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) 1/4 — 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.
- (b) x = 5 — Area = length × width, so x(x + 3) = 40, which rearranges to x² + 3x − 40 = 0. This factorises as (x + 8)(x − 5) = 0: the two numbers in the brackets must multiply to −40 and add to +3, and the pair 8 and −5 does both. This gives x = −8 or x = 5. Since x is a length, it cannot be negative, so x = 5. A candidate who gives both solutions without rejecting the negative one, which cannot be a length, answers x = 5 or x = −8. A candidate who picks the wrong factor pair of 40, such as 10 and −4 instead of 8 and −5, gets (x + 10)(x − 4) = 0 and answers x = 4. A candidate who rejects the wrong root, keeping the negative solution instead of the positive one, answers x = −8.
- (a) £20,000 — Method: add the parts of the ratio, divide the profit by the number of parts to find the value of one part, then multiply by the parts belonging to the partner asked about. Working: 3 + 5 + 4 = 12 parts, £48,000 ÷ 12 = £4,000 for one part, and that partner has 5 parts, so 5 × £4,000 = £20,000. Answer: £20,000. The distractors: £4,000 is the value of one part; £12,000 is the 3-part share; £16,000 is the 4-part share, which is what a candidate reaches by assuming the biggest share must go with the number written last in the ratio.
- (c) 1 — Method: the instruction states that a curved surface counts as a face, so count the surfaces of the sphere on that basis. Working: a sphere has exactly one continuous curved surface and no flat surfaces at all. A student who answers 0 has ignored the instruction and refused to count the curved surface. A student who answers 2 has confused the sphere with a cylinder, which has two flat circular faces. A student who answers 3 has imagined extra hidden surfaces that do not exist. Answer: 1 face.
- (a) 3,860 — The probability a bulb works correctly is the complement of being defective: 1 − 0.035 = 0.965. Expected number working correctly = 0.965 × 4,000 = 3,860. Using the probability of being defective instead of its complement gives 4,000 × 0.035 = 140, the expected number of DEFECTIVE bulbs, not working ones. Shifting the decimal point in the complement, using 0.0965 instead of 0.965, gives 4,000 × 0.0965 = 386. Assuming every bulb works, ignoring the 0.035 probability altogether, gives the full batch of 4,000.
- (a) 12 mm — Method: the range is the highest value minus the lowest value. Working: the highest rainfall is 15 mm and the lowest is 3 mm, so the range is 15 − 3 = 12 mm. Giving 15 mm alone states the highest value, not the range. Giving 3 mm alone states the lowest value, not the range. Sorting the six values, 3, 5, 9, 11, 12 and 15, and averaging the middle two, (9 + 11) ÷ 2 = 10 mm, finds the median, a completely different statistic. The range always needs BOTH the highest and the lowest value — never just one of them.
- (c) 32 — Method: a power tells you how many times to multiply the base by itself. Working: 2⁵ = 2 × 2 × 2 × 2 × 2 = 32. Answer: 32. (10 comes from multiplying the base by the power, 2 × 5, instead of using repeated multiplication. 25 comes from swapping the base and the power and working out 5² instead. 16 comes from using one fewer 2, effectively working out 2⁴.)
- (b) y = 1/x — A graph with two separate curved branches must have a value of x that is excluded from it, creating a break in the curve. For y = 1/x, x = 0 is excluded, so the graph splits into one branch where x and y are both positive, and one where x and y are both negative. y = x² is a single U-shaped curve with no break, even though it has two 'arms'. y = x³ is one continuous curve that passes through the origin without any break. y = 2x + 1 is a single straight line; even though it passes through regions where x and y share the same sign, it is one continuous line, not two separate branches.
- (c) 1.6 — The multiplier from tins to pounds is found by dividing the cost by the number of tins: £8 ÷ 5 = 1.6. 0.625 comes from dividing the wrong way round, tins ÷ cost (5 ÷ 8), which is the multiplier from pounds back to tins. 3 comes from subtracting the number of tins from the cost (8 − 5), instead of dividing. 40 comes from multiplying the cost by the number of tins (8 × 5), instead of dividing.
- (b) £5.00 — Perimeter of the kite = 25 + 25 + 40 + 40 = 130 cm = 1.3 m. The frame is sold only in whole metres, so 2 m must be bought. Cost = 2 × £2.50 = £5.00. A student who buys the exact 1.3 m instead of rounding up to whole metres gets 1.3 × £2.50 = £3.25. A student who never converts the perimeter from centimetres to metres and costs 130 × £2.50 gets £325.00.9 m, which rounds up to 3 whole metres, costing 3 × £2.50 = £7.50.
- (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.
- (d) 3,200,000 — 1 million = 1,000,000, so 3.2 million = 3.2 × 1,000,000 = 3,200,000. A candidate who moves the decimal point one place too many gets 32,000,000. A candidate who moves it one place too few gets 320,000. A candidate who writes the .2 as extra thousands instead of hundred-thousands gets 3,002,000.
- (b) 6x + 6 — Perimeter = 2[(2x + 5) + (x − 2)] = 2(3x + 3) = 6x + 6. A candidate who adds the length and width but forgets to double for the perimeter gets 3x + 3. A candidate who makes a sign error and adds 2 instead of subtracting it before doubling gets 2[(2x + 5) + (x + 2)] = 6x + 14. A candidate who multiplies the length and width instead of adding them, confusing the perimeter formula with the area formula, and then doubles that product, gets 2(2x + 5)(x − 2) = 4x² + 2x − 20.
- (c) 3 hours — Time = distance ÷ speed, so 210 ÷ 70 = 3 hours. Working out 210 − 70 = 140 subtracts the speed from the distance instead of dividing. Working out 210 + 70 = 280 adds the two values together, which does not give a time. Working out 210 × 70 = 14700 multiplies distance and speed together instead of dividing. The journey takes 3 hours.
- (c) The midpoint of AB — M is a single point lying exactly halfway along the straight line AB, so 'the midpoint of AB' correctly describes it. 'The vertex of AB' is wrong because a vertex is a corner point where two edges or lines meet, which does not apply to a point on a single straight line. 'A perpendicular of AB' is wrong because perpendicular describes two lines meeting at right angles, not a single point. 'A plane of AB' is wrong because a plane is a flat two-dimensional surface, not a point.
- (b) 63 — 10% of 180 = 18, so 5% = 9. 35% = (3 × 18) + 9 = 54 + 9 = 63. A candidate who uses 25% instead of 35% gets 45. A candidate who doubles 35% to get 70% by mistake gets 126. A candidate who subtracts 35 from 180 instead of finding a percentage gets 145.
- (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.
- (a) 200 seconds — Method: change the minutes into seconds using 1 minute = 60 seconds, then add on the loose seconds. Working: 3 × 60 = 180, and 180 + 20 = 200. Answer: 200 seconds. The distractors: 320 seconds comes from taking a minute as 100 seconds, giving 300 + 20; 180 seconds comes from converting the 3 minutes and forgetting the extra 20 seconds; 23 seconds comes from adding 3 and 20 without converting the minutes at all.
- (d) −1.7 — Since the numbers have different signs, find the difference between their sizes: 4.5 − 2.8 = 1.7, then keep the sign of the number further from zero. So −4.5 + 2.8 = −1.7. A candidate who drops the negative sign gets 1.7. A candidate who adds the magnitudes instead of finding the difference gets −(4.5 + 2.8) = −7.3. A candidate who takes the smaller digit from the larger in the tenths column, doing 8 − 5 = 3 instead of borrowing to make 15 − 8 = 7, gets 2.3 and so −2.3.
- (a) 675 ml — How much a jug holds is a volume, and volumes of similar solids scale with the cube of the length scale factor. The length scale factor is 12 ÷ 8 = 1.5, so the volume scale factor is 1.5 × 1.5 × 1.5 = 3.375. The larger jug holds 200 × 3.375 = 675 ml. Multiplying the scale factor by 3 instead of raising it to the power 3 is the mistake to guard against here.
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.