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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GCSE Higher sample Paper 2 (calculator)
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- 1.Simplify 2³ × 2⁴, giving your answer as a single power of 2.
- 2.y = 5 − 2x. Work out the value of x when y = 11.
- 3.A vehicle is worth £18000. Its value decreases by 15% each year. Using the recurrence V_{n+1} = 0.85V_n, with V_0 = 18000, find the first whole number of years after which the vehicle's value drops below £10000.
- 4.A hiker starts by walking on a bearing of 245°. She turns clockwise through 50°, then turns clockwise through a further 15°, and continues walking in a straight line. What bearing is she now walking on?
- 5.A bag contains beads that are exactly one of red, white or black. The probability of taking a red bead is 3/10 and the probability of taking a white bead is 1/4. Work out the probability of taking a bead that is red or white.
- 6.A factory makes 4,000 light bulbs a day. In a random sample of 80 of one day's bulbs, 3 were faulty. Work out an estimate for the number of faulty bulbs the factory makes in a day.
- 7.Work out (−3) × 4 + 2 × (−5)
- 8.The 2nd term of a geometric sequence is 6 and the 4th term is 24. Given that all the terms of the sequence are positive, work out the common ratio.
- 9.Two mathematically similar picture frames have perimeters 30 cm and 45 cm. The area of the smaller frame is 40 cm². Work out the area of the larger frame.
- 10.A party hat is in the shape of a cone with a base radius of 6 cm and a slant height of 10 cm. Work out the curved surface area of the party hat. Use π = 3.14 and the formula curved surface area = πrl.
- 11.A biased six-sided dice is rolled once. The probability that it lands on 6 is 0.25. The other five scores are all equally likely. Work out the probability that it lands on 3.
- 12.√700 lies between which two consecutive integers?
- 13.Solve the simultaneous equations 5x − 2y = 16 and 3x + 2y = 16. Work out the value of x.
- 14.The number of tickets a group can afford is inversely proportional to the price per ticket. At £4 per ticket, the group can afford 12 tickets. Work out how many tickets the group can afford at £6 per ticket.
- 15.A carpenter measures two triangular braces. Brace 1 has a 55° angle, a 65° angle, and, opposite the 55° angle, a side of 8 cm. Brace 2 has matching 55° and 65° angles and, opposite its 55° angle, a side of 8 cm. The carpenter says that because the equal 8 cm side is not between the two equal angles in either brace, he cannot be sure the two braces are the same size. Is the carpenter correct?
- 16.A zip-line runs from a platform 25 m high to a landing point 5 m high, over a horizontal distance of 40 m. Work out the gradient of the zip-line.
- 17.Two numbers are in the ratio 3 : 7. Their sum is 60. Work out the positive difference between the two numbers.
- 18.A shape is reflected in the line x = 1, and the image is then reflected in the line x = 5. Which single transformation is equivalent to this combination, for every point?
- 19.The nth term of a geometric sequence is given by aₙ = 3 × 2ⁿ⁻¹. Work out the position of the first term that is greater than 1,000.
- 20.The nth term of a sequence is n² + 3. Work out the first term of the sequence that is greater than 50.
Answer key
- (d) 2⁷ — When multiplying powers of the same base, the indices add: 3 + 4 = 7, so 2³ × 2⁴ = 2⁷. Multiplying the indices instead of adding them gives 3 × 4 = 12, so 2¹². Subtracting the indices instead of adding them gives 4 − 3 = 1, so 2¹. Multiplying the bases together as well as adding the indices gives 2 × 2 = 4, so 4⁷.
- (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) 4 years — Apply the recurrence repeatedly. V_1 = 0.85 × 18000 = 15300. V_2 = 0.85 × 15300 = 13005. V_3 = 0.85 × 13005 = 11054.25. V_4 = 0.85 × 11054.25 = 9396.1125. V_3 = £11054.25 is still above £10000, but V_4 = £9396.11 has dropped below it, so the answer is 4 years. Stopping at V_3 and calling it '3 years' misreads £11054.25 as already below £10000, or comes from wrongly modelling the fall as a flat £2700 a year (15% of the original value each time, without compounding), which crosses £10000 a year too early. Continuing one extra step to V_5 = 0.85 × 9396.1125 = 7986.70 and calling it '5 years' overshoots, since the value had already dropped below £10000 at V_4. Doubling the percentage decrease to 30% by mistake gives V_1 = 0.7 × 18000 = 12600, then V_2 = 0.7 × 12600 = 8820, which is already below £10000 after only 2 years — the wrong rate crosses the threshold too fast.
- (d) 310 — Method: since both turns are clockwise, add both angles to the starting bearing. Working: 245° + 50° = 295°; 295° + 15° = 310°. A student who answers 295 has only added the first turn and forgotten the second one. A student who answers 180 has subtracted both turns instead of adding them. A student who answers 320 has added the two turns as 75° instead of 65° by misreading the second turn. Answer: 310°.
- (c) 11/20 — 'Red or white' combines two mutually exclusive events, so add their probabilities: 3/10 = 6/20 and 1/4 = 5/20, giving 6/20 + 5/20 = 11/20. Multiplying the two probabilities instead of adding them, 3/10 × 1/4, gives 3/40, which would be the probability of red and white together, not red or white — and a bead can't be both colours. Subtracting the sum from 1, 1 − 11/20 = 9/20, gives the probability of the bead being black instead of red or white. Converting 1/4 as 4/20 instead of 5/20 (dividing 20 by 4 but forgetting to scale the numerator) gives 6/20 + 4/20 = 1/2.
- (a) 150 bulbs — Method: assume the proportion faulty in a random sample is the proportion faulty in the whole day's output, and scale the sample up to the population. Working: the sample of 80 has to be scaled up to 4,000 bulbs, and 4,000 ÷ 80 = 50, so the day's output is 50 sample-sized batches. Each batch is expected to contain the same 3 faulty bulbs, so the estimate is 3 × 50 = 150. Answer: 150 bulbs, and it is an estimate, because another sample of 80 would probably contain a different number of faulty bulbs. The distractors: 50 bulbs is the scale factor 4,000 ÷ 80 written down as though it were the answer, so it reports how many batches there are rather than how many faulty bulbs; 120 bulbs comes from reading 3 out of 80 as 3%, then taking 0.03 × 4,000 = 120, but 3 out of 80 is 3.75%; 240 bulbs comes from 3 × 80 = 240, multiplying the faulty bulbs by the size of the sample instead of by the scale factor, which uses the 80 twice and the 4,000 not at all.
- (a) −22 — Method: both multiplications are carried out before the addition, and a positive multiplied by a negative is negative. Working: (−3) × 4 = −12 and 2 × (−5) = −10, so the calculation becomes −12 + (−10) = −22. Answer: −22. The distractors: 22 comes from ignoring the minus signs and working out 3 × 4 + 2 × 5 = 22; 50 comes from working from left to right with no priority at all, giving −12 + 2 = −10 and then −10 × (−5) = 50; −2 comes from taking 2 × (−5) as +10, so that −12 + 10 = −2.
- (c) 2 — For a geometric sequence, a₄ = a₂ × r². So r² = a₄ ÷ a₂ = 24 ÷ 6 = 4, giving r = 2 or r = −2. Since all the terms are positive, r = 2. 4 comes from computing a₄ ÷ a₂ = 4 and stopping there, without taking the square root to find r. 18 comes from subtracting instead of dividing: a₄ − a₂ = 24 − 6 = 18, taken directly as the common ratio. −2 is the other square root of 4, but it is rejected because all the sequence's terms are given as positive.
- (c) 90 cm² — The perimeter ratio is 30 : 45, which simplifies to 2 : 3, so the larger frame is 1.5 times the perimeter of the smaller one. Areas scale with the square of this length scale factor: 1.5² = 2.25. 40 × 2.25 = 90, so the larger frame has an area of 90 cm². Giving 60 cm² uses the scale factor, 1.5, without squaring it (40 × 1.5 = 60). Giving 135 cm² cubes the scale factor, 1.5³ = 3.375, as if area scaled like a volume (40 × 3.375 = 135). Giving 2.25 cm² is the squared scale factor on its own, without multiplying by the smaller frame's area of 40 cm².
- (a) 188.4 cm² — Curved surface area of a cone = πrl. With r = 6 cm, l = 10 cm and π = 3.14, curved surface area = 3.14 × 6 × 10 = 188.4 cm². A student who uses the cylinder's curved surface area formula, 2πrl, instead of the cone's gets 2 × 3.14 × 6 × 10 = 376.8 cm². A student who uses the circle-area formula πr² instead of πrl gets 3.14 × 36 = 113.04 cm². A student who multiplies r × l but leaves out π entirely gets 6 × 10 = 60 cm².
- (b) 0.150 — The six scores are exhaustive, so their probabilities sum to 1. The probability of not landing on 6 is 1 − 0.25 = 0.75, and this is shared equally between the other five scores, so each has probability 0.75 ÷ 5 = 0.150. Giving 0.750 as the answer stops after finding the probability of not landing on 6, without sharing it out between the five remaining scores. Dividing 0.75 by 6 instead of by 5 gives 0.125, wrongly including the score of 6 among the equally likely scores. Ignoring the bias completely and dividing 1 by all six scores gives 1 ÷ 6 = 0.167.
- (d) 26 and 27 — Find the two consecutive perfect squares either side of 700: 26² = 676 and 27² = 729. Since 676 < 700 < 729, √700 lies between 26 and 27. Answering 7 and 8 comes from stripping the two zeros off 700 and using the 7 itself as the size of the root, instead of comparing 700 with the perfect squares around it — dividing the number under the root by 100 divides the root by 10, so the digits do not simply carry across. Answering 25 and 26 comes from checking 25² = 625, seeing that it is less than 700, and stopping there without also checking the square directly above it. Answering 35 and 36 comes from halving 700 to 350 and then treating that halved value as if it were ten times the true root, drifting into the thirties instead of the twenties.
- (b) 4 — Adding the two equations: the y-terms, −2y and +2y, cancel, and the x-terms combine to 5x + 3x = 8x; the right-hand sides add to 16 + 16 = 32. This gives 8x = 32, so x = 4. A candidate who adds only one of the right-hand sides, instead of both, would get 8x = 16, so x = 2. A candidate who divides 32 by 4 instead of 8 would get x = 8. A candidate who subtracts the equations instead of adding them, getting 2x − 4y = 0, and then wrongly assumes y = 0, would get x = 0.
- (d) 8 — The product of price and number of tickets is constant: k = 4 × 12 = 48. At £6 per ticket, the number of tickets is 48 ÷ 6 = 8. Getting 18 comes from treating price and tickets as directly proportional and working out 12 × 6 ÷ 4 instead of dividing k by the new price. Getting 12 assumes the number of tickets does not change when the price changes. Getting 6 comes from writing down the new price instead of working out the number of tickets.
- (d) No — third angle is also fixed — Since both braces have angles of 55° and 65°, their third angles must both be 60°, because angles in a triangle sum to 180°. All three angles now match, so the braces have the same shape. Both 8 cm sides lie in the same position relative to those angles — opposite the 55° angle in each brace — so one matching pair of corresponding sides fixes the size as well, exactly as ASA or AAS would. The braces are therefore guaranteed to be congruent and the carpenter is incorrect: 'No — third angle is also fixed' is correct. 'Yes — side must be included' is wrong because the side does not have to lie physically between the two named angles; once the third angle is fixed, a corresponding equal side anywhere is enough. 'No — any two angles enough alone' is wrong because two equal angles with no side length at all would only show the triangles are similar, not congruent. 'Yes — third angle may differ' is wrong because the third angle is fixed at 60° by the angle sum and cannot vary.
- (b) −0.5 — Method: gradient = change in height ÷ horizontal distance, and the height decreases so the change is negative. Working: change in height = 5 − 25 = −20, horizontal distance = 40, so gradient = −20 ÷ 40 = −0.5. Answer: the gradient is −0.5. 0.5 comes from dropping the negative sign, ignoring that the zip-line descends. 2 comes from inverting the gradient, dividing the horizontal distance by the drop in height instead of the other way round. −0.8 comes from dividing the drop by the platform height, 25, instead of by the horizontal distance, 40.
- (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) Translation by the vector (8, 0) — Method: reflecting twice in two parallel vertical lines is always equivalent to a single translation, at right angles to the lines, of twice the distance between them. Working: the two lines are 5 − 1 = 4 units apart, so the translation is 2 × 4 = 8 units in the positive x-direction. Answer: translation by the vector (8, 0). Using just the gap itself, without doubling it, gives (4, 0); translating in the negative x-direction, from the second line back towards the first, gives (−8, 0); and describing the combination as a single reflection in the line halfway between them, x = 3, confuses this combination with the effect of a single reflection — two reflections in parallel lines are always equivalent to a translation, never to another reflection. Always double the gap between the lines, and translate in the direction from the first line towards the second.
- (b) 10 — Check n = 9: a₉ = 3 × 2⁸ = 3 × 256 = 768, below 1,000. Check n = 10: a₁₀ = 3 × 2⁹ = 3 × 512 = 1,536, above 1,000. So the first term greater than 1,000 is at position n = 10. Answering 9 comes from forgetting the −1 shift and using the formula as 3 × 2ⁿ instead of 3 × 2ⁿ⁻¹: checking 3 × 2⁹ = 1,536 (which exceeds 1,000) but then reporting the position as n = 9, the exponent used, instead of n = 10 — wrong, because the exponent in the real formula is n − 1, not n. Answering 11 comes from going one term too far: correctly finding that a₁₀ already exceeds 1,000, but then checking one position further and reporting n = 11 instead of stopping at the first position that already works — wrong, because n = 11 is not the FIRST term greater than 1,000. Answering 1,536 gives the VALUE of the term (a₁₀ itself) rather than its position — wrong, because the question asks which term it is (the value of n), not what that term is worth.
- (a) 52 — Test successive terms: n=6 gives 6²+3=39, which is not greater than 50. n=7 gives 7²+3=52, which is greater than 50, so the first term greater than 50 is 52. A candidate who stops at n=6, before checking whether 39 actually exceeds 50, would give 39. A candidate who solves n²>50 instead of n²+3>50, ignoring the +3 in the search, would find n=8 is the first value with n²>50 (since 7²=49) and compute 8²+3=67. A candidate who computes n² by doubling n instead of squaring it would compute 2×7+3=17.
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.