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%.
CalculatorGCSE Higher
GCSE Higher sample Paper 2 (calculator)
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- 1.A window display has red baubles, gold baubles and green baubles in the ratio 4 : 5 : 6. What fraction of the baubles are not green?
- 2.A taxi charges a £3 fixed fee plus £2 for each mile travelled. Write an expression, in pounds, for the total cost of a journey of n miles.
- 3.A metal alloy is made from copper and tin in the ratio 7:3. Work out the mass of tin in 250 g of the alloy.
- 4.Triangle PQR has vertices P(0, 0), Q(9, 0) and R(0, 4). Work out the area of triangle PQR.
- 5.In a class of 32 pupils, 19 study geography and 15 study history. 8 study both subjects. Work out n(G ∪ H), the number of pupils who study geography or history or both.
- 6.The seven members of Team A took 20, 21, 22, 23, 24, 25 and 40 seconds to finish a task. The seven members of Team B took 20, 30, 32, 34, 36, 38 and 40 seconds. Tomás says that because the two teams have the same range, their times are spread out in the same way. Is Tomás right? Give a reason for your answer.
- 7.A jacket normally costs £65. In a sale it is reduced by 20%, and the shop then takes a further £5 off at the till. Work out the final price.
- 8.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.
- 9.A cup of tea is cooling. Its temperature, in °C, is plotted against time, in minutes, since it was poured. At t = 4 minutes, the gradient of the tangent to the graph is −3.2. What does this tell you about the tea at t = 4 minutes?
- 10.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.
- 11.A box contains 9 red balls and 11 green balls. Two balls are taken out at random, one after the other, without being replaced. Given that both balls taken out are the same colour, work out the probability that both balls are red.
- 12.Given that 5³ = 125 and 6³ = 216, use a midpoint test to estimate ∛130 to 1 decimal place.
- 13.Solve x² + 7x = 0.
- 14.The number of members of a running club increases from 45 to 54. Work out the percentage increase.
- 15.A parallelogram has an area of 96 cm² and a base of 8 cm. Work out the perpendicular height of the parallelogram.
- 16.The equation x³ + 4x − 9 = 0 is to be solved by iteration. Work out which one of these iterative formulas comes from a correct rearrangement of that equation.
- 17.6 identical taps fill a paddling pool in 20 minutes. Each tap fills at the same steady rate. Work out how long 3 of these taps would take to fill the same pool.
- 18.A sector of a circle has radius 12 cm and angle 45°. Using π = 3.14, work out the area of the sector.
- 19.A model rocket's height is given by h = −5t² + 20t, where h is in metres and t is in seconds. A student says the graph of h against t is n-shaped. Which statement gives the correct verdict on the SHAPE and the reason that settles it from the equation?
- 20.Describe the single transformation that maps the graph of y = x² onto the graph of y = x² + 3.y = x²y = x² + 3
Answer key
- (b) 3/5 — Method: add all three parts for the total, add together the parts that are not green, then write this over the total. Working: total parts = 4 + 5 + 6 = 15. Not green = 4 + 5 = 9. Fraction = 9/15 = 3/5. Answer: 3/5. 2/5 comes from finding the fraction that IS green (6/15 = 2/5) instead of not green. 4/15 comes from only counting the red baubles as 'not green' and forgetting the gold ones. 9/10 comes from adding only two of the three ratio parts to find the total (4+6=10), missing out the gold part, while still using 9 for the numerator.
- (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.
- (b) 75 g — Method: split the total mass into the number of parts shown by the ratio, then find the mass of tin. Working: the ratio 7:3 has 7 + 3 = 10 parts, so one part is 250 ÷ 10 = 25 g, and the mass of tin is 3 × 25 = 75 g. So the alloy contains 75 g of tin. Distractor 175 g is the mass of copper, not tin. Distractor 125 g comes from splitting the alloy into two equal halves, ignoring the ratio. Distractor 25 g is the value of one part, found correctly but never multiplied by 3.
- (a) 18 — PQ lies along the x-axis with length 9, and PR lies along the y-axis with length 4, and these two sides meet at right angles at P, so they can be used as the base and height of the triangle. Area = 1/2 × base × height = 1/2 × 9 × 4 = 18. 36 comes from multiplying the base and height but forgetting to halve the result. 13 comes from adding the two lengths, 9 + 4, instead of multiplying them. 26 comes from the perimeter-style calculation 2 × (9 + 4) instead of the triangle area formula.
- (b) 26 — Method: n(G ∪ H) = n(G) + n(H) − n(G ∩ H), taking off the overlap once so the pupils who study both subjects are not counted twice. Working: 19 + 15 − 8 = 26. Answer: 26. Watch out: adding 19 and 15 without taking off the overlap gives 34, which counts the 8 pupils who study both subjects twice. Taking the 8 off both totals before adding, 19 − 8 + 15 − 8 = 18, counts only the pupils who study exactly one of the two subjects and leaves out the 8 who study both. And writing down 11, which is 19 − 8, gives the number who study geography only, not the number who study geography or history or both.
- (d) No, the range uses only the fastest and slowest time — Method: check what the range is built from, then look at what it leaves out. Working: both teams have a fastest time of 20 seconds and a slowest of 40 seconds, so both ranges are 40 − 20 = 20 seconds and Tomás has that part right. But the range is calculated from those two values alone. Six of Team A's seven times lie between 20 and 25 seconds, with a single time far out at 40; Team B is the other way round, with six of its seven times at 30 seconds or more and a single time far out at 20. So Team A bunches at the fast end and Team B at the slow end. The two patterns are quite different, and the range cannot see the difference because the five middle times never enter the calculation. Answer: no, because the range uses only the fastest and slowest time. The distractors: comparing the means answers a different question, since a mean measures position rather than spread, and two sets with the same spread can have different means; saying that equal ranges mean equal spread is the very assumption that fails here; saying that seven times each forces the spreads to match confuses the size of a data set with how its values are arranged inside it.
- (c) £47.00 — First apply the 20% reduction: £65 × 0.8 = £52.00. Then take off the further £5: £52.00 − £5 = £47.00. Treating the 20% as a flat £20 rather than a percentage of the price, £65 − £20 − £5, gives £40.00. Applying the 20% reduction correctly but forgetting to take off the extra £5 leaves £52.00. Taking off the £5 first and then applying the 20% reduction to the smaller amount, (£65 − £5) × 0.8, gives £48.00.
- (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) Falling at 3.2°C per minute — The gradient of a tangent gives the instantaneous rate of change, in °C per minute here, not a temperature and not a total change. The negative sign means the temperature is falling, not rising, so the tea is cooling at a rate of 3.2°C per minute at the instant t = 4. Reading the sign the wrong way round gives 'rising at 3.2°C per minute', which would mean the tea is heating up. Treating −3.2 as a total drop since the tea was poured confuses a rate with an accumulated change, which would need the temperatures at two different times, not the gradient at one instant. Treating −3.2 as the temperature reading itself confuses the gradient, a rate of change, with the y-value on the graph. Always check whether a number is a rate, a total, or a single reading before you use it.
- (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.
- (d) 36/91 — Method: P(both red | same colour) = P(both red) ÷ P(same colour), where P(same colour) = P(both red) + P(both green). Working: P(both red) = 9/20 × 8/19 = 72/380 = 18/95. P(both green) = 11/20 × 10/19 = 110/380 = 11/38. P(same colour) = 18/95 + 11/38 = 36/190 + 55/190 = 91/190. P(both red | same colour) = (36/190) ÷ (91/190) = 36/91. Answer: 36/91. Watch out: stopping at 18/95 gives P(both red) itself, without dividing by the probability that the colours matched at all. Working out 55/91 finds the same-colour probability for green instead of red — check which colour's count you are putting on top. And 9/20 is just the chance the first ball drawn is red, which ignores the second draw and the without-replacement condition completely.
- (b) 5.1 — ∛130 lies between 5 and 6, since 125 < 130 < 216, and closer to 5 because 130 is much nearer 125 than 216. To pin down the first decimal place, test the midpoint of the tenth, 5.05: 5.05³ = 5.05 × 5.05 × 5.05 ≈ 128.79. Since 130 is greater than 128.79, ∛130 lies above 5.05, so it rounds to 5.1 rather than 5.0. Rounding down to 5.0, on the assumption that a value close to the lower bound 125 must round down, ignores that 5.05³ is already less than 130. Estimating 5.2 overshoots the true root: 5.2³ = 140.608, which is well above 130, so ∛130 cannot round to 5.2. Taking 6.0, the upper of the two whole numbers the root lies between, ignores that 130 is far nearer to 5³ = 125 than to 6³ = 216, so the root sits just above 5, not just below 6.
- (c) x = 0 or x = −7 — Factorising: x² + 7x = x(x + 7) = 0, so x = 0 or x + 7 = 0, giving x = 0 or x = −7. A candidate who divides both sides of the original equation by x, which loses the solution x = 0, gets only x = −7. A candidate who makes a sign error solving x + 7 = 0 gets x = 0 or x = 7. A candidate who misreads the coefficient and doubles it gets x = 0 or x = −14.
- (c) 20% — Method: percentage increase = (increase ÷ original) × 100. Working: the increase is 54 − 45 = 9, and 9 ÷ 45 = 0.2, so the percentage increase is 0.2 × 100 = 20. Answer: 20%. The distractors: 9% comes from writing the actual increase as a percentage; 16.7% comes from dividing by the new value 54 instead of the original 45; 120% is the multiplier 1.2 written as a change rather than the change itself.
- (a) 12 cm — Area of a parallelogram = base × perpendicular height, so height = area ÷ base = 96 ÷ 8 = 12 cm. A student who adds the area and base instead of dividing gets 96 + 8 = 104 cm. A student who multiplies the area and base instead of dividing gets 96 × 8 = 768 cm. A student who uses the triangle area formula, area = 1/2 × base × height, instead of the parallelogram formula solves 96 = 1/2 × 8 × h and gets h = 24 cm.
- (c) xₙ₊₁ = ∛(9 − 4xₙ) — Method: a formula xₙ₊₁ = f(xₙ) is a correct rearrangement when the equation x = f(x) turns back into the equation you started with, so rearrange x³ + 4x − 9 = 0 by making the cube the subject. Working: x³ + 4x − 9 = 0 gives x³ = 9 − 4x, because the 4x and the 9 each change sign as they cross the equals sign; taking the cube root of both sides gives x = ∛(9 − 4x), which is the formula xₙ₊₁ = ∛(9 − 4xₙ). Answer: xₙ₊₁ = ∛(9 − 4xₙ). The distractors: ∛(9 + 4xₙ) comes from writing x³ = 9 + 4x, moving the 4x across the equals sign without changing its sign; (9 + xₙ³)/4 comes from making the linear term the subject but keeping the sign of the cube, writing 4x = 9 + x³ when the equation gives 4x = 9 − x³; ∛(9 − 4xₙ³) cubes the x in the linear term as well, changing a term the original equation never cubed.
- (c) 40 minutes — Method: this is inverse proportion — fewer taps means longer, not shorter — so the number of taps × the time taken stays constant. Working: 6 × 20 = 120, and with 3 taps the time is 120 ÷ 3 = 40 minutes. So 3 taps take 40 minutes. Distractor 10 minutes comes from treating it as direct proportion instead of inverse, working out 20 × 3 ÷ 6. Distractor 30 minutes comes from halving the number of taps and adding half the original time, 20 + 10, instead of doubling the time. Distractor 17 minutes comes from subtracting the number of taps removed, 3, directly from the original time, 20.
- (c) 56.52 cm² — Sector area = (45 ÷ 360) × 3.14 × 12² = 0.125 × 3.14 × 144 = 56.52 cm². (4.71 cm² comes from forgetting to square the radius; 452.16 cm² comes from finding the area of the whole circle and forgetting the angle fraction; 9.42 cm² comes from using the arc length formula instead of the sector area formula.)
- (a) Yes — the t² coefficient is negative, giving an n-shape. — The coefficient of t² in h = −5t² + 20t is −5, which is negative, so the graph is n-shaped with a maximum point — this matches the physical story of the rocket rising then falling, but the shape itself is decided by the negative coefficient of t², not by the story alone. Saying the shape comes from the story rather than the coefficient gets the reasoning backwards — the algebra determines the shape, and the story happens to agree with it. Saying it is U-shaped because height starts by increasing confuses the early part of the curve with its overall shape; a U-shaped curve would mean the rocket's height eventually increases again forever, which does not happen here. Saying it is n-shaped only because the rocket lands treats a consequence of the shape as if it were the cause.
- (a) A translation by vector (0, 3) — y = x² + 3 adds a constant outside the squaring, so it is a vertical translation of y = x² — every point moves the same distance parallel to the y-axis, with no change in x. The vector is (0, 3), matching the +3. A vector of (3, 0) confuses this with a horizontal shift; (0, −3) has the right axis but the wrong sign, since the graph moves up, not down; a stretch changes the shape of the curve, which adding a constant term never does.
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.