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) 17 — Roots and powers are worked out first: √25 = 5 and 4² = 16. Division comes next: 12 ÷ 3 = 4. Then addition and subtraction, left to right: 5 + 16 − 4 = 17. A candidate who treated 4² as 4 × 2 = 8, multiplying the base by the exponent instead of squaring it, worked out 5 + 8 − 4 = 9. A candidate who did not evaluate the root and used 25 itself worked out 25 + 16 − 4 = 37. A candidate who ignored the priority of division and worked through 5 + 16 − 12 ÷ 3 strictly left to right got 5 + 16 = 21, then 21 − 12 = 9, then 9 ÷ 3 = 3.
- (c) 3c + 2d — Method: 'triple c' is 3c, 'double d' is 2d, and 'add' joins the two separate terms with a plus sign. Working: 3c + 2d. Answer: 3c + 2d. 2c + 3d comes from swapping which letter gets tripled and which gets doubled. 6cd comes from multiplying the two terms together instead of adding them, and also multiplying the coefficients (3 × 2 = 6). 5(c + d) comes from adding the coefficients (3 + 2 = 5) and applying that single number to both letters together, as if c and d always came as a pair.
- (b) 1:3 — Let the number who cycle be 1 part. Since 3 times as many pupils walk, the number who walk is 3 parts. The ratio of cycle to walk is 1:3. (3:1 comes from writing the ratio the wrong way round, walk to cycle. 1:4 comes from treating “3 times as many” as “3 more”, so walk = cycle + 3, giving parts 1 and 4. 3:4 comes from comparing the number who walk with the total number of pupils, 3 parts out of 4 parts altogether, instead of comparing cycle with walk.)
- (d) (−2, 0) — A 180° rotation about a centre (a, b) maps (x, y) to (2a − x, 2b − y). Here that gives (2 × 1 − 4, 2 × 1 − 2) = (−2, 0). A pupil who rotates about the origin instead of (1, 1) gets (−4, −2). A pupil who adds the centre's coordinates instead of applying the rotation formula gets (4 + 1, 2 + 1) = (5, 3). A pupil who just subtracts the centre's coordinates from E's, without doubling and reversing, gets (4 − 1, 2 − 1) = (3, 1). The correct image is (−2, 0).
- (b) 70 — Saloon, estate and hatchback are exhaustive, so their probabilities sum to 1: the probability of a hatchback is 1 − 0.28 − 0.37 = 0.35. The number of hatchbacks is 0.35 × 200 = 70. Treating the SUM of the other two probabilities, 0.28 + 0.37 = 0.65, as the probability of a hatchback instead of its complement gives 0.65 × 200 = 130. Multiplying the correct probability, 0.35, by 100 instead of the 200 cars actually surveyed gives 35. Averaging the two given probabilities, (0.28 + 0.37) ÷ 2 = 0.325, instead of subtracting them from 1, and then multiplying by 200 gives 65.
- (c) Because every pupil has an equal chance of being picked — Method: whether a sample represents its population is decided by the selection method, not by the size of the sample, so ask whether the method gives every member of the population the same chance of being chosen. Working: the names are drawn at random from a list of all 10,000 pupils, so each pupil has the same chance, 500 out of 10,000, of being drawn, and no group of pupils is more likely to appear than any other; that is what keeps bias out of the sample. Answer: because every pupil has an equal chance of being picked. The distractors: the reply about 5% treats the sampling fraction as the test of fairness, but a badly chosen 5% is still biased and a well chosen 1% is not; the reply about 500 being large enough makes size the test instead, which is the same mistake in another form, since a large sample drawn from one school would still misrepresent the city; the reply about the most willing pupils describes self-selection, which hands the choice of who is in the sample to the pupils who feel most strongly about the question.
- (a) 0.55, 58%, 3/5 — Converting all three to decimals: 3/5 = 0.6, 0.55 stays as 0.55, and 58% = 0.58. In order from smallest to largest, this is 0.55, then 58%, then 3/5. Writing the numbers in the reverse order, largest to smallest, gives 3/5, 58%, 0.55. Misconverting 3/5 as 0.5 instead of 0.6 makes it appear smaller than both other values, giving the order 3/5, 0.55, 58%. Misconverting 58% as 0.058 instead of 0.58, by moving the decimal point two extra places, makes it appear smallest of the three, giving the order 58%, 0.55, 3/5.
- (b) −3 — Method: the gradient of a straight line is the change in y divided by the change in x, with the two coordinates taken in the same order in the numerator as in the denominator. Working: going from (−1, 5) to (3, −7), the change in y is −7 − 5 = −12 and the change in x is 3 − (−1) = 4, so the gradient is −12 ÷ 4 = −3. Answer: −3. The distractors: 3 comes from subtracting the y-coordinates in one order and the x-coordinates in the other, giving 12 ÷ 4; −1/3 comes from dividing the change in x by the change in y instead of the other way round, giving 4 ÷ (−12); −6 comes from working out 3 − (−1) as 3 − 1 = 2, so that the change in y is divided by 2 rather than by 4.
- (b) 8:12 — Method: two ratios are equivalent when one is obtained from the other by multiplying, or dividing, both parts by the same number. Working: multiplying both parts of 2:3 by 4 gives 2 × 4 = 8 and 3 × 4 = 12, and the check runs the other way too, since the highest common factor of 8 and 12 is 4 and dividing both parts by 4 returns 2:3. Answer: 8:12. The distractors: 8:3 comes from multiplying only the first part by 4 and leaving the second part alone; 12:8 comes from multiplying both parts by 4 correctly but then writing the two parts the wrong way round; 4:5 comes from adding 2 to each part instead of multiplying, and adding the same amount to both parts changes the ratio.
- (d) positive — Moving from A to B, the x-coordinate increases from 1 to 4 and the y-coordinate also increases from 1 to 7, so the line rises as it goes from left to right, which means the gradient is positive. "negative" would need y to decrease as x increases, which is not the case here. "zero" would need the y-coordinate to stay the same, but it changes from 1 to 7. "cannot be determined" is wrong because the coordinates of both points are known, so the direction of the line can always be found.
- (d) 1/8 — Method: a run of flips of a fair coin gives equally likely sequences of heads and tails, so count the sequences that match and divide by how many sequences there are. Working: each flip lands two ways and no flip affects another, so three flips give 2 × 2 × 2 = 8 equally likely sequences: HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. Only HHH has a head at every flip, so 1 sequence of the 8 matches. Answer: the probability is 1/8. The distractors: 1/4 comes from treating 'three heads', 'two heads', 'one head' and 'no heads' as four equally likely results, which they are not, since one sequence gives three heads and three sequences give two; 1/6 comes from taking the number of sequences to be 2 + 2 + 2 = 6, adding the two ways each flip can land instead of multiplying them; 1/2 comes from reading the first flip only and giving the probability of a head on one flip, without combining it with the other two.
- (c) 0.94 — Method: a fraction is written as a decimal by making the denominator a power of ten, because the decimal places record tenths, hundredths and thousandths. Working: 50 × 2 = 100, so the numerator must also be multiplied by 2, giving 47 × 2 = 94 and the equivalent fraction 94/100; 94 hundredths is written with two digits after the decimal point. Answer: 0.94. The distractors: 0.47 comes from treating the denominator as though it were already 100 and writing the digits of the numerator straight after the point; 4.7 comes from dividing 47 by 10 instead of by 50; 0.094 comes from multiplying the denominator by 20 to reach 1000 but the numerator by only 2, giving 94/1000.
- (d) 28 cm — Method: rate = 12 cm ÷ 3 min = 4 cm per minute. Depth after 7 minutes = 4 × 7 = 28 cm. Distractor origins: 84 cm multiplies the given depth by 7 directly, without first finding the rate per minute (12 × 7 = 84); 24 cm simply doubles the given depth instead of scaling correctly by the ratio of times; 16 cm combines the numbers with subtraction and addition (12 − 3 + 7 = 16) instead of finding a rate.
- (b) 3 : 5 — The highest common factor of 27 and 45 is 9. Divide both parts by 9: 27 ÷ 9 = 3 and 45 ÷ 9 = 5, giving 3 : 5. Giving 9 : 15 divides by 3, a common factor but not the highest one, so it is not yet in simplest form. Giving 5 : 3 swaps the order of the two parts. Giving 27 : 45 has not been simplified at all.
- (c) 67° — Angles on a straight line add up to 180°. Add the two known angles: 48° + 65° = 113°. Subtract from 180°: 180° − 113° = 67°.
- (a) 2/3 — Total parts = 6 + 4 + 5 = 15. Sopranos and altos together are not tenors: 6 + 4 = 10 parts, so the fraction is 10/15, which simplifies to 2/3. 1/3 comes from finding the fraction of tenors instead of the fraction that is not tenors. 2/5 comes from counting only the sopranos as not tenors and leaving the altos out. 4/9 comes from leaving sopranos out of the total, 4 + 5 = 9, and then using only the altos as the fraction that is not tenors.
- (c) 16 km/h — Method: total distance = 12 + 8 = 20 km. Total time = 30 + 30 + 15 = 75 minutes = 1.25 hours. Average speed = total distance ÷ total time = 20 ÷ 1.25 = 16 km/h. Distractor origins: 20 km/h gives the total distance without ever dividing by the total time; 24 km/h uses only the speed of the first stage (12 km in 30 minutes), ignoring the second stage and the rest; 28 km/h averages the two separate stage speeds, 24 km/h and 32 km/h, instead of using total distance over total time.
- (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.)
- (c) 3 or −5 — Method: a point a fixed distance from another can lie on either side of it, so move the given distance in each direction from the starting point. Working: moving 4 units to the right gives −1 + 4 = 3, and moving 4 units to the left gives −1 − 4 = −5. Answer: 3 or −5. The distractors: 5 or −3 comes from starting at 1 instead of −1, giving 1 + 4 and 1 − 4; 3 only comes from moving to the right and forgetting that the point could lie to the left as well; 4 or −4 comes from measuring the distance from zero instead of from point A, which just repeats the given distance.
- (d) £270 — Method: the £90 is a difference between two shares, so turn it into a number of parts before finding the value of one part. Working: the largest share is 3 parts and the smallest is 1 part, so the difference is 3 − 1 = 2 parts and 2 parts are worth £90; one part = £90 ÷ 2 = £45; the whole amount is 1 + 2 + 3 = 6 parts, so 6 × £45 = £270. Answer: £270. The distractors: £540 comes from treating the £90 as the value of one part and multiplying it by the 6 parts; £180 comes from finding the £45 correctly but adding only the 1-part and 3-part shares and forgetting the middle share; £135 comes from multiplying £45 by 3 and giving the largest share instead of the total.
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.