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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GCSE Higher sample Paper 2 (calculator)
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- 1.A country's population is 8,340,000, and it is estimated that 1,950,000 of them live in the capital city. By rounding each number to 1 significant figure, work out an estimate for the number of people who do not live in the capital city.
- 2.To rearrange the formula y = 3x − 8 to make x the subject, Priya writes: 'Add 8 to both sides to get y + 8 = 3x, then divide both sides by 3 to get x = y + 8 ÷ 3.' Work out the correct expression for x.y = 3x − 8
- 3.A statue exerts a downward force of 68.4 N on its base, which has an area of 2.85 m². Work out the pressure that the statue exerts on its base, in N/m².
- 4.From an external point P, two tangents PA and PB touch a circle with centre O at the points A and B. Angle APB = 46°. C is a point on the major arc AB, the arc on the opposite side of AB from P. Work out the size of angle ACB.
- 5.A biased spinner is spun 200 times. It lands on red 70 times, on blue 50 times, and on green 80 times. Using these results, work out the expected number of times the spinner does NOT land on red, in 500 spins of the same spinner.
- 6.Grace asked 12 children how many brothers and sisters they have. Her results, in order, were 0, 0, 1, 1, 1, 1, 2, 2, 3, 3, 4, 6. Work out the median number of brothers and sisters.
- 7.Round 0.006852 to 2 significant figures.
- 8.The diagram shows the graph of a function. Which equation could it represent?
- 9.Noah runs 2 km in 10 minutes. Work out how long he takes to run 5 km at the same speed.
- 10.A solid has 10 vertices and 15 edges. Using the formula F + V − E = 2, work out how many faces it has.
- 11.A pupil must choose 2 different subjects at random from these five: maths, physics, biology, chemistry and history. Work out the probability that the two subjects chosen are maths and biology.
- 12.Which of these is written correctly in standard form?
- 13.Which of these values of x is a solution of x² + 2x − 15 = 0?
- 14.There are 400 students at a school. 25% of them have a brother, 40% have a sister and 15% have both a brother and a sister. Work out how many of the students have neither a brother nor a sister.
- 15.A triangular plot of land has sides AB = 13 m and AC = 10 m, with angle BAC = 72° between them. Work out the perimeter of the plot. Give your answer to 1 decimal place.
- 16.A market trader builds a display of oranges in a rectangular block. The block is n oranges wide and 4 oranges longer than it is wide. The number of oranges needed is 5 when n = 1, 12 when n = 2, 21 when n = 3, 32 when n = 4, and 45 when n = 5. Work out an expression, in terms of n, for the number of oranges needed for a display of width n.
- 17.A concrete mix is made from gravel and cement in the ratio 5 : 1 by mass. The mass of cement is c kg and the total mass of the mix is m kg. Write down a formula for m in terms of c.
- 18.In pentagon PQRST, which of the following correctly names the interior angle at vertex R, using standard three-letter angle notation?
- 19.A student says that (x + 4)² is equivalent to x² + 16. For which value of x do the two expressions give the SAME result, making it look (misleadingly) like the student could be right?
- 20.(2x + 3)(x + a) ≡ 2x² + 11x + 12 is an identity. Work out the value of a.
Answer key
- (d) 6,000,000 — Method: round each number to 1 significant figure, then subtract. Working: 8,340,000 rounds to 8,000,000 (1 s.f.); 1,950,000 rounds to 2,000,000 (1 s.f.); 8,000,000 − 2,000,000 = 6,000,000. Answer: 6,000,000. 6,390,000 is the exact difference, found without rounding the numbers first. 8,000,000 comes from rounding the population correctly but forgetting to subtract the capital's population at all. 6,300,000 comes from rounding 8,340,000 to the nearest hundred thousand, 8,300,000, instead of to 1 significant figure, then subtracting the correctly rounded 2,000,000.
- (d) x = (y + 8) / 3 — Method: add 8 to both sides, then divide the whole side by 3. Working: y = 3x − 8, so adding 8 to both sides gives y + 8 = 3x, then dividing both sides by 3 gives x = (y + 8) / 3. The value x = y + 8 / 3 is Priya's version, which comes from dividing only the 8 by 3 instead of dividing the whole expression y + 8 by 3. The value x = (y − 8) / 3 comes from a sign error, keeping −8 instead of moving it to +8. The value x = 3(y + 8) comes from multiplying by 3 instead of dividing.
- (a) 24 N/m² — Pressure = force ÷ area. 68.4 ÷ 2.85 = 24 N/m². 194.94 N/m² comes from multiplying the force by the area instead of dividing (68.4 × 2.85). 65.55 N/m² comes from subtracting the area from the force (68.4 − 2.85) instead of dividing. 0.04 N/m² comes from dividing the area by the force instead of the force by the area (2.85 ÷ 68.4).
- (a) 67° — PA and PB are tangents from the same point, so OAPB is a kite with right angles at A and B, by the tangent–radius theorem. The kite's angles sum to 360°, so angle AOB = 360° − 90° − 90° − 46° = 134°. C is on the major arc AB, so by the angle at the centre theorem, angle ACB = 134° ÷ 2 = 67°. Doubling angle APB instead of first finding angle AOB from the kite gives 46° + 46° = 92°. Using angle APB directly as if it were the centre angle, then halving it, gives 46° ÷ 2 = 23°. Finding angle AOB = 134° correctly but forgetting to halve it for the angle at the circumference gives 134°. Halve angle AOB, once you've found it properly, and you get 67°.
- (c) 325 — Method: first find the relative frequency of NOT landing on red from the 200 spins, then scale that up to 500 spins. Working: non-red results = 50 + 80 = 130, out of 200 spins, so P(not red) = 130 ÷ 200 = 0.65. Expected non-red results in 500 spins = 500 × 0.65 = 325. Answer: 325. Watch out: writing down 175 finds the expected number of RED results instead, 70 ÷ 200 × 500 = 175, answering the opposite of what was asked. Writing down 250 assumes landing red or not landing red must be a fair 50-50 split, but the spinner is biased and the actual results do not split evenly. And writing down 130 stops after finding how many of the 200 spins were non-red and forgets to scale that figure up to the 500 spins asked for.
- (a) 1.5 — Method: with an even number of values the median is the mean of the two middle values, which for 12 values are the 6th and the 7th once the data are in order. Working: the results are already in order, and 12 ÷ 2 = 6, so the middle pair are the 6th value, 1, and the 7th value, 2; the median is (1 + 2) ÷ 2 = 1.5. Answer: 1.5 brothers and sisters. The distractors: 1 comes from reading the 6th value and stopping there instead of averaging the middle pair; 2 comes from working out the mean, 24 ÷ 12, instead of the median; 6 comes from working out the range, 6 − 0, which measures spread rather than centre.
- (a) 0.0069 — Leading zeros are not significant, so the significant figures in 0.006852 start at 6: 6, 8, 5, 2. Rounding to 2 significant figures means keeping 6 and 8, and looking at the next digit, 5, to decide whether to round up. Since 5 rounds up, the second significant figure increases from 8 to 9: 0.006852 rounds to 0.0069. A candidate who rounded to 1 significant figure instead of 2 wrote 0.007. A candidate who rounded to 3 significant figures instead of 2 wrote 0.00685. A candidate who did not round up despite the next digit being 5 wrote 0.0068.
- (b) $y = x^3 - 4x$ — Method: count how many times the curve crosses the x-axis and check whether it is a cubic (an S-shape with up to three crossing points) rather than a lower power, and note which way it runs overall from bottom-left to top-right or the reverse. Working: the curve crosses the x-axis at three points, x = −2, 0 and 2, and runs from bottom-left to top-right, which matches y = x³ − 4x = x(x − 2)(x + 2). Answer: y = x³ − 4x. Distractor refutation: y = −x³ + 4x comes from a sign error on every term, which would flip the curve so it ran from top-left to bottom-right instead. y = x³ + 4x comes from a sign error on the x term only, which removes two of the three crossing points, since x(x² + 4) has only x = 0 as a real root. y = x² − 4x comes from dropping the cubic term altogether, giving a parabola with only two crossing points instead of an S-shaped curve with three.
- (d) 25 minutes — Method: find the time for one kilometre, then multiply by the number of kilometres — the unitary method with a rate. Working: 10 ÷ 2 = 5 minutes per km, and 5 × 5 = 25. Answer: 25 minutes. The distractors: 20 minutes comes from multiplying the 10 minutes by 2, the distance in the given rate, instead of by the scale factor 2.5; 50 minutes comes from multiplying 10 by 5, treating the 10 minutes as the time for a single kilometre; 15 minutes comes from adding the 5 km on to the 10 minutes, adding quantities that are not the same kind.
- (a) 7 — Rearranging F + V − E = 2 gives F = 2 − V + E = 2 − 10 + 15 = 7. A candidate who works out E − V without the +2, giving 15 − 10, answers 5. A candidate who rearranges with a sign error, working out 2 + V − E = 2 + 10 − 15 = −3 and then drops the negative sign, answers 3. A candidate who adds all three numbers together, V + E + 2 = 10 + 15 + 2, answers 27, having used the wrong operation entirely. The correct number of faces is 7.
- (b) 1/10 — Method: list every possible pair of subjects systematically, so that no pair is missed and no pair is counted twice, then compare the number of successful pairs with the size of the list. Working: pairing maths with each of the other four gives 4 pairs, physics with each subject after it gives 3, biology gives 2 and chemistry gives 1, so there are 4 + 3 + 2 + 1 = 10 pairs. Exactly one of them is maths with biology. Answer: the probability is 1/10. The distractors: 1/5 comes from counting maths-then-biology and biology-then-maths as two separate successes while still dividing by the 10 unordered pairs; 1/15 comes from a list that also pairs each subject with itself, giving 15 entries instead of 10; 1/4 comes from working out the second step alone, that 1 of the 4 subjects left after maths is biology, without allowing for the chance that maths is picked at all.
- (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.
- (a) x = 3 — Method: factorise x² + 2x − 15 as (x + 5)(x − 3), since 5 × (−3) = −15 and 5 + (−3) = 2. Setting each bracket equal to zero gives x + 5 = 0 or x − 3 = 0, so x = −5 or x = 3. Only x = 3 is offered here. Distractor origins: x = −3 reverses the sign of the factor pair, treating the bracket (x − 3) as giving x = −3 instead of x = 3; x = 5 takes the number from the other factor, (x + 5), but with the wrong sign, giving x = 5 instead of x = −5; x = 15 takes the constant term of the original expression as if it were a root, without factorising at all.
- (c) 200 students — Method: find the percentage who have a brother or a sister, taking care that the students with both are not counted twice, then take that percentage from 100% and apply the result to the 400 students. Working: 25% + 40% = 65%, but the 15% with both has been counted in each of those figures, so 65% − 15% = 50% have a brother or a sister; that leaves 100% − 50% = 50%, and 50% of 400 = 200. Answer: 200 students. The distractors: 260 students is 65% of 400, the number with a brother or a sister when the 15% overlap is counted twice; 140 students comes from taking that same uncorrected 65% away from the 400; 300 students comes from subtracting only the 25% with a brother and ignoring the sisters altogether.
- (c) 36.7 m — Method: use the cosine rule to find the missing side BC, then add all three sides for the perimeter. Working: BC² = 13² + 10² − 2 × 13 × 10 × cos 72°, so BC = 13.7 m, and perimeter = 13 + 10 + 13.7 = 36.7 m. Forgetting the negative sign in the cosine rule (using + instead of −) gives BC = 18.7 m and a perimeter of 41.7 m; leaving AC out of the total and only adding AB and BC gives 26.7 m; and adding AB twice instead of AB and AC gives 39.7 m. Always add all three named sides once the missing one has been found.
- (b) n² + 4n — First differences of 5, 12, 21, 32, 45 are 7, 9, 11, 13. Second differences are 2, 2, 2, so a = 2 ÷ 2 = 1. Subtracting n² (1, 4, 9, 16, 25) from the totals (5, 12, 21, 32, 45) leaves 4, 8, 12, 16, 20, the linear expression 4n. So the number of oranges is n² + 4n. Using the second difference itself as the coefficient of n², without halving it, gives 2n² + 4n. Finding a = 1 correctly but dropping the linear remainder 4n entirely leaves n². Treating the first first difference (7) as a constant common difference and building a linear formula 5 + 7(n − 1) = 7n − 2 fits only the first two totals, and gives 19 for n = 3 instead of 21.
- (c) m = 6c — Method: the whole is the sum of the parts in the ratio, and the cement is 1 part, so one part weighs c kg. Working: the mix has 5 + 1 = 6 parts, each of mass c kg, so the total mass is 6 × c, giving m = 6c. Answer: m = 6c. The distractors: m = 5c uses the 5 gravel parts as the multiplier and forgets that the cement is in the mix too, so it gives the mass of the gravel and not the total; m = c + 5 comes from reading the ratio as '5 more than' and adding, which treats a number of parts as a mass in kilograms; m = c/6 turns the relationship upside down, as though the total were shared into the cement rather than the cement multiplied up to the total.
- (a) ∠QRS — Method: the middle letter in three-letter angle notation is always the vertex of the angle, and the outer two letters are the neighbouring vertices along the shape's sides. Working: at vertex R, the two adjacent vertices along the pentagon are Q and S, so the interior angle is written ∠QRS, with R in the middle. Options: ∠PQR names the angle at Q, not R, since Q is the middle letter there; ∠RST puts R first rather than in the middle, so it actually names the angle at S; ∠TRP does have R in the middle, but T and P are not the vertices adjacent to R along the pentagon's sides, so it does not describe R's interior angle. Answer: ∠QRS.
- (c) x = 0 — Expand (x + 4)² correctly: (x + 4)² = x² + 8x + 16. This equals x² + 16 only when 8x is zero, i.e. when x = 0 — at every other value of x the two expressions differ by 8x. Choosing x = 4 confuses the constant inside the bracket with the value of x that makes the expressions match. Choosing x = −4 makes the same confusion but with the sign flipped. Choosing x = 8 mistakes the coefficient of the middle term, 8x, for the value of x itself.
- (d) a = 4 — Expand the left-hand side: (2x + 3)(x + a) = 2x² + 2ax + 3x + 3a = 2x² + (2a + 3)x + 3a. For this to match 2x² + 11x + 12 for every value of x, the x-coefficients must be equal and the constants must be equal: 2a + 3 = 11 and 3a = 12. Both give a = 4, so a = 4. Writing a = 12 comes from the constant-term equation 3a = 12: reading it as saying a itself is 12, rather than dividing both sides by 3. Writing a = 8 comes from the x-coefficient equation 2a + 3 = 11: working out 11 − 3 = 8 correctly but then stopping, without dividing by the 2 in front of a. Writing a = −4 comes from rearranging 2a + 3 = 11 the wrong way round, as 2a = 3 − 11 = −8, which gives a = −4 instead of a = 4.
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.