Sample paper · GCSE Foundation · grades 1–5
GCSE Foundation sample Paper 1 (non-calculator)
The real Paper 1 is 1.5 hour 30 minutes and 80 marks, non-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 no calculator-only items. AO1 / AO2 / AO3 at this tier: 50% / 25% / 25%.
Non-calculatorGCSE Foundation
Answer key: GCSE Foundation sample Paper 1 (non-calculator)
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- (d) 0.3 — Converting the fractions to decimals, 1/4 = 0.25 and 2/5 = 0.4, so any decimal between 0.25 and 0.4 is a valid answer, and 0.3 fits. Confusing 1/4 with 1/5 and converting it as 0.2 instead of 0.25 gives a value below the true lower bound. Confusing 2/5 with 1/2 and converting it as 0.5 instead of 0.4 gives a value above the true upper bound. Converting the fractions correctly but choosing a decimal above the true upper bound of 0.4 instead of between the two values gives 0.45.
- (a) 5 — Method: apply the machine's operation to the input, and treat zero as an input like any other. Working: the machine gives 0 + 5, and counting on 5 from zero leaves the 5 unchanged. Answer: 5. The distractors: 0 comes from assuming that an input of 0 must give an output of 0, which is true for a machine that multiplies but not for one that adds; −5 comes from subtracting 5 instead of adding it; 6 comes from treating the input as 1 rather than 0 and working out 1 + 5.
- (d) 3:8 — Convert 2 hours to minutes: 2 hours = 120 minutes. The ratio is 45 : 120. The highest common factor of 45 and 120 is 15. Divide both parts by 15: 45 ÷ 15 = 3 and 120 ÷ 15 = 8, giving 3 : 8. Leaving the hours unconverted gives 45 : 2 — the units on each side are different, so this does not compare like with like. Dividing by 5 instead of 15 gives 9 : 24, which still shares a common factor of 3, so it is not fully simplified. Swapping the order gives 8 : 3, hours to minutes instead of minutes to hours.
- (a) A smaller circle — Since the cutting plane is parallel to the circular base, the cross-section is also a circle, but smaller than the base because the cone narrows as it rises towards the apex, so 'a smaller circle' is correct. 'A triangle' wrongly describes the outline seen from the side of the cone, not a horizontal cross-section. 'An ellipse' would only result from a cut made at an angle to the base, not one parallel to it. 'The same size circle as the base' wrongly ignores that the cone tapers, so any parallel cross-section above the base must be smaller.
- (d) 18 — Method: a frequency tree can be filled from either end, because a pair of branches adds up to the number above it and the matching end branches add up to the total for that outcome. Working: the children number 180 − 108 = 72. Of the 153 finishers, 99 were adults, so the children who finished number 153 − 99 = 54. Along the children's branch, 72 − 54 = 18 children did not finish. Answer: 18 children did not finish. The distractors: 27 is 180 − 153 and counts everyone who did not finish, adults included; 9 is 108 − 99 and counts only the adults who did not finish; 54 is the number of children who did finish, which is the branch next to the one asked for.
- (c) 28 — Method: multiply the number of whole symbols by the value of one symbol, then add the value of any half symbol shown. Working: 3 whole symbols represent 3 × 8 = 24 cars. The half symbol represents 4 cars. Total cars sold in March = 24 + 4 = 28. Leaving out the half symbol, 3 × 8 = 24, undercounts by exactly the value of that half symbol. Treating the half symbol as if it were a full symbol, 4 × 8 = 32, overcounts because it doubles the value the half symbol is worth. Giving 3.5 reports the number of symbols shown, not the number of cars they represent — the key still needs to be applied. Always apply the key to every symbol shown, including a half symbol, rather than reading off the symbol count itself.
- (b) £13.48 — Method: an amount of money is written to the nearest penny, which is 2 decimal places, so the calculator display has to be rounded to 2 decimal places. Working: 53.90 ÷ 4 = 13.475, and the digit in the third decimal place is 5, so the penny digit goes up from 7 to 8. Answer: £13.48. The distractors: £13.47 comes from chopping the third decimal place off instead of rounding with it; £13.50 comes from rounding to the nearest 10p rather than to the nearest penny; £13.40 comes from cutting the display short at 1 decimal place, which is both the wrong degree of accuracy and a truncation rather than a rounding.
- (a) 216 — The cube numbers are formed by cubing 1, 2, 3, 4, 5, ..., so the next one is 6³ = 216. A candidate who squares 6 instead of cubing it gets 6² = 36. A candidate who assumes the sequence doubles each time gets 125 × 2 = 250. A candidate who uses 5 × 6² instead of 6 × 6 × 6 gets 180.
- (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) ∠QRS — Method: the middle letter in three-letter angle notation is always the vertex of the angle, and the outer two letters are the neighbouring vertices along the shape's sides. Working: at vertex R, the two adjacent vertices along the pentagon are Q and S, so the interior angle is written ∠QRS, with R in the middle. Options: ∠PQR names the angle at Q, not R, since Q is the middle letter there; ∠RST puts R first rather than in the middle, so it actually names the angle at S; ∠TRP does have R in the middle, but T and P are not the vertices adjacent to R along the pentagon's sides, so it does not describe R's interior angle. Answer: ∠QRS.
- (c) 1/36 — Method: the two dice do not affect each other, so the probability that both results happen is the product of the two separate probabilities. Working: a 1 on the red dice has probability 1/6 and a 2 on the blue dice has probability 1/6. Multiplying gives 1 × 1 = 1 on the top and 6 × 6 = 36 on the bottom. Answer: the probability is 1/36. The distractors: 1/18 comes from allowing either order, counting a 2 on the red dice with a 1 on the blue dice as well, although the stem names which dice shows which score; 1/3 comes from adding 1/6 and 1/6 instead of multiplying; 1/6 comes from giving the probability for one dice only and treating the other as making no difference.
- (c) 3:4 — The jacket costs 3/7 of £84, which is 3 × (84 ÷ 7) = 3 × 12 = £36. The bag then costs the rest of the money, £84 − £36 = £48. The ratio of the jacket to the bag is 36:48, which simplifies to 3:4. Writing the fraction spent on the jacket, 3/7, directly as the ratio, without working out that the bag's share is the remaining 4/7, gives 3:7. Giving the ratio the wrong way round, bag to jacket instead of jacket to bag, gives 4:3. Assuming the jacket and bag cost the same, ignoring the fraction given, gives 1:1.
- (c) y = 200 − 8x — Method: in a linear model the amount present at the start is the constant term and the steady rate of change is the gradient, which is negative because the amount is falling. Working: at x = 0 minutes there are 200 litres, so the constant term is 200; 8 litres are lost every minute, so after x minutes 8x litres have gone and the amount left is y = 200 − 8x. Answer: y = 200 − 8x. The distractors: y = 8x + 200 treats the draining as filling, so the butt would gain 8 litres a minute; y = 200x − 8 swaps the two numbers over, using the starting 200 litres as the rate per minute and the 8 litres per minute as the starting amount; y = −8x − 200 makes the starting amount negative as well as the rate, so the butt would begin 200 litres in deficit.
- (c) 21% — Method: an increase of 10% is a multiplier of 1.1, and two successive increases are found by multiplying the multipliers. Working: 1.1 × 1.1 = 1.21, so the rent is 121% of the original, which is an increase of 21%. Answer: 21%. The distractors: 20% comes from adding the two percentages, which ignores that the second 10% is taken of a larger amount; 121% is the multiplier written as the change rather than the change itself; 11% comes from slipping in the multiplication and getting 1.11 instead of 1.21.
- (b) Parallel, since AB and DC are both horizontal lines — A and B both have y-coordinate 1, so AB is horizontal; D and C both have y-coordinate 4, so DC is horizontal too. Two horizontal lines are always parallel, so AB and DC are parallel. "Not parallel, since AB and DC have different lengths" is wrong twice over: AB runs from x = 1 to x = 5 and DC from x = 2 to x = 6, so both are in fact 4 units long, and in any case length has no bearing on whether two lines are parallel — only direction does. "Not parallel, since AB is horizontal and DC is vertical" misreads the coordinates of D and C, which share the y-coordinate 4 and so give a horizontal line, not a vertical one. "Parallel, since A and D have the same x-coordinate" rests on a false claim: A has x-coordinate 1 and D has x-coordinate 2, so they do not share an x-coordinate — and even if they did, it would say nothing about whether AB is parallel to DC.
- (d) 1,000 — Method: round each number to 1 significant figure, then multiply the rounded values. Working: 49 rounds to 50 and 21 rounds to 20, and 50 × 20 = 1,000 because 5 × 2 = 10 and the two rounded numbers carry one zero each. Answer: 1,000. The distractors: 800 comes from rounding 49 down to 40 instead of to the nearest ten; 1,500 comes from rounding 21 up to 30 rather than down to 20; 1,029 is the exact product 49 × 21, worked out in full when the question asks for an estimate.
- (d) 6 — Method: write an expression for each person's savings after w weeks, and set them equal. Working: 40 + 6w = 10 + 11w. Subtract 6w from both sides: 40 = 10 + 5w. Subtract 10: 30 = 5w, so w = 6. Answer: 6 weeks. 0 comes from setting only the weekly amounts equal, 6w = 11w, and ignoring the different starting amounts entirely. 10 comes from adding the two starting amounts and dividing by the difference in weekly amounts, (40 + 10) ÷ (11 − 6), instead of forming and solving the correct equation. 1.76 comes from adding the two weekly amounts instead of subtracting them when rearranging, (40 − 10) ÷ (11 + 6).
- (d) 75% — Method: change a fraction to a percentage by dividing the numerator by the denominator and multiplying by 100. Working: 3 ÷ 4 = 0.75, and 0.75 × 100 = 75. Answer: 75%. The distractors: 34% comes from reading the digits 3 and 4 straight off as a percentage; 0.75% comes from dividing but forgetting to multiply by 100; 133% comes from inverting the fraction and working out 4 ÷ 3 instead.
- (d) 600 — Method: the number of seats is the number of rows multiplied by the number of seats in each row, so round each number to 1 significant figure and then multiply the rounded values, which is quick because a product of two multiples of ten is found by multiplying the non-zero digits and attaching the zeros. Working: 21 rounds to 20 and 29 rounds to 30; 2 × 3 = 6, and 20 and 30 carry one zero each, so two zeros follow the 6. Answer: about 600 seats. The distractors: 50 comes from adding the two rounded numbers instead of multiplying them, 20 + 30; 60 comes from multiplying 20 by the 3 of 30 and forgetting the zero in 30; 6,000 comes from attaching three zeros to 2 × 3 when 20 and 30 provide only two between them.
- (d) 7/4 — A part-to-part ratio a : b gives the fraction a/b when the first quantity is written as a fraction of the second, so 7 : 4 gives 7/4. Writing 4/7 puts the parts the wrong way round — blue as a fraction of red, not red as a fraction of blue. Writing 7/11 uses the total number of counters, 7 + 4 = 11, as the denominator instead of the number of blue counters — that is red as a fraction of the whole bag, not red as a fraction of blue. Writing 11/7 has both the wrong denominator and the parts inverted.
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.