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.Postage on a parcel is calculated as £4.60, correct to the nearest 20p. Which of these could not be the actual cost of the postage?
- 2.Which of these rules generates the sequence 6, 11, 16, 21, …?
- 3.The population of a village is 1200. It is predicted to grow by 5% next year. Work out the predicted population after 1 year, to the nearest whole number.
- 4.p is the column vector with top number 5 and bottom number 2. q is the column vector with top number 1 and bottom number 6. Work out p + q, giving your answer as a column vector in the form (top, bottom).
- 5.A bag is known to contain red, blue and green counters in equal numbers, so the theoretical probability of taking each colour is 1/3. In 90 trials with replacement, red was taken 38 times, blue was taken 26 times and green was taken 26 times. Which colour is over-represented compared with its theoretical probability?
- 6.The seven members of Team A took 20, 21, 22, 23, 24, 25 and 40 seconds to finish a task. The seven members of Team B took 20, 30, 32, 34, 36, 38 and 40 seconds. Tomás says that because the two teams have the same range, their times are spread out in the same way. Is Tomás right? Give a reason for your answer.
- 7.A number, n, is a multiple of both 6 and 9. Work out the smallest possible value of n that is greater than 20.
- 8.A distance-time graph shows a car travelling at a constant speed, covering 90 km in 2 hours. Work out the car's speed in km/h.
- 9.A model railway is built at a scale where the ratio of model length to real length is 1 : 76. Given that a real carriage is 20.6 metres long, work out the length of the model carriage in centimetres, to 1 decimal place.
- 10.A tangent touches a circle with centre O at the point P. Q is a point on the tangent. Write down the circle fact that tells you the size of angle OPQ.
- 11.A basketball player has a free-throw success probability of 0.75. She wants to expect to score 60 successful free throws. Work out how many free throws she needs to attempt.
- 12.Write these numbers in order, starting with the smallest: 3, −1, 0, −5
- 13.Tom buys 2 notebooks and 6 pencils on Monday, priced at p pence each for a notebook and q pence each for a pencil. On Tuesday he buys 3 more notebooks and 1 more pencil. Write a simplified expression, in pence, for the total amount Tom has spent on notebooks and pencils.
- 14.Two mathematically similar jugs have heights 8 cm and 12 cm. The smaller jug holds 200 ml when it is full. Work out how much the larger jug holds when it is full.
- 15.Which of these correctly compares a 'line' with a 'line segment'?
- 16.Write 5,000,000 in standard form.
- 17.Which of these equations gives a graph shaped like a U, symmetrical about a vertical line?
- 18.Two mathematically similar flags are made in different sizes. The length scale factor from the smaller flag to the larger flag is 5. Work out the area scale factor from the smaller flag to the larger flag.
- 19.A café orders 340 bread rolls at 24p each and 85 cakes at £1.35 each. Work out the total cost of the order.
- 20.The ratio of the number of red beads to blue beads in a bracelet is 3 : 7. There are 21 blue beads. Work out the number of red beads.
Answer key
- (d) £4.80 — Method: find the error interval, then check which value falls outside it. Working: half of 20p is 10p, so the actual cost, c, satisfies £4.50 ≤ c < £4.70. £4.80 is above £4.70, so it could not be the actual cost. Answer: £4.80. (£4.50 is a genuine possible cost — it sits at the included lower boundary. £4.65 is a genuine possible cost, below the £4.70 upper boundary. £4.55 is a genuine possible cost, well inside the interval.)
- (c) 5n + 1 — Method: find the common difference, then use it as the coefficient of n in the position-to-term rule, and find the constant by checking against the first term. Working: the common difference is 5, so the rule has the form 5n + c. Using the 1st term: 5(1) + c = 6, so c = 1. The rule is 5n + 1. Answer: 5n + 1. 5n − 1 uses the correct coefficient but the wrong sign for the constant. 6n comes from using the first term as the coefficient of n instead of the common difference — it matches the 1st term by coincidence but fails from the 2nd term onward. n + 5 swaps the coefficient and the constant around, using the common difference as the constant instead of the coefficient of n.
- (a) 1260 — To increase by 5%, multiply by 1.05 (100% + 5%). 1200 × 1.05 = 1260. 60 comes from working out only the increase (1200 × 0.05) and forgetting to add it to the original population. 1205 comes from adding 5 directly to 1200 instead of 5% of 1200. 1800 comes from multiplying by 1.5, using 50% instead of 5%.
- (d) (6, 8) — Method: add column vectors by adding the top numbers together and the bottom numbers together. Working: top numbers 5 + 1 = 6; bottom numbers 2 + 6 = 8. Answer: p + q = (6, 8). A candidate who subtracts instead of adds gets (4, −4). A candidate who swaps the top and bottom of the correct answer gets (8, 6). A candidate who only adds the top numbers and leaves the bottom number of p unchanged gets (6, 2).
- (c) Red — Theoretical probability is 1/3 ≈ 0.333 for each colour. Red's relative frequency is 38/90 ≈ 0.422, above 1/3, so red is over-represented. Blue's relative frequency is 26/90 ≈ 0.289, below 1/3, so blue is under-represented, not over. Green's relative frequency is also 26/90 ≈ 0.289, below 1/3 for the same reason. Since red's relative frequency clearly exceeds 1/3, it is not true that none of the colours are over-represented.
- (d) No, the range uses only the fastest and slowest time — Method: check what the range is built from, then look at what it leaves out. Working: both teams have a fastest time of 20 seconds and a slowest of 40 seconds, so both ranges are 40 − 20 = 20 seconds and Tomás has that part right. But the range is calculated from those two values alone. Six of Team A's seven times lie between 20 and 25 seconds, with a single time far out at 40; Team B is the other way round, with six of its seven times at 30 seconds or more and a single time far out at 20. So Team A bunches at the fast end and Team B at the slow end. The two patterns are quite different, and the range cannot see the difference because the five middle times never enter the calculation. Answer: no, because the range uses only the fastest and slowest time. The distractors: comparing the means answers a different question, since a mean measures position rather than spread, and two sets with the same spread can have different means; saying that equal ranges mean equal spread is the very assumption that fails here; saying that seven times each forces the spreads to match confuses the size of a data set with how its values are arranged inside it.
- (a) 36 — Method: find the lowest common multiple of 6 and 9, then move up the list of common multiples until one is greater than 20. Working: the common multiples of 6 and 9 are 18, 36, 54 …. 18 is not greater than 20, so the next one, 36, is the smallest value of n that is greater than 20. 18 is the lowest common multiple itself, but it fails the 'greater than 20' condition. 54 is the common multiple after 36, one step too far. 27 is a multiple of 9 but not of 6, since 27 ÷ 6 is not a whole number. Answer: 36.
- (c) 45 km/h — Method: speed = distance ÷ time = 90 ÷ 2 = 45 km/h. Distractor origins: 180 km/h multiplies distance by time instead of dividing (90 × 2 = 180); 88 km/h subtracts the time from the distance (90 − 2 = 88); 92 km/h adds the time to the distance (90 + 2 = 92).
- (c) 27.1 cm — The model length is 20.6 ÷ 76 = 0.271052... metres. Converting to centimetres by multiplying by 100 gives 27.1052..., which rounds to 27.1 cm. Forgetting to convert metres to centimetres leaves the answer as 0.271052... metres, which rounds to 0.3 cm if the unit is simply relabelled. Multiplying by 1000 instead of 100 when converting metres to centimetres gives 271.052..., which rounds to 271.1 cm. Multiplying by 76 instead of dividing, 20.6 × 76 = 1565.6, uses the scale factor the wrong way round — that would be the real length if the model were 20.6 units long, not the other way round.
- (d) The angle between a tangent and a radius is 90° — OP is the radius drawn to the point of contact P, and a circle theorem states that a tangent always meets that radius at a right angle, so angle OPQ = 90°. A tangent is not parallel to the radius it touches — at the point of contact it is perpendicular to that radius, not parallel to it. A tangent does not pass through the centre — a straight line through the centre that also touches the circle at one point would have to be a diameter, which is a different line entirely. The angle-in-a-semicircle theorem needs a triangle drawn inside the circle with a diameter as its longest side; there is no such triangle here, just a tangent and a radius.
- (b) 80 — To find the number of attempts needed, divide the target number of successes by the probability of success: 60 ÷ 0.75 = 80. Writing 45 is wrong because 60 × 0.75 = 45 multiplies instead of dividing — that is the number of successes expected from 60 attempts, not the number of attempts needed for 60 successes. Writing 240 is wrong because 60 ÷ 0.25 = 240 uses 0.25, the probability of MISSING, instead of 0.75, the probability of scoring. Writing 90 is wrong because it comes from misremembering 0.75 as 2/3 and dividing by that instead: 60 ÷ (2/3) = 90. She needs to attempt 80 free throws.
- (d) −5, −1, 0, 3 — Method: order the numbers by their position on a number line, smallest (furthest left) first. Working: both −5 and −1 lie to the left of 0, and 3 lies to the right of 0. Of the two negatives, −5 is 5 units from zero and −1 is 1 unit from zero, so −5 is further left. Answer: −5, −1, 0, 3. The distractors: 3, 0, −1, −5 is the correct order written the wrong way round, starting with the largest; −1, −5, 0, 3 comes from ordering the two negatives by the size of their digits, so that −1 is treated as the smaller; 0, −1, −5, 3 comes from believing that zero is the smallest number there is and then listing the negatives by their digits.
- (a) 5p + 7q — Tom's total spend combines like terms: (2p + 6q) + (3p + q) = 5p + 7q, adding the notebook terms (2p + 3p = 5p) and the pencil terms (6q + q = 7q) separately. Answering 12pq adds every coefficient together (2 + 6 + 3 + 1 = 12) and multiplies the letters, combining unlike terms as though notebooks and pencils were the same item. Answering 5p + 6q correctly combines the notebook terms but forgets to add Tuesday's extra pencil to the 6q. Answering 2p + 7q correctly combines the pencil terms but forgets to add Tuesday's 3 extra notebooks to the 2p. The total amount Tom has spent is 5p + 7q pence.
- (a) 675 ml — How much a jug holds is a volume, and volumes of similar solids scale with the cube of the length scale factor. The length scale factor is 12 ÷ 8 = 1.5, so the volume scale factor is 1.5 × 1.5 × 1.5 = 3.375. The larger jug holds 200 × 3.375 = 675 ml. Multiplying the scale factor by 3 instead of raising it to the power 3 is the mistake to guard against here.
- (c) A line has no endpoints; a segment has two — A line extends without end in both directions, whereas a line segment is the part of a line between two specific fixed endpoints, so 'a line has no endpoints; a segment has two' is correct. 'A line has two endpoints; a segment has none' reverses these two definitions, so it is wrong. 'A line segment is always curved' is wrong because a line segment is straight, not curved, and does not extend infinitely. 'A line segment is a closed shape' is wrong because a line segment is a straight length between two points, not a polygon.
- (c) 5 × 10⁶ — Method: standard form is written as A × 10ⁿ, where A is at least 1 and less than 10 and n counts the places the decimal point moves. Working: the digits of 5,000,000 give a coefficient of A = 5, and the decimal point travels from the end of 5,000,000 until it sits just after the 5, a move of 6 places, so n = 6. Answer: 5 × 10⁶. The distractors: 50 × 10⁵ comes from stopping before the coefficient has been brought into range, and 50 is not less than 10, so it is not standard form; 5 × 10⁷ comes from counting the seven digits of 5,000,000 instead of the six places the decimal point moves; 5 × 10⁻⁶ comes from making the index negative because the decimal point was carried to the left, when a negative index belongs to a number smaller than 1.
- (a) y = x² + 2 — A quadratic equation, with x², gives a graph shaped like a symmetrical U (or an upside-down U). Of the four equations, y = x² + 2 is quadratic, so it gives this U-shape. y = x³ + 2 is cubic, giving an S-shaped curve rather than a symmetrical U. y = 2x + 2 is linear, giving a straight line. y = 2/x is reciprocal, giving two separate curved branches rather than a single U-shape.
- (a) 25 — Method: for similar shapes, the area scale factor is the length scale factor squared. Working: 5² = 25. Answer: 25. 5 comes from using the length scale factor itself as if it were the area scale factor, without squaring it. 10 comes from doubling the length scale factor instead of squaring it. 125 comes from cubing the length scale factor, which would give the volume scale factor, not the area scale factor.
- (c) £196.35 — Method: convert both prices to pounds, multiply each by its quantity, then add the two totals. Working: 340 rolls at £0.24 each = £81.60; 85 cakes at £1.35 each = £114.75; £81.60 + £114.75 = £196.35. Answer: £196.35. £81.60 comes from working out the cost of the rolls only and forgetting to add the cost of the cakes. £114.75 comes from working out the cost of the cakes only and forgetting to add the cost of the rolls. £122.91 comes from converting 24p to £0.024 instead of £0.24, a place value error of a factor of 10 in the price of the rolls, before adding the correctly worked out cost of the cakes.
- (b) 9 — Method: divide the known quantity by its ratio number to find the value of one part, then multiply by the other ratio number. Working: 21 ÷ 7 = 3 (value of one part). Red beads = 3 × 3 = 9. Wrong options: 49 comes from dividing by the wrong ratio number (21 ÷ 3 × 7); 24 comes from adding the ratio number for red (3) to 21 instead of scaling; 14 comes from subtracting the ratio number for blue (7) from 21 instead of scaling.
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.