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.The decimal 0.272727... repeats the block 27 for ever. Write 0.27 recurring as a fraction in its simplest form.
- 2.Meera writes the statement 3(x + 4) = 3x + 12. Which of these correctly describes what she has written, with a reason?
- 3.The number of euros, e, received is directly proportional to the number of pounds, p, exchanged. Exchanging £40 gives 46 euros. Work out how many euros are received for £65, giving your answer to the nearest euro.
- 4.A rhombus-shaped floor tile has sides of length 12 cm and an interior angle of 55°. Work out the area of the tile. Give your answer to 1 decimal place.
- 5.A ferry company finds that on 20% of days the sea is rough. If the sea is rough, the probability that a crossing is delayed is 0.75. If the sea is calm, the probability that a crossing is delayed is 0.1. Given that a crossing was delayed, work out the probability that the sea was rough that day.
- 6.A recruitment agency in Manchester compares the weekly pay, in pounds, of seven employees at two branches. Branch A: £480, £495, £500, £505, £510, £515, £1,200. Branch B: £480, £490, £500, £510, £520, £530, £540. An advert for the agency claims 'Branch A pays more on average.' Decide whether this claim is fairly supported, using an appropriate average, and choose the correct conclusion.
- 7.A weather app records the temperature at three points in one day: 6 °C at noon, −2 °C at midnight, and −7 °C just before dawn. Work out the difference between the highest and lowest of these three temperatures.
- 8.The point A(−6, 8) lies on the circle x² + y² = 100, whose centre is the origin O. The tangent to the circle at A crosses the y-axis at the point B. Work out the length of OB.
- 9.Line P passes through the origin and the point (3, 21). Line Q passes through the origin and the point (5, 45). Both lines represent the distance, in metres, run by an athlete against the time, in seconds. Work out the gradient of the steeper line.
- 10.A tangent to a circle touches the circle at exactly one point, P. Work out the size of the angle between the tangent and the radius drawn to P.
- 11.A charity raffle has three types of ticket: winning, near-miss and losing, and every ticket is exactly one of these. The probability that a ticket is winning is 1/8 and the probability that it is a near-miss is 1/4. Work out the probability that a ticket is losing.
- 12.A minibus can carry 16 passengers. A school is taking 179 pupils on a trip. By rounding 179 to the nearest 10, work out an estimate for the number of minibuses needed, given that the school cannot hire part of a minibus.
- 13.Kai writes the expression 8 − 3x + 5x² and says it has three terms. Ryan says it only has two terms because a number on its own is not a term. Work out the correct number of terms in the expression.
- 14.Two mathematically similar tins have heights 8 cm and 12 cm. Jayden says that the volume of the larger tin is 1.5 times the volume of the smaller tin. Give a reason why Jayden is incorrect, and work out the correct volume scale factor.
- 15.Triangle PQR has angle P = 90°. The side PQ = 9 cm and the hypotenuse QR = 15 cm. Work out the length of PR.
- 16.f(x) = x³ − 3x² − 4. Work out the pair of consecutive integers between which the solution of f(x) = 0 lies.y = x
- 17.A shop's profit, P pounds, is in direct proportion to the number of items sold, n. When 15 items are sold, the profit is £45. Write down the ratio n : P in its simplest form for this shop.
- 18.Triangle ABC is similar to triangle PQR. Side AB is 8 cm and corresponds to side PQ, which is 12 cm. Side AC is 6 cm and corresponds to side PR. Work out the length of PR.
- 19.The graph of W = 840 ÷ L shows how the width, W metres, of a rectangular field of area 840 m² depends on its length, L metres. Use the relationship to work out the width when the length is 35 m, and hence the perimeter of the field.
- 20.The equation x² + 2x − 5 = 0 can be solved using the iterative formula xₙ₊₁ = 5/(xₙ + 2). The starting value is x₀ = 1, so x₁ is the value after the formula has been used once. Work out x₂ correct to 2 decimal places.
Answer key
- (c) 3/11 — Method: let a letter stand for the recurring decimal, multiply by the power of ten that shifts exactly one repeating block past the point, subtract the original equation so that the recurring tail cancels, then solve and cancel. Working: let x = 0.272727...; the repeating block is two digits long, so multiply by 100 to give 100x = 27.272727...; subtracting gives 99x = 27, so x = 27/99; the highest common factor of 27 and 99 is 9, and 27 ÷ 9 = 3 with 99 ÷ 9 = 11. Answer: 3/11. The distractors: 27/100 comes from writing the repeating block over 100 instead of over 99, forgetting that subtracting x leaves 99x rather than 100x; 3/10 comes from rounding the decimal to one place and converting 0.3; 2/9 comes from treating only the 2 as recurring and converting 0.222... instead.
- (a) An identity, true for every value of x — Expanding the bracket on the left gives 3x + 12, which matches the right-hand side exactly, so the statement is true for every value of x — this makes it an identity. A candidate who reasons that any statement with an equals sign must be an equation picks that option, missing that an equation is only true for particular value(s) of x, not all of them. A candidate who confuses an identity with a formula, because both relate two expressions, picks the formula option — but a formula connects two different quantities, such as area and side length, not two equivalent forms of the same expression. A candidate who assumes it can be solved for a single value of x, as with a normal equation, picks that option, not realising there is no single solution here.
- (b) 75 — The exchange rate is constant: k = 46 ÷ 40 = 1.15 euros per pound. For £65, the number of euros is 1.15 × 65 = 74.75, which rounds to 75 euros. Getting 74 comes from rounding 74.75 down instead of to the nearest whole number. Getting 57 comes from using the reciprocal rate (40 ÷ 46) instead of 46 ÷ 40. Getting 71 comes from adding the difference between 65 and 40 (25) onto 46 instead of using the proportional rate.
- (d) 118.0 cm² — Method: a rhombus is made of two congruent triangles either side of a diagonal, each with area 1/2 × 12 × 12 × sin 55°, so the whole rhombus has area 12 × 12 × sin 55° (side² × sin of the interior angle). Working: area = 12² × sin 55° = 118.0 cm². Stopping at one triangle's area, 1/2 × 12² × sin 55°, and forgetting to double it gives 59.0 cm²; using cos 55° instead of sin 55° gives 82.6 cm²; and multiplying the two sides together with no trig term at all gives 144.0 cm². Splitting the rhombus into its two triangles is the safest way to see why the 1/2 disappears from the whole-shape formula.
- (c) 15/23 — Method: find P(rough and delayed) and the overall P(delayed) using the tree, then divide. Working: P(rough and delayed) = 0.2 × 0.75 = 0.15. P(calm and delayed) = 0.8 × 0.1 = 0.08. P(delayed) = 0.15 + 0.08 = 0.23. P(rough | delayed) = 0.15 ÷ 0.23 = 15/23. Answer: 15/23. Watch out: leaving the answer as 0.15 (3/20) gives P(rough and delayed) itself, without dividing by the overall probability that a crossing is delayed. Giving 0.75 (3/4) is the probability you were told to start with — that a crossing is delayed GIVEN the sea is rough — which is the reverse of what's being asked. And 0.2 (1/5) is just the original probability that the sea is rough, before you take the fact that the crossing was delayed into account.
- (c) No — median £505 at A vs £510 at B. — Branch A's seven wages in order are £480, £495, £500, £505, £510, £515 and £1,200, so the median, the 4th value, is £505. Branch B's in order are £480, £490, £500, £510, £520, £530 and £540, so the median is £510. Since £505 is lower than £510, the median wage is not higher at Branch A, so the claim is not fairly supported. Choosing 'Yes — mean £600.71 at A vs £510 at B' uses the mean: 480 + 495 + 500 + 505 + 510 + 515 + 1200 = 4205, and 4205 ÷ 7 = 600.71, a figure pulled upward by the £1,200 outlier that does not represent a typical wage. Choosing 'Yes — median £515 at A vs £510 at B' miscounts the middle position, taking the 6th wage, £515, instead of the correct 4th value, £505. Choosing 'Yes — highest wage £1,200 at A vs £540 at B' compares the highest wage at each branch rather than a measure of the typical, or average, wage.
- (a) 13 °C — Method: subtract the lowest temperature from the highest temperature to find the difference. Working: the highest temperature is 6 °C and the lowest is −7 °C. Difference = 6 − (−7) = 6 + 7 = 13. Answer: 13 °C. 8 °C comes from using −2 °C as the lowest temperature instead of −7 °C: 6 − (−2) = 8. 5 °C comes from finding the difference between the two negative temperatures instead of the highest and lowest: −2 − (−7) = 5. −1 °C comes from adding the highest and lowest temperatures instead of subtracting: 6 + (−7) = −1.
- (a) 12.5 — Method: the tangent at A is perpendicular to the radius OA, so find the gradient of OA, take its negative reciprocal, write the equation of the tangent and find where it meets the y-axis; the length of OB is then the distance of that crossing from the origin. Working: OA runs from (0, 0) to (−6, 8), so its gradient is 8 ÷ (−6), which cancels to −4/3; the negative reciprocal of −4/3 is 3/4. Substituting into y − 8 = 3/4(x + 6) gives y = 0.75x + 4.5 + 8, so y = 0.75x + 12.5 and B is (0, 12.5). The length OB is therefore 12.5. Answer: 12.5. The distractors: 10 is the radius of the circle, quoted on the assumption that the tangent always meets an axis one radius from the centre, which is only true when the radius itself lies along that axis; 8 is the y-coordinate of A, quoted by treating the tangent as horizontal so that it keeps the height of A; 3.5 comes from turning the gradient of OA upside down without changing its sign, which gives y = −0.75x + 3.5.
- (b) 9 m/s — The gradient of line P is 21 ÷ 3 = 7, so P has a rate of 7 m/s. The gradient of line Q is 45 ÷ 5 = 9, so Q has a rate of 9 m/s. Because 9 is greater than 7, line Q is the steeper line, with gradient 9 m/s. Taking line P's gradient instead of Q's gives 7 m/s, the less steep line. Subtracting the two lines' coordinates directly, (45 − 21) ÷ (5 − 3) = 24 ÷ 2 = 12 m/s, mixes points from different lines rather than using one line's own two points. Adding the two gradients, 7 + 9 = 16 m/s, treats 'steeper' as a total rather than a comparison.
- (c) 90° — A tangent to a circle always meets the radius drawn to the point of contact at a right angle, so the angle between the tangent and the radius at P is 90°. 180° confuses the tangent with the diameter through P, as if the radius continued in a straight line into the tangent. 45° halves the true angle by mistake. 60° comes from confusing this fact with the angle of an equilateral triangle.
- (a) 5/8 — Winning, near-miss and losing are exhaustive, so the three probabilities sum to 1. Writing 1/4 as 2/8 so every fraction has the same denominator, 1 − 1/8 − 2/8 = 8/8 − 1/8 − 2/8 = 5/8. Subtracting only the winning probability and forgetting the near-miss probability gives 1 − 1/8 = 7/8. Subtracting only the near-miss probability and forgetting the winning probability gives 1 − 1/4 = 3/4. Adding the two given probabilities and stopping there gives 1/8 + 2/8 = 3/8, the probability that a ticket is winning or a near-miss, not the probability that it is losing.
- (b) 12 — Method: round the number of pupils to the nearest 10, divide by the number of passengers each minibus can carry, then round up because a part-full minibus still needs a whole vehicle. Working: 179 rounds to 180 (nearest 10); 180 ÷ 16 = 11.25; 11 minibuses only carry 176 passengers, so a 12th minibus is needed for the rest. Answer: 12. 11 comes from rounding 11.25 to the nearest whole number in the usual way, without checking that the leftover pupils still need transporting. 10 comes from rounding 179 down to 170 instead of to the nearest 10, 180. 180 comes from stopping after rounding the number of pupils, without dividing by the number of passengers each minibus carries at all.
- (b) 3 — Method: a term is any part of an expression separated from the rest by a + or − sign, including a number on its own. Working: 8 − 3x + 5x² splits at the + and − signs into 8, −3x and 5x² — that is three separate terms. Answer: 3. Ryan's answer of 2 comes from wrongly excluding the number 8, thinking a term must contain a letter. 5 comes from miscounting by treating the coefficients and powers as separate terms as well as the letters. 1 comes from treating the whole expression as a single term because it is written without brackets.
- (a) 3.375 — Method: the volume scale factor between similar shapes is the length scale factor cubed, not the length scale factor itself. Working: the length scale factor is 12 ÷ 8 = 1.5, and 1.5³ = 3.375. Answer: 3.375. Jayden's answer, 1.5, is only the LENGTH scale factor — he never cubed it to get the volume scale factor. 2.25 comes from squaring the length scale factor instead of cubing it, which would give the area scale factor. 4.5 comes from multiplying the length scale factor by 3 (the number of dimensions) instead of cubing it.
- (c) 12 cm — By Pythagoras' Theorem, QR² = PQ² + PR², so PR² = QR² − PQ² = 15² − 9² = 225 − 81 = 144. Square root: √144 = 12 cm.
- (d) 3 and 4 — f(3) = 3³ − 3 × 3² − 4 = 27 − 27 − 4 = −4, and f(4) = 4³ − 3 × 4² − 4 = 64 − 48 − 4 = 12. Since f(3) is negative and f(4) is positive, there is a change of sign, so the solution lies between 3 and 4. A sign slip when expanding −3x², treating it as +3x², gives f(3) = 27 + 27 − 4 = 50, a positive value; taken with the correctly negative f(2) = 8 − 12 − 4 = −8, this reads as a change of sign between 2 and 3. The same sign slip applied at x = 1 gives f(1) = 1 + 3 − 4 = 0, read as the point the sign changes, against f(0) = 0 + 0 − 4 = −4, giving 0 and 1. Shifting the correctly found interval up by one integer, an indexing slip, reports the change as lying between 4 and 5 instead of 3 and 4.
- (c) 1 : 3 — n : P = 15 : 45. Dividing both parts by their highest common factor, 15, gives 1 : 3. Inverting the ratio, 3 : 1, swaps profit and number of items. Dividing only the n-part by 15, getting 1, but leaving the P-part as 45 gives 1 : 45 — only one side has been simplified. Dividing only the P-part by 15, getting 3, but leaving the n-part as 15 gives 15 : 3, the opposite partial mistake.
- (c) 9 cm — The scale factor from triangle ABC to triangle PQR is PQ ÷ AB = 12 ÷ 8 = 1.5. So PR = AC × 1.5 = 6 × 1.5 = 9 cm. (4 cm comes from using the scale factor upside down, AB ÷ PQ; 10 cm comes from adding the difference between AB and PQ, 4 cm, to AC instead of scaling; 16 cm comes from multiplying PQ by the ratio AB : AC instead of scaling AC by the correct scale factor.)
- (b) 118 m — Width = 840 ÷ 35 = 24 m. Perimeter = 2 × (length + width) = 2 × (35 + 24) = 2 × 59 = 118 m. The option 59 m gives the sum of the length and width but forgets to double it for the perimeter. The option 70 m doubles only the length (2 × 35 = 70) and leaves out the width entirely. The option 48 m doubles only the width (2 × 24 = 48) and leaves out the length entirely.
- (a) 1.36 — Method: put the starting value into the right-hand side to get x₁, feed that value back in to get x₂, and round only once the second value has been found. Working: x₁ = 5 ÷ (1 + 2) = 5 ÷ 3 = 1.66666…; x₂ = 5 ÷ (1.66666… + 2) = 5 ÷ 3.66666… = 1.36363…. The digit in the third decimal place is 3, so x₂ = 1.36 correct to 2 decimal places. Answer: 1.36. The distractors: 1.67 is x₁, the value after a single use of the formula, given by a candidate who counts the starting value itself as x₁; 1.49 is x₃ = 1.48648…, one use of the formula too many; 1.37 comes from writing x₁ down as 1.66, truncating the display instead of keeping it in full, and then working out 5 ÷ 3.66 = 1.36612…, which rounds up to 1.37.
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.