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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- (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.
- (b) y = 2x − 1 — Method: in a rule that multiplies and then adds, the multiplier is the step in the outputs for each step of 1 in the input, and the number added on is the output when the input is 0. Working: the inputs 0, 1, 2 rise in ones while the outputs −1, 1, 3 rise by 2 each time, so the input is multiplied by 2; an input of 0 gives 2 × 0 = 0 and the output must be −1, so 1 is subtracted. Answer: y = 2x − 1, checked against the last pair by 2 × 2 − 1 = 3. The distractors: y = 2x + 1 comes from finding the multiplier 2 correctly and then reading the output at an input of 0 as +1 instead of −1; y = x − 1 comes from taking the multiplier as 1 because the inputs go up in ones, instead of using the step in the outputs; y = 3x − 1 comes from reading the largest output, 3, as the multiplier.
- (a) 20% — Method: percentage increase = increase ÷ original amount × 100. Working: the increase is 84 − 70 = 14 marks, and 14 ÷ 70 = 0.2, so 0.2 × 100 = 20. Answer: an increase of 20%. The distractors: 14% comes from quoting the 14 mark increase as though marks and per cent were the same thing; 17% comes from dividing the 14 by the new mean 84 instead of by the original 70, which gives 17% to the nearest per cent; 120% is the new mean written as a percentage of the old one, which is the whole of the new mean rather than the increase.
- (c) 80 cm² — Method: a composite shape made of two rectangles that do not overlap has an area equal to the sum of the two rectangle areas, so work out each area and add them. Working: the first rectangle has area 12 × 4 = 48 cm² and the second has area 4 × 8 = 32 cm², so the total is 48 + 32 = 80. Answer: 80 cm². The distractors: 48 cm² comes from working out the larger rectangle only and treating it as the whole shape; 32 cm² comes from working out the smaller rectangle only; 28 cm comes from adding the four given lengths, 12 + 4 + 4 + 8, which is a distance rather than an area and so carries a plain centimetre unit.
- (c) 90 — Method: over many future trials the expected number of successes is the number of trials multiplied by the probability of a success. Working: the sections are equal, so each is equally likely and P(red) is 3 out of 8. Over 240 spins the expected number of reds is 240 × 3 ÷ 8 = 90. Answer: about 90 of the spins would be expected to land on red. The distractors: 150 uses the 5 sections that are not red, 240 × 5 ÷ 8 = 150, which is the expected number of spins that do not land on red; 30 is 240 ÷ 8 and is the expected count for one single section; 80 comes from dividing by the number of red sections instead of by the number of sections, 240 ÷ 3 = 80.
- (c) 22 kg — Method: multiply the mean by the number of parcels to rebuild the total mass, then subtract the masses that are known. Working: four parcels with a mean mass of 17 kg have a total mass of 17 × 4 = 68 kg; the three known parcels total 12 + 16 + 18 = 46 kg; so the fourth parcel has mass 68 − 46 = 22 kg. Answer: 22 kg. The distractors: 68 kg comes from stopping at the total mass of all four parcels; 17 kg comes from assuming the missing parcel must have the mean mass; 5 kg comes from multiplying the mean by 3, the number of parcels whose mass is given, leaving 51 − 46 = 5.
- (c) 1,000 m² — Method: round each length to 1 significant figure, then use area of a rectangle = length × width on the rounded lengths. Working: 19.6 m rounds to 20 m and 48.3 m rounds to 50 m, so the estimate is 20 × 50 = 1,000 and the area is about 1,000 m². Answer: 1,000 m². The distractors: 800 m² comes from rounding 48.3 down to 40 when the digit after its first significant figure is 8 and sends it up to 50, giving 20 × 40 = 800; 140 m² is the perimeter of the rounded rectangle, 2 × 20 + 2 × 50 = 140, not its area; 70 m² comes from adding the rounded lengths, 20 + 50 = 70, instead of multiplying them.
- (a) y = 2 — Method: make x the subject of the simpler equation, substitute it into the other equation and then read off the letter the question asks for. Working: x − y = 1 gives x = y + 1, so 2x + 3y = 12 becomes 2(y + 1) + 3y = 12, that is 2y + 2 + 3y = 12, so 5y = 10 and y = 2. Answer: y = 2, and the matching value x = 3 checks in 2 × 3 + 3 × 2 = 12. The distractors: y = 3 comes from solving the pair correctly and then writing down the value of x; y = 2.2 comes from expanding 2(y + 1) as 2y + 1, which leaves 5y = 11; y = 10 comes from rearranging x − y = 1 as x = 1 − y, which turns the first equation into 2 + y = 12.
- (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) Isosceles trapezium: base angles are equal — WX is parallel to ZY and the two non-parallel sides WZ and XY are equal in length, so WXYZ is an isosceles trapezium. In an isosceles trapezium the two angles at each of the parallel sides are equal, so angle W = angle X (and angle Z = angle Y). A student who answers with the parallelogram property has quoted a fact that is true of a parallelogram, but WZ and XY are given as non-parallel so WXYZ is not a parallelogram — and in a parallelogram "opposite angles" pairs W with Y, not W with X. A student who answers with the kite property has again taken a true fact about the wrong shape: a kite's equal sides are two pairs of adjacent sides, not the two non-parallel sides of a trapezium. A student who answers with the rhombus property has used something no rhombus has — a rhombus has two pairs of parallel sides and only two pairs of equal angles; all four angles are equal only in a square.
- (d) 315 — The relative frequency from the survey is 42 ÷ 120 = 0.35, so the expected number who prefer paper bags among 900 shoppers is 0.35 × 900 = 315. Giving 42 as the answer reuses the original survey count without scaling it up to 900 shoppers at all. Finding the expected number who prefer PLASTIC bags instead of paper, using the relative frequency 78 ÷ 120 = 0.65, gives 0.65 × 900 = 585. Using 1000 shoppers instead of the 900 actually stated gives 0.35 × 1000 = 350.
- (d) 10π cm — Method: for a circle, circumference = π × diameter. Working: circumference = π × 10 = 10π cm. Answer: 10π cm. 20π cm comes from using the formula 2 × π × radius but plugging in the diameter as if it were the radius: 2 × π × 10 = 20π. 25π cm comes from confusing circumference with area, using π × radius² with radius 5: π × 5² = 25π. 5π cm comes from halving the diameter and then multiplying by π, as if circumference were π × radius: π × 5 = 5π.
- (c) t = 1 second and t = 5 seconds — The ball is at ground level when h = 0, which happens when either bracket is zero. t − 1 = 0 gives t = 1, and 5 − t = 0 gives t = 5. The option with t = −1 makes a sign error solving the first bracket. The option with t = −5 makes a sign error solving the second bracket, treating 5 − t = 0 as if it gave a negative solution. The option with t = 4 seconds is an arithmetic slip in solving 5 − t = 0.
- (b) 20 km — The scale 2 cm : 5 km means each 1 cm on the map represents 5 ÷ 2 = 2.5 km in real life. The footpath is 8 cm on the map, so its real length is 8 × 2.5 = 20 km. 40 km comes from multiplying 8 by 5 directly, ignoring that the scale's '2 cm' has to be divided out first: 8 × 5 = 40. 3.2 km comes from dividing 8 by 2.5 instead of multiplying: 8 ÷ 2.5 = 3.2. 5 km comes from multiplying 2.5 by the scale's '2' instead of by the footpath's 8 cm: 2.5 × 2 = 5.
- (c) minor segment — A chord splits a circle into two segments; the smaller of the two is called the minor segment and the larger one the major segment. "major segment" names the larger region, the opposite of what is asked for. "sector" is a different region altogether, enclosed by two radii and an arc, not by a chord. "arc" is a curved length along the circumference, not an enclosed region at all.
- (d) 36 — Method: the fraction is acting as an operator on the whole class, so one third of the class equals 12; the operation has to be reversed, and the inverse of dividing by 3 is multiplying by 3. Working: 1/3 × (number of pupils) = 12, so the number of pupils = 12 × 3 = 36. Answer: 36 pupils. The distractors: 4 comes from applying the operator instead of reversing it, working out 12 ÷ 3 = 4; 18 comes from reading the 12 girls as two thirds of the class, giving 12 ÷ 2 × 3 = 18; 24 comes from working out the number of boys, the other two thirds, as 2 × 12 = 24 and giving that instead of the size of the class.
- (d) 3 — The expression has three terms separated by + and − signs: 4x², −7x and 3. A candidate who thinks a number on its own does not count as a term says there are 2 terms, missing the 3. A candidate who miscounts the minus sign as creating an extra term says there are 4. A candidate who splits each term into its coefficient and letter part separately (4, x², 7, x, 3) says there are 5.
- (c) 2.5 — x : y = 2 : 5 means that for every matching pair of values, y ÷ x = 5 ÷ 2 = 2.5. So y = 2.5x, and comparing with y = kx gives k = 2.5. Dividing the other way round, 2 ÷ 5 = 0.4, gives x in terms of y — that is the constant for x = 0.4y, not for y = kx. Taking the y-part of the ratio on its own, 5, reads one number off the ratio instead of dividing the y-part by the x-part; 5 would only be right if the x-part were 1. Subtracting the two parts, 5 − 2 = 3, treats the ratio as a difference, but a ratio compares two quantities by multiplication, not by subtraction. The constant is k = 2.5.
- (c) 8,000,000,000 — Method: a power of ten is 1 followed by that many zeros, so 8 × 10⁹ is 8 multiplied by 1 followed by nine zeros. Working: 10⁹ = 1,000,000,000, and 8 × 1,000,000,000 puts an 8 in front of those nine zeros. Answer: 8,000,000,000. The distractors: 800,000,000 comes from writing nine digits altogether instead of nine zeros after the 8; 720 comes from reading 10⁹ as 10 × 9 = 90 and then working out 8 × 90; 0.000000008 comes from treating the index as negative and carrying the decimal point nine places to the left.
- (a) 22p — First convert the hourly rate to pence: £13.20 = 1320p. Then convert from per hour to per minute by dividing by 60, since there are 60 minutes in an hour: 1320 ÷ 60 = 22p per minute. Working out 1320 × 60 = 79200p multiplies by 60 instead of dividing, going the wrong way between per hour and per minute. Working out 13.20 × 10 = 132p converts pounds to pence using the wrong power of ten, and dividing that by 60 carries the error through to give 2.2p. Working out 13.20 × 100 = 1320p converts the currency correctly but stops there, leaving the rate as pence per hour rather than completing the second conversion to pence per minute. The rate of pay is 22p per minute.
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.