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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- (c) 3.05 kg — There are 1000 g in 1 kg, so 450 g = 450 ÷ 1000 = 0.45 kg. The total mass is 0.45 + 2.6 = 3.05 kg. 7.1 kg divides the grams by 100 instead of 1000, 2.645 kg divides them by 10 000, and 452.6 kg adds the two numbers without converting the grams at all.
- (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) y is always 4 times x. — Check the multiplier for each pair: 12 ÷ 3 = 4, 20 ÷ 5 = 4, 32 ÷ 8 = 4 — the same multiplier every time, so y is always 4 times x. 'x is always 4 times y' comes from writing the multiplier the wrong way round. 'y is always x plus 9' only works for the first pair (3 + 9 = 12); it gives 14 for x = 5 and 17 for x = 8, not 20 and 32. 'y is always double x, plus 6' also only works for the first pair (3 × 2 + 6 = 12); it gives 16 for x = 5, not 20.
- (d) 5 — A square-based pyramid has four vertices at the corners of the square base, plus one more vertex at the apex where the four triangular faces meet: 4 + 1 = 5 vertices. Choosing 8 counts the edges instead of the vertices. Choosing 4 counts only the base corners and forgets the apex at the top. Choosing 6 is the vertex count of a triangular prism, not a square-based pyramid.
- (c) 3/8 — Method: write out every result of the three coins as a string of three letters, H for heads and T for tails, count the results that match the description and divide by how many results the list holds. Working: each coin lands two ways and no coin affects another, so the list holds 2 × 2 × 2 = 8 equally likely results. Exactly two heads means one coin lands on tails and the other two on heads, so the results are HHT, HTH and THH — 3 of the 8. Answer: the probability is 3/8. The distractors: 4/8 comes from reading 'exactly two heads' as 'at least two heads' and counting HHH as well; 2/8 comes from a list made without a system, in which HHT and THH are written down and HTH, the result with the tail between the two heads, is missed; 6/8 comes from counting 3 × 2 = 6 ways of picking which two of the three coins show heads, which counts every pair of coins twice, once in each order.
- (a) The means are equal, and Class B's marks are the more consistent because its range is smaller. — Method: comparing two distributions needs two things — a measure of average and a measure of spread — and each must be put into the context of the question. Working: both classes have a mean mark of 70, so on average the two classes scored the same; the range measures spread, and Class A's range of 30 marks is three times Class B's range of 10 marks, so Class B's marks sit closely around the mean while Class A's are far more spread out. Answer: the means are equal, and Class B's marks are the more consistent because its range is smaller. The distractors: the reply crediting Class A with more consistency reverses the meaning of the range, treating a larger range as tighter data when a larger range means more spread; the reply that Class A's mean mark is higher compares the wrong pair of figures, reading the range of 30 as an average; the reply that Class B's mean mark is higher reads the spread correctly but its claim about the means is false, since both means are 70.
- (c) 2 — First write 1 1/2 as an improper fraction, 3/2. To divide by 3/4, multiply by its reciprocal, 4/3: 3/2 × 4/3 = 12/6 = 2. Dropping the whole number and dividing only the fractional part, 1/2 ÷ 3/4 = 1/2 × 4/3, gives 2/3. Multiplying by 3/4 directly instead of using its reciprocal, 3/2 × 3/4, gives 9/8. Using the reciprocal of the first fraction instead of the second, 2/3 × 3/4, gives 1/2.
- (b) x = 0 — Method: the first coordinate of a point measures how far to the left or right of the origin it is, so a point that is neither left nor right of the origin has a first coordinate of zero. Working: the y-axis is the vertical line through the origin; every point on it lies directly above or directly below the origin with no sideways movement at all, so its first coordinate is zero while its second coordinate may take any value. Answer: x = 0. The distractors: y = 0 is the condition for lying on the x-axis, the other axis; x = 1 comes from confusing the y-axis with the vertical line one unit to the right of it; x = y is the condition for the diagonal through the origin, which meets the y-axis at the origin only.
- (a) £9.00 — Rate of pay = total pay ÷ number of hours. £54 ÷ 6 = £9.00 per hour. £324.00 comes from multiplying the pay by the hours instead of dividing (£54 × 6). £0.11 comes from dividing the hours by the pay instead of the pay by the hours (6 ÷ 54). £48.00 comes from subtracting the hours from the pay (£54 − 6) instead of dividing.
- (b) an arc of the same radius, centred at B — The perpendicular bisector construction needs two arcs of equal radius, one centred at each end of the segment — after the arc from A, the next step is an arc of exactly the same radius from B, so the two arcs cross at two points; the line through those two crossing points is the perpendicular bisector. "an arc of the same radius, centred at the midpoint of AB" is not possible yet, since the midpoint is only found once both arcs are drawn — it is the RESULT of the construction, not a step in it. "a smaller arc, centred at A again" gives two different-sized arcs from the same point, which never cross to give the bisector. "a straight line joining the ends of the first arc" only connects points on one arc, and locates nothing.
- (b) 26 — Method: n(G ∪ H) = n(G) + n(H) − n(G ∩ H), taking off the overlap once so the pupils who study both subjects are not counted twice. Working: 19 + 15 − 8 = 26. Answer: 26. Watch out: adding 19 and 15 without taking off the overlap gives 34, which counts the 8 pupils who study both subjects twice. Taking the 8 off both totals before adding, 19 − 8 + 15 − 8 = 18, counts only the pupils who study exactly one of the two subjects and leaves out the 8 who study both. And writing down 11, which is 19 − 8, gives the number who study geography only, not the number who study geography or history or both.
- (b) 5 — Method: list the factors of each number and pick the largest value that appears in both lists. Working: the factors of 15 are 1, 3, 5 and 15; the factors of 25 are 1, 5 and 25. The values in both lists are 1 and 5, and the larger of those is 5. Answer: 5. The distractors: 3 comes from choosing a factor of 15 without checking that it also divides 25; 15 comes from assuming that the smaller of the two numbers is always a factor of the larger one; 75 is the lowest common multiple of 15 and 25, given by taking the highest power of each prime instead of the lowest.
- (a) x = 10 or x = −10 — Rearranging, x² = 100. Taking the square root of both sides gives x = ±10, i.e. x = 10 or x = −10. A candidate who forgets the negative root gives only x = 10. A candidate who halves 100 instead of taking its square root gets x = 50. A candidate who applies the ± sign to 100 itself instead of to its square root gets x = 100 or x = −100.
- (d) 75 pages — Method: find the number of pages printed in one minute, then scale up to 10 minutes. Working: 45 ÷ 6 = 7.5 pages per minute, so 7.5 × 10 = 75 pages. Answer: 75 pages. 27 pages comes from using the ratio upside down, 45 × 6 ÷ 10, instead of finding the rate per minute first. 55 pages comes from simply adding the extra minutes, 10, onto the original number of pages, 45. 70 pages comes from rounding the rate down to 7 pages per minute before multiplying by 10, instead of using the exact rate of 7.5.
- (d) 4 — Method: use the definition of a rectangle — a quadrilateral in which every interior angle is a right angle. Working: a quadrilateral has four interior angles, and in a rectangle all four of them measure 90°, which is consistent with the 360° angle sum because 4 × 90° = 360°. Answer: 4. The distractors: 2 comes from applying the rule that opposite angles are equal and concluding that only one pair of corners is square; 1 comes from thinking that a single square corner is enough to make a shape a rectangle; 0 comes from confusing a rectangle with a general parallelogram, which has no right angles unless it is a rectangle.
- (a) 10π cm — The circumference of a circle is found from C = 2 × π × r. With a radius of 5 cm, this gives C = 2 × π × 5 = 10π cm. Using the radius directly in the formula without doubling it gives 5π cm, missing the factor of 2. Using the formula for area, π × r², instead of circumference gives 25π cm, which is also the wrong units for a length. Doubling the radius to get a diameter of 10 and then applying the circumference formula a second time gives 20π cm, doubling the answer that is already correct.
- (d) m³n — m × m × m × n means three m's multiplied together, then multiplied by n, which is written as m³n. Putting the power of 3 on n instead of m, mn³, puts the power on the wrong letter. Writing 3mn treats the repeated multiplication as if it were repeated addition, as if it meant 3 × m × n. Writing m³ + n adds the n instead of multiplying it in.
- (d) £7060 — A 6% decrease each year means the value becomes 100% − 6% = 94% of the previous year's value, and 94% = 0.94, so the multiplier is 0.94. Apply it once for each of the 3 years: £8500 × 0.94 = £7990 after 1 year, £7990 × 0.94 = £7510.60 after 2 years, £7510.60 × 0.94 = £7059.96 after 3 years, which rounds to £7060 to the nearest pound. (£6970 comes from using simple depreciation instead of compound, taking 6% of the original £8500 three times: £8500 − 3 × £510 = £6970. £7990 is the value after only 1 year, forgetting the remaining 2 years. £7511 is the value after only 2 years, £8500 × 0.94² = £7510.60, forgetting the third year.)
- (a) 1/2, 3/5, 2/3, 5/6 — Convert all four fractions to a common denominator of 30: 2/3 is 20/30, 3/5 is 18/30, 5/6 is 25/30, and 1/2 is 15/30. Ordering by these numerators, smallest to largest, gives 15/30, 18/30, 20/30, 25/30, which is 1/2, 3/5, 2/3, 5/6. Ordering by the size of the numerator in the original fractions, 1, 2, 3, 5, rather than converting to a common denominator, gives the wrong order 1/2, 2/3, 3/5, 5/6, because it ignores that the denominators are different. Ordering largest to smallest instead of smallest to largest, as the question asks, gives 5/6, 2/3, 3/5, 1/2. Using the rule "the bigger the denominator, the smaller the fraction" to place the last two, so that 5/6 is put below 2/3 because 6 is bigger than 3, gives 1/2, 3/5, 5/6, 2/3 — that rule only holds when the numerators are the same, and here 20/30 really is smaller than 25/30. So the correct order, smallest to largest, is 1/2, 3/5, 2/3, 5/6.
- (c) 120 km/h — Method: for a fixed distance the average speed multiplied by the time is constant, and that constant is the distance, so divide the distance by the new time. Working: speed × time = 240, so in 2 hours the speed needed is 240 ÷ 2 = 120 km/h. Answer: 120 km/h. The distractors: 80 km/h is the average speed of the original journey, 240 ÷ 3, which answers for the 3-hour timing rather than the 2-hour one; 160 km/h comes from halving the 3 hours to 1.5 hours and working out 240 ÷ 1.5, instead of using the 2 hours the question gives; 480 km/h comes from multiplying the distance by the 2 hours rather than dividing by it.
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.