taking the reciprocal gives −4/3. Taking the reciprocal of the radius's gradient without negating it gives 3/4.
- (c) x = 0 or x = 3 — Method: take out the common factor 5x: 5x(x − 3) = 0, so 5x = 0 or x − 3 = 0, giving x = 0 or x = 3. Distractor origins: x = 0 or x = 15 comes from forgetting to divide the second term by the common factor 5x correctly, leaving 15 instead of 3; x = 3 loses the solution x = 0 by dividing both sides by x; x = 5 or x = 3 mistakes the coefficient 5 itself for one of the solutions.
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.