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) 1 × 10⁷ seconds — Method: the time for a journey is the distance divided by the speed, so round each number to 1 significant figure and then divide; dividing numbers in standard form means dividing the coefficients and subtracting the indices. Working: 3.1 × 10¹⁵ rounds to 3 × 10¹⁵ and 2.998 × 10⁸ rounds to 3 × 10⁸; 3 ÷ 3 = 1 for the coefficients, and 15 − 8 = 7 for the indices. Answer: about 1 × 10⁷ seconds. The distractors: 3 × 10⁷ seconds comes from subtracting the indices correctly but leaving the coefficient as 3 instead of dividing 3 by 3; 1 × 10⁻⁷ seconds comes from dividing the speed by the distance instead of the distance by the speed; 9 × 10²³ seconds comes from multiplying the two quantities instead of dividing them, since 3 × 3 = 9 and 15 + 8 = 23.
- (d) Isla is right, because 5 + 1 = 6 — Method: put the input through the machine yourself, then compare what comes out with the output claimed, and check that the reason given with the verdict is itself true. Working: the machine adds 1 to whatever is put in, so an input of 5 gives 5 + 1 = 6, which is exactly the output claimed. Answer: Isla is right, and the reason is that 5 + 1 = 6. The distractors: the statement that the machine adds 1 to its output describes a different machine — this one adds 1 to what goes in — so the verdict is right but the reason is false; 5 − 1 = 4 comes from running the machine backwards and subtracting instead of adding; 5 × 1 = 5 comes from reading ‘adds 1’ as ‘multiplies by 1’, which leaves the input unchanged.
- (d) 5/4 — Work out the number of free seats: 54 − 24 = 30. Form the fraction 30/24; both numbers share a factor of 6, so 30 ÷ 6 = 5 and 24 ÷ 6 = 4, giving 5/4. 4/5 comes from writing the fraction the wrong way round, as reserved over free (24/30). 5/9 comes from comparing the free seats with the total number of seats (30/54), instead of with the reserved seats. 5/6 comes from miscalculating 54 − 24 as 20 instead of 30, then forming 20/24.
- (d) SSS, using shared side QS — PQ equals RQ and PS equals RS are two given pairs of equal sides, and QS is common to both triangles, so QS equals itself and gives a third pair of equal sides. Three pairs of equal sides is exactly the SSS condition, so 'SSS, using shared side QS' is correct. 'SAS, using the angle at Q' is wrong because no angle is given anywhere in this question; angle PQS and angle RQS are not stated to be equal, and assuming they are would be assuming the very thing being proved. 'Only two pairs of sides — not enough' is wrong because it forgets that the shared side QS is itself a third pair of equal sides. 'Cannot prove — no angle given' is wrong because SSS is one of the four basic congruence conditions and specifically requires no angle at all.
- (b) 41/160 — In total, 27 + 14 = 41 of the 160 employees cycle to work, so the probability is 41/160 (41 and 160 share no common factor, so this is already in its simplest form). Writing 27/160 is wrong because it only counts the full-time cyclists and leaves out the 14 part-time cyclists. Writing 41/90 is wrong because it uses the full-time total (90) as the denominator instead of the whole survey (160). Writing 1/5 is wrong because it only uses the part-time branch, simplifying 14/70 to 1/5 and ignoring the full-time cyclists completely. The probability is 41/160.
- (c) The frequency of that category — Method: a bar chart for categorical data has one bar for each category, and the vertical scale on which the bars are measured is a count. Working: a bar drawn twice as tall as another tells you that twice as many items of data fell into its category, so the height measures how many items of data belong to that one category, which is exactly what a frequency is. Answer: the height of each bar is the frequency of that category — a count of items of data. The distractors: the number of different categories comes from reading the vertical scale as though it counted the bars, which is shown along the horizontal axis instead; the total of all the data comes from treating one bar as though it stood for the whole data set rather than for one category; the mean of all the data comes from confusing a bar chart with a measure of average, which no single bar can show.
- (d) 36 — To find 3/8 of 96, divide by the denominator and multiply by the numerator: 96 ÷ 8 = 12, and 12 × 3 = 36. Dividing 96 by the numerator instead of the denominator, 96 ÷ 3 = 32, uses the wrong part of the fraction. Using 3/4 instead of 3/8, perhaps misreading the denominator, gives 96 × 3/4 = 72. Finding 96 ÷ 8 = 12 correctly but forgetting to multiply by the numerator 3 leaves just 12 as the final answer.
- (a) x < 7 — Subtract 5 from both sides: x < 12 − 5, so x < 7. A candidate who adds 5 instead of subtracting gets x < 17. A candidate who subtracts the wrong way round gets x < −7. A candidate who correctly finds 7 but wrongly flips the inequality (as if dividing by a negative had happened) writes x > 7.
- (a) 20 cm — Method: match the measurement you are given to its own part of the ratio, use it to find the value of one part, then multiply by the parts belonging to the measurement asked for. Working: the length is the second measurement listed, so it matches 4 parts and one part = 16 ÷ 4 = 4 cm; the height is 5 parts, so 5 × 4 = 20. Answer: 20 cm. The distractors: 12 cm is the width, which is the 3-part measurement; 4 cm is the value of one part only; 80 cm comes from multiplying the 16 cm by 5 without first dividing by the 4 parts the length is worth.
- (b) 68° — Method: two properties are needed. Angle A and angle D are co-interior angles between the parallel sides AB and DC, so they add up to 180°; and because the trapezium is isosceles, the two angles on the side AB are equal, so angle B = angle A. Working: angle A = 180° − 112° = 68°, and angle B = angle A = 68°. Answer: 68°. The distractors: 112° comes from assuming that angles B and D are equal, which is the property of a parallelogram, not of a trapezium; 90° comes from assuming that the angles on the other parallel side must be right angles; 248° comes from using the 360° angle sum of a quadrilateral and taking away only the one angle that is given.
- (b) 37 — Method: on a frequency tree each pair of branches adds back up to the number it came from, so work out the failures on each branch and then add the two end branches. Working: on the men's branch 70 − 45 = 25 men failed. On the women's branch 50 − 38 = 12 women failed. Adding the two failing end branches gives 25 + 12 = 37. Answer: 37 of the 120 people failed the test. The distractors: 83 comes from adding the two passing branches, 45 + 38 = 83, and so reads the tree for the wrong outcome; 25 is the men's failing branch on its own, with the women never added; 12 is the women's failing branch on its own, with the men never added.
- (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.
- (d) 4n + 2 — Method: find the common difference, then find the constant by adjusting the first term. Working: 10 − 6 = 4, 14 − 10 = 4, so the common difference is 4 and the coefficient of n is 4. The constant is the first term minus the common difference: 6 − 4 = 2. Answer: the nth term is 4n + 2. 4n + 6 comes from using the first term, 6, as the constant without subtracting the common difference. 4n − 2 comes from working out the constant the wrong way round, as the common difference minus the first term (4 − 6 = −2) instead of the first term minus the common difference. 6n + 4 comes from swapping the common difference and the first term.
- (a) 135 minutes — Method: to change hours into minutes, multiply by 60. Working: 2.25 × 60 = 135 minutes. So the film runs for 135 minutes. Distractor 145 minutes comes from reading the '.25' as 25 minutes instead of a quarter of an hour, giving 2 hours 25 minutes. Distractor 225 minutes comes from multiplying by 100 instead of 60. Distractor 150 minutes comes from rounding 2.25 hours to 2.5 hours before converting.
- (b) circumference — The distance all the way around the outside edge of a circle is called the circumference. The diameter is a straight line across the circle through the centre, so it is a length through the circle, not around it. The radius is a straight line from the centre to the edge, again a length across, not around. The area is the amount of surface inside the circle, a region, not a length at all.
- (c) £7.05 — Tom is correct — Method: multiply the cost of one jar by the number of jars, then compare the total to £7. Working: 3 × £2.35 = £7.05, and since £7.05 is more than £7, Tom's claim is correct. Answer: £7.05 — Tom is correct. "£7.05 — Tom is incorrect" comes from reaching the right total but reading the 05 after the point as making the amount less than £7. "£6.95 — Tom is incorrect" comes from working out 3 × 35p as 3 × 30p + 5p = 95p, multiplying only the tens digit, and adding it to 3 × £2 = £6. "£7.20 — Tom is correct" comes from rounding £2.35 up to £2.40 before multiplying: 3 × £2.40 = £7.20.
- (b) 1215 — Method: generate the terms one at a time with the term-to-term rule and compare each with 1000 as you go, stopping at the first one that passes it. Working: the terms are 5, then 5 × 3 = 15, then 45, then 135, then 405, and 405 × 3 = 1215; 405 is still below 1000 while 1215 is above it. Answer: 1215. The distractors: 405 comes from stopping at the last term that is still below 1000 instead of giving the first one above it; 3645 comes from carrying on one term too far, past the first term that passes 1000; 2187 comes from using the multiplier 3 as the first term as well, generating 3, 9, 27, 81, 243, 729, 2187 instead of the sequence described.
- (b) 1.5 km — Multiply the map length by the scale factor: 6 × 25000 = 150000 cm. Convert to kilometres, using 100 cm = 1 m and 1000 m = 1 km, so 100000 cm = 1 km: 150000 ÷ 100000 = 1.5 km. (1500 km comes from converting only as far as metres, 150000 ÷ 100 = 1500 m, and then writing kilometres on the end. 15 km comes from dividing by 10000 instead of 100000. 0.15 km comes from dividing by 1000000 instead of 100000.)
- (b) No — their possible jump lengths do not overlap — Method: each recorded jump stands for the lengths within half of 0.1 m, that is 0.05 m, of the figure recorded, and Priya is right only if the two ranges overlap. Working: Priya's jump is at least 3.8 − 0.05 = 3.75 m and below 3.85 m, because a jump of 3.85 m would have been recorded as 3.9 m; Nadia's jump is at least 3.85 m and below 3.9 + 0.05 = 3.95 m. Every length Priya could have jumped is below 3.85 m and every length Nadia could have jumped is at least 3.85 m, so Nadia jumped further whatever the exact lengths were. Answer: No — their possible jump lengths do not overlap. The distractors: the reason that a recorded jump is exactly the length jumped reaches the same verdict by treating a rounded record as exact, which is the idea this question tests; both jumps being 3.85 m would put 3.85 m inside Priya's range, when a jump of that length is recorded as 3.9 m; Priya jumping up to 3.9 m goes a whole 0.1 m above her record instead of half of it.
- (b) 460 — To increase by 15%, multiply by 1.15 (100% + 15%). 400 × 1.15 = 460. 60 comes from working out only the increase, 400 × 0.15 = 60, and forgetting to add it to the original number. 415 comes from adding 15 directly to 400 instead of 15% of 400. 340 comes from multiplying by 0.85, decreasing instead of increasing: 400 × 0.85 = 340.
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.