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)
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- 1.Work out √144 − 2 × 3 + √25
- 2.Write down the gradient of the line with equation y = 6 + 4x.
- 3.A rope of length A is 1.5 times as long as a rope of length B. Write the ratio A : B in its simplest form.
- 4.Write down the exact value of tan 0°.
- 5.The probability that a machine completes a task successfully is 0.85. Work out the probability that the machine does not complete the task successfully.
- 6.The mean mass of four parcels is 17 kg. Three of the parcels have masses 12 kg, 16 kg and 18 kg. Work out the mass of the fourth parcel.
- 7.3/5 of a 5/6 litre bottle of juice is poured out. Work out the exact volume poured out, in litres.
- 8.Which of these is an equation, rather than an expression, a formula or an identity?
- 9.A hosepipe fills a paddling pool at a constant rate. The volume of water in the pool increases by 15 litres every 3 minutes. Write down the gradient of the graph of volume against time.
- 10.Two straight roads cross at a junction. The angle between them, measured with a protractor, is 118°. What is the size of the angle vertically opposite to it?
- 11.A frequency tree records the results of 160 patients who took a new medicine. It splits them into those who reported side effects and those who did not. 15% of the patients reported side effects. Work out how many of the 160 patients did not report side effects.
- 12.A digital timer truncates every time to 1 decimal place. It shows a swimmer's time for one length as 12.3 seconds. Using t for the swimmer's actual time in seconds, write down the error interval for t.
- 13.A straight line passes through the points (−1, 2) and (3, 14). Work out the equation of the line.
- 14.Two mathematically similar cylinders have heights in the ratio 3 : 4. Write the ratio of their volumes in its simplest form.
- 15.A cylinder has a volume of 942 cm³ and a height of 12 cm. Using π = 3.14, work out the radius of the cylinder.
- 16.A spreadsheet shows that 812 − 397 = 315. Work out an estimate for 812 − 397, by rounding each number to the nearest 100, to check whether the spreadsheet's answer is reasonable.
- 17.Which expression means the same as a × a × a?
- 18.Ollie says that the ratio 6 : 15 is equivalent to the ratio 2 : 5. Is he correct? Give a reason for your answer.
- 19.Work out an estimate for 312 × 19, by rounding each number to 1 significant figure.
- 20.A laptop priced at £520 is first increased by 15%, and then the new price is decreased by 20%. Work out the final price of the laptop.
Answer key
- (b) 11 — Method: roots and the multiplication are worked out before the addition and subtraction, and what is left is then worked through from left to right. Working: √144 = 12, √25 = 5 and 2 × 3 = 6, so the calculation becomes 12 − 6 + 5, which gives 6 + 5 = 11. Answer: 11. The distractors: 1 comes from carrying out the addition before the subtraction, giving 12 − (6 + 5) = 12 − 11 = 1; 35 comes from working from left to right with no priority, giving 12 − 2 = 10, then 10 × 3 = 30 and 30 + 5 = 35; 7 comes from combining the two roots as √(144 + 25) = √169 = 13 and then subtracting the product, giving 13 − 6 = 7.
- (d) 4 — Method: compare the equation with the general form y = mx + c after writing the x term first, then read off the coefficient of x. Working: y = 6 + 4x can be written as y = 4x + 6, so comparing with y = mx + c gives m = 4. Answer: the gradient is 4. The value 6 comes from reading off the constant, written first in the equation, and treating it as the gradient because it comes before the x term. The value 10 comes from adding the two numbers in the equation, 6 + 4, instead of reading off the coefficient of x. The value 24 comes from multiplying the two numbers in the equation, 6 × 4, instead of reading off the coefficient of x.
- (c) 3 : 2 — Method: 'A is k times B' means that if B is taken as 1 part then A is k parts, and a ratio in its simplest form is written with whole numbers that have no common factor. Working: taking B as 1, A is 1.5, so A : B = 1.5 : 1; multiplying both parts by 2 clears the decimal and gives 3 : 2, and 3 and 2 share no factor. Answer: 3 : 2. The distractors: 2 : 3 comes from writing B before A, reversing the order the question asks for; 1.5 : 1 is the right relationship but is not in its simplest form, because a ratio in its simplest form uses whole numbers; 3 : 5 comes from comparing A with the combined length rather than with B, since A : (A + B) = 1.5 : 2.5 = 3 : 5.
- (a) 0 — tan 0° = 0, because there is no opposite side to consider when the angle itself is 0° — one of the exact values you need to know. 1 is the exact value of tan 45°, not tan 0°. √3 is the exact value of tan 60°. 1/√3 is the exact value of tan 30°.
- (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.
- (c) 22 kg — Method: multiply the mean by the number of parcels to rebuild the total mass, then subtract the masses that are known. Working: four parcels with a mean mass of 17 kg have a total mass of 17 × 4 = 68 kg; the three known parcels total 12 + 16 + 18 = 46 kg; so the fourth parcel has mass 68 − 46 = 22 kg. Answer: 22 kg. The distractors: 68 kg comes from stopping at the total mass of all four parcels; 17 kg comes from assuming the missing parcel must have the mean mass; 5 kg comes from multiplying the mean by 3, the number of parcels whose mass is given, leaving 51 − 46 = 5.
- (a) 1/2 — To find a fraction of an amount, multiply the fractions together: 3/5 × 5/6 = 15/30, which simplifies to 1/2 litre. Adding the fractions instead of multiplying them, using a common denominator of 30, gives 18/30 + 25/30 = 43/30, a value greater than the whole bottle. Dividing by 5/6 instead of multiplying by it, using its reciprocal 6/5, gives 3/5 × 6/5 = 18/25. Multiplying 5/6 by itself instead of by 3/5 gives 25/36.
- (c) 3x + 5 = 17 — An equation contains an equals sign and is true only for particular value(s) of the unknown — solving 3x + 5 = 17 gives the single value x = 4. 3x + 5 is an expression: it has no equals sign, so it cannot be solved, only simplified or evaluated. A = πr² is a formula: it shows the general relationship between different quantities (area and radius), rather than asking for one unknown value. 3(x + 5) ≡ 3x + 15 is an identity: the ≡ sign shows it is true for every value of x, not just one. The equation is 3x + 5 = 17.
- (d) 5 — The gradient equals the amount gained divided by the time taken: 15 ÷ 3 = 5 litres per minute.
- (a) 118° — When two straight lines cross, the angles that are vertically opposite each other are always equal. So the angle vertically opposite 118° is also 118°. A candidate who instead finds the angle next to it on the straight line, using 180° − 118° = 62°, has found the adjacent angle, not the vertically opposite one. A candidate who answers 180° has confused the rule with angles on a straight line. A candidate who doubles the angle, giving 236°, has applied no valid angle rule at all. The vertically opposite angle is 118°.
- (d) 136 — 15% of 160 = 0.15 × 160 = 24 patients reported side effects, so 160 − 24 = 136 did not. Stopping after finding the number who reported side effects, 24, answers the wrong question — it is not the number who did NOT report them. Misreading '15%' as a raw count of 15 patients, rather than a percentage, gives 160 − 15 = 145. Subtracting 15% of 160 twice, 160 − 24 − 24 = 112, double-counts the side-effect group.
- (c) 12.3 ≤ t < 12.4 — Method: truncating cuts the later digits off instead of rounding them, so nothing is ever pushed upwards. The displayed value is therefore the smallest the time can be, and the time can run up to, but not reach, the next value the display can show. Working: the display reads 12.3, so the actual time is at least 12.3 seconds; as soon as the time reaches 12.3 + 0.1 = 12.4 seconds the display would read 12.4, so 12.4 is not included. Answer: 12.3 ≤ t < 12.4. The distractors: 12.25 ≤ t < 12.35 is the interval for a time rounded to 1 decimal place, and this display does not round; 12.3 < t ≤ 12.4 excludes the one value the display certainly allows and includes the one it rules out; 12.3 ≤ t ≤ 12.4 treats 12.4 seconds as possible, but at 12.4 seconds the display would no longer read 12.3.
- (c) y = 3x + 5 — Method: find the gradient from the two points, then substitute one point into y = mx + c to find c. Working: gradient = (14 − 2) ÷ (3 − (−1)) = 12 ÷ 4 = 3. Using the point (3, 14): 14 = 3(3) + c, so 14 = 9 + c, giving c = 5. Answer: y = 3x + 5. y = 6x − 4 comes from mishandling the negative x-coordinate, treating 3 − (−1) as 3 − 1 = 2, so the gradient becomes 12 ÷ 2 = 6, and then c = 14 − 18 = −4. y = 3x + 23 comes from a sign error isolating c, adding 9 to 14 instead of subtracting it: c = 14 + 9 = 23. y = x/3 + 13 comes from dividing the change in x by the change in y instead of the other way round, giving a gradient of 4 ÷ 12 = 1/3, and then c = 14 − 1 = 13.
- (c) 27 : 64 — For similar solids, the ratio of volumes is the ratio of lengths cubed: 3³ : 4³ = 27 : 64. 3 : 4 comes from using the height ratio itself as the volume ratio, without cubing it at all. 9 : 16 comes from squaring each part instead of cubing (3² : 4²) — squaring is the rule for area, not volume. 27 : 4 comes from cubing only the first part of the ratio (3³ = 27), and leaving the second part uncubed.
- (c) 5 cm — Volume of a cylinder = πr²h, so r² = V ÷ (πh) = 942 ÷ (3.14 × 12) = 942 ÷ 37.68 = 25, and r = √25 = 5 cm. A pupil who finds r² = 25 but forgets to take the square root gives 25 cm. A pupil who forgets to divide by π, using r² = 942 ÷ 12 = 78.5, gets r = √78.5 ≈ 8.9 cm. A pupil who forgets to divide by the height, using r² = 942 ÷ 3.14 = 300, gets r = √300 ≈ 17.3 cm. The correct radius is 5 cm.
- (b) 400 — Method: round each number to the nearest 100, then subtract the rounded values. Working: 812 rounds to 800 (nearest 100) and 397 rounds to 400 (nearest 100). 800 − 400 = 400. Answer: 400. 500 comes from rounding 397 down to 300 instead of up to the nearest 100, 400. 300 comes from rounding 812 down to 700 instead of up to the nearest 100, 800. 415 is the exact value of 812 − 397, found without rounding first, so it is not an estimate — the spreadsheet's answer of 315 is too far from the estimate of 400 to be correct.
- (c) a³ — a × a × a means a multiplied by itself three times, which is written using powers as a³ — the small 3 shows how many times a is multiplied by itself. Writing 3a instead uses the 3 as a coefficient, as if the expression meant a + a + a (three lots of a added together) rather than three a's multiplied together. Writing a² only accounts for two of the three a's being multiplied together, missing one factor. Writing 3 + a treats the repeated multiplication as an addition of 3 and a, which has no connection to the original expression. The expression that means a × a × a is a³.
- (a) Yes — dividing both parts of 6 : 15 by 3 gives 2 : 5 — Method: divide both parts of the ratio by their highest common factor and compare. Working: the highest common factor of 6 and 15 is 3. 6 ÷ 3 = 2 and 15 ÷ 3 = 5, giving 2 : 5, so Ollie is correct. Wrong options: '6 : 15 simplifies to 3 : 5' divides incorrectly, giving the wrong simplified ratio; 'cannot simplify a ratio that does not start with an even number' states a false rule — any ratio can be simplified if its parts share a common factor; '15 ÷ 6 is not a whole number, so it cannot be simplified' wrongly tries to divide one part by the other instead of finding a common factor.
- (c) 6000 — Method: round each number to 1 significant figure, then multiply the rounded values. Working: 312 rounds to 300 (1 s.f.) and 19 rounds to 20 (1 s.f.). 300 × 20 = 6000. Answer: 6000. 5928 is the exact value of 312 × 19, found by multiplying without rounding first, which is not an estimate. 600 comes from rounding 19 down to 2 instead of to 20, losing a zero from its place value. 6200 comes from rounding 312 to the nearest 10, 310, instead of to 1 significant figure, 300, then multiplying by the correctly rounded 20.
- (c) £478.40 — Method: apply the percentage increase, then apply the percentage decrease to the new price. Working: after the increase, the laptop costs £520 × 1.15. Multiplying this result by 0.80 gives the final price, £478.40. Answer: £478.40. £494 comes from combining the two percentages into a single net change (15% − 20% = −5%) and applying it directly, £520 × 0.95 = £494, instead of applying the two changes one after the other. £416 comes from applying only the 20% decrease to the original price, £520 × 0.80 = £416, forgetting the increase entirely. £598 comes from applying only the 15% increase and stopping there, forgetting to apply the decrease 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.