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.Simplify x⁷ × x⁴, giving your answer as a single power of x.
- 2.The formula for the perimeter of a rectangle is P = 2(l + w). One rectangle has l = 9 cm and w = 4 cm. A second rectangle has l = 6 cm and w = 5 cm. Work out how many cm greater the perimeter of the first rectangle is than the perimeter of the second.
- 3.The rent on a flat increases by 10% one year and by a further 10% the following year. Work out the overall percentage increase over the two years.
- 4.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.
- 5.A bag contains 4 red counters and 6 blue counters. A counter is taken at random, its colour noted, and it is put back in the bag before a second counter is taken at random. Work out the probability that the two counters are different colours. Give your answer as a fraction in its simplest form.
- 6.A sports centre in Ipswich has 2,000 members. It wants to know what its members think of its opening hours, so it asks a random sample of 100 of them. Write down what the population is in this survey.
- 7.A lift has a safe working load of 500 kg. Four people get in the lift and the lift's display records their total mass as 493 kg, correct to the nearest kg. Decide whether the four people are definitely within the safe working load.
- 8.The curve y = x² − 3x. Use the chord between x = 1 and x = 4 to estimate the gradient of the curve at x = 2.5.y = x² − 3x
- 9.Write £3.60 : £2.40 as a ratio in its simplest form.
- 10.A rectangular garden bed measures 5 m by 3 m. A semicircular flower border is attached along one of the 3 m-wide ends, on the outside of the rectangle, so that the 3 m side is the diameter of the semicircle. Work out the total perimeter of the combined shape (the two long sides, the one remaining short side, and the curved edge). Use π = 3.14.
- 11.A box contains 9 red balls and 11 green balls. Two balls are taken out at random, one after the other, without being replaced. Given that both balls taken out are the same colour, work out the probability that both balls are red.
- 12.In standard form, 2,000 is written as 2 × 10ⁿ. Write down the value of n.
- 13.A rectangular garden has width w metres and length (w + 3) metres. A gardener writes its perimeter as 2w + 3. Which statement corrects the gardener's mistake?
- 14.y is directly proportional to x. When x = 8, y = 20. Work out the value of y when x = 12.
- 15.Rectangle ABCD has vertices A(−1, 2), B(4, 2), C(4, 5) and D(−1, 5). The rectangle is translated so that C moves to (0, −1). Write down the coordinates of the image of A.
- 16.A rule turns each input x into an output y. The inputs are x = 1, 2, 3, 4 and the matching outputs are y = −5, −3, −1 and one missing value. Work out the missing value of y.
- 17.y is directly proportional to x², and x is positive. When x = 4, y = 32. Construct the equation connecting x and y, then work out the value of x when y = 200.
- 18.A garden is made of two triangular flower beds, ABC and ACD, joined along the edge AC. In triangle ABC, angle ABC = 90°, AB = 5 m and BC = 12 m. In triangle ACD, AD = 9 m and angle CAD = 40°, the angle between AC and AD. Work out the area of flower bed ACD. Give your answer to 1 decimal place.
- 19.The graph of y = f(x) passes through the point (0, 4). Work out the y-coordinate of the point where the graph of y = −f(x) + 5 crosses the y-axis.
- 20.Which expression is equivalent to 3(x + 4) − 2(x − 1)?
Answer key
- (a) x¹¹ — Method: when multiplying powers of the same base, add the indices. Working: 7 + 4 = 11, so x⁷ × x⁴ = x¹¹. x²⁸ comes from multiplying the indices, 7 × 4 = 28, instead of adding them. x³ comes from working out 7 − 4 = 3, which is the rule for dividing powers, not multiplying them. 11x comes from adding the indices to make 11 but then treating x as a coefficient instead of a power. Answer: x¹¹.
- (a) 4 — Rectangle 1: P = 2(9 + 4) = 2 × 13 = 26 cm. Rectangle 2: P = 2(6 + 5) = 2 × 11 = 22 cm. Difference = 26 − 22 = 4 cm. 2 comes from finding the difference between (l + w) for each rectangle, 13 − 11 = 2, but forgetting to double it. 26 comes from giving rectangle 1's whole perimeter instead of the difference between the two. 6 comes from doubling only the change in length, 2 × (9 − 6), and ignoring that the width also changed.
- (c) 21% — Method: an increase of 10% is a multiplier of 1.1, and two successive increases are found by multiplying the multipliers. Working: 1.1 × 1.1 = 1.21, so the rent is 121% of the original, which is an increase of 21%. Answer: 21%. The distractors: 20% comes from adding the two percentages, which ignores that the second 10% is taken of a larger amount; 121% is the multiplier written as the change rather than the change itself; 11% comes from slipping in the multiplication and getting 1.11 instead of 1.21.
- (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.
- (a) 12/25 — P(red then blue) = 4/10 × 6/10 = 24/100. P(blue then red) is the same, 6/10 × 4/10 = 24/100. Since either order counts as different colours, add them: 24/100 + 24/100 = 48/100 = 12/25. Working out only one order, red-then-blue, and forgetting blue-then-red, gives 24/100 = 6/25. Working out the probability that the two counters are the SAME colour instead — 4/10 × 4/10 + 6/10 × 6/10 = 16/100 + 36/100 = 52/100 = 13/25 — answers a different question. Using 6/9 for the second draw, as if the first counter were not replaced, gives 2 × (4/10 × 6/9) = 48/90 = 8/15.
- (b) All 2,000 members of the sports centre. — Method: in a survey, the population is the whole group the survey is trying to find out about, and the sample is the smaller group actually asked. Working: this survey wants to know what the sports centre's members think, so the population is every one of the 2,000 members — whether or not they were personally asked. Saying the population is the 100 members who were asked names the sample, not the population; the sample is drawn FROM the population, so it is smaller than it, not the same as it. Saying the population is everybody who lives in Ipswich widens the group far beyond who the survey is actually about — plenty of Ipswich residents are not members of the sports centre at all, so they are outside this survey altogether. Saying the population is the members who say they are unhappy confuses the population with a result of the survey: whether a member turns out to be happy or unhappy is something the survey finds out, not part of the definition of who is being studied. The population is always the whole group the question is about, before any sampling or any results come in.
- (c) Yes — the greatest possible total is 493.5 kg, under 500 kg — 493 kg correct to the nearest kg means the true total mass, m, satisfies 492.5 kg ≤ m < 493.5 kg. The greatest possible total is 493.5 kg, which is under the 500 kg safe working load, so the four people are definitely within it. 'The true total could be as high as 498 kg' comes from treating 'nearest kg' as an error of ±5 kg instead of ±0.5 kg. 'Cannot be decided without the exact total' overlooks that the error interval already gives the greatest possible total, so the decision can be made without knowing the exact figure. '493 kg is only an estimate, so it may be over 500 kg' ignores that the error interval is bounded — the true total cannot exceed 493.5 kg, well under 500 kg.
- (c) 2 — At x = 1, y = 1² − 3(1) = 1 − 3 = −2. At x = 4, y = 4² − 3(4) = 16 − 12 = 4. Gradient of the chord = change in y ÷ change in x = (4 − (−2)) ÷ (4 − 1) = 6 ÷ 3 = 2. Writing down the change in y, 6, and stopping there without dividing by the change in x gives 6. Losing the negative sign on y = −2 at x = 1 and treating it as +2 gives (4 − 2) ÷ (4 − 1) = 2 ÷ 3 = 2/3. Dividing the wrong way round, change in x ÷ change in y, gives (4 − 1) ÷ (4 − (−2)) = 3 ÷ 6 = 1/2.
- (d) 3 : 2 — Convert both amounts to pence: £3.60 = 360p and £2.40 = 240p, giving the ratio 360 : 240. Divide both parts by their highest common factor, 120, to get 3 : 2. Giving 360 : 240 has not been simplified at all. Giving 2 : 3 swaps the order. Giving 36 : 24 has been divided by 10, which is a common factor but not the highest one, so it is not yet in simplest form.
- (c) 17.71 m — The perimeter is the two long sides (5 m + 5 m = 10 m) plus the one remaining short side (3 m) plus the curved semicircular edge. The 3 m side is the diameter of the semicircle, so its radius is 3 ÷ 2 = 1.5 m and the curved length is half the circumference: (1/2) × 2 × 3.14 × 1.5 = 4.71 m. Total: 10 + 3 + 4.71 = 17.71 m.
- (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.
- (c) 3 — Method: the index counts how many times the coefficient has been multiplied by 10, which is the number of places the decimal point moves from the end of the number to just after the first significant digit. Working: 2,000 = 2 × 1,000, and 1,000 = 10 × 10 × 10, which is three tens. Answer: 3. The distractors: 4 comes from counting the four digits of 2,000 rather than the three places the decimal point moves; 2 comes from copying the coefficient 2 into the index; −3 comes from making the index negative, which would describe a number smaller than 1 rather than two thousand.
- (a) It is 2(w + (w + 3)) = 4w + 6, not 2w + 3. — The perimeter of a rectangle is twice the width plus twice the length: 2 × w + 2 × (w + 3) = 2w + 2w + 6 = 4w + 6, so the gardener's 2w + 3 is wrong. Writing w + (w + 3) = 2w + 3 forgets to double the sides at all, only adding one width and one length once. Writing 4(w + 3) = 4w + 12 wrongly treats all four sides as equal to the length, as if the garden were a square. Writing 3w + 6 comes from doubling the length correctly but adding the width only once instead of doubling it too.
- (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.
- (d) (−5, −4) — Method: every vertex of a translated shape moves by the same vector, so find that vector from the one vertex whose image is given, then apply it to A. Working: C(4, 5) moves to (0, −1), so across 0 − 4 = −4 and up −1 − 5 = −6, giving the vector $\binom{-4}{-6}$. Applying it to A(−1, 2): −1 − 4 = −5 and 2 − 6 = −4. Answer: the image of A is (−5, −4). Working the vector out as object minus image gives 4 to the right and 6 up, which applied to A gives (3, 8). Getting the horizontal movement right but reversing the vertical one gives (−5, 8). Treating (0, −1) as the image of every vertex ignores that a translation carries each vertex to a different place.
- (d) 1 — Method: find the step in the outputs for each step of 1 in the input, write the rule from that step and from one pair of values, then apply the rule to the last input. Working: the outputs −5, −3, −1 rise by 2 while x rises in ones, so x is multiplied by 2; at x = 1, 2 × 1 = 2 while y = −5, so 7 is subtracted, giving y = 2x − 7; at x = 4 the rule gives 2 × 4 = 8 and 8 − 7 = 1. Answer: y = 1. The distractors: 3 comes from carrying the outputs on one step too far, to x = 5; 0 comes from assuming the outputs −5, −3, −1 carry on by adding 1 rather than by adding 2; 8 comes from doubling the input and forgetting to subtract the 7.
- (a) 10 — Since y is directly proportional to x², y = kx². Using x = 4, y = 32: 32 = k × 16, so k = 32 ÷ 16 = 2. The equation is y = 2x². When y = 200: x² = 200 ÷ 2 = 100, so x = 10, taking the positive root because the question states that x is positive. Stopping at x² = 100 without taking the square root leaves 100, not the value of x itself. Treating the relationship as if y were proportional to x itself gives k = 32 ÷ 4 = 8 and then x = 200 ÷ 8 = 25, which is a different relationship entirely. Multiplying by k instead of dividing when rearranging gives x² = 200 × 2 = 400 and x = 20, which is not this equation rearranged correctly. The value of x when y = 200 is 10.
- (c) 37.6 m² — Triangle ABC has a right angle at B, so use Pythagoras' theorem to find AC: AC² = AB² + BC² = 5² + 12² = 25 + 144 = 169, so AC = 13 m. In triangle ACD, use Area = 1/2 × AC × AD × sin(angle CAD) = 1/2 × 13 × 9 × sin 40° = 58.5 × 0.6428 = 37.6 m² (1 d.p.). Adding AB and BC to get AC = 17 m instead of applying Pythagoras gives 1/2 × 17 × 9 × sin 40° = 49.2 m². Using cos 40° instead of sin 40° gives 1/2 × 13 × 9 × cos 40° = 44.8 m². Substituting AB = 5 m directly instead of finding AC first gives 1/2 × 5 × 9 × sin 40° = 14.5 m².
- (c) 1 — At x = 0, f(0) = 4. Applying the transformations in order — reflect in the x-axis first, then translate up by 5 — gives −f(0) + 5 = −4 + 5 = 1. Applying the translation but forgetting the reflection gives f(0) + 5 = 9. Applying the reflection to the whole expression, including the +5, gives −f(0) − 5 = −9. Applying the reflection but forgetting the translation gives −f(0) = −4.
- (a) x + 14 — Expand each bracket separately: 3(x + 4) = 3x + 12, and −2(x − 1) = −2x + 2 (multiply −2 by both x and −1). Combine: 3x + 12 − 2x + 2 = x + 14. Writing x + 10 comes from taking −2(x − 1) as −2x − 2, not flipping the sign of the −1 inside the bracket. Writing 5x + 10 comes from treating the second bracket as +2(x − 1) instead of subtracting it, so the x-terms are added rather than subtracted. Writing x + 13 comes from only multiplying the 2 by the x, and carrying the −1 across unmultiplied.
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.