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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- (a) 7 — Dividing by 0.1 is the same as multiplying by 10, so 0.7 ÷ 0.1 = 7. Meera's answer of 0.07 comes from dividing 0.7 by 10 instead of by 0.1, the wrong way round. A candidate who confuses 0.1 with 0.01 multiplies by 100 instead of 10 and gets 70. A candidate who thinks dividing by a number less than 1 does not change the value gets 0.7.
- (b) x ≥ 5 — Expand the bracket: 2(3x − 1) = 6x − 2, so the inequality is 6x − 2 ≥ 4x + 8. Subtract 4x from both sides and add 2 to both sides: 2x ≥ 10. Divide both sides by 2: x ≥ 5. A candidate who only multiplies the 3x by 2 and forgets to multiply the −1 gets 6x − 1 ≥ 4x + 8, leading to x ≥ 4.5. A candidate who adds 4x instead of subtracting it gets 10x ≥ 10, leading to x ≥ 1. A candidate who multiplies by 2 instead of dividing gets x ≥ 20.
- (b) 24 — Method: let one part of the ratio be worth x, write both groups in terms of x, and use the fact that the two groups end up equal. Working: the boys are 3x and the girls are 4x; after the 6 boys join, 3x + 6 = 4x, so x = 6; the girls are 4 parts, so 4 × 6 = 24. Answer: 24 girls. The distractors: 18 is the number of boys before the 6 join, which is 3 × 6; 30 comes from adding the 6 new members to the girls as well as to the boys; 42 is the total number of members in the choir before the 6 boys join, the 18 boys and the girls together.
- (c) 12.4 cm — Using the cosine rule, BC² = AB² + AC² − 2 × AB × AC × cos(A) = 9² + 6² − 2 × 9 × 6 × cos(110°) = 81 + 36 − 108 × cos(110°). Since cos(110°) ≈ −0.34202, 108 × cos(110°) ≈ −36.94, so BC² ≈ 117 + 36.94 = 153.94. Taking the square root, BC ≈ 12.4072, which rounds to 12.4 cm. 8.9 cm comes from treating cos(110°) as if it were positive (using +0.342 instead of −0.342), which wrongly subtracts instead of adds and gives BC² ≈ 80.06. 153.9 cm is BC² itself, rounded, with the square root never taken. 11.6 cm comes from leaving out the factor of 2 in the formula, computing BC² = 81 + 36 − 9 × 6 × cos(110°) ≈ 135.47 instead.
- (d) 75% — Method: count how many cards are not hearts, write that count over the total number of cards, cancel the fraction down and then turn it into a percentage. Working: 52 − 13 = 39 cards are not hearts, so the probability is 39/52; dividing the numerator and the denominator by 13 gives 3/4, and 3/4 = 0.75, so 0.75 × 100 = 75. Answer: 75%, three quarters of the way along the 0 to 1 scale. The distractors: 25% comes from giving the probability that the card is a heart, 13 out of 52, which cancels to 1/4; 50% comes from reading 'not a heart' as 'not a red card' and halving the pack; 39% comes from writing the count of 39 cards straight down as the percentage without comparing it with the 52 cards in the pack.
- (a) The means are equal, and Class B's marks are the more consistent because its range is smaller. — Method: comparing two distributions needs two things — a measure of average and a measure of spread — and each must be put into the context of the question. Working: both classes have a mean mark of 70, so on average the two classes scored the same; the range measures spread, and Class A's range of 30 marks is three times Class B's range of 10 marks, so Class B's marks sit closely around the mean while Class A's are far more spread out. Answer: the means are equal, and Class B's marks are the more consistent because its range is smaller. The distractors: the reply crediting Class A with more consistency reverses the meaning of the range, treating a larger range as tighter data when a larger range means more spread; the reply that Class A's mean mark is higher compares the wrong pair of figures, reading the range of 30 as an average; the reply that Class B's mean mark is higher reads the spread correctly but its claim about the means is false, since both means are 70.
- (b) 750,000,000 — Method: multiplying by 10⁸ moves the decimal point eight places to the right, and every empty place is filled with a zero. Working: 10⁸ = 100,000,000, and moving the decimal point in 7.5 eight places to the right gives 7.5 × 100,000,000. Answer: 750,000,000. The distractors: 7,500,000,000 comes from removing the decimal point first to make 75 and then writing eight zeros after it, which carries the digits one place too far; 600 comes from reading 10⁸ as 10 × 8 = 80 and working out 7.5 × 80; 0.000000075 comes from moving the decimal point eight places to the left, as though the index were negative.
- (c) (3, 9) — Reflecting in the horizontal line y = 2 keeps the x-coordinate the same and maps y to 2 × 2 − y = 4 − (−5) = 9, so S = (3, 9). A candidate who uses k − y instead of 2k − y gets 2 − (−5) = 7, giving (3, 7). A candidate who reflects in the x-axis instead of the line y = 2, simply changing the sign of y, gets (3, 5). A candidate who also changes the sign of the x-coordinate, as if reflecting in both axes, gets (−3, 9).
- (d) 2:3:5 — The highest common factor of 12, 18 and 30 is 6. Divide every part by 6: 12 ÷ 6 = 2, 18 ÷ 6 = 3, 30 ÷ 6 = 5, giving 2 : 3 : 5. Dividing by 2 instead of 6 gives 6 : 9 : 15, which still shares a common factor of 3, so it is not fully simplified. Dividing by 3 instead of 6 gives 4 : 6 : 10, which still shares a common factor of 2, so it is not fully simplified either. Swapping the first two parts gives 3 : 2 : 5, the parts in the wrong order.
- (b) 2.8 m — Use Pythagoras' Theorem: the plank is the hypotenuse (3.5 m) of a right-angled triangle formed with the wall and the ground (2.1 m). height² = 3.5² − 2.1² = 12.25 − 4.41 = 7.84. height = √7.84 = 2.8 m. A student who subtracts the two given lengths directly instead of using Pythagoras gets 3.5 − 2.1 = 1.4 m. A student who doubles the distance from the wall by mistake gets 2.1 × 2 = 4.2 m.
- (b) £0.30 profit for the stall — The stall keeps the £1.50 entry fee whatever happens, and expects to pay out prize × probability of winning = £6 × 0.2 = £1.20 on average. So its expected profit per game is £1.50 − £1.20 = £0.30. Reporting the expected pay-out of £1.20 itself as the profit forgets that the stall also keeps the entry fee. Assuming the player always wins gives an expected cost of £6 − £1.50 = £4.50, treated as a loss for the stall. Using the probability of NOT winning, 0.8, to find the expected pay-out gives £6 × 0.8 = £4.80, and £1.50 − £4.80 = −£3.30, a £3.30 loss.
- (c) −1 — Method: find each cube root separately, keeping its sign, and then add the two results. Working: (−3) × (−3) × (−3) = −27, so ∛(−27) = −3, and 2 × 2 × 2 = 8, so ∛8 = 2. Adding gives −3 + 2 = −1. Answer: −1. The distractors: 5 comes from taking the cube root of a negative number as positive, giving 3 + 2; −5 comes from reading the minus sign as applying to the whole sum and working out −(3 + 2); −6 comes from multiplying the two roots, −3 × 2, instead of adding them.
- (c) y = (5/12)x − 169/12 — The gradient of the radius to (5, −12) is (−12 − 0) ÷ (5 − 0) = −12/5. A tangent is perpendicular to the radius at that point, so its gradient is the negative reciprocal, 5/12. Using y − y₁ = m(x − x₁) with (5, −12): y + 12 = (5/12)(x − 5), which gives y = (5/12)x − 169/12. y = −(12/5)x comes from using the radius's own gradient, −12/5, instead of turning it into the perpendicular gradient, and building the line through the origin (as the radius itself does). y = −(5/12)x − 119/12 comes from taking the reciprocal of −12/5 correctly as a size but keeping the wrong sign, using −5/12 instead of 5/12. y = (5/12)x − 25/12 comes from using the correct gradient 5/12 but building the line through (5, 0) instead of (5, −12) — dropping the point's y-coordinate.
- (b) 12 — The difference between the parts of the ratio is 5 − 2 = 3 parts, and this is worth 18. Divide to find one part: 18 ÷ 3 = 6. Cats have 2 parts: 2 × 6 = 12. (30 is the number of dogs, using 5 parts instead of 2. 6 is the value of one part — the number of cats is 2 lots of this, not just one. 9 comes from dividing 18 by 2 and stopping there, instead of dividing by the difference in parts, 3, and then multiplying by 2.)
- (d) Wrong - the given angle is not the included angle — Sides AB and BC meet at vertex B, so the included angle needed for SAS is angle B, not angle A — the information given is SSA. SSA does not prove congruence: with AB = 10 cm, BC = 7 cm and angle A = 40° there are two different triangles that fit, one with angle C ≈ 74.6° and one with angle C ≈ 105.4°, so Sam's triangles need not be the same shape and size at all. Sam is not correct just because two sides and an angle are equal, since the angle must be the one INCLUDED between those two sides. The condition is not ASA either, because ASA needs two angles, and only one angle is given here. It is also not true that nothing matches — the stated lengths and angle DO match between the two triangles; the problem is which angle was given, not whether the values agree.
- (c) 50 — Method: the edge of the track is a circle centred on the origin, so it has equation x² + y² = r²; use the floodlight whose coordinates are given to find r², then substitute x = 0 to find where that circle meets the positive y-axis. Working: 30² + 40² = 900 + 1600 = 2500, so the edge of the track is x² + y² = 2500. Substituting x = 0 gives k² = 2500, so k = √2500 = 50, taking the positive root because k is positive. Answer: k = 50. The distractors: 2500 is r² used as though it were the radius, which would place the second floodlight fifty times too far from the centre; 70 adds the two coordinates, 30 + 40, instead of adding their squares; 40 takes the y-coordinate of the first floodlight to be the radius, which would only be true if that floodlight stood on the y-axis itself.
- (c) 30 — Method: use y = kx and find k from the given pair of values, then substitute x = 12. Working: k = 20 ÷ 8 = 2.5, so y = 2.5 × 12 = 30. Answer: 30. 24 comes from treating the relationship as additive, adding the increase in x (12 − 8 = 4) straight onto y (20 + 4 = 24), instead of multiplying by k. 14.5 comes from finding k correctly (2.5) but then adding it to x instead of multiplying (12 + 2.5 = 14.5). 4.8 comes from finding k upside down, 8 ÷ 20 = 0.4, and multiplying by x: 12 × 0.4 = 4.8.
- (a) $\binom{-5}{-4}$ — Method: one translation followed by another is a single translation, and the two vectors are added: top to top, bottom to bottom. Working: across, 4 − 9 = −5; up, −7 + 3 = −4. Answer: $\binom{-5}{-4}$. Subtracting the second vector instead of adding it gives 13 on top and −7 − 3 = −10 underneath. Adding the top numbers correctly but subtracting the bottom ones gives −10 underneath with −5 on top. Adding 4 and 9 as though both were positive and then keeping the minus sign of the larger gives −13 on top.
- (a) 120 m — The gradient of the second line is (0 − V)/(9 − 6) = −V/3, and this equals −8, so V = 24. The distance from t = 0 to t = 6 is the area of a trapezium with parallel sides 4 and 24 and width 6: 1/2 × (4 + 24) × 6 = 84. The distance from t = 6 to t = 9 is the area of a triangle with base 3 and height 24: 1/2 × 3 × 24 = 36. The total distance is 84 + 36 = 120 m. Using V = 8, treating the gradient's number as the missing velocity itself rather than solving −V/3 = −8 for V, gives a trapezium area of 1/2 × (4 + 8) × 6 = 36 and a triangle area of 1/2 × 3 × 8 = 12, a total of 48 m. Leaving out the 1/2 in the trapezium formula, (4 + 24) × 6 = 168, plus the correct triangle of 36, gives 204 m. Using the full 6 seconds as the triangle's base instead of the 3 seconds the second line actually lasts, 1/2 × 6 × 24 = 72, plus the correct trapezium of 84, gives 156 m.
- (b) (3, 3) and (−1, −5) — Set the two expressions for y equal: x² − 6 = 2x − 3, which rearranges to x² − 2x − 3 = 0. Factorise: (x − 3)(x + 1) = 0, so x = 3 or x = −1. Substitute into the linear equation, y = 2x − 3: x = 3 gives y = 3; x = −1 gives y = −5. The points are (3, 3) and (−1, −5). Distractor routes: (3, 3) and (1, −1) comes from mis-factorising x² − 2x − 3 as (x − 3)(x − 1), giving a second root of 1 instead of −1. (3, 9) and (−1, 1) comes from finding the y-coordinate from y = x² instead of substituting back into the given line equation y = 2x − 3. (3, 3) alone stops after finding only the first root of the quadratic.
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.