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%.
CalculatorGCSE Foundation
GCSE Foundation sample Paper 2 (calculator)
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- 1.Work out how many factors 36 has.
- 2.A sequence begins at 60, and each term after that is found by subtracting 7 from the term before it. Work out the 5th term of the sequence.
- 3.Write the ratio 3/4 : 1/2 as a ratio of whole numbers in its simplest form.
- 4.Two straight roads cross at a junction. The angle between them, measured with a protractor, is 118°. What is the size of the angle vertically opposite to it?
- 5.A table shows how two Year 10 classes did in a spelling test. In class A, 18 out of 24 pupils passed. In class B, 21 out of 30 pupils passed. Which class had the greater proportion of pupils passing?
- 6.There are 10,000 pupils in a city. A researcher picks a sample of 500 of them by drawing names at random from a list of every pupil in the city. Give the reason why this method gives a representative sample.
- 7.Work out an estimate for 8,900 ÷ 29, by rounding each number to 1 significant figure.
- 8.Factorise fully 5x + 5y − 5
- 9.A cyclist rides 30 km in 1 hour 30 minutes. Work out the average speed of the cyclist in km/h.
- 10.Point A has coordinates (2, 5). Point A is translated to point B using the column vector with top number 6 and bottom number −3. Write down the coordinates of point B.
- 11.Two ordinary fair dice are rolled and the two scores are added together. Work out the probability that the total is a prime number.
- 12.Work out the highest common factor of 12 and 18.
- 13.A number machine multiplies its input by 3 and then adds 7. The output is 1. Work out the input.
- 14.A straight-line graph shows the cost, C pounds, of hiring a bike for h hours. The line passes through the points (1, 12) and (4, 27). Which of these statements about the line is true?
- 15.Shape S has a vertex at (9, 6). It is enlarged by a scale factor of 1/3, centre the origin. Work out the coordinates of the image of this vertex.
- 16.The number 36 can be written as 2² × 3², and the number 84 can be written as 2² × 3 × 7. Work out the highest common factor of 36 and 84.
- 17.Make b the subject of the formula a = 5b.
- 18.A box holds 60 crayons. 25 of the crayons are broken and the rest are unbroken. Write the number of unbroken crayons as a fraction of the number of broken crayons. Give your answer in its simplest form.
- 19.Work out (4 × 10⁻³) × (2 × 10⁵). Give your answer in standard form.
- 20.A map has a scale of 1 : 25 000. Two towns are 5 km apart in real life. Work out the distance between the towns on the map, in centimetres.
Answer key
- (d) 9 — List all the factors of 36 in pairs that multiply to give 36: 1 × 36, 2 × 18, 3 × 12, 4 × 9, and 6 × 6. This gives the factors 1, 2, 3, 4, 6, 9, 12, 18 and 36 — nine factors in total, with 6 counted only once even though it appears in a pair with itself. Forgetting that 36 is itself a factor of 36 and leaving it off the list gives 8. Counting the number of factor pairs, five of them, rather than the number of individual factors gives 5. Treating the repeated pair 6 × 6 as two separate factors, 6 and 6 again, gives 10 instead of 9. So 36 has 9 factors.
- (c) 32 — The terms are 60, 53, 46, 39, 32 — each found by subtracting 7 from the term before, so the 5th term is 32. Subtracting 7 five times from the first term instead of four times, 60 − 7 × 5 = 25, treats the first term as if it came before the sequence starts. Adding 7 four times instead of subtracting, 60 + 7 × 4 = 88, uses the wrong operation. Stopping one term early gives the 4th term, 39.
- (b) 3:2 — Write both fractions over a common denominator of 4: 3/4 stays as 3/4, and 1/2 = 2/4. Comparing the numerators gives the ratio 3 : 2. Getting 2 : 3 swaps the two parts round. Getting 3 : 1 comes from using the numerator of the first fraction and the original numerator of the second fraction (1) without converting to a common denominator. Getting 2 : 1 comes from using only the denominators, 4 and 2, and simplifying those instead of the numerators.
- (a) 118° — When two straight lines cross, the angles that are vertically opposite each other are always equal. So the angle vertically opposite 118° is also 118°. A candidate who instead finds the angle next to it on the straight line, using 180° − 118° = 62°, has found the adjacent angle, not the vertically opposite one. A candidate who answers 180° has confused the rule with angles on a straight line. A candidate who doubles the angle, giving 236°, has applied no valid angle rule at all. The vertically opposite angle is 118°.
- (a) Class A — Class A: 18/24 = 0.75 = 75%. Class B: 21/30 = 0.7 = 70%. Since 75% > 70%, class A had the greater proportion passing, even though fewer pupils passed there in total. Choosing class B compares the raw numbers of pupils who passed (21 > 18) rather than the proportions. The two proportions are not equal — 0.75 and 0.7 are different values, so the two classes did not have the same pass rate. The class sizes being different does not prevent a comparison: converting each to a proportion makes the two classes directly comparable, so the answer can be determined.
- (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) 300 — Method: round each number to 1 significant figure, then divide the rounded values. Working: 8,900 rounds to 9,000 and 29 rounds to 30; cancelling a zero from each gives 900 ÷ 3. Answer: 300. The distractors: 450 comes from rounding 29 down to 20 instead of to the nearest ten, giving 9,000 ÷ 20; 3,000 comes from rounding 29 to 3 rather than to 30, a place-value slip that divides by a number ten times too small; 307 is the exact quotient rounded to the nearest whole number, worked out in full when the question asks for an estimate.
- (c) 5(x + y − 1) — Method: take out the highest common factor of all three terms and divide every term by it, the number term included. Working: the highest common factor of 5x, 5y and −5 is 5; dividing gives 5x ÷ 5 = x, 5y ÷ 5 = y and −5 ÷ 5 = −1, so the bracket holds x + y − 1. Answer: 5(x + y − 1), which multiplies back out to 5x + 5y − 5. The distractors: 5(x + y + 1) comes from dividing −5 by 5 and losing the minus sign; 5(x + y − 5) comes from dividing only the terms containing a letter by 5 and carrying the −5 into the bracket unchanged; 5(xy − 1) comes from collecting the unlike terms 5x and 5y as 5xy before factorising.
- (d) 20 km/h — Method: average speed = total distance ÷ total time, with the time written in hours. Working: 1 hour 30 minutes = 1.5 hours, and 30 ÷ 1.5 = 20. Answer: 20 km/h. The distractors: 45 km/h comes from multiplying 30 by 1.5 instead of dividing; 15 km/h comes from dividing by 2, as if the ride had taken 2 hours; 30 km/h comes from dividing by the whole hour only and ignoring the extra 30 minutes.
- (a) (8, 2) — Method: translating a point by a column vector means adding the vector's top number to the x-coordinate and its bottom number to the y-coordinate. Working: x-coordinate 2 + 6 = 8; y-coordinate 5 + (−3) = 2. Answer: B = (8, 2). A candidate who ignores the negative sign and adds 3 instead of −3 gets (8, 8). A candidate who translates in the reverse direction, subtracting the vector from A instead of adding it, gets (−4, 8). A candidate who swaps the vector's top and bottom numbers before adding gets (−1, 11).
- (d) 5/12 — The possible totals are 2 to 12. The prime totals are 2, 3, 5, 7 and 11, with 1, 2, 4, 6 and 2 ways to make them, giving 1 + 2 + 4 + 6 + 2 = 15 outcomes out of 36, so the probability is 15/36 = 5/12. Choosing 7/18 comes from forgetting that a total of 2 counts as prime, giving 14/36 = 7/18 instead. Choosing 19/36 comes from wrongly including a total of 9 as a prime number, adding its 4 ways to the correct 15 to get 19/36. Choosing 13/36 comes from forgetting the 2 ways of making a total of 11, giving 13/36 instead of 15/36.
- (b) 6 — List the factors of each number: the factors of 12 are 1, 2, 3, 4, 6 and 12; the factors of 18 are 1, 2, 3, 6, 9 and 18. The common factors are 1, 2, 3 and 6, and the highest of these is 6. Picking 2, a common factor but not the largest, gives an answer that is too small. Picking 3, also a common factor but still not the largest, gives another answer that is too small. Working out the lowest common multiple instead of the highest common factor gives 36. So the highest common factor of 12 and 18 is 6.
- (d) −2 — Method: run the machine backwards, undoing the operations in the opposite order and swapping each one for its inverse. Working: the machine added 7 last, so take 7 off the output: 1 − 7 = −6; before that the machine had multiplied by 3, so divide: −6 ÷ 3, and a negative divided by a positive stays negative. Answer: −2, which checks because 3 × (−2) + 7 = −6 + 7 = 1. The distractors: 2 comes from dividing 6 by 3 and losing the minus sign; −6 comes from taking the 7 off and stopping there, never undoing the multiplication; −18 comes from multiplying −6 by 3 instead of dividing by 3.
- (a) The hourly rate is £5 and the fixed fee is £7 — The gradient is (27 − 12) ÷ (4 − 1) = 15 ÷ 3 = £5, the hourly rate. Using the point (1, 12): 12 = 5 × 1 + fee, so the fee is 12 − 5 = £7. That gives 'The hourly rate is £5 and the fixed fee is £7'. Swapping the two figures gives the statement with £7 as the rate and £5 as the fee, which has them the wrong way round. Taking the C-value of the first point, £12, as the fixed fee ignores that 1 hour of hire is already included in that £12. Using 15, the change in C, as the hourly rate without dividing by the change in h (3 hours) gives the statement claiming a £15 hourly rate.
- (c) (3, 2) — For an enlargement centred on the origin, multiply every coordinate by the scale factor: (9 × 1/3, 6 × 1/3) = (3, 2). A pupil who multiplies by 3 instead of by 1/3 gets (27, 18). A pupil who subtracts a third of each coordinate instead of scaling by a third gets (9 − 3, 6 − 2) = (6, 4). A pupil who applies the scale factor to the x-coordinate only gets (3, 6). The correct image is (3, 2).
- (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) b = a/5 — Writing 5b means 5 × b, so b has been multiplied by 5. The inverse of multiplying by 5 is dividing by 5, and dividing both sides by 5 gives a/5 = b, which is written b = a/5. Multiplying both sides by 5 instead gives 5a, which applies the operation a second time; treating the 5 as though it were added gives a − 5; writing 5/a turns the fraction upside down.
- (d) 7/5 — The number of unbroken crayons is 60 − 25 = 35. The comparison is with the broken crayons, so the broken crayons are the denominator: 35/25. Both parts divide by 5: 35 ÷ 5 = 7 and 25 ÷ 5 = 5. The fraction is 7/5, which is greater than 1 because there are more unbroken crayons than broken ones.
- (a) 8 × 10² — Method: to multiply numbers written in standard form, multiply the coefficients and add the indices. Working: 4 × 2 = 8 for the coefficients, and −3 + 5 = 2 for the indices; 8 already lies between 1 and 10, so no adjustment is needed. Answer: 8 × 10². The distractors: 6 × 10² comes from adding the coefficients, 4 + 2, instead of multiplying them; 8 × 10⁸ comes from ignoring the minus sign and adding 3 + 5; 8 × 10⁻¹⁵ comes from multiplying the indices, −3 × 5, instead of adding them.
- (c) 20 cm — Convert 5 km to centimetres: 5 km = 5000 m = 500 000 cm. Divide by the scale factor: 500 000 ÷ 25 000 = 20, giving 20 cm. Converting only as far as metres, 5000 ÷ 25 000 = 0.2, gives 0.2 cm — the conversion to centimetres was never finished. Dropping a zero in the division gives 2 cm, ten times too small. Misreading the scale as 1 : 2500 instead of 1 : 25 000 gives 500 000 ÷ 2500 = 200 cm, ten times too big.
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.