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) 2.5 × 10⁷ — Method: place the decimal point so that the coefficient is at least 1 and less than 10, then count the places it has moved. Working: the digits give a coefficient of 2.5, and the decimal point travels from the end of 25,000,000 until it sits between the 2 and the 5, a move of 7 places. Answer: 2.5 × 10⁷. The distractors: 25 × 10⁶ is the same area but not in standard form, because the coefficient must be less than 10; 2.5 × 10⁸ comes from counting the eight digits of 25,000,000 instead of the seven places the decimal point moves; 2.5 × 10⁻⁷ comes from making the index negative because the decimal point was carried to the left.
- (d) The graph never crosses either axis — Since x ≠ 0, there is no point on the graph where x = 0, so it cannot cross the y-axis; likewise 1/x is never equal to 0 for any x, so it cannot cross the x-axis either — the graph never touches either axis. A candidate who forgets the restriction x ≠ 0 might think the graph behaves like other graphs and passes through the origin, (0, 0). A candidate who correctly rules out the x-axis but forgets that x = 0 is also excluded might say the graph crosses the y-axis but never the x-axis. A candidate who only pictures the branch where x and y are both positive might say the graph has only one branch, in quadrant 1, forgetting the second branch where x and y are both negative.
- (b) 75 cm² — Areas of similar shapes are in the ratio of the squares of their lengths. Squaring both parts of 3 : 5 gives an area ratio of 9 : 25, so the larger area is 25/9 of the smaller one. Working with the smaller area: 27 ÷ 9 = 3, and 3 × 25 = 75. The area of the larger rectangle is 75 cm².
- (a) 8 — A cylinder has two flat circular faces (the top and the base) and one curved surface, so only its 2 flat faces are wrapped. A cuboid has 6 faces and all of them are flat, so all 6 are wrapped. Adding these: 2 + 6 = 8, so 8 is correct. 9 comes from wrongly counting the cylinder's curved surface as a flat face. 6 comes from counting only the cuboid and forgetting the cylinder's two flat circular faces. 3 comes from counting only the cylinder and including its curved surface in that total.
- (d) 3/5 — 'Chosen from the boys' restricts the sample space to the 45 boys, of whom 27 walk. So the probability is 27/45 = 3/5. Using all 80 students as the denominator instead of just the boys gives 27/80, which is the probability that a student chosen from everyone is a boy who walks — not what was asked. Using the 18 boys who do NOT walk (45 − 27) as the numerator instead of the 27 who do gives 18/45 = 2/5. Dividing the 27 walking boys by the number of girls (80 − 45 = 35) instead of by the number of boys gives 27/35.
- (d) 20 — 0.4x + 1 = 9, so 0.4x = 9 − 1 = 8, and 8 ÷ 0.4 = 20, so 20 sessions are needed. Choosing 23 divides 9 by 0.4 without first subtracting the 1: 9 ÷ 0.4 = 22.5, rounded up to 23. Choosing 25 subtracts the wrong way, adding the 1 instead of taking it away: 9 + 1 = 10, and 10 ÷ 0.4 = 25. Choosing 2 misplaces the decimal point in the gradient, dividing by 4 instead of by 0.4: 8 ÷ 4 = 2.
- (d) 1/4 — Method: count the digits after the decimal point to choose a denominator that is a power of ten, then cancel the fraction by the highest common factor of the numerator and the denominator. Working: 0.25 has two digits after the point, so 0.25 is 25 hundredths and can be written as 25/100; the highest common factor of 25 and 100 is 25, and 25 ÷ 25 = 1 with 100 ÷ 25 = 4. Answer: 1/4. The distractors: 2/5 comes from reading the two digits after the point as a numerator and a denominator and writing 2 over 5; 1/25 comes from putting 1 over the digits after the point, as though 0.25 meant one twenty-fifth; 1/40 comes from miscounting the decimal places and using a denominator of 1000, giving 25/1000, which cancels by 25 to 1/40.
- (d) −5 — In y = mx + c, the y-intercept is c, the constant term, not the coefficient of x. Here c = −5, so the y-intercept is −5. 4 is Beth's value, the coefficient of x (the gradient), not the intercept. 5 comes from taking the constant term but dropping its negative sign. −4 comes from taking the gradient but giving it the sign of the intercept.
- (d) 75 pages — Method: find the number of pages printed in one minute, then scale up to 10 minutes. Working: 45 ÷ 6 = 7.5 pages per minute, so 7.5 × 10 = 75 pages. Answer: 75 pages. 27 pages comes from using the ratio upside down, 45 × 6 ÷ 10, instead of finding the rate per minute first. 55 pages comes from simply adding the extra minutes, 10, onto the original number of pages, 45. 70 pages comes from rounding the rate down to 7 pages per minute before multiplying by 10, instead of using the exact rate of 7.5.
- (c) tan θ = opposite/adjacent — The tangent ratio is defined as tan θ = opposite/adjacent, so this is the ratio that connects the opposite and adjacent sides. "sin θ = opposite/adjacent" wrongly labels this ratio as sine, when sine actually connects the opposite side and the hypotenuse. "cos θ = opposite/adjacent" wrongly labels it as cosine, when cosine connects the adjacent side and the hypotenuse. "tan θ = adjacent/opposite" uses the correct ratio name but has the opposite and adjacent sides the wrong way round.
- (b) 3 pupils — Method: an expected frequency is the probability multiplied by the number of trials, so multiply the probability by the number of pupils. Working: 30 × 1/10 means finding one tenth of 30, and 30 ÷ 10 = 3. Answer: 3 pupils would be expected to have a nut allergy. The distractors: 27 pupils comes from working out how many are expected NOT to have the allergy, 30 − 3, instead of how many are; 10 pupils comes from reading the 10 in the fraction 1/10 as the number of pupils; 1 pupil comes from reading the numerator of the fraction as the expected number.
- (b) 0.6 — Method: convert the fraction to an equivalent fraction with denominator 10, then read off the decimal. Working: 3/5 = 6/10 (multiplying numerator and denominator by 2) = 0.6. Answer: 0.6. 0.35 comes from combining the digits 3 and 5 directly after the decimal point instead of converting the fraction. 0.53 comes from writing the numerator and denominator digits in the wrong order. 1.67 comes from flipping the fraction to 5/3 before converting to a decimal.
- (b) (−4, 3) — Method: in a rectangle whose sides are parallel to the axes only two different x-coordinates and two different y-coordinates appear, and each of them is shared by a pair of vertices, so the missing vertex takes the x-coordinate and the y-coordinate that so far appear only once. Working: the x-coordinates given are −4, 2 and 2, so 2 is already used twice and −4 is used once; the y-coordinates given are −1, −1 and 3, so −1 is already used twice and 3 is used once; the fourth vertex therefore has x = −4 and y = 3. Answer: (−4, 3). The distractors: (3, −4) comes from picking the two unpaired coordinates correctly and then writing them in the wrong order; (−4, −5) comes from matching the 4-unit vertical side but measuring it downwards from (−4, −1) instead of upwards; (8, 3) comes from carrying on round the shape with the horizontal step used earlier, adding 6 to the x-coordinate of (2, 3) instead of closing the rectangle.
- (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.
- (a) Every square is a rectangle — Method: test each statement against the definitions. A rectangle is a quadrilateral with four right angles and opposite sides equal; a square is a quadrilateral with four right angles and all four sides equal; a rhombus is a quadrilateral with all four sides equal. Working: a square has four right angles and its opposite sides are equal, so every square meets the definition of a rectangle and the statement that every square is a rectangle is true. Answer: every square is a rectangle. The distractors: the claim that every rectangle is a square reverses the inclusion, and fails for any rectangle whose length and width differ; the claim that every rhombus is a rectangle treats four equal sides as enough, and drops the right-angle condition — a tilted rhombus has no right angles; the claim that every rectangle is a rhombus reads 'opposite sides equal' as though it meant 'all four sides equal'.
- (a) 6 × 10² metres — Method: distance = speed × time, so multiply the coefficients and add the indices. Working: 3 × 2 = 6 for the coefficients, and 8 + (−6) = 2 for the indices; 6 lies between 1 and 10, so the coefficient needs no adjustment. Answer: 6 × 10² metres, which is 600 metres. The distractors: 5 × 10² metres comes from adding the coefficients, 3 + 2, instead of multiplying them; 6 × 10¹⁴ metres comes from subtracting the indices, 8 − (−6), which is the rule for dividing rather than for multiplying; 6 × 10⁻⁴⁸ metres comes from multiplying the indices, 8 × (−6), instead of adding them.
- (c) 3x + 4y — Collect the x terms: 5x − 2x = 3x. Collect the y terms: 3y + y = 4y. So 5x + 3y − 2x + y = 3x + 4y. A candidate who subtracts the y terms instead of adding them (3y − y) gets 3x + 2y. A candidate who adds 2x instead of subtracting it (5x + 2x) gets 7x + 4y. A candidate who wrongly combines the x and y terms into a single term gets 6xy.
- (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) −5 — Method: the bracket is worked out first, then the multiplication and the division, which stand as separate parts, and the subtraction that joins them is carried out last; subtracting a negative is the same as adding. Working: (−3) + 5 = 2, so the product is 2 × (−4) = −8; the division gives (−6) ÷ 2 = −3; joining them gives −8 − (−3) = −8 + 3 = −5. Answer: −5. The distractors: −11 comes from taking away 3 instead of taking away −3, giving −8 − 3 = −11; −1 comes from working from left to right once the bracket is done, giving −8 − (−6) = −2 and then −2 ÷ 2 = −1; 11 comes from treating the first product as positive because it was worked out from a bracket, giving 8 − (−3) = 11.
- (d) 25 — Gradient = (60 − 20) ÷ (15 − 5) = 40 ÷ 10 = 4 litres per minute. Since the butt is empty at t = 0, V = 4t. Setting V = 100 gives t = 100 ÷ 4 = 25 minutes.
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.