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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- (c) 3:4 — The jacket costs 3/7 of £84, which is 3 × (84 ÷ 7) = 3 × 12 = £36. The bag then costs the rest of the money, £84 − £36 = £48. The ratio of the jacket to the bag is 36:48, which simplifies to 3:4. Writing the fraction spent on the jacket, 3/7, directly as the ratio, without working out that the bag's share is the remaining 4/7, gives 3:7. Giving the ratio the wrong way round, bag to jacket instead of jacket to bag, gives 4:3. Assuming the jacket and bag cost the same, ignoring the fraction given, gives 1:1.
- (b) 6 — Reflecting y = f(x) in the x-axis turns the minimum point (2, −3) into a maximum point at (2, 3), since −f(x) negates every y-value: −(−3) = 3. Adding a then gives 3 + a = 9, so a = 9 − 3 = 6. Forgetting the reflection and using the original y-value of −3 gives −3 + a = 9, so a = 12 — this ignores that −f(x) changes the sign of the y-coordinate before a is added. Writing a = −12 comes from subtracting in the wrong order, working out 9 − (−3) as −3 − 9 instead. Writing a = −6 comes from taking the negative of the correct answer, as if the final value of a needed to be reflected too, on top of the turning point.
- (b) −3 — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, with both changes taken in the same direction along the line. Working: going from (1, 20) to (5, 8) the change in y is 8 − 20 = −12 and the change in x is 5 − 1 = 4, so the gradient is −12 ÷ 4 = −3. Answer: −3, and the negative sign is expected because the line falls from left to right. The distractors: 3 comes from subtracting the smaller y from the larger, 20 − 8 = 12, while still taking the x values from left to right, which loses the minus sign that says the line falls; −12 is the change in y left undivided by the change in x of 4; −1/3 comes from dividing the change in x by the change in y, 4 ÷ (−12), turning the gradient upside down.
- (d) 6√3 cm — Method: AB lies alongside the 30° angle at A and AC is the hypotenuse, so the ratio needed is cosine: cos 30° = AB ÷ AC. Working: the exact value of cos 30° is √3/2, so AB = 12 × √3 ÷ 2, and half of 12 is 6. Answer: AB = 6√3 cm, which is about 10.4 cm. Using sine by mistake gives 12 × 1/2 = 6 cm, which is the length of BC rather than AB. Using tan 30° = 1/√3 gives 12 ÷ √3, which is 4√3 cm. Remembering cos 30° as √3 rather than as √3 halved gives 12√3 cm, longer than the hypotenuse and so impossible.
- (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) A person's shoe size and their favourite colour — A person's shoe size is not linked to which colour they prefer, so these two show no correlation. The other three pairs are all genuinely correlated: distance travelled and fuel used rise together, which is positive correlation; hours of revision and test score generally rise together, which is also positive correlation; and as outdoor temperature rises, fewer woolly hats are sold, which is negative correlation. Negative correlation is still a real relationship between two variables — it is not the same thing as no relationship at all, so the temperature and hats pair is not the answer to this question.
- (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.
- (d) £16.70 — The mileage charge is 2.20 × 6 = £13.20. Adding the fixed fee: £13.20 + £3.50 = £16.70. A candidate who forgets the fixed fee gives just the mileage charge, £13.20. A candidate who adds the fixed fee to the per-mile rate before multiplying by the number of miles, (3.50 + 2.20) × 6, gets £34.20. A candidate who rounds £2.20 down to £2 gets 2 × 6 + 3.50 = £15.50.
- (c) £338.69 — To increase by 12% each year, multiply by 1.12 twice. £270 × 1.12 × 1.12 = £338.688, which rounds to £338.69 (nearest penny, since the third decimal place is 8). £334.80 comes from treating the two 12% increases as a single flat 24% increase applied once instead of compounding: £270 × 1.24 = £334.80. £302.40 comes from applying the 12% increase only once, for 1 year instead of 2: £270 × 1.12 = £302.40. £338.68 comes from rounding £338.688 down to the nearest penny instead of up.
- (d) −2 — Method: the scale factor is the ratio of the image vector to the object vector, both measured FROM THE CENTRE of enlargement, keeping every sign. Working: the vector from the centre (2, 1) to P(2, 5) is (0, 4); the vector from the centre to P′(2, −7) is (0, −8). The scale factor is −8 ÷ 4 = −2. Answer: −2. Measure both vectors from the CENTRE, not from the origin, divide the IMAGE vector by the OBJECT vector and not the other way round, and keep the negative sign: a negative scale factor is not the same size as its positive counterpart with the sign dropped.
- (d) 1/8 — Method: a run of flips of a fair coin gives equally likely sequences of heads and tails, so count the sequences that match and divide by how many sequences there are. Working: each flip lands two ways and no flip affects another, so three flips give 2 × 2 × 2 = 8 equally likely sequences: HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. Only HHH has a head at every flip, so 1 sequence of the 8 matches. Answer: the probability is 1/8. The distractors: 1/4 comes from treating 'three heads', 'two heads', 'one head' and 'no heads' as four equally likely results, which they are not, since one sequence gives three heads and three sequences give two; 1/6 comes from taking the number of sequences to be 2 + 2 + 2 = 6, adding the two ways each flip can land instead of multiplying them; 1/2 comes from reading the first flip only and giving the probability of a head on one flip, without combining it with the other two.
- (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.
- (d) −3 — Substitute y = 11 into y = 5 − 2x, giving 11 = 5 − 2x. Subtracting 5 from both sides gives 6 = −2x, so x = 6 ÷ (−2) = −3. A candidate who mishandles the negative sign when rearranging, treating the equation as 6 = 2x, gets x = 3. A candidate who correctly finds −2x = 6 but forgets to divide by 2 at all gets x = 6. A candidate who adds 5 and 11 instead of subtracting, getting 2x = 16, gets x = 8.
- (d) 48 — Find the rate first: 18 ÷ 3 = 6 bottles per minute. Then apply it to the new time: 6 × 8 = 48 bottles. Working out 18 + (8 − 3) = 23 adds the extra 5 minutes onto the number of bottles instead of scaling proportionally. Working out 18 × 8 = 144 multiplies the given number of bottles by the new number of minutes without finding the rate first. Writing 18 keeps the count the same, not realising it must change with the time. In 8 minutes the machine fills 48 bottles.
- (b) 10.5 — Method: multiply the drawing length by the scale factor to get the real length. Working: 7 cm × 1.5 = 10.5 m. A student who answers 8.5 has added the scale factor to the drawing length instead of multiplying (7 + 1.5). A student who answers 14 has rounded the scale factor up to 2 before multiplying. A student who answers 3.5 has divided the drawing length by 2 instead of multiplying it by 1.5. Answer: 10.5 m.
- (a) Subtracting 4x gives 3 = 10, which is never true. — Method: try to solve the equation as normal and see what happens. Working: subtract 4x from both sides: 4x + 3 − 4x = 4x + 10 − 4x, giving 3 = 10. This statement is false for every value of x, so the equation has no solution. Answer: subtracting 4x gives 3 = 10, which is never true. "x would have to be negative" invents a constraint on x that the equation never states. "It's true for every x" confuses this equation with an identity, where both sides would simplify to the same expression. "x = 7" misreads the false statement 3 = 10 as something to solve for x, rather than recognising it means no solution exists.
- (a) 5:3 — Divide both prices by their highest common factor, 3: 15 ÷ 3 = 5 and 9 ÷ 3 = 3, giving the ratio 5:3. Choosing 3:5 comes from writing the ratio the wrong way round, as child price to adult price. Choosing 2:3 comes from using the difference between the two prices (15 − 9 = 6) as the first part of the ratio instead of the adult price, then simplifying 6:9 by dividing by 3. Choosing 5:8 comes from comparing the adult price with the total cost of both tickets (£15 out of £24) instead of comparing it with the child price.
- (b) (12, 9) — Apply the transformations in the order given: first translate, then enlarge. Translating (3, 5) by the vector (1, −2) gives (3 + 1, 5 − 2) = (4, 3). Enlarging this by scale factor 3 about the origin multiplies both coordinates by 3: (4 × 3, 3 × 3) = (12, 9). Enlarging first and translating afterwards reverses the order and gives (3 × 3 + 1, 5 × 3 − 2) = (10, 13), a different point because the two transformations do not commute. Enlarging the original point by scale factor 3 while forgetting to translate it at all gives (3 × 3, 5 × 3) = (9, 15). Reversing the signs of the translation vector before applying it gives (3 − 1, 5 + 2) = (2, 7), which then enlarges to (2 × 3, 7 × 3) = (6, 21).
- (c) 17.7 cm — Method: the iteration converges on x, the depth of the box, which is also the side of each square cut away; a square is removed from both ends of the 20 cm side, so the longer side of the base is 20 − 2x. Run the formula three times, then carry out that subtraction. Working: x₁ = (16 × 1² − 1³ + 50) ÷ 60 = 65 ÷ 60 = 1.08333…; x₂ = 67.50636… ÷ 60 = 1.12510…; x₃ = 68.82959… ÷ 60 = 1.14715…. The longer side of the base is 20 − 2 × 1.14715… = 17.70568…, which is 17.7 cm correct to 1 decimal place. Answer: 17.7 cm. The distractors: 1.1 cm is x₃ itself rounded, the depth of the box, given by a candidate who runs the iteration correctly and then stops before the step the question actually asks for; 18.9 cm is 20 − 1.14715…, taking away one square instead of two and forgetting that a corner is cut from each end of that side; 9.7 cm is 12 − 2 × 1.14715…, the shorter side of the base, which measures the wrong edge of the sheet.
- (d) x = 4 and x = −4 — Substituting y = 3 gives x² + 9 = 25, which simplifies to x² = 16, so x = 4 or x = −4. Choosing 'x = 3 and x = −3' uses the given value y = 3 as if it were the x-coordinate. Choosing 'x = 4' alone finds the positive square root of 16 but forgets the negative root. Choosing 'x = 5 and x = −5' skips subtracting 3² = 9 from 25 and takes the square root of 25 directly.
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.