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.Work out 35% of 180.
(a)45
(b)63
(c)145
(d)126
2.A company's profit was £2000 in its first year. Each following year, the profit increases by £800. Work out the first year in which the profit is more than £7000.
(a)9
(b)6
(c)7
(d)8
3.Which of these ratios is equivalent to 5 : 4?
(a)15 : 12
(b)4 : 5
(c)15 : 16
(d)9 : 8
4.A solid pyramid has a square base of side 10 cm and four identical triangular faces. Its apex is directly above the centre of the base, at a vertical height of 12 cm, and the slant height of each triangular face — the distance from the apex to the midpoint of a base edge — is 13 cm. Work out the total surface area of the pyramid.
(a)340 cm²
(b)360 cm²
(c)620 cm²
(d)260 cm²
5.On Amelia's walk to school there are two sets of traffic lights. The probability that the first set is green when she reaches it is 0.4. The probability that the second set is green when she reaches it is 0.5. The two sets work independently of each other. Work out the probability that both sets are green.
(a)0.2
(b)0.1
(c)0.45
(d)0.9
6.A bar chart shows the number of books borrowed from a school library on each of five days. The bar heights are: Monday 12, Tuesday 18, Wednesday 9, Thursday 15 and Friday 21. Work out the total number of books borrowed over the five days.
(a)66
(b)21
(c)15
(d)75
7.Write 90 as a product of its prime factors.
(a)2 × 3² × 5
(b)2 × 45
(c)2 × 3 × 15
(d)3 × 3 × 10
8.The diagram shows part of the graph of a reciprocal function of the form $y = \frac{k}{x}$ , passing through the labelled point. Work out the value of k.
(a)5
(b)6
(c)9
(d)1.5
9.£120 is shared between three cousins in the ratio 3:4:5. Work out the largest share.
(a)£10
(b)£50
(c)£30
(d)£40
10.A cyclist rides around a perfectly circular track with centre O. At every moment of the ride, the distance from O to the cyclist is exactly the same. Which term names this fixed distance?
(a)radius
(b)diameter
(c)circumference
(d)chord
11.A spinner is divided into sectors of 144°, 90° and 126°, coloured purple, orange and grey in that order. The spinner is spun 300 times. Work out how many times you would expect it to land on purple.
(a)100
(b)120
(c)75
(d)43
12.Write down the symbol that makes this statement true: −9 ___ −4
(a)≥
(b)>
(c)<
(d)=
13.A geometric sequence begins 162, 54, 18, 6, ... Work out the next term in the sequence.
(a)2
(b)3
(c)−6
(d)18
14.In a car park there are 35 cars and 15 vans. Write the ratio of the number of vans to the number of cars in its simplest form.
(a)3 : 10
(b)7 : 3
(c)3 : 7
(d)15 : 35
15.A vertical flagpole casts a shadow 6 m long at the same time as a 1.2 m vertical post casts a shadow 1.8 m long. The flagpole and the post are both vertical, and the triangles formed by each object, its shadow, and the sun's ray are similar. Work out the height of the flagpole.
(a)5.4 m
(b)4 m
(c)9 m
(d)7.2 m
16.Work out 3/8 of 96.
(a)72
(b)12
(c)32
(d)36
17.The mean of four numbers is 12.5. Three of the numbers are 8, 15 and 19. Form an equation using n for the fourth number, and solve it to find n.
(a)−4.5
(b)50
(c)29.5
(d)8
18.Ravi has 5 pears. He has 4 times as many apples as pears. Write the number of pears as a fraction of the number of apples.
(a)4/1
(b)5/9
(c)4/5
(d)1/4
19.Work out the lowest common multiple of 4 and 6.
(a)12
(b)2
(c)6
(d)24
20.y is always the same multiple of x. When x = 6, y = 15. Work out the value of y when x = 10.
(a)4
(b)15
(c)25
(d)19

Answer key

(b) 63 — 10% of 180 = 18, so 5% = 9. 35% = (3 × 18) + 9 = 54 + 9 = 63. A candidate who uses 25% instead of 35% gets 45. A candidate who doubles 35% to get 70% by mistake gets 126. A candidate who subtracts 35 from 180 instead of finding a percentage gets 145.
(d) 8 — Method: write the nth term of the sequence, 2000 + 800(n − 1), and find the smallest whole n for which it is greater than 7000. Working: 2000 + 800(n − 1) > 7000, so 800(n − 1) > 5000, giving n − 1 > 6.25. Since n − 1 must be a whole number, the smallest value is 7, so n = 8. Check: year 8's total is 2000 + 800 × 7 = 7600, which is more than £7000, while year 7's total is 2000 + 800 × 6 = 6800, which is not. Answer: year 8. 7 comes from rounding 6.25 to the nearest whole number, 6, and then adding 1, instead of rounding up to the next whole number before adding 1. 6 comes from using 6.25 rounded down to 6 as the year number directly, without adding the 1 needed to convert from the number of increases to the year number. 9 comes from adding one extra year beyond the year that already satisfies the condition.
(a) 15 : 12 — Multiply both parts of the ratio 5 : 4 by the same number, 3, to get an equivalent ratio: 5 × 3 = 15 and 4 × 3 = 12, giving 15 : 12. Giving 15 : 16 multiplies the two parts by different scale factors (×3 and ×4), which changes the ratio. Giving 9 : 8 adds 4 to each part instead of multiplying, which also changes the ratio. Giving 4 : 5 swaps the order of the two parts.
(b) 360 cm² — Method: the total surface area is the square base plus the four triangular faces, and the height used for a triangular face is the slant height of 13 cm, not the vertical height of 12 cm. Working: the base is 10 × 10 = 100 cm²; one triangular face is (10 × 13) ÷ 2 = 65 cm², so four faces give 4 × 65 = 260 cm²; the total is 100 + 260 = 360. Answer: 360 cm². The distractors: 260 cm² comes from adding the four triangular faces and leaving out the base; 340 cm² comes from using the vertical height of 12 cm as the height of each triangle, 100 + 4 × 60; 620 cm² comes from working out each face as 10 × 13 without halving, 100 + 4 × 130.
(a) 0.2 — Method: for independent events, the probability that both happen is the product of the two probabilities. Working: the first set is green with probability 0.4 and the second with probability 0.5, so the calculation is 0.4 × 0.5. Since 4 × 5 = 20 and the two factors carry one decimal place each, the product carries two. Answer: the probability is 0.2. The distractors: 0.9 comes from adding 0.4 and 0.5 instead of multiplying them; 0.45 comes from averaging the two probabilities; 0.1 comes from subtracting 0.4 from 0.5, treating the question as a difference.
(d) 75 — Method: to find a total from a bar chart, add together the height of every bar. Working: 12 + 18 + 9 + 15 + 21 = 75 books. Leaving out Wednesday's bar by mistake, 12 + 18 + 15 + 21 = 66, misses one day out of the total. Giving 21 states only Friday's bar, the tallest one, not the total of all five days. Dividing the total by the number of days, 75 ÷ 5 = 15, finds the mean number of books per day, not the total borrowed. Add up every single bar — do not stop at the biggest one, and do not divide once you have added them all.
(a) 2 × 3² × 5 — Method: divide repeatedly by the smallest prime number until only prime factors remain. Working: 90 ÷ 2 = 45, 45 ÷ 3 = 15, 15 ÷ 3 = 5, and 5 is prime, so 90 = 2 × 3 × 3 × 5, written as 2 × 3² × 5. 2 × 3 × 15 stops before the 15 is broken down into 3 × 5, so it is not fully factorised. 3 × 3 × 10 stops before the 10 is broken down into 2 × 5. 2 × 45 stops after only one division. Answer: 2 × 3² × 5.
(b) 6 — Method: for any point that lies on y = k/x, the value of k is found by multiplying the x-coordinate and the y-coordinate together, since k = x × y. Working: k = 2 × 3 = 6. Answer: k = 6. Distractor refutation: 1.5 comes from dividing the y-coordinate by the x-coordinate instead of multiplying them. 5 comes from adding the two coordinates instead of multiplying them. 9 comes from misreading the point's x-coordinate as 3 instead of 2, then multiplying 3 × 3.
(b) £50 — Method: add the parts of the ratio, divide the amount by the number of parts to find the value of one part, then multiply by the parts in the largest share. Working: 3 + 4 + 5 = 12 parts, £120 ÷ 12 = £10 for one part, and the largest share is 5 parts, so 5 × £10 = £50. Answer: £50. The distractors: £10 is the value of one part only; £30 is the 3-part share, which is the smallest one; £40 is the 4-part share, the middle one.
(a) radius — The distance from the centre O to any point on the circle is the radius. The diameter is the distance right across the circle, through the centre, from one side to the other — twice the radius, not the same measurement. The circumference is the total distance around the circle, not from the centre. A chord is a line joining two points on the circle that does not have to pass through the centre at all.
(b) 120 — Purple's probability is its angle out of the full circle: 144° ÷ 360° = 2/5. Expected number on purple = 2/5 × 300 = 120. Assuming the three colours are equally likely, ignoring their different angles, gives 300 ÷ 3 = 100. Using orange's angle, 90°, instead of purple's 144° gives a probability of 1/4, so 300 × 1/4 = 75. Treating 144 as a percentage instead of finding its fraction of 360° gives 300 × 0.144 = 43.2, which rounds to 43.
(c) < — Method: compare the two negative numbers by their position on a number line, not by the size of the digit. Working: −9 is nine places to the left of zero and −4 is four places to the left of zero, so −9 is further from zero in the negative direction, which makes it the smaller number. Answer: −9 < −4, so the symbol is <. > comes from comparing 9 and 4 as if both numbers were positive, ignoring the negative signs. = comes from treating both numbers as equal simply because they are both negative. ≥ combines > (false) with = (also false).
(a) 2 — The common ratio is 54 ÷ 162 = 1/3, confirmed by 18 ÷ 54 = 1/3 and 6 ÷ 18 = 1/3. The next term is 6 × 1/3 = 2. A candidate who wrongly treats the sequence as arithmetic and subtracts the last difference (18 − 6 = 12) again gets 6 − 12 = −6. A candidate who multiplies by 3 instead of dividing, effectively reversing the sequence, gets 18. A candidate who guesses the ratio is 1/2 gets 6 × 1/2 = 3.
(c) 3 : 7 — Method: a ratio 'of A to B' is written with A first, and it is simplified by dividing both parts by their highest common factor. Working: vans are named first, so the ratio starts as 15 : 35; the highest common factor of 15 and 35 is 5, and 15 ÷ 5 = 3 while 35 ÷ 5 = 7. Answer: 3 : 7. The distractors: 7 : 3 comes from writing the two numbers in the order they appear in the question, cars before vans, instead of the order asked for; 3 : 10 comes from comparing the vans with the total number of vehicles, 15 : 50, a part-to-whole ratio in place of the part-to-part ratio asked for; 15 : 35 is the right comparison left unsimplified, and the question asks for the simplest form.
(b) 4 m — The two triangles are similar, so height ÷ shadow is the same for both objects. For the post, height ÷ shadow = 1.2 ÷ 1.8. Flagpole height = 6 × (1.2 ÷ 1.8) = 4 m. A student who multiplies the flagpole's shadow by the post's height without applying the ratio gets 6 × 1.2 = 7.2 m. A student who assumes a constant DIFFERENCE between shadow and height instead of a constant ratio finds the post's shadow minus height, 1.8 − 1.2 = 0.6 m, then subtracts that from the flagpole's shadow to get 6 − 0.6 = 5.4 m.
(d) 36 — To find 3/8 of 96, divide by the denominator and multiply by the numerator: 96 ÷ 8 = 12, and 12 × 3 = 36. Dividing 96 by the numerator instead of the denominator, 96 ÷ 3 = 32, uses the wrong part of the fraction. Using 3/4 instead of 3/8, perhaps misreading the denominator, gives 96 × 3/4 = 72. Finding 96 ÷ 8 = 12 correctly but forgetting to multiply by the numerator 3 leaves just 12 as the final answer.
(d) 8 — Since the mean of the four numbers is 12.5, their total is 4 × 12.5 = 50. The three known numbers add up to 8 + 15 + 19 = 42, so n = 50 − 42 = 8. A candidate who multiplies the mean by 3 instead of 4 gets a total of 37.5, giving n = 37.5 − 42 = −4.5. A candidate who forgets to subtract the three known numbers and gives the total itself as n states n = 50. A candidate who subtracts the mean from the total of the three known numbers instead of the other way round gets n = 42 − 12.5 = 29.5.
(d) 1/4 — Method: work out both amounts first, then put the amount named before the word 'of' on top and the amount named after it on the bottom. Working: there are 4 × 5 = 20 apples, so the fraction is 5/20; dividing the numerator and the denominator by 5 gives 1/4. Answer: 1/4. The distractors: 4/1 comes from quoting the multiplier given in the question, which compares the apples with the pears rather than the pears with the apples; 4/5 comes from writing the 20 apples over the 25 fruits altogether, 20/25, a part-to-whole fraction; 5/9 comes from reading '4 times as many' as '4 more than', giving 5 + 4 = 9 apples and the fraction 5/9.
(a) 12 — List the multiples of each number: multiples of 4 are 4, 8, 12, 16, 20, 24; multiples of 6 are 6, 12, 18, 24. The lowest number that appears in both lists is 12. Picking 24, a common multiple but not the lowest one, gives an answer that is too big. Picking 6, the larger of the two original numbers rather than a common multiple, ignores that the lowest common multiple must appear in both lists. Working out the highest common factor instead of the lowest common multiple gives 2. So the lowest common multiple of 4 and 6 is 12.
(c) 25 — Find the constant multiplier from the given pair: 15 ÷ 6 = 2.5, so y is always 2.5 times x. When x = 10, y = 10 × 2.5 = 25. 19 comes from assuming an additive relationship instead of a multiplicative one — adding the difference 15 − 6 = 9 onto 10. 4 comes from using the multiplier the wrong way round (6 ÷ 15 = 0.4) and then multiplying by 10. 15 comes from simply repeating the given value of y, without applying the multiplier to the new value of x at all.

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.