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) 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.
- (b) 1215 — Method: generate the terms one at a time with the term-to-term rule and compare each with 1000 as you go, stopping at the first one that passes it. Working: the terms are 5, then 5 × 3 = 15, then 45, then 135, then 405, and 405 × 3 = 1215; 405 is still below 1000 while 1215 is above it. Answer: 1215. The distractors: 405 comes from stopping at the last term that is still below 1000 instead of giving the first one above it; 3645 comes from carrying on one term too far, past the first term that passes 1000; 2187 comes from using the multiplier 3 as the first term as well, generating 3, 9, 27, 81, 243, 729, 2187 instead of the sequence described.
- (b) 460 — To increase by 15%, multiply by 1.15 (100% + 15%). 400 × 1.15 = 460. 60 comes from working out only the increase, 400 × 0.15 = 60, and forgetting to add it to the original number. 415 comes from adding 15 directly to 400 instead of 15% of 400. 340 comes from multiplying by 0.85, decreasing instead of increasing: 400 × 0.85 = 340.
- (d) SSS - three sides equal — All three pairs of corresponding sides are equal in length (5 cm, 7 cm and 9 cm in both triangles), so the triangles are congruent by SSS. SAS needs an angle to be given as well as two sides, but no angle is given here. ASA needs two angles and the side between them, but no angles are given at all. RHS needs a right angle and the hypotenuse, but no angle is stated to be a right angle.
- (b) 0.4 — The two events are independent, so multiply the two probabilities: 0.8 × 0.5 = 0.4. Choosing 1.3 comes from adding the two probabilities instead of multiplying, 0.8 + 0.5 = 1.3. Choosing 0.3 comes from subtracting the two probabilities, 0.8 − 0.5 = 0.3, instead of multiplying them. Choosing 0.1 comes from using the probability that Ben is NOT on time, 1 − 0.8 = 0.2, and multiplying that by Freya's probability instead, 0.2 × 0.5 = 0.1.
- (d) 19 kg — Method: estimate the mean of grouped data by multiplying each class's midpoint by its frequency, adding the four totals, then dividing by the total frequency. Working: the midpoints are 5, 15, 25 and 35 kg. The weighted totals are 11 × 5 = 55, 5 × 15 = 75, 5 × 25 = 125 and 9 × 35 = 315, which add to 570. Dividing by the 30 dogs gives an estimate of 570 ÷ 30 = 19 kg. Giving 5 kg reads off the midpoint of the modal class, 0 < m ≤ 10, the class with the most dogs — but the class with the most dogs is not where the mean falls, and neither is a substitute for actually calculating it. Giving 20 kg averages the four midpoints, (5 + 15 + 25 + 35) ÷ 4, treating every class as equally likely and ignoring that far more dogs are in the lightest and heaviest classes than in the middle two. Giving 570 kg stops after finding the correct weighted total and forgets the final division by the 30 dogs. Always weight each midpoint by its own frequency, and always finish by dividing by the total frequency, not the number of classes.
- (c) 1/4 — A negative index means the reciprocal of the positive power, so 2⁻² = 1 ÷ 2² = 1/4. Treating the negative sign as making the answer negative instead gives −(2²) = −4. Ignoring the negative sign altogether gives just 2² = 4. Finding the reciprocal correctly but then also applying a negative sign gives −1/4.
- (c) 3x + 4y — Collect the x terms: 5x − 2x = 3x. Collect the y terms: 3y + y = 4y. So 5x + 3y − 2x + y = 3x + 4y. A candidate who subtracts the y terms instead of adding them (3y − y) gets 3x + 2y. A candidate who adds 2x instead of subtracting it (5x + 2x) gets 7x + 4y. A candidate who wrongly combines the x and y terms into a single term gets 6xy.
- (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) SSA — Method: recall the four basic congruence criteria for triangles and compare them with each option. Working: the four basic conditions are SSS, SAS, ASA and RHS. SSA (two sides and a non-included angle) is not one of them, because knowing two sides and an angle that is not between them does not always fix a unique triangle. Options: SSS, ASA and RHS are each genuine congruence conditions, so none of them is the correct answer to 'which is NOT'. Answer: SSA.
- (c) 0.5 — Method: count the favourable outcomes, write them over the total number of equally likely outcomes and then divide to turn the fraction into a decimal. Working: the odd scores are 1, 3 and 5, which is 3 of the 6 equally likely scores, so the probability is 3/6, and 3 ÷ 6 = 0.5. Answer: 0.5, the middle of the 0 to 1 probability scale. The distractors: 0.33 comes from listing only 3 and 5 as odd and working out 2 ÷ 6; 0.17 comes from giving the probability of one particular odd score, 1 ÷ 6; 0.3 comes from writing '3 out of 6' as 0.3, reading the 3 straight off as tenths instead of dividing.
- (c) 8 hours 35 minutes — From 21:35 to midnight is 2 hours 25 minutes, and from midnight to 06:10 is a further 6 hours 10 minutes, giving a total of 8 hours 35 minutes. Misreading the departure time as 22:35 instead of 21:35 loses an hour from the calculation and gives 7 hours 35 minutes. Misreading the departure time as 20:35 instead of 21:35 gains an hour and gives 9 hours 35 minutes. Subtracting the times as if both fell on the same day, without crossing midnight, gives 15 hours 25 minutes.
- (b) y = 7 — Method: one of the equations already gives the value of x, so that value is substituted into the other equation. Working: replacing x with 2 in x + y = 9 gives 2 + y = 9, so y = 9 − 2 = 7. Answer: y = 7, and the pair x = 2 with y = 7 satisfies both equations. The distractors: y = 11 comes from rearranging x + y = 9 as y = 9 + x and adding instead of subtracting; y = 9 comes from ignoring the x term and reading the total as the value of y; y = 4.5 comes from halving 9, as though x and y had to be equal.
- (a) 10/7 — Put Saturday's distance over Sunday's distance: 17.5/12.25. Multiply both numbers by 100 to clear the decimals: 1750/1225. Divide both by their highest common factor, 175: 1750÷175 = 10, 1225÷175 = 7, giving 10/7. (7/10 comes from writing the distances the wrong way round. 3/7 comes from finding the difference, 17.5 − 12.25 = 5.25 km, and writing it as a fraction of Sunday's distance, 5.25/12.25. 10/17 comes from comparing Saturday's distance to the total distance ridden, 17.5/29.75.)
- (b) 270° — Bearings are measured clockwise from north (000°). Due north is 000°, due east is 090°, due south is 180° and due west is 270°. Choosing 090° gives the bearing for due east, not west. Choosing 180° gives the bearing for due south. Choosing 000° gives the bearing for due north.
- (c) 12 — Method: division and multiplication have equal priority, so they are carried out in the order they are written, from left to right; a negative divided by a negative is positive. Working: (−36) ÷ (−6) = 6, and then 6 × 2 = 12. Answer: 12. The distractors: 3 comes from carrying out the multiplication first, (−6) × 2 = −12 followed by (−36) ÷ (−12) = 3; −12 comes from treating a negative divided by a negative as negative, giving −6 and then −6 × 2 = −12; 6 comes from stopping at the division and never carrying out the multiplication by 2.
- (a) 50 — Method: in ab², only the b is squared, so square b first, then multiply by a. Working: b² = 5² = 25, then a × b² = 2 × 25 = 50. Answer: 50. 100 comes from squaring the product ab instead of just b: (2 × 5)² = 100. 20 comes from squaring a instead of b: a² × b = 4 × 5 = 20. 10 comes from ignoring the square altogether and working out a × b = 2 × 5 = 10.
- (c) 70 cm — Method: convert the real length to centimetres, then divide by the scale factor. Working: 84 m = 8400 cm. 8400 ÷ 120 = 70 cm. Wrong options: 0.7 cm comes from dividing 84 by 120 without converting metres to centimetres; 1,008,000 cm comes from multiplying instead of dividing (8400 × 120); 7 cm comes from converting 84 m to 840 cm (using ×10 instead of ×100) before dividing.
- (c) 24 — Method: round each number to the nearest whole number, then multiply the rounded values. Working: 6.4 rounds to 6 (nearest whole number) and 3.9 rounds to 4 (nearest whole number). 6 × 4 = 24. Answer: 24. 18 comes from rounding 3.9 down to 3 instead of up to the nearest whole number, 4, giving 6 × 3. 28 comes from rounding 6.4 up to 7 instead of down to the nearest whole number, 6, giving 7 × 4. 25 is the exact value of 6.4 × 3.9, which is 24.96, rounded to the nearest whole number after multiplying, rather than estimated by rounding first.
- (a) 1/9 — The ratio copper : tin is 9:1, so write tin over copper: 1/9. (9/1 comes from writing the ratio the wrong way round, copper over tin. 1/10 comes from comparing the tin to the total mass of the alloy, 1 part out of 10. 9/10 comes from comparing the copper to the total mass of the alloy, 9 parts out of 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.