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) 7/8 — Method: write the decimal over the power of ten that matches the number of digits after the point, then divide the numerator and the denominator by their highest common factor. Working: 0.875 has three digits after the point, so it is 875 thousandths and can be written as 875/1000; the highest common factor of 875 and 1000 is 125, and 875 ÷ 125 = 7 with 1000 ÷ 125 = 8. Answer: 7/8. The distractors: 8/7 comes from cancelling correctly but writing the two parts the wrong way round; 9/10 comes from rounding 0.875 to one decimal place as 0.9 before converting; 7/80 comes from counting four decimal places instead of three and using a denominator of 10000, giving 875/10000.
- (d) x = −2, x = 6; y-intercept (0, 12) — Reflecting y = f(x) in the x-axis, to get y = −f(x), negates every y-value but leaves every x-value fixed. The x-intercepts happen where y = 0, and −0 = 0, so they are unaffected: y = −f(x) still crosses the x-axis at x = −2 and x = 6. The y-intercept is the value at x = 0: f(0) = −12, so −f(0) = 12, giving the point (0, 12) — the sign flips because the y-intercept is a nonzero y-value, unlike the roots. Writing 'x = 2, x = −6; y-intercept (0, −12)' comes from confusing −f(x) with f(−x) — reflecting in the y-axis instead of the x-axis, which negates the x-values of the intercepts (turning −2 into 2 and 6 into −6) but leaves f(0) unchanged, since f(−0) = f(0) = −12. Writing 'x = −2, x = 6; y-intercept (0, −12)' comes from forgetting that −f(x) is a reflection at all, and assumes both intercepts stay exactly as they were. Writing 'x = 2, x = −6; y-intercept (0, 12)' correctly negates the y-intercept but wrongly negates the x-intercepts too, as if a reflection in the x-axis also flipped the sign of every x-value.
- (a) 4 hours — This is inverse proportion: more pumps take less time. Multiply the original numbers to find the total pump-hours needed: 2 × 10 = 20 pump-hours. Divide by the new number of pumps: 20 ÷ 5 = 4 hours. Working out 10 × 5 ÷ 2 = 25 hours treats it as direct proportion, as if more pumps needed more time. Stopping at 20 gives the total pump-hours, not the number of hours. Working out 10 − (5 − 2) = 7 hours subtracts the extra number of pumps straight from the number of hours, treating pumps and hours as the same kind of quantity. 5 pumps take 4 hours.
- (c) cos 45° — cos 45° = √2/2, and √2 cannot be written as an exact fraction, so this value is irrational. tan 0° = 0, sin 90° = 1 and sin 30° = 1/2 are all rational, since each can be written as an exact whole number or fraction.
- (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.
- (c) The data show a link only; a third factor may affect both — Method: a study of this kind measures two quantities and reports how they change together; deciding that one of them produces the other is a further claim, and it needs evidence that the measurements alone cannot give. Working: the study shows that more coffee goes with better concentration, which is a positive correlation; but a third factor that was never measured, such as how motivated someone is, could raise both the coffee drinking and the concentration, and the concentration could equally be what leads to the extra coffee. Answer: the data show a link only, because a third factor may be affecting both quantities, so no claim about cause can be made. The distractors: calling the conclusion safe because the correlation is positive treats the direction of a correlation as proof of cause, which no direction can give; calling it wrong because the correlation is negative misreads the direction of the relationship, since the study reports both quantities rising together; saying the two quantities are not linked denies the correlation the study actually found, when what fails is only the claim about cause.
- (b) −0.7 < −0.25 — Method: compare the two negative decimals by their distance from zero on a number line. Working: −0.7 is 0.7 away from zero and −0.25 is 0.25 away from zero, so −0.7 is further from zero in the negative direction, making it the smaller number. Answer: −0.7 < −0.25 is true. "−0.7 > −0.25" comes from comparing 0.7 and 0.25 as if both numbers were positive, ignoring the negative signs. "−0.7 = −0.25" comes from assuming the two numbers are equal because they are both negative decimals. "−0.7 ≥ −0.25" combines the false statement "−0.7 > −0.25" with the false statement "−0.7 = −0.25".
- (d) 168 m — Split the area under the graph into two parts. The rising section (0 to 8 s) is a triangle: area = 0.5 × 8 × 12 = 48. The constant section, from 8 s to 18 s (10 s), is a rectangle: area = 12 × 10 = 120. Total distance = 48 + 120 = 168 m. Leaving out the 0.5 doubles the triangle, giving 8 × 12 + 120 = 216 m; ignoring the triangle section altogether gives only 120 m; averaging the start and final speeds over the whole 18 seconds, (0 + 12) ÷ 2 = 6, then 6 × 18 = 108 m, wrongly treats the tram as accelerating the whole time, when the graph is flat for the last 10 seconds.
- (d) 24 — Method: split 60 into 3 + 7 = 10 equal parts, find the value of one part, then use the difference in ratio parts. Working: 60 ÷ 10 = 6, so the numbers are 3 × 6 = 18 and 7 × 6 = 42, and their difference is 42 − 18 = 24. Answer: 24. 4 comes from finding the difference between the ratio numbers, 7 − 3, but forgetting to multiply by the value of one part. 60 comes from adding the two numbers back together instead of subtracting, which just repeats the given sum. 80 comes from dividing 60 by the first ratio number, 3, instead of by the total number of parts, 10, giving a part value of 20 and a difference of 7 × 20 − 3 × 20 = 80.
- (c) 10 cm — The diagonals of a rhombus bisect each other at right angles, splitting it into four congruent right-angled triangles with legs 8 cm (half of 16 cm) and 6 cm (half of 12 cm). By Pythagoras' Theorem, side² = 8² + 6² = 64 + 36 = 100. Square root: √100 = 10 cm.
- (b) 10 — Method: A′ means everything in the universal set that is NOT in A, so n(A′) = n(universal set) − n(A). Working: the universal set has 15 elements. A = {3, 6, 9, 12, 15}, so n(A) = 5. n(A′) = 15 − 5 = 10. Answer: 10. Watch out: writing down 5 gives n(A) itself, the size of the multiples-of-3 set, which is the opposite of its complement. Writing down 11 comes from missing 15 off the list of multiples of 3, treating A as only {3, 6, 9, 12}, so A is undercounted as 4 and A′ is overstated as 15 − 4. And writing down 12 comes from only listing the multiples of 3 up to 9 — 3, 6 and 9 — and missing that 12 and 15 also belong to A, undercounting A as 3 rather than 5.
- (a) 5 — Method: the number of glasses is the amount in the jug divided by the amount one glass holds. Write the mixed number as an improper fraction, then divide by multiplying by the reciprocal. Working: 3 1/3 = (3 × 3 + 1)/3 = 10/3, and 10/3 ÷ 2/3 = 10/3 × 3/2 = 30/6 = 5. Answer: 5. The distractors: 2 comes from writing 3 1/3 as 4/3, adding the whole number to the numerator instead of multiplying it by the denominator first, and then dividing 4/3 by 2/3; 20/9 comes from multiplying by 2/3 instead of dividing by it; 5/3 comes from dividing by 2 rather than by 2/3, as though each glass held 2 litres.
- (c) y = −x + 30 — Method: in a linear model the value at the start is the constant term and the steady rate of change is the gradient, which is negative when the quantity is falling. Working: at x = 0 the candle is 30 cm tall, so the constant term is 30; it loses 1 cm every hour, so the gradient is −1 and the height after x hours is y = −x + 30. Answer: y = −x + 30. The distractors: y = x + 30 comes from taking the rate as +1 and making the candle grow rather than shrink; y = 30x − 1 comes from building the model correctly as 30 − 1x and then copying it into the form y = mx + c with the two numbers left where they stood, so the starting height 30 ends up multiplying x and the hourly 1 is left behind as the constant being taken away; y = −30x + 1 comes from reading the two numbers the other way round, taking 30 cm per hour as the rate and 1 cm as the starting height, which burns 30 cm an hour from a candle only 1 cm tall.
- (b) 25 m/s — Convert km/h to m/s by multiplying by 1000 (km to m) and dividing by 3600 (hours to seconds): 90 × 1000 ÷ 3600 = 25 m/s. Working out 90 ÷ 60 = 1.5 converts using 60, as if going from hours to minutes rather than to seconds. Working out 90 × 3.6 = 324 multiplies by 3.6 instead of dividing by it, going the wrong way between the units. Working out 90 × 1000 = 90000 converts kilometres to metres but forgets to convert hours to seconds at all. The train's speed is 25 m/s.
- (d) 6.4 cm — The scale factor from the larger sign to the smaller sign is 4 ÷ 9, so the smaller side is 14.4 × 4 ÷ 9 = 6.4 cm. The distractor 32.4 cm comes from using the ratio the wrong way round, 14.4 × 9 ÷ 4 = 32.4. The distractor 9.4 cm comes from subtracting the difference between the ratio numbers (9 − 4 = 5) from the given length, 14.4 − 5 = 9.4. The distractor 3.6 cm comes from dividing the given length by 4 only, 14.4 ÷ 4 = 3.6, without also using the other ratio number.
- (c) y = 3x + 2 — Method: divide the change in the outputs by the change in the inputs to find the multiplier, then put one pair of values into the rule to find the number added on. Working: the output rises by 11 − 5 = 6 while the input rises by 3 − 1 = 2, so the multiplier is 6 ÷ 2 = 3; with an input of 1, 3 × 1 = 3 and the output is 5, so 2 is added. Answer: y = 3x + 2, checked against the second pair by 3 × 3 + 2 = 11. The distractors: y = 3x − 2 comes from finding the multiplier 3 and then subtracting the 2 instead of adding it; y = 2x + 3 comes from swapping the multiplier and the number added on; y = x + 4 comes from assuming the input is multiplied by 1 and using 5 − 1 = 4 as the number added on.
- (b) €230.00 — Multiply the amount in pounds by the exchange rate: 200 × 1.15 = 230, so £200 = €230.00. Working out 200 + 1.15 = 201.15 treats the exchange rate as an amount to add rather than a multiplier. Working out 200 × 0.15 = 30 finds only the extra amount earned for every pound and forgets to add it back to the original £200. Working out 200 × 11.5 = 2300.00 misplaces the decimal point in the exchange rate, multiplying by 11.5 instead of 1.15. £200 converts to €230.00.
- (c) minor segment — A chord splits a circle into two segments; the smaller of the two is called the minor segment and the larger one the major segment. "major segment" names the larger region, the opposite of what is asked for. "sector" is a different region altogether, enclosed by two radii and an arc, not by a chord. "arc" is a curved length along the circumference, not an enclosed region at all.
- (b) 8 — Method: a circle centred on the origin has equation x² + y² = r², where r is the radius, so the number on the right-hand side is the square of the radius and not the radius itself. Working: comparing x² + y² = 64 with x² + y² = r² gives r² = 64, so r = √64 = 8. Answer: the radius is 8. The distractors: 64 is r² read straight off the equation as though the right-hand side were the radius, which is the commonest error on this form; 32 comes from halving 64, treating the right-hand side as a diameter that has to be halved; 16 is the diameter, 2 × 8, quoted in place of the radius.
- (c) 2.61 — The limit L satisfies L = 12 ÷ (L + 2), so L(L + 2) = 12, giving L² + 2L − 12 = 0 and L = (−2 + √52) ÷ 2 = 2.6056, which is 2.61 to 2 decimal places (the early iterates 4, 2, 3, 2.4, 2.73, ... oscillate around this value before settling). Choosing 3.00 reads off x₃, one of the early oscillating values, before the sequence has settled close to the limit. Choosing 4.00 reads off x₁, the very first iterate, not the value the sequence approaches. Choosing 6.00 comes from writing the limit equation as L = 12 ÷ 2, leaving L itself out of the denominator.
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.