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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Answer key: GCSE Foundation 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.
- (a) 5 — A factor of the whole expression must divide every term exactly. 15x + 20 = 5(3x + 4), so 5 is a factor. Distractor origins: 3 divides 15x exactly but does not divide 20 exactly; 4 divides 20 exactly but does not divide 15x exactly; 10 also divides 20 exactly but does not divide 15x exactly, so it is a factor of only one term, not of the whole expression.
- (b) 20 — The gradient is the change in p divided by the change in t: (90 − 30) ÷ (5 − 2) = 60 ÷ 3 = 20. Choosing 0.05 comes from dividing the change in t by the change in p, the wrong way round (3 ÷ 60). Choosing −20 comes from subtracting the coordinates in the wrong order for one part of the calculation, for example (30 − 90) ÷ (5 − 2), giving a negative value. Choosing 18 comes from dividing the second p-coordinate by the second t-coordinate directly (90 ÷ 5) instead of using the change between the two points.
- (b) 2 cm — Method: a diameter is made of two radii end to end, so going back from a diameter to a radius undoes that doubling, which gives radius = diameter ÷ 2. Working: the diameter is 4 cm, so the radius is 4 ÷ 2 = 2 cm. Answer: 2 cm. The distractors: 8 cm comes from multiplying by 2 instead of dividing by it, the relationship applied in the wrong direction; 1 cm comes from halving twice, once to reach the radius and then once more as though a second halving were called for; 0.5 cm comes from writing the division upside down as 2 ÷ 4 rather than 4 ÷ 2.
- (b) 0.15 — Method: success and failure are the only two outcomes of the task, so they form an exhaustive set and their probabilities add to 1; subtract the given probability from 1. Working: writing 1 as 1.00 so that both numbers have two decimal places gives 1.00 − 0.85; exchanging once, the hundredths give 10 − 5 = 5 and the tenths, now 9, give 9 − 8 = 1. Answer: 0.15, close to the left-hand end of the 0 to 1 scale because the machine nearly always succeeds. The distractors: 0.85 comes from giving back the probability of success instead of the probability of failure; 0.25 comes from taking each decimal column from 10 on its own in 1.00 − 0.85, writing 10 − 5 = 5 in the hundredths and 10 − 8 = 2 in the tenths instead of reducing the tenths to 9 after the exchange; 0.5 comes from assuming that success and failure must be equally likely because there are only two outcomes.
- (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) 25 — Reverse the operations in reverse order: undo the subtraction by adding 6, then undo the division by multiplying by 5. −1 + 6 = 5, so the number divided by 5 equals 5, and 5 × 5 = 25 — checking, 25 ÷ 5 − 6 = 5 − 6 = −1. A candidate who subtracted 6 again instead of adding worked out −1 − 6 = −7, then −7 × 5 = −35. A candidate who multiplied by 5 before undoing the subtraction, doing the inverse operations in the wrong order, worked out −1 × 5 = −5, then −5 + 6 = 1. A candidate who multiplied by 5 but forgot to undo the subtraction at all worked out −1 × 5 = −5 and stopped there.
- (a) 3 — Method: substitute the given value into the formula and carry out the subtraction in the order the formula is written. Working: replacing m with 8 gives L = 8 − 5, and 8 − 5 = 3. Answer: 3. The distractors: 13 comes from adding 5 to 8 instead of subtracting it; 40 comes from reading m − 5 as a multiplication and working out 8 × 5; −3 comes from subtracting the wrong way round and working out 5 − 8.
- (d) £7060 — A 6% decrease each year means the value becomes 100% − 6% = 94% of the previous year's value, and 94% = 0.94, so the multiplier is 0.94. Apply it once for each of the 3 years: £8500 × 0.94 = £7990 after 1 year, £7990 × 0.94 = £7510.60 after 2 years, £7510.60 × 0.94 = £7059.96 after 3 years, which rounds to £7060 to the nearest pound. (£6970 comes from using simple depreciation instead of compound, taking 6% of the original £8500 three times: £8500 − 3 × £510 = £6970. £7990 is the value after only 1 year, forgetting the remaining 2 years. £7511 is the value after only 2 years, £8500 × 0.94² = £7510.60, forgetting the third year.)
- (a) 28 cm — Method: the perimeter is the total distance round the outside, and a rhombus has four sides of equal length, so the perimeter is 4 × the side length. Working: 4 × 7 = 28. Answer: 28 cm. The distractors: 14 cm comes from 2 × 7, adding only one pair of sides and forgetting that a rhombus has two pairs; 49 cm comes from working out 7 × 7, which is the calculation for the area of a square rather than a distance round the outside; 11 cm comes from adding the side length to the number of sides, 7 + 4, instead of multiplying them.
- (d) 1/20 — Method: add the counts on the faulty end branches, divide by the total number of items in the experiment, then cancel. Working: the faulty items number 15 + 5 = 20, and 400 items were checked, so the probability is 20/400. Dividing the top and the bottom by 20 gives 1/20. Answer: the probability is 1/20. The distractors: 3/50 is 15/250 and comes from dividing machine A's faults by machine A's output, which is that machine's own fault rate rather than the probability for the whole batch; 1/30 is 5/150 and does the same on machine B's branch; 19/20 is 380/400 and gives the probability that the item picked is not faulty.
- (a) 187.5 g — Method: the smallest possible actual mass is half the rounding unit below the given value. Working: half of 25 g is 12.5 g, so the smallest possible mass is 200 − 12.5 = 187.5 g. Answer: 187.5 g. (175 g comes from subtracting the whole rounding unit, 25, instead of half of it. 200 g comes from giving the rounded value itself rather than the lower bound. 212.5 g comes from adding the half unit instead of subtracting it, giving the upper bound.)
- (d) 110 — Method: form the equation 15 + 0.08m = 23.80, where m is the number of minutes, then solve for m. Working: subtract the fixed fee: 0.08m = 23.80 − 15 = 8.80. Divide by the cost per minute: m = 8.80 ÷ 0.08 = 110. Answer: 110 minutes. 1.1 comes from using 8 instead of 0.08 as the cost per minute, forgetting to convert pence to pounds. 297.5 comes from dividing the whole bill by the cost per minute without subtracting the fixed fee first, 23.80 ÷ 0.08. 485 comes from adding the fixed fee to the bill instead of subtracting it, before dividing by the cost per minute, (23.80 + 15) ÷ 0.08.
- (c) 10 g — Method: mass = density × volume. Working: 2 × 5 = 10, in grams because the density is in grams per cubic centimetre. Answer: 10 g. The distractors: 2.5 g comes from dividing the volume by the density, 5 ÷ 2, instead of multiplying; 0.4 g comes from dividing the density by the volume, 2 ÷ 5; 7 g comes from adding the density and the volume.
- (b) It is the longest of the three chords — Method: in any circle the length of a chord is decided by how far the chord lies from the centre, because a chord passing nearer the centre cuts further across the circle. Working: the chord through O lies at a distance of zero from the centre, and no chord can lie closer than that, so no chord of the circle can be longer than it; a chord through the centre is a diameter. The other two chords lie at some distance greater than zero, so each of them falls short of that maximum. Answer: It is the longest of the three chords. The distractors: It is the shortest of the three chords comes from reversing the rule and picturing a chord near the centre as a short line tucked inside; It is the same length as the other two chords comes from carrying the fact that all radii of a circle are equal across to chords, which are not all equal; It is half the length of each of the other two chords comes from confusing a chord through the centre with a radius, which really is half a diameter.
- (d) £33.00 — The profit is 3/8 of £24 = (£24 ÷ 8) × 3 = £3 × 3 = £9.00. Selling price = £24 + £9.00 = £33.00. A candidate who gives the profit instead of the selling price gets £9.00. A candidate who subtracts the profit instead of adding it gets £24 − £9 = £15.00. A candidate who works out one eighth of £24 and adds that on, forgetting to multiply by the numerator 3, gets £24 + £3 = £27.00.
- (d) 7 — Method: set up the equation 2.50 + 1.80m = 15.10, then subtract the booking fee and divide by the cost per mile. Working: 1.80m = 15.10 − 2.50 = 12.60; m = 12.60 ÷ 1.80 = 7. Answer: 7 miles. 8.39 comes from dividing the whole £15.10 by £1.80 without first subtracting the booking fee: 15.10 ÷ 1.80 ≈ 8.39. 5.32 comes from swapping the two amounts round, subtracting £1.80 and dividing by £2.50: (15.10 − 1.80) ÷ 2.50 ≈ 5.32. 9.78 comes from adding the booking fee instead of subtracting it: (15.10 + 2.50) ÷ 1.80 ≈ 9.78.
- (b) Provider Y — £21.00 against Provider X's £23.00 — Provider X's gradient is (35 − 15) ÷ 100 = 0.2, so cost = 15 + 0.2 × 40 = 15 + 8 = £23.00. Provider Y's gradient is (45 − 5) ÷ 100 = 0.4, so cost = 5 + 0.4 × 40 = 5 + 16 = £21.00. £21.00 is less than £23.00, so Provider Y is cheaper: 'Provider Y — £21.00 against Provider X's £23.00'. Getting both costs right but naming Provider X as cheaper compares the two numbers the wrong way round — £23.00 is more than £21.00, not less. Comparing only the fixed fees, £15.00 and £5.00, ignores the cost of the 40 gigabytes actually used. Reading off the costs at 100 gigabytes, £35.00 and £45.00, directly from the graph answers a different usage from the 40 gigabytes the question asks about.
- (a) 12 — Compare the powers of each prime that appears in both factorisations. In 2² × 3² and 2² × 3 × 7, the prime 2 appears with power 2 in both, and the prime 3 appears with power 2 in one and only power 1 in the other — take the lower power, 3¹. Multiplying the shared primes at their lower powers, 2² × 3, gives 12. Using power 1 for both primes instead of comparing the powers properly, 2 × 3, gives 6, which misses that 2 is common at power 2, not power 1. Multiplying the primes at their higher powers and including 7, which only appears in 84, gives 2² × 3² × 7, which comes to 252 — this is the lowest common multiple, not the highest common factor. Only spotting that 3 is a common prime and overlooking that 2 is common as well gives 3. So the highest common factor of 36 and 84 is 12.
- (c) 30 km — First convert 1 hour 15 minutes to hours: 15 minutes is a quarter of an hour, so the time is 1.25 hours. Then multiply by the speed: 24 × 1.25 = 30 km. Reading 15 minutes as 0.15 hours (writing the minutes after the decimal point instead of as a fraction of 60) gives 24 × 1.15 = 27.6 km. Working out 24 × 75 = 1800 multiplies by the number of minutes instead of converting to hours first. Working out 24 ÷ 1.25 = 19.2 divides by the time instead of multiplying. The cyclist travels 30 km.
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.