Sample paper · GCSE Foundation · grades 1–5

GCSE Foundation 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 Foundation qualification — number 25%, algebra 20%, ratio, proportion and rates of change 25%, geometry and measures 15%, probability 7.5%, statistics 7.5% — with a calculator allowed. AO1 / AO2 / AO3 at this tier: 50% / 25% / 25%.

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GCSE Foundation sample Paper 2 (calculator)

Maths · GCSE Foundation · Paper 2 · Calculator · 20 questions

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1.A number, y, is equal to 8.2 when rounded to 1 decimal place. Write down the error interval for y.
(a)8.15 ≤ y < 8.25
(b)8.15 < y < 8.25
(c)8.1 ≤ y < 8.2
(d)8.2 ≤ y < 8.3
2.Rosie is asked whether 4(x + 3) = 4x + 12 is an equation or an identity. Which statement is correct? Give a reason for your answer.
(a)An identity, because it is true for every value of x
(b)An identity, because the brackets can be expanded
(c)An equation, because solving it gives x = 3
(d)A formula, because it works out 4x + 12 from x
3.The ratio of y to x is always 3 : 4. Write down which equation could represent this relationship.
(a)y = 3x + 4
(b)y = 3/(4x)
(c)y = 4x/3
(d)y = 3x/4
4.Work out the exact value of cos 0° − sin 30°.
(a)1/2
(b)−1/2
(c)3/2
(d)1
5.A game has two independent stages. P(win stage 1) = 0.4, and P(win stage 2) = 0.25. Work out the probability that Ffion wins at least one of the two stages.
(a)0.1
(b)0.55
(c)0.45
(d)0.65
6.A garden centre's sales, in thousands of pounds, at the end of each quarter last year were: quarter 1 — 18, quarter 2 — 34, quarter 3 — 30, quarter 4 — 22. Work out the increase in sales from quarter 1 to the quarter with the highest sales.
(a)£16,000
(b)£12,000
(c)£34,000
(d)£16
7.Which of these units would be most appropriate to use when stating the mass of a small car?
(a)Kilograms
(b)Litres
(c)Grams
(d)Metres
8.The first five square numbers are 1, 4, 9, 16, 25. Write down the next square number in the sequence.
(a)36
(b)30
(c)35
(d)49
9.A shop sells ribbon by the metre. 2 m costs £3.00, 4 m costs £6.00, and 7 m costs £10.50. Does this data show that the cost is directly proportional to the length of ribbon bought? Choose the correct verdict and reason.
(a)Yes — the cost per metre is £1.50 each time
(b)Yes — the cost increases as the length increases
(c)No — 10.50 ÷ 7 gives 1.05, unlike the other rates
(d)No — the cost must double whenever length doubles
10.A right-angled triangle has its two shorter sides equal to 8 cm and 15 cm. Work out the value of cos θ, where θ is the angle opposite the 8 cm side. Give your answer as a fraction.
(a)17/15
(b)8/17
(c)15/17
(d)8/15
11.A survey of 60 readers records: 21 read only fiction, 17 read only non-fiction, 14 read both, and the rest read neither. Work out the probability that a randomly chosen reader reads exactly one of the two types, giving your answer as a fraction in its simplest form.
(a)13/15
(b)19/26
(c)7/30
(d)19/30
12.A piece of ribbon is 2.4 metres long. Kim cuts 40 cm from it. Work out the length of ribbon left, in millimetres.
(a)2800 mm
(b)200 mm
(c)2000 mm
(d)2360 mm
13.A company charges for hiring chairs for a day. Hiring 1 chair costs £12, hiring 2 chairs costs £19, hiring 3 chairs costs £26, and hiring 4 chairs costs £33. Work out an expression, in terms of n, for the cost in pounds of hiring n chairs for a day.
(a)7n
(b)12n + 7
(c)7n + 12
(d)7n + 5
14.A solution is made by dissolving 50 g of sugar in 200 g of water. Work out the concentration of sugar in the solution as a percentage.
(a)20%
(b)25%
(c)80%
(d)0.2%
15.A raised garden bed is a prism whose cross-section is a triangle with a base of 1.2 m and a perpendicular height of 0.8 m. The bed is 3 m long. Work out the volume of soil needed to fill it, in m³.
(a)3.6 m³
(b)2.88 m³
(c)1.44 m³
(d)0.48 m³
16.Work out 1/2 of 1/4 of 80.
(a)20
(b)10
(c)60
(d)40
17.A distance-time graph shows a car travelling at a constant speed, covering 90 km in 2 hours. Work out the car's speed in km/h.
(a)88 km/h
(b)180 km/h
(c)45 km/h
(d)92 km/h
18.The ratio of cats to dogs in a shelter is 3:4. There are 18 cats. Work out the total number of animals in the shelter.
(a)42
(b)126
(c)25
(d)24
19.Light travels at 2.998 × 10⁸ metres per second. A distant object in space is 3.1 × 10¹⁵ metres from Earth. Work out an estimate for the number of seconds light takes to travel from the object to Earth, by rounding each number to 1 significant figure.
(a)9 × 10²³ seconds
(b)3 × 10⁷ seconds
(c)1 × 10⁻⁷ seconds
(d)1 × 10⁷ seconds
20.Two quantities x and y are in direct proportion. When x = 8, the value of y is 20. Work out the value of y when x = 14.
(a)26
(b)5.6
(c)28
(d)35

Answer key

(a) 8.15 ≤ y < 8.25 — Rounding to 1 decimal place means the error interval spans half of 0.1, so 0.05, either side of 8.2: 8.2 − 0.05 = 8.15 and 8.2 + 0.05 = 8.25. The lower bound uses ≤ because 8.15 itself rounds to 8.2, but the upper bound uses < because 8.25 would round up to 8.3. So the error interval is 8.15 ≤ y < 8.25. A candidate who used the wrong rounding band gave 8.1 ≤ y < 8.2. A candidate who used a strict inequality at both ends wrote 8.15 < y < 8.25, wrongly excluding 8.15 itself. A candidate who added the full 0.1 instead of half of it wrote 8.2 ≤ y < 8.3.
(a) An identity, because it is true for every value of x — Expanding the brackets multiplies both terms inside by 4, giving 4x + 12, which is exactly the right-hand side. The two sides are therefore equal whatever x is, and a statement true for every value of the letter is an identity. An equation is true only for particular values, and trying to solve this one leads to 0 = 0, which places no restriction on x at all. Being able to expand brackets is not what makes a statement an identity, since any equation with brackets can be expanded. Nor is it a formula: a formula links two different quantities, and only one letter appears here.
(d) y = 3x/4 — y : x = 3 : 4 means y/x = 3/4. Rearranging to make y the subject gives y = (3/4)x = 3x/4. A student who mixes up which quantity goes on top gets y = 4x/3. A student who treats the ratio numbers as the coefficient and constant of a linear equation instead of a proportional relationship gets y = 3x + 4. A student who mistakes the relationship for inverse proportion gets y = 3/(4x).
(a) 1/2 — cos 0° = 1 and sin 30° = 1/2, so cos 0° − sin 30° = 1 − 1/2 = 1/2. 1 comes from writing down cos 0° alone and forgetting to subtract sin 30°. 3/2 comes from adding the two values instead of subtracting. −1/2 comes from working out sin 30° − cos 0°, the two terms the wrong way round.
(b) 0.55 — Method: it is easier to find the probability that Ffion wins NEITHER stage, then subtract that from 1. Working: P(lose stage 1) = 1 − 0.4 = 0.6, and P(lose stage 2) = 1 − 0.25 = 0.75. P(neither) = 0.6 × 0.75 = 0.45. P(at least one) = 1 − 0.45 = 0.55. Answer: 0.55. Watch out: adding the two win probabilities, 0.4 + 0.25 = 0.65, treats winning both as impossible and overcounts — that is not how independent probabilities combine. Multiplying the two win probabilities, 0.4 × 0.25 = 0.1, gives the probability of winning BOTH stages, not at least one. And stopping at 0.45, the probability of winning neither stage, forgets the final step of subtracting from 1.
(a) £16,000 — Method: first find the quarter with the highest sales figure, then subtract quarter 1's sales from it — remembering that every figure is given in THOUSANDS of pounds. Working: the highest sales figure is quarter 2, at £34,000 (34 thousand pounds). The increase from quarter 1 is £34,000 − £18,000 = £16,000. Giving £34,000 reads off the highest sales figure on its own, without subtracting quarter 1's sales — that is the highest quarter's total, not the increase. Giving £12,000 uses quarter 3's sales, 30, the SECOND-highest figure, instead of quarter 2's 34, the actual highest — 30 − 18 = 12, but quarter 3 is not the quarter with the highest sales. Giving £16 gets the subtraction right, 34 − 18 = 16, but forgets that every figure in the question is in thousands of pounds, so the increase is £16,000, not £16. Always identify the correct quarter FIRST, and always check the units the numbers are given in before writing your final answer.
(a) Kilograms — A small car has a mass of roughly 1000–1500 kilograms, so kilograms gives a sensible number to work with — neither too small nor too large. Grams would give an enormous number, over a million, since grams suit much lighter objects. Litres measures capacity, not mass, so it is the wrong type of unit for a car's mass entirely. Metres measures length, not mass, so it is also the wrong type of unit.
(a) 36 — Square numbers are n² for n = 1, 2, 3, ...; the fifth term, 25, is 5². The sixth square number is 6² = 36. A candidate who mislabels 25 as the sixth square number would compute 7² = 49 instead. A candidate who repeats an earlier difference between terms (5, from 4 to 9) rather than the correct next difference (11, since the differences are the odd numbers 3, 5, 7, 9, 11) would reach 30. A candidate who instead adds 10 would reach 35.
(a) Yes — the cost per metre is £1.50 each time — Direct proportion holds if the cost per metre is the same every time. Check each pair: 3.00 ÷ 2 = 1.50, 6.00 ÷ 4 = 1.50, and 10.50 ÷ 7 = 1.50. All three give the same rate, £1.50 per metre, so the data does show direct proportion. Saying only that the cost increases as the length increases is not enough on its own — many non-proportional relationships also increase, so this reason does not prove proportion. Misreading 10.50 ÷ 7 as 1.05 by misplacing the decimal point gives a false mismatch that is not actually there. Requiring every length to be a double of another confuses a special case (doubling) with the general test, which is that the rate itself stays constant. The data does show direct proportion, at £1.50 per metre.
(c) 15/17 — Method: cos θ = adjacent ÷ hypotenuse, so find the hypotenuse with Pythagoras' theorem first and then decide which short side is next to θ. Working: the hypotenuse is √(8² + 15²) = √(64 + 225) = √289 = 17 cm. The angle θ is opposite the 8 cm side, so the side next to it is the 15 cm side, and cos θ = 15 ÷ 17. Answer: 15/17. The distractors: 8/17 is sin θ, opposite over hypotenuse, used in place of the cosine; 8/15 is tan θ, opposite over adjacent; 17/15 comes from writing the cosine ratio upside down, as hypotenuse over adjacent.
(d) 19/30 — Exactly one means only fiction or only non-fiction, not both: 21 + 17 = 38 out of the 60 readers, which simplifies to 19/30. Including the 14 who read both as well gives 21 + 17 + 14 = 52, so 52/60 = 13/15 — that is at least one, not exactly one. Using only the both-count, 14, as the numerator gives 14/60 = 7/30, the probability of reading both, not exactly one. Using 52, the number who read at least one type, as the denominator instead of the full 60 readers surveyed gives 38/52 = 19/26.
(c) 2000 mm — Put both lengths into the same unit first. There are 1000 mm in a metre, so the ribbon is 2.4 × 1000 = 2400 mm, and there are 10 mm in a centimetre, so the piece cut off is 40 × 10 = 400 mm. The length left is 2400 − 400 = 2000 mm. 2360 mm subtracts 40 mm instead of 400 mm, 200 mm works in centimetres and then labels the result as millimetres, and 2800 mm adds the piece that was cut off instead of subtracting it.
(d) 7n + 5 — Method: find the rate charged per extra chair, then find the fixed part of the cost that fits hiring 1 chair. Working: the cost rises by £7 for each extra chair (19 − 12 = 7, 26 − 19 = 7, 33 − 26 = 7), so the cost has the form 7n + c. Substituting n = 1: 7(1) + c = 12, so c = 5. Answer: the cost in pounds is 7n + 5. The value 7n comes from ignoring the fixed part of the charge entirely. The value 7n + 12 comes from using the cost of 1 chair as the fixed part directly, without subtracting the per-chair rate first. The value 12n + 7 comes from swapping the roles of the cost of hiring 1 chair, £12, and the rate per extra chair, £7 — using the total for one chair as the coefficient of n and the rate as the fixed part.
(a) 20% — Method: a percentage concentration compares the sugar with the whole solution, so add the two masses to get the mass of solution and then scale the ratio of sugar to solution to a denominator of 100. Working: the solution has a mass of 200 + 50 = 250 g; sugar:solution = 50:250, and scaling to per 100 gives 50 ÷ 250 × 100 = 20, so the ratio is 20:100. Answer: 20%. The distractors: 25% comes from comparing the sugar with the 200 g of water, 50:200, rather than with the whole solution; 80% is the percentage of the solution that is water, 200:250, which answers for the wrong part of the mixture; 0.2% comes from working out 50 ÷ 250 = 0.2 and writing that decimal down as a percentage without multiplying by 100.
(c) 1.44 m³ — The cross-section is a triangle, so its area = base × height ÷ 2. Base × height = 1.2 × 0.8 = 0.96 m², and half of that is 0.96 ÷ 2 = 0.48 m². The volume of the prism = cross-sectional area × length = 0.48 × 3 = 1.44 m³. A pupil who forgets to halve when finding the triangle's area gets 1.2 × 0.8 × 3 = 2.88 m³. A pupil who ignores the height altogether, treating the cross-section as base × length, gets 1.2 × 3 = 3.6 m³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length of the bed stops at 0.48 m³. The correct volume of soil is 1.44 m³.
(b) 10 — First find 1/4 of 80, which is 20, then find 1/2 of that: 20 ÷ 2 = 10. Adding the two fractions together instead of applying them one after the other, 1/2 + 1/4 = 3/4, and finding 3/4 of 80 gives 60. Finding 1/4 of 80 = 20 correctly but stopping before applying the second fraction leaves 20 as the final answer. Finding 1/2 of 80 = 40 first but forgetting to then find 1/4 of that leaves 40 as the final answer.
(c) 45 km/h — Method: speed = distance ÷ time = 90 ÷ 2 = 45 km/h. Distractor origins: 180 km/h multiplies distance by time instead of dividing (90 × 2 = 180); 88 km/h subtracts the time from the distance (90 − 2 = 88); 92 km/h adds the time to the distance (90 + 2 = 92).
(a) 42 — Find the value of one part: 18 ÷ 3 = 6. Find the number of dogs: 4 × 6 = 24. Add the cats and the dogs to find the total: 18 + 24 = 42. (24 is the number of dogs only, not the total number of animals. 126 comes from multiplying the number of cats by the total number of parts, 18 × 7, instead of finding one part first. 25 comes from adding the ratio numbers 3 and 4 directly to the number of cats.)
(d) 1 × 10⁷ seconds — Method: the time for a journey is the distance divided by the speed, so round each number to 1 significant figure and then divide; dividing numbers in standard form means dividing the coefficients and subtracting the indices. Working: 3.1 × 10¹⁵ rounds to 3 × 10¹⁵ and 2.998 × 10⁸ rounds to 3 × 10⁸; 3 ÷ 3 = 1 for the coefficients, and 15 − 8 = 7 for the indices. Answer: about 1 × 10⁷ seconds. The distractors: 3 × 10⁷ seconds comes from subtracting the indices correctly but leaving the coefficient as 3 instead of dividing 3 by 3; 1 × 10⁻⁷ seconds comes from dividing the speed by the distance instead of the distance by the speed; 9 × 10²³ seconds comes from multiplying the two quantities instead of dividing them, since 3 × 3 = 9 and 15 + 8 = 23.
(d) 35 — Method: in direct proportion the ratio y : x is the same for every pair, so find the constant and substitute the new value of x. Working: k = 20 ÷ 8 = 2.5, so y = 2.5x; when x = 14, y = 2.5 × 14 = 35. Answer: 35. The distractors: 26 comes from additive thinking — x rises by 6, so 6 is added to y — which would keep the difference constant rather than the ratio; 28 comes from rounding the constant 2.5 down to 2 and working out 2 × 14, which loses the half in the constant; 5.6 comes from using the constant upside down, 8 ÷ 20 = 0.4, and working out 0.4 × 14.

How the 20 questions are shared out

Number — 5 questions (25% of the qualification)
Algebra — 4 questions (20% of the qualification)
Ratio, proportion and rates of change — 5 questions (25% of the qualification)
Geometry and measures — 3 questions (15% 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.