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.Write 0.45 as a fraction in its simplest form.
- 2.Make x the subject of the formula y = 3(x + 2).
- 3.A plumber's charge, C pounds, for a job lasting h hours is shown on a straight-line graph. The line passes through the points (1, 45) and (3, 85). Work out what the gradient of this line represents, in context.
- 4.A water tank holds 2.5 m³. Work out this volume in cm³.
- 5.The probability that Ben catches his bus on time is 0.8. The probability that Freya catches her bus on time is 0.5. The two events are independent. Work out the probability that both Ben and Freya catch their bus on time.
- 6.A charity shop in Bath holds 2,000 books. Volunteer A checks a random sample of 50 books and finds 35 paperbacks. Volunteer B checks a different random sample of 50 books and finds 31 paperbacks. Work out the estimate each sample gives for the whole stock, and write down what the shop should do next.
- 7.Amelia has 49 boxes of apples with 21 apples in each box. Work out an estimate for the total number of apples, by rounding each number to 1 significant figure.
- 8.Solve 2x − 2 = 2
- 9.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.
- 10.Write down which one of these statements about quadrilaterals is true.
- 11.A frequency tree records the results of 160 patients who took a new medicine. It splits them into those who reported side effects and those who did not. 15% of the patients reported side effects. Work out how many of the 160 patients did not report side effects.
- 12.Zoe worked out 3/8 + 1/8 by adding the numerators and the denominators, and got 4/16. Work out the correct value of 3/8 + 1/8, giving your answer in its simplest form.
- 13.Which of these values of x is a solution of x² + 2x − 15 = 0?
- 14.y is directly proportional to x. When x = 5, y = 18. Work out the value of y when x = 15.
- 15.A rhombus has all four sides equal in length. Write down how many pairs of parallel sides a rhombus has.
- 16.A supermarket sells apples at £1.85 per kg. Anna buys 3.6 kg of apples. Estimate the cost by rounding each number to 1 significant figure before multiplying. Work out Anna's estimate.
- 17.Which expression means 'y squared, multiplied by 3'?
- 18.A plan is drawn to a scale of 1 : 25. A real fence is 10 m long. Work out the length of the fence on the plan, in centimetres.
- 19.A charity raffle sells 240 tickets at £1.85 each. 40% of the money raised is given to a local hospital. Work out how much money the hospital receives.
- 20.In a test, Amelia answered 18 of the 24 questions correctly. Work out the percentage of the questions she answered correctly.
Answer key
- (a) 9/20 — Method: write the decimal over 100 using its two decimal places, then simplify. Working: 0.45 = 45/100 = 9/20 (dividing both numerator and denominator by 5). Answer: 9/20. 9/100 comes from dividing only the numerator by 5 and leaving the denominator as 100. 9/2 comes from writing the decimal over 10 instead of 100, as if there were only one decimal place, then simplifying 45/10. 9/200 comes from writing the decimal over 1000 instead of 100, as if there were three decimal places, then simplifying 45/1000.
- (b) x = y/3 − 2 — The bracket containing x has been multiplied by 3, so divide both sides by 3 first, which gives y/3 = x + 2. Subtracting 2 from both sides then leaves y/3 − 2 = x, so x = y/3 − 2. Taking the 2 away before dividing gives (y − 2)/3, which divides the 2 by 3 as well, although the 2 was never divided in the original formula. Adding 2 rather than subtracting it gives y/3 + 2, and multiplying by 3 instead of dividing gives 3y − 2.
- (a) The plumber charges £20 for each extra hour worked — The gradient is the change in C divided by the change in h: (85 − 45) ÷ (3 − 1) = 40 ÷ 2 = £20. On this graph the gradient represents the extra amount charged for each extra hour worked, so the answer is 'The plumber charges £20 for each extra hour worked'. Not dividing by the change in h gives 40, the option that reads 'charges £40 for each extra hour worked' — that is the total change in cost between the two points, not the rate per hour. Reading off the C-value of the first point, 45, gives the option using £45 — that is the cost of a job lasting 1 hour, not the rate. Treating the gradient as a flat total charge regardless of the time taken misunderstands what a straight-line graph through two different h-values shows: the cost does depend on h, so it cannot be a single fixed total.
- (b) 2 500 000 cm³ — Method: a volume conversion uses the length factor three times, once for each dimension. Working: 1 m = 100 cm, so a cube of side 1 m is a cube of side 100 cm, and 100 × 100 × 100 = 1 000 000, giving 1 m³ = 1 000 000 cm³. Then 2.5 × 1 000 000 = 2 500 000. Answer: 2 500 000 cm³. Using the plain length factor 100 gives 250 cm³. Using 1000, which is the factor that turns cubic metres into litres, gives 2500 cm³. Using 10 000, which is the factor that belongs to square metres and square centimetres, gives 25 000 cm³.
- (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.
- (a) 1,400 and 1,240, so combine the samples for one estimate — Method: scale each sample up to the whole stock, then use the fact that a larger sample gives a more reliable estimate than a smaller one. Working: the first sample gives 35 ÷ 50 = 0.7 and 0.7 × 2,000 = 1,400 paperbacks; the second gives 31 ÷ 50 = 0.62 and 0.62 × 2,000 = 1,240 paperbacks. Two random samples of the same size are expected to differ a little, so neither estimate is wrong. Putting the two together gives 35 + 31 = 66 paperbacks in 100 books, and 66 ÷ 100 = 0.66 with 0.66 × 2,000 = 1,320, an estimate resting on twice as many books as either volunteer checked. Answer: 1,400 and 1,240, so combine the samples for one estimate. The distractors: keeping 1,400 because it is larger picks an estimate by its size, when both samples held 50 books and neither has a stronger claim; saying a volunteer must have miscounted assumes two random samples ought to agree exactly, which is precisely what random sampling does not promise; 1,750 and 1,550 come from 35 × 50 = 1,750 and 31 × 50 = 1,550, multiplying each count by the size of the sample instead of scaling by 2,000 ÷ 50.
- (d) 1,000 — Method: round each number to 1 significant figure, then multiply the rounded values. Working: 49 rounds to 50 and 21 rounds to 20, and 50 × 20 = 1,000 because 5 × 2 = 10 and the two rounded numbers carry one zero each. Answer: 1,000. The distractors: 800 comes from rounding 49 down to 40 instead of to the nearest ten; 1,500 comes from rounding 21 up to 30 rather than down to 20; 1,029 is the exact product 49 × 21, worked out in full when the question asks for an estimate.
- (a) x = 2 — Method: undo the subtraction first, then undo the multiplication, doing the same to both sides each time. Working: adding 2 to both sides gives 2x = 4, and dividing both sides by 2 gives x = 2. Answer: x = 2. The distractors: x = 4 comes from stopping at 2x = 4 and writing 4 as the value of x; x = 0 comes from subtracting 2 from both sides instead of adding it, giving 2x = 0; x = 8 comes from multiplying 4 by 2 instead of dividing by 2.
- (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.
- (a) Every square is a rectangle — Method: test each statement against the definitions. A rectangle is a quadrilateral with four right angles and opposite sides equal; a square is a quadrilateral with four right angles and all four sides equal; a rhombus is a quadrilateral with all four sides equal. Working: a square has four right angles and its opposite sides are equal, so every square meets the definition of a rectangle and the statement that every square is a rectangle is true. Answer: every square is a rectangle. The distractors: the claim that every rectangle is a square reverses the inclusion, and fails for any rectangle whose length and width differ; the claim that every rhombus is a rectangle treats four equal sides as enough, and drops the right-angle condition — a tilted rhombus has no right angles; the claim that every rectangle is a rhombus reads 'opposite sides equal' as though it meant 'all four sides equal'.
- (d) 136 — 15% of 160 = 0.15 × 160 = 24 patients reported side effects, so 160 − 24 = 136 did not. Stopping after finding the number who reported side effects, 24, answers the wrong question — it is not the number who did NOT report them. Misreading '15%' as a raw count of 15 patients, rather than a percentage, gives 160 − 15 = 145. Subtracting 15% of 160 twice, 160 − 24 − 24 = 112, double-counts the side-effect group.
- (d) 1/2 — Method: since the fractions already share a denominator, add only the numerators and keep the denominator the same. Working: 3/8 + 1/8 = 4/8 = 1/2. Answer: 1/2. Zoe's method, adding the denominators too, gives 4/16 = 1/4. 4/9 comes from adding the numerators correctly to get 4, but then building the denominator by adding the 8 of the first fraction to the 1 of the second (8 + 1 = 9), mixing a denominator with a numerator. 3/8 comes from ignoring the second fraction and simply restating the first one.
- (a) x = 3 — Method: factorise x² + 2x − 15 as (x + 5)(x − 3), since 5 × (−3) = −15 and 5 + (−3) = 2. Setting each bracket equal to zero gives x + 5 = 0 or x − 3 = 0, so x = −5 or x = 3. Only x = 3 is offered here. Distractor origins: x = −3 reverses the sign of the factor pair, treating the bracket (x − 3) as giving x = −3 instead of x = 3; x = 5 takes the number from the other factor, (x + 5), but with the wrong sign, giving x = 5 instead of x = −5; x = 15 takes the constant term of the original expression as if it were a root, without factorising at all.
- (c) 54 — Method: y = kx, so k = y ÷ x. Working: k = 18 ÷ 5 = 3.6. At x = 15: y = 3.6 × 15 = 54. Wrong options: 28 comes from adding the change in x (10) onto y instead of scaling; 6 comes from treating the relationship as inverse proportion (k = 5 × 18 = 90, then y = 90 ÷ 15 = 6); 60 comes from rounding the constant up to 4 instead of using 3.6.
- (b) 2 — A rhombus is a special parallelogram, so opposite sides are parallel — that is two pairs of parallel sides. A quadrilateral with only one pair of parallel sides is a trapezium, not a rhombus, so 1 is wrong. A kite has 0 pairs of parallel sides, not a rhombus. There are only two pairs of opposite sides in a quadrilateral altogether, so 4 is not possible. The correct answer is 2.
- (b) £8 — Rounding to 1 significant figure: £1.85 rounds to £2, and 3.6 kg rounds to 4 kg. The estimate is £2 × 4 = £8. A candidate who used the unrounded values instead of estimating worked out 1.85 × 3.6 = £6.66. A candidate who rounded only the mass and used the exact price worked out 1.85 × 4 = £7.40. A candidate who rounded the price to the nearest 10p instead of 1 significant figure worked out 1.9 × 4 = £7.60.
- (a) 3y² — 'y squared, multiplied by 3' means the square is applied to y only, and the result is then multiplied by 3, written as 3y². Writing y³ mistakes the multiplication by 3 for an extra factor of y, adding to the power instead of using a coefficient. Writing (3y)² squares the whole of 3y, including the 3, which gives 9y² rather than 3y² — the square should apply to y alone. Writing 3 + y² adds the 3 instead of multiplying by it. The expression for 'y squared, multiplied by 3' is 3y².
- (d) 40 cm — Method: convert the real length to centimetres, then divide by the scale factor. Working: 10 m = 1000 cm. 1000 ÷ 25 = 40 cm. Wrong options: 0.4 cm comes from dividing 10 by 25 without converting metres to centimetres first; 25000 cm comes from multiplying instead of dividing (1000 × 25); 4 cm comes from converting 10 m to 100 cm (using ×10 instead of ×100) before dividing.
- (a) £177.60 — Total raised = 240 × £1.85 = £444.00. The hospital receives 40% of this: £444.00 × 0.4 = £177.60. A candidate who works out the remaining 60% instead of the 40% given away gets £266.40. A candidate who forgets to find the percentage and gives the full total gets £444.00. A candidate who halves 40% by mistake and uses 20% gets £88.80.
- (a) 75% — Method: to express one quantity as a percentage of another, divide the part by the whole and multiply by 100. Working: 18 ÷ 24 = 0.75, and 0.75 × 100 = 75. Answer: 75%. The distractors: 25% is the percentage she got wrong, 6 out of 24; 133% comes from dividing the whole by the part, 24 ÷ 18; 18% comes from writing the number of correct answers with a percent sign.
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.