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) £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.
- (c) 19 — Method: set up the equation 15 + 9n = 186, then subtract the deposit and divide by the cost per student. Working: 9n = 186 − 15 = 171; n = 171 ÷ 9 = 19. Answer: 19 students. 20.67 comes from dividing the whole £186 by £9 without first subtracting the deposit: 186 ÷ 9 ≈ 20.67. 11.8 comes from swapping the two amounts round, subtracting £9 and dividing by £15: (186 − 9) ÷ 15 = 11.8. 22.33 comes from adding the deposit instead of subtracting it: (186 + 15) ÷ 9 ≈ 22.33.
- (a) 3/4 — Method: use tan = opposite ÷ adjacent in the right-angled triangle to find tan(angle B), then use the fact that corresponding angles in similar shapes are equal, so they have equal trigonometric ratios. Working: for angle B, the opposite side is AC = 3 cm and the adjacent side is BC = 4 cm, so tan(angle B) = 3/4. Angle Q corresponds to angle B, so angle Q = angle B and tan(angle Q) = 3/4. Answer: 3/4. 4/3 comes from writing the ratio upside down, adjacent ÷ opposite, giving the reciprocal instead of the tangent. 3/7 and 4/7 come from treating 3 and 4 as if they were parts of a total of 3 + 4 = 7, which is how a ratio is shared, not how a trigonometric ratio is formed.
- (d) 56.52 cm — Circumference = πd. Substitute d = 18: circumference = 3.14 × 18 = 56.52 cm.
- (c) 120 — 27 of the 45 sweets are orange (45 − 18 = 27), so the probability of taking an orange sweet is 27/45 = 3/5, and 200 × 3/5 = 120. Writing 80 is wrong because 200 × 18/45 = 80 uses the LEMON sweets' fraction instead of orange. Writing 182 is wrong because it takes the 18 lemon sweets away from the 200 repeats (200 − 18 = 182), applying the ‘the rest are orange’ subtraction to the number of goes instead of to the 45 sweets in the bag. Writing 27 is wrong because it is simply the number of orange sweets in the bag — it has not been scaled up to account for the 200 repeats. The expected number of times an orange sweet is taken is 120.
- (c) 140 — Method: the mean shares the total equally between the items, so add the values and then divide by how many there are. Working: the total is 120 + 200 + 160 + 80 = 560 pages and there are 4 books, so the mean is 560 ÷ 4 = 140 pages. Answer: 140. The distractors: 560 comes from stopping at the total number of pages and never dividing by 4; 280 comes from dividing the total by 2 instead of by the 4 books; 120 comes from working out the range, 200 − 80, which measures spread instead of centre.
- (c) 87 litres — Work out how much water is drained: 8 × 6 = 48 litres. Subtract this from the starting amount: 120 − 48 = 72 litres. Then add the 15 litres from the hose: 72 + 15 = 87 litres. Subtracting the 15 litres instead of adding it, as though the hose also removed water, gives 120 − 48 − 15 = 57 litres. Stopping after the drain step, without adding the hose water back in, leaves the working at 72 litres. Adding the rate and the time instead of multiplying them, 8 + 6 = 14 litres drained, and then working from there gives 120 − 14 + 15 = 121 litres. So 87 litres of water is left in the tank.
- (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) 3/2 — Find each average speed: car = 180 ÷ 3 = 60 mph; lorry = 160 ÷ 4 = 40 mph. Put the car's speed over the lorry's speed: 60/40. Divide both numbers by their highest common factor, 20: 60÷20 = 3, 40÷20 = 2, giving 3/2. (2/3 comes from writing the speeds the wrong way round. 9/8 comes from comparing the distances travelled, 180/160, without working out the speeds. 3/4 comes from comparing the times taken, 3/4, instead of the speeds.)
- (a) (3, 9) — Two column vectors are parallel only when one is the other multiplied by a single scalar — both numbers scaled by the same factor with the same sign. r has top:bottom ratio 2:6, which simplifies to 1:3. (3, 9) has ratio 3:9, also 1:3, so (3, 9) = 1.5 × r: it is parallel to r. (6, 2) swaps the top and bottom numbers of r, giving ratio 6:2 (3:1), which is not the same ratio. (2, 3) halves only the bottom number of r, giving ratio 2:3, not 1:3. (−2, 6) changes the sign of only the top number of r, giving ratio −2:6, which no longer matches 2:6 throughout.
- (c) 170 — 68% of the 250 packets germinated: 250 × 0.68 = 170. Using the complement, the 32% that did NOT germinate, gives 250 × 0.32 = 80. Shifting the decimal point, using 0.068 instead of 0.68, gives 250 × 0.068 = 17. Rounding 68% up to 70% before multiplying gives 250 × 0.70 = 175.
- (a) 3/13 — Vegetables are 9 of the 9 + 4 = 13 parts, so vegetables are 9/13 of the plot. Potatoes are a third of the vegetable section, so potatoes are 1/3 of 9/13, which is 9/39, simplifying to 3/13, of the whole plot. 9/13 comes from stopping after finding the fraction of the plot that is vegetables, without taking the further third for potatoes. 1/3 gives the fraction of the vegetable section that is potatoes, not the fraction of the whole plot. 4/39 comes from taking a third of the flowers' fraction, 4/13, instead of the vegetables' fraction.
- (d) x² − 49 — Area = length × width = (x + 7)(x − 7). Expanding: x × x = x², x × (−7) = −7x, 7 × x = 7x, 7 × (−7) = −49. The two middle terms −7x and 7x cancel, leaving x² − 49. A candidate who misremembers the difference-of-two-squares result as a sum gets x² + 49. A candidate who makes a sign error and treats both middle terms as −7x instead of cancelling gets x² − 14x − 49. A candidate who confuses area with perimeter and simply adds the length and width gets 2x.
- (c) 7800 kg/m³ — Method: build the conversion factor from the two unit changes separately — one for the mass, one for the volume. Working: 1 kg = 1000 g, so the mass figure is divided by 1000; 1 m = 100 cm, so 1 m³ = 100 × 100 × 100 = 1000000 cm³ and the volume figure is multiplied by 1000000. The density figure is therefore multiplied by 1000000 ÷ 1000 = 1000, giving 7.8 × 1000 = 7800. So the density of the metal is 7800 kg/m³. Distractor 780 kg/m³ comes from multiplying by 100 instead of 1000. Distractor 78000 kg/m³ comes from multiplying by 10000, an extra zero. Distractor 7.8 kg/m³ comes from not converting the units at all.
- (d) 1 — An isosceles trapezium has one line of symmetry, running through the midpoints of the two parallel sides. 0 would be true for a scalene trapezium, whose non-parallel sides are unequal, but this trapezium's non-parallel sides are equal, so it is symmetrical. 2 is the number of lines of symmetry of a rectangle, not a trapezium. 4 comes from wrongly applying the 'number of sides equals number of lines of symmetry' rule, which only holds for REGULAR polygons — a trapezium is not regular.
- (a) 5 — Method: 10% is 10/100, which cancels to 1/10, so finding 10% of an amount means dividing it by 10. Working: 10% = 10/100 = 1/10, and 50 ÷ 10 = 5. Answer: 5 questions. The distractors: 10 comes from reading the 10% as a count of 10 questions rather than as an operator acting on the 50; 45 comes from working out how many questions Freya answers correctly, the other 90% of the quiz, instead of how many she gets wrong; 500 comes from multiplying by 10 instead of dividing by 10.
- (c) 4x − 3 — Subtract each term inside the bracket: 5x − (x + 3) = 5x − x − 3 = 4x − 3. Writing 4x + 3 comes from subtracting the x but not the 3, keeping its sign positive. Writing 6x − 3 comes from adding the x term, 5x + x = 6x, instead of subtracting it. Writing 5x − 3 comes from ignoring the x term inside the bracket and only subtracting the 3.
- (b) 12 — Method: in an equation of direct proportion the value of x is substituted and multiplied by the constant. Working: y = 3x with x = 4 gives y = 3 × 4 = 12. Answer: 12. The distractors: 7 comes from adding the 3 and the 4 instead of multiplying them, reading 3x as 3 + x; 34 comes from writing the 3 and the 4 side by side, treating 3x as the digits of a two-digit number rather than as a product; 1 comes from working out 4 − 3, which turns the constant into an amount to be taken away.
- (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.
- (c) 1 : 27 — The edge lengths are in the ratio 2 : 6, which simplifies to 1 : 3. Volumes scale with the cube of the length ratio, so the volume ratio is 1³ : 3³ = 1 : 27. Giving 1 : 3 uses the length ratio without cubing it. Giving 1 : 9 squares the length ratio, which is the rule for areas, instead of cubing it, which is the rule for volumes. Giving 27 : 1 has the ratio the right way round for larger to smaller, not smaller to larger as the question asks.
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.