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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- (b) £1,000, so £900 is not enough — Method: round each number to 1 significant figure, multiply to estimate the total cost, then compare the estimate with the money available. Working: 187 rounds to 200 and £4.85 rounds to £5, so the estimate is 200 × 5 = 1,000, and £1,000 is more than the £900 the school has. Answer: £1,000, so £900 is not enough. The distractors: £800 comes from cutting £4.85 down to £4 instead of rounding it up to £5, giving 200 × 4 = 800, and that estimate wrongly suggests the money stretches; £935 comes from rounding the price only and keeping 187 lunches, giving 187 × 5 = 935; £950 comes from rounding 187 to the nearest 10 rather than to 1 significant figure, giving 190 × 5 = 950.
- (b) y = 84 − 14x — Method: the distance still to go is the whole route minus the distance already covered, and at a steady speed the distance covered after x hours is the speed multiplied by x. Working: after x hours Freya has cycled 14x km, so the distance remaining is the 84 km route take away 14x, giving y = 84 − 14x, a straight line with intercept 84 and gradient −14. Answer: y = 84 − 14x. The distractors: y = 84 + 14x comes from adding the distance cycled to the length of the route rather than taking it away, so the ride would grow longer the further she goes; y = 14x − 84 comes from carrying out the subtraction the wrong way round, which is negative for the whole of the ride; y = 14 − 84x comes from swapping the two numbers over, treating 84 km as the hourly speed and 14 km as the length of the route.
- (d) 20 — Gradient = change in T ÷ change in t = (140 − 60) ÷ (6 − 2) = 80 ÷ 4 = 20. A student who subtracts in the wrong order gets −20. A student who divides 80 by 2 instead of 4 gets 40. A student who wrongly treats the line as passing through the origin and uses the point (2, 60) on its own gets 60 ÷ 2 = 30.
- (d) 7.7 m — The horizontal distance is adjacent to the 50° angle and the zip wire is the hypotenuse, so horizontal distance = 12 × cos 50° = 12 × 0.6428... = 7.71...≈ 7.7 m. "9.2 m" uses the sine ratio instead of cosine, 12 × sin 50° = 9.19...≈ 9.2 m, which actually finds the vertical drop of the zip wire, not the horizontal distance. "15.7 m" comes from dividing by the sine ratio instead of multiplying by the cosine ratio, 12 ÷ sin 50° = 15.66...≈ 15.7 m, both the wrong operation and the wrong ratio. "12.0 m" simply uses the length of the zip wire itself as the horizontal distance, ignoring the angle of 50° altogether.
- (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.
- (b) Treat them as outliers and check them before deciding — Method: a point lying a long way from the pattern the rest of the data make is called an outlier, and an outlier is investigated before anything is done with it, because it may be an error in the data or it may be a genuine but unusual case. Working: 48 of the 50 points lie close to the rising line of best fit, so the trend is set by those 48; the two remaining points do not follow it, so they are identified as outliers and checked — a mistake in measuring or recording would be corrected, while a genuine reading would be kept and reported. Answer: treat them as outliers and check them before deciding what to do with them. The distractors: deleting them at once assumes that every point far from the line must be an error, which throws away real data; moving the line so that it passes through them assumes a line of best fit must touch particular points, when it is drawn to follow all 50; taking them as proof that there is no correlation lets two points overturn the pattern that the other 48 agree on.
- (d) 1 × 10⁻³ — Method: for a number smaller than 1 the index is negative, and it counts the places the decimal point moves to the right to leave a coefficient between 1 and 10. Working: the only significant digit is 1, so the coefficient is 1; the decimal point in 0.001 moves three places to the right to reach that 1, so the index is −3. Answer: 1 × 10⁻³. The distractors: 1 × 10³ comes from taking the index as positive, which describes one thousand grams rather than one thousandth of a gram; 0.1 × 10⁻² is the same mass written with a coefficient of 0.1, which is smaller than 1 and so is not standard form; 1 × 10⁻⁴ comes from counting the zero in front of the decimal point as well as the places after it.
- (c) x = 0 or x = −7 — Factorising: x² + 7x = x(x + 7) = 0, so x = 0 or x + 7 = 0, giving x = 0 or x = −7. A candidate who divides both sides of the original equation by x, which loses the solution x = 0, gets only x = −7. A candidate who makes a sign error solving x + 7 = 0 gets x = 0 or x = 7. A candidate who misreads the coefficient and doubles it gets x = 0 or x = −14.
- (c) 7.6 — Density = mass ÷ volume, so 356.5 ÷ 47 = 7.585..., which rounds to 7.6 g/cm³ (1 d.p.). (0.1 comes from dividing the volume by the mass instead of the mass by the volume, the wrong way round. 7.5 comes from rounding 7.585 down instead of up to 1 decimal place. 403.5 comes from adding the mass and the volume instead of dividing.)
- (c) 32 cm — Method: a perimeter is lengths added together, so it scales by the length scale factor itself, which is 4 ÷ 3 going from the smaller triangle to the larger one — not by its square. Working: 24 ÷ 3 = 8, and 8 × 4 = 32. Answer: 32 cm. The distractors: 18 cm comes from multiplying by 3 ÷ 4, scaling from the larger triangle down to the smaller one; 25 cm comes from adding the difference between the parts of the ratio, 4 − 3 = 1, to the perimeter; 8 cm comes from dividing by 3 and stopping there, before multiplying by 4.
- (c) 5/8 — Method: the sections are all the same size, so every section is equally likely; count the sections that are not red and write that count over the total number of sections. Working: 8 − 3 = 5 sections are not red, and there are 8 sections altogether. Answer: 5/8, a value between 0 and 1 and a little above the halfway point of the scale. The distractors: 3/8 comes from giving the probability that the spinner does land on red; 5/11 comes from adding the 3 red sections to the 8 sections to make a total of 11 instead of using the 8 sections that exist; 1/2 comes from assuming that 'red' and 'not red' must be equally likely because there are only two possibilities.
- (a) 10 °C — Method: work out the coldest and warmest of the four temperatures, then subtract to find the difference. Working: the coldest temperature is Sam's, −9 °C, and the warmest is Alex's, 1 °C. The difference is 1 − (−9) = 1 + 9 = 10. Answer: 10 °C. 8 °C comes from working out 1 − 9 = −8 and reporting 8, dropping the negative sign on −9 instead of turning the subtraction into an addition. 3 °C comes from comparing the wrong pair, Sam's −9 °C and Ben's −6 °C, instead of the coldest and the warmest: −6 − (−9) = 3. 7 °C comes from comparing Ben's −6 °C with Alex's 1 °C, mistakenly treating Ben's reading as the coldest instead of Sam's.
- (c) 7 — To reverse the rule, subtract the constant then divide by the coefficient: 30−2=28, then 28÷4=7, so n=7. A candidate who adds the constant instead of subtracting it, a sign error when rearranging, would compute (30+2)÷4=32÷4=8. A candidate who subtracts the constant correctly but then forgets to divide by the coefficient would stop at 30−2=28. A candidate who treats 4n+2 as a single term 6n, adding the coefficient and constant together, would compute 30÷6=5.
- (d) £120 — Method: add the parts of the ratio, divide the prize by the number of parts to find the value of one part, then multiply by the parts in the share asked for. Working: 2 + 3 + 4 = 9 parts, £360 ÷ 9 = £40 for one part, and the share asked for is 3 parts, so 3 × £40 = £120. Answer: £120. The distractors: £40 is the value of one part rather than of a 3-part share; £80 is the 2-part share; £160 is the 4-part share, which is what a candidate reaches by picking the largest share instead of the one the question names.
- (a) 60° — Method: the angles of a triangle add up to 180°, so add the three expressions, solve for x and then substitute back into the expression for angle B. Working: 2x + 3x + 4x = 9x, so 9x = 180 and x = 20. Angle B is 3x, so angle B = 3 × 20 = 60. Answer: 60°. The distractors: 20° is the value of x, from stopping as soon as the equation is solved instead of substituting back; 120° comes from using 360° as the angle sum, which gives x = 40 and 3x = 120; 80° is 4x, the angle at C, from substituting into the wrong expression.
- (d) 1050 ml — Convert both amounts to millilitres: 1.8 litres = 1800 ml and 3/4 litre = 750 ml. Subtracting gives 1800 − 750 = 1050 ml. Confusing 3/4 with 75% and converting it as 75 ml instead of 750 ml gives 1800 − 75 = 1725 ml. Adding the amount poured out instead of subtracting it gives 1800 + 750 = 2550 ml. Misreading 1.8 litres as 0.8 litres, losing the whole litre, gives 800 − 750 = 50 ml.
- (d) −3 — Method: rewrite the equation in the form y = mx + c, then read off the gradient. Working: y = 4 − 3x can be written as y = −3x + 4, so comparing with y = mx + c gives m = −3. Answer: the gradient is −3. The value 3 comes from ignoring the negative sign on the x term. The value 4 comes from reading off the y-intercept instead of the gradient. The value −4 comes from a sign error, applying the negative sign to the intercept instead of the gradient.
- (d) 1200 cm — Method: multiply the drawing length by the scale factor. Working: 4 × 300 = 1200 cm. Wrong options: 0.013 cm comes from dividing instead of multiplying (4 ÷ 300); 304 cm comes from adding the scale factor to the drawing length instead of multiplying; 600 cm comes from using half the scale factor (150) instead of 300.
- (b) £300 — The increased price is 110% of the original, so the original price = £330 ÷ 1.1 = £300. A candidate who finds 10% of £330 and subtracts it, wrongly treating £330 as the original, gets £330 − £33 = £297. A candidate who adds 10% of £330 again instead of reversing the increase gets £330 + £33 = £363. A candidate who divides by 0.1 instead of 1.1 gets £3,300.
- (a) £54 — Method: for direct proportion, wage = rate × hours. Working: £9 × 6 = £54. Wrong options: £15 comes from adding the rate and the hours instead of multiplying (£9 + 6); £63 comes from using 7 hours instead of 6; £1.50 comes from dividing the rate by the hours instead of multiplying (£9 ÷ 6).
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.