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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- (d) 24 km — Method: a fraction acts as an operator, so finding 8/12 of a distance means dividing by the denominator and multiplying by the numerator. Working: 36 ÷ 12 = 3, so one twelfth of the walk is 3 km, and eight twelfths is 3 × 8 = 24 km. Answer: 24 km. The distractors: 12 km comes from working out the part of the walk still left, the other four twelfths, instead of the part already completed; 288 km comes from multiplying by the numerator without dividing by the denominator, giving 36 × 8 = 288; 4.5 km comes from dividing by the numerator instead of multiplying by it, giving 36 ÷ 8 = 4.5.
- (a) 2 m/s² — Method: acceleration = change in speed ÷ time = (17 − 5) ÷ 6 = 12 ÷ 6 = 2 m/s². Distractor origins: 12 m/s² stops after finding the change in speed and forgets to divide by the time; 22 m/s² adds the two speeds instead of subtracting them, and also forgets to divide by time (5 + 17 = 22); 72 m/s² multiplies the change in speed by the time instead of dividing (12 × 6 = 72).
- (a) £11.70 — Rate of pay = total pay ÷ hours worked, so £105.30 ÷ 9 = £11.70 per hour. Working out £105.30 − 9 = £96.30 subtracts the number of hours from the total pay instead of dividing. Working out £105.30 × 9 = £947.70 multiplies total pay by hours worked instead of dividing. Misplacing the decimal point in the correct answer gives £117.00 instead of £11.70. Maya's rate of pay is £11.70 per hour.
- (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.
- (a) 1200 — Method: take the estimate from the larger sample, because an unbiased relative frequency tends towards the true probability as the sample grows, then multiply by the number of bulbs made in a week. Working: Inspector B tested 500 bulbs, far more than Inspector A's 40, so use B's relative frequency: 30 ÷ 500 = 0.06. A week's production is 4000 × 5 = 20000 bulbs. The expected number of faulty bulbs is 20000 × 0.06 = 1200. Answer: about 1200 faulty bulbs a week. The distractors: 2000 uses Inspector A's estimate, 4 ÷ 40 = 0.1, giving 20000 × 0.1 = 2000, and so rests on a sample of only 40 bulbs; 1600 comes from averaging the two estimates of 0.1 and 0.06 to get 0.08, and 20000 × 0.08 = 1600, which gives the small sample equal weight with the large one; 240 uses the right estimate but stops at a single day, 4000 × 0.06 = 240.
- (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) 0.0065 — Leading zeros are never significant, so counting from the first non-zero digit, the first two significant figures of 0.006482 are 6 and 4. Look at the next digit along, 8, to decide whether the second figure rounds up: since 8 is 5 or more, the 4 rounds up to 5, giving 0.0065. Rounding to 2 decimal places instead of 2 significant figures gives 0.01, which answers a different question. Wrongly counting one of the leading zeros as a significant figure and stopping one figure short gives 0.006. Keeping an extra digit, as in 0.00648, gives 3 significant figures rather than 2.
- (b) 8 kg — The cost above the flat £5 charge is 17 − 5 = £12. At £2 per kg, this covers 12 ÷ 2 = 6 kg above the first 2 kg, so the total weight is 2 + 6 = 8 kg. Dividing the full £17 by £2 per kg without first taking off the £5 flat charge gives 17 ÷ 2 = 8.5 kg. Taking off the £5 flat charge and dividing by £2 per kg, but forgetting to add back the 2 kg that the flat charge covers, gives 12 ÷ 2 = 6 kg. Taking off £2 instead of £5 as the flat charge, (17 − 2) ÷ 2 = 7.5 kg, swaps which number is the fixed fee.
- (b) 6 litres — Method: find the volume of the cuboid in cm³, then change cm³ into litres using 1 litre = 1000 cm³. Working: 30 × 20 × 10 = 6000 cm³, and 6000 ÷ 1000 = 6. Answer: 6 litres. The distractors: 60 litres comes from using 1 litre = 100 cm³; 600 litres comes from using 1 litre = 10 cm³; 0.6 litres comes from using 1 litre = 10 000 cm³.
- (b) 10 — Corresponding angles are equal, so 5x + 4 = 3x + 24. Subtracting 3x from both sides gives 2x + 4 = 24, then subtracting 4 gives 2x = 20, so x = 10. 14 comes from adding the constants, 4 + 24, instead of subtracting them when rearranging. 19 comes from treating the angles as co-interior (summing to 180°): 5x + 4 + 3x + 24 = 180 gives 8x = 152, so x = 19. 20 correctly reaches 2x = 20 but stops without dividing by 2.
- (a) 195 — The relative frequency from the trial is 52 ÷ 80 = 0.65, and the expected number of point-up landings in 300 drops is 0.65 × 300 = 195. Giving 52 as the answer reuses the original count from the 80-drop trial without scaling it up to 300 drops at all. Misreading 52 out of 80 as 52% and finding 52% of 300 gives 156. Finding the expected number of point-DOWN landings instead of point-up, using the relative frequency 28 ÷ 80 = 0.35, gives 0.35 × 300 = 105.
- (a) 12 — Compare the powers of each prime that appears in both factorisations. In 2² × 3² and 2² × 3 × 7, the prime 2 appears with power 2 in both, and the prime 3 appears with power 2 in one and only power 1 in the other — take the lower power, 3¹. Multiplying the shared primes at their lower powers, 2² × 3, gives 12. Using power 1 for both primes instead of comparing the powers properly, 2 × 3, gives 6, which misses that 2 is common at power 2, not power 1. Multiplying the primes at their higher powers and including 7, which only appears in 84, gives 2² × 3² × 7, which comes to 252 — this is the lowest common multiple, not the highest common factor. Only spotting that 3 is a common prime and overlooking that 2 is common as well gives 3. So the highest common factor of 36 and 84 is 12.
- (b) 9 — Subtract 14 from both sides: 4s ≤ 36. Divide both sides by 4: s ≤ 9, so the greatest number of tickets is 9. A candidate who forgets the £14 coach cost solves 4s ≤ 50, getting s ≤ 12.5, rounded down to 12. A candidate who adds the £14 instead of subtracting it solves 4s ≤ 64, getting s = 16. A candidate who miscalculates 50 − 14 as 32 solves 4s ≤ 32, getting s = 8.
- (b) 16 — Method: substitute x into each equation separately, then subtract the smaller value from the larger. Working: y = 3 × 4 = 12 and z = 7 × 4 = 28, so the difference is 28 − 12 = 16. Answer: 16. The distractors: 4 comes from subtracting the constants, 7 − 3, which is the difference between the two gradients rather than the difference between the values at x = 4; 40 comes from adding the two values, 12 + 28, instead of subtracting them; 28 is the value of z on its own, given instead of being compared with the value of y.
- (d) (−4, −3) — The combined translation is the sum of the two column vectors, added component by component: top numbers 3 + (−7) = −4, bottom numbers −5 + 2 = −3, giving (−4, −3). (10, −7) subtracts the second vector from the first instead of adding them. (−4, 3) gets the top number right but makes a sign error on the bottom, treating −5 + 2 as +3. (4, −3) gets the bottom number right but makes a sign error on the top, treating 3 + (−7) as +4.
- (b) 8/5 — The reciprocal of a fraction a/b is b/a — swap the numerator and denominator. So the reciprocal of 5/8 is 8/5. A candidate who wrote 5/8 gave the original fraction, not its reciprocal. A candidate who worked out 1 − 5/8 = 3/8 confused 'reciprocal' with subtracting from 1. A candidate who wrote −8/5 inverted the fraction correctly but introduced an unnecessary sign change.
- (d) 3 — Method: add the constant term to both sides first, then divide by the coefficient of x. Working: 3x = 4 + 5 = 9; x = 9 ÷ 3 = 3. Answer: x = 3. −1/3 comes from a sign error when moving the 5, subtracting instead of adding: 3x = 4 − 5 = −1, then x = −1/3. 6 comes from subtracting the coefficient 3 instead of dividing by it: 9 − 3 = 6. 9 comes from correctly finding 3x = 9 but forgetting to divide by 3.
- (b) 8 — Method: the gradient of a straight line is the change in the vertical value divided by the change in the horizontal value between two points on the line. Working: from the origin (0, 0) to (5, 40) the vertical change is 40 − 0 = 40 and the horizontal change is 5 − 0 = 5, so the gradient is 40 ÷ 5 = 8, which here means a cost of £8 for each litre. Answer: 8. The distractors: 0.125 comes from dividing the horizontal change by the vertical change, 5 ÷ 40, which gives litres per pound instead of the gradient; 40 comes from reading off the vertical value of the point and calling it the gradient, ignoring the 5 litres it took to reach that cost; 35 comes from subtracting the two coordinates, 40 − 5, instead of dividing them.
- (c) 90 — Multiply the number of choices for each course: 5 × 6 × 3 = 90. 14 comes from adding the three numbers instead of multiplying them. 30 comes from multiplying only the starters and mains, 5 × 6, and forgetting the dessert. 18 comes from multiplying only the mains and desserts, 6 × 3, and forgetting the starter.
- (a) 12 — Speed × time is constant: k = 20 × 15 = 300. At 25 pages per minute, the time is 300 ÷ 25 = 12 minutes. Getting 18.75 comes from treating speed and time as directly proportional and working out 15 × 25 ÷ 20 instead of dividing k by the new speed. Getting 20 comes from adding the increase in speed (25 − 20 = 5) onto the time (15 + 5 = 20). Getting 10 comes from subtracting that same increase in speed from the time (15 − 5 = 10).
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.