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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- (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.
- (b) x = (y − 3)/5 — To make x the subject of y = 5x + 3, first subtract 3 from both sides to get y − 3 = 5x, then divide both sides by 5: x = (y − 3)/5. Writing x = (y + 3)/5 keeps the division correct but does not change the sign of the 3 when moving it across. Writing x = y/5 − 3 divides only the y term by 5 and leaves the 3 as a separate subtraction, instead of subtracting first and dividing the whole expression. Writing x = 5(y − 3) applies the correct order of subtracting 3 first, but then multiplies by 5 instead of dividing — the inverse of 5x is division, not multiplication. The correct rearrangement is x = (y − 3)/5.
- (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.
- (d) 21.2 m — Method: the cable is the hypotenuse of a right-angled triangle whose vertical side is the drop from the roof to the bracket and whose horizontal side is 15 m, so use Pythagoras' theorem. Working: the drop is 20 − 5 = 15 m, so c² = 15² + 15² = 225 + 225 = 450 and c = √450 = 21.213…, which is 21.2 m to 1 decimal place. Answer: 21.2 m. The distractors: 25.0 m comes from using the whole 20 m height of the roof as the vertical side and forgetting that the bracket is already 5 m up; 30.0 m comes from adding the two sides of the triangle, 15 + 15, instead of using Pythagoras' theorem; 15.0 m is the horizontal distance on its own, which would be the length of the cable only if it ran level.
- (a) 4/9 — There are 3 × 6 = 18 equally likely outcomes. The product is a multiple of 5 whenever Spinner C shows 5, whatever Spinner D shows, which is 6 outcomes, or whenever Spinner D shows 5 and Spinner C shows 2 or 3, which is 2 more outcomes. This gives 6 + 2 = 8 outcomes, so the probability is 8/18 = 4/9. Choosing 1/3 comes from only counting the 6 outcomes where Spinner C shows 5 and forgetting the 2 outcomes where Spinner D shows 5 instead. Choosing 1/9 comes from only counting the 2 outcomes where Spinner D shows 5 and forgetting the 6 outcomes where Spinner C shows 5. Choosing 1/2 comes from counting 9 outcomes instead of 8, by listing the case where Spinner C shows 5 and Spinner D shows 5 twice over.
- (d) Testing destroys bulbs, so testing all leaves none to sell. — Method: testing every item in a population instead of a sample is a census — sensible only when testing does not use up or destroy what is being tested. Working: here, testing a bulb to find its lifespan destroys it, so testing all 50,000 bulbs would leave nothing left to sell — a sample lets the company estimate the typical lifespan without destroying its whole stock. Extra electricity used in testing is not the real reason a census is avoided here — it is the destruction of the product that matters. Saying a sample is always more accurate than a full census is the wrong way round: a census, if it could be carried out, gives the exact figure for the whole population — it is testing being destructive, not a lack of accuracy, that rules it out here. There is no law against testing every item a company makes — nothing in the question suggests that. When testing destroys the item being tested, sampling is necessary, not just convenient.
- (c) 31 — Method: work out each power separately, remembering that any non-zero base raised to the power 0 is 1 and a base raised to the power 1 is itself, then add the three values. Working: 5⁰ = 1, 5¹ = 5 and 5² = 25, so the total is 1 + 5 + 25 = 31. Answer: 31. The distractors: 30 comes from taking 5⁰ as 0 instead of 1; 35 comes from taking 5⁰ as 5, treating a zero index as leaving the base unchanged; 125 comes from adding the indices first, as though the three terms were being multiplied, and working out 5³.
- (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.
- (a) £20,000 — Method: add the parts of the ratio, divide the profit by the number of parts to find the value of one part, then multiply by the parts belonging to the partner asked about. Working: 3 + 5 + 4 = 12 parts, £48,000 ÷ 12 = £4,000 for one part, and that partner has 5 parts, so 5 × £4,000 = £20,000. Answer: £20,000. The distractors: £4,000 is the value of one part; £12,000 is the 3-part share; £16,000 is the 4-part share, which is what a candidate reaches by assuming the biggest share must go with the number written last in the ratio.
- (d) 0 — A vertex is a sharp corner point where edges meet. A cylinder has two curved, circular edges but no sharp corner points at all, so it has 0 vertices, making that the correct answer. '2' wrongly treats the two circular edges themselves as vertices, but an edge is not the same as a vertex. '4' overcounts by treating each circular edge as if it had two end-vertices, which does not apply to a continuous curved edge. '1' wrongly imagines the curved surface itself forming a single corner point, which it does not.
- (d) 3/5 — 'Chosen from the boys' restricts the sample space to the 45 boys, of whom 27 walk. So the probability is 27/45 = 3/5. Using all 80 students as the denominator instead of just the boys gives 27/80, which is the probability that a student chosen from everyone is a boy who walks — not what was asked. Using the 18 boys who do NOT walk (45 − 27) as the numerator instead of the 27 who do gives 18/45 = 2/5. Dividing the 27 walking boys by the number of girls (80 − 45 = 35) instead of by the number of boys gives 27/35.
- (b) −18 — Method: the multiplication is carried out before the addition, and a negative multiplied by a positive is negative. Working: (−4) × 3 = −12, so the calculation becomes (−6) + (−12) = −18. Answer: −18. The distractors: −30 comes from adding first and multiplying afterwards, giving (−6 + −4) × 3 = −10 × 3 = −30; 6 comes from treating (−4) × 3 as +12 on the grounds that a minus sign makes a product positive, giving −6 + 12 = 6; 18 comes from ignoring both minus signs and working out 6 + 4 × 3 = 18.
- (d) 5 — Method: running a machine backwards means undoing its operation, and the operation that undoes multiplying by 3 is dividing by 3. Working: the input multiplied by 3 gives 15, so the input is 15 ÷ 3. Answer: 5, which checks because 3 × 5 = 15. The distractors: 45 comes from running the machine forwards on the output, 15 × 3; 12 comes from subtracting 3 from 15 instead of dividing; 18 comes from adding 3 to 15, which would undo a machine that subtracts rather than one that multiplies.
- (d) 48 — Find the rate first: 18 ÷ 3 = 6 bottles per minute. Then apply it to the new time: 6 × 8 = 48 bottles. Working out 18 + (8 − 3) = 23 adds the extra 5 minutes onto the number of bottles instead of scaling proportionally. Working out 18 × 8 = 144 multiplies the given number of bottles by the new number of minutes without finding the rate first. Writing 18 keeps the count the same, not realising it must change with the time. In 8 minutes the machine fills 48 bottles.
- (c) 2 hours 15 minutes — Divide the total minutes by 60: 135 ÷ 60 = 2 remainder 15, so that is 2 hours 15 minutes. Treating 100 minutes as one hour, a metric-style mistake, gives 135 − 100 = 35, so 1 hour 35 minutes. Subtracting 60 twice to reach the 15 minutes left over, but losing count and recording only one of the two hours removed, gives 1 hour 15 minutes. Writing 135 ÷ 60 = 2.25 and then reading the '25' as minutes, instead of converting the 0.25 of an hour into 15 minutes, gives 2 hours 25 minutes.
- (d) 1/16 — Method: a negative index means the reciprocal of the power, so 2⁻⁴ is 1 divided by 2⁴. Working: 2⁴ = 2 × 2 × 2 × 2 = 16, so the value is 1/16. Answer: 1/16. The distractors: −16 comes from reading the negative index as a minus sign on the result; 16 comes from ignoring the minus sign and working out 2⁴; 1/8 comes from multiplying the base by the index, 2 × 4, and writing 1 over that product.
- (b) 7 — Method: compare the equation with y = mx + c, where m is the gradient. Working: in y = 7x − 2, the coefficient of x is 7. Answer: the gradient is 7. −2 comes from confusing the gradient with the y-intercept. −7 comes from a sign error, treating the coefficient of x as negative. 2 comes from using the y-intercept but dropping its sign.
- (a) 70 km/h — Method: average speed for a whole journey is the total distance divided by the total time, not the mean of the separate speeds. Working: 3 × 80 = 240 km and 1 × 40 = 40 km, giving 280 km in 3 + 1 = 4 hours, so 280 ÷ 4 = 70. Answer: 70 km/h. The distractors: 60 km/h comes from taking the mean of 80 and 40, which would only be right if equal times were spent at each speed; 280 km/h is the total distance written with a speed unit, from forgetting to divide by the total time; 80 km/h comes from quoting the speed of the longer leg as the average for the whole journey.
- (c) £384.00 — Adding 20% VAT means multiplying the price by 1.2: £320 × 1.2 = £384.00. Treating the 20% as a flat £20 rather than a percentage of the price, £320 + £20, gives £340.00. Working out the VAT amount alone, £320 × 0.2 = £64.00, and stopping there without adding it back to the original price gives just the VAT, not the total price. Misplacing the decimal point and using 2% instead of 20%, £320 × 1.02, gives £326.40.
- (c) 5 m — The scale 1 : 100 means 1 cm on the map represents 100 cm in real life. The path is 5 cm on the map, so the real length is 5 × 100 = 500 cm. Convert to metres: 500 cm = 5 m. Dividing instead of multiplying, 5 ÷ 100 = 0.05, gives 0.05 m — the scale must be used to make the real object bigger than the map, not smaller. Leaving the answer as 500 without converting to metres and calling it 500 m mistakes centimetres for metres. Using a scale of 1 : 1000 instead of the given 1 : 100 gives 5 × 1000 = 5000 cm = 50 m, ten times too large.
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.