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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- (a) 5 — Method: each index is worked out first, and subtracting a negative number is the same as adding the positive. Working: (−2)³ = (−2) × (−2) × (−2) = −8 and (−3)² = (−3) × (−3) = 9, while − (−4) becomes + 4, so the calculation becomes −8 + 9 + 4 = 5. Answer: 5. The distractors: −13 comes from taking (−3)² as −9, giving −8 − 9 + 4 = −13; −3 comes from reading − (−4) as − 4, giving −8 + 9 − 4 = −3; 21 comes from treating every power of a negative number as positive, so that (−2)³ is taken as 8 and the calculation becomes 8 + 9 + 4 = 21.
- (a) 36 — Square numbers are n² for n = 1, 2, 3, ...; the fifth term, 25, is 5². The sixth square number is 6² = 36. A candidate who mislabels 25 as the sixth square number would compute 7² = 49 instead. A candidate who repeats an earlier difference between terms (5, from 4 to 9) rather than the correct next difference (11, since the differences are the odd numbers 3, 5, 7, 9, 11) would reach 30. A candidate who instead adds 10 would reach 35.
- (a) 200 seconds — Method: change the minutes into seconds using 1 minute = 60 seconds, then add on the loose seconds. Working: 3 × 60 = 180, and 180 + 20 = 200. Answer: 200 seconds. The distractors: 320 seconds comes from taking a minute as 100 seconds, giving 300 + 20; 180 seconds comes from converting the 3 minutes and forgetting the extra 20 seconds; 23 seconds comes from adding 3 and 20 without converting the minutes at all.
- (c) 21 cm — Method: the perimeter is the distance all the way round the edge, so every side is counted once and the three side lengths are added. Working: the two equal sides give 8 + 8 = 16 cm, and the third side adds 5 cm to that. Answer: the perimeter is 21 cm. The distractors: 16 cm comes from adding the two 8 cm sides and handing that total in before the third side has been included; 13 cm comes from adding one 8 cm side to the 5 cm side, as though the badge carried only the two different lengths printed on it rather than three sides; 24 cm comes from taking all three sides to be 8 cm and working out 3 × 8, which would be the perimeter only if the badge were equilateral.
- (c) 80 — Since 180 calls are 0.75 of all the technical support calls, the technical support total is 180 ÷ 0.75 = 240. The billing calls make up the rest of the 320 calls, so 320 − 240 = 80. Choosing 240 comes from stopping after finding the technical support total and forgetting the question asks for the billing calls, which are the rest. Choosing 185 comes from multiplying 180 × 0.75 = 135 instead of dividing, then working out 320 − 135 = 185. Choosing 140 comes from using 180 directly as the whole technical support total, ignoring the probability altogether, then working out 320 − 180 = 140.
- (b) No — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. — Method: the median is the middle value of the data in order of size, so the data must be sorted before any position is read off. Working: Noah's list 9, 3, 7, 1, 5 is not in order; sorted it becomes 1, 3, 5, 7, 9, and with 5 values the middle position is the third, which now holds 5 rather than 7. Noah has read the third value of the unsorted list. Answer: no — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. The distractors: the reply giving 3 as the median sorts the data correctly but then reads the value in the second place instead of the third; the reply that 7 is the third number he wrote accepts a position in the unsorted list, which is exactly the mistake the question is about; the reply using the mean claims a value of 7 for it, but the mean is 25 ÷ 5 = 5, so that reasoning is false as well.
- (c) 2/3 — Method: multiply the numerators together and the denominators together, then divide both parts of the result by their highest common factor. Working: 3 × 14 = 42 and 7 × 9 = 63, giving 42/63; the highest common factor of 42 and 63 is 21, and 42 ÷ 21 = 2 with 63 ÷ 21 = 3. Answer: 2/3. The distractors: 17/16 comes from adding the numerators and adding the denominators, giving (3 + 14)/(7 + 9); 27/98 comes from turning the second fraction upside down and multiplying, which divides instead of multiplying and gives 3/7 × 9/14; 2/21 comes from cancelling the 7 into the 14 in the numerator but leaving the 7 in the denominator, giving 6/63.
- (b) 3 + 2m — Method: add the fixed fee to the cost of the miles travelled. Working: the miles cost 2 × m = 2m pounds, and the fixed fee is £3, so the total is 3 + 2m. Answer: 3 + 2m. 2 + 3m comes from swapping which number is the fixed fee and which is the per-mile rate. 5 + m comes from adding the fixed fee and the rate together into one number, 3 + 2 = 5, before adding the miles, instead of multiplying the rate by the number of miles. 5m comes from multiplying that combined total by the miles, treating the whole £5 as if it were a per-mile rate.
- (c) 27 : 64 — For similar solids, the ratio of volumes is the ratio of lengths cubed: 3³ : 4³ = 27 : 64. 3 : 4 comes from using the height ratio itself as the volume ratio, without cubing it at all. 9 : 16 comes from squaring each part instead of cubing (3² : 4²) — squaring is the rule for area, not volume. 27 : 4 comes from cubing only the first part of the ratio (3³ = 27), and leaving the second part uncubed.
- (a) Infinitely many — Method: a radius is any straight line from the centre of a circle to a point on the circle, so counting the radii means counting the possible end points, and those end points are the points of the circle itself. Working: a circle is a continuous curve, so between any two points on it there is always another point; the supply of end points therefore never runs out, and each end point gives one radius. Every one of those radii is the same length, because equal distance from the centre is what makes the curve a circle in the first place. Answer: Infinitely many. The distractors: Exactly one comes from treating a radius as a single fixed line, because a textbook diagram usually shows only one drawn in; Exactly two comes from picturing a diameter and counting the two halves it is made of; Exactly four comes from picturing the circle cut into quarters by two perpendicular diameters and counting the four radii that appear in that picture.
- (c) The green bag (7/24) — Method: fractions can only be ordered once they share a denominator, so rewrite all four over the lowest common denominator and compare the numerators. Working: the lowest common denominator of 12, 8, 24 and 3 is 24, and scaling gives 5/12 = 10/24, 3/8 = 9/24, 7/24 stays as it is, and 1/3 = 8/24; the numerators are then 10, 9, 7 and 8. Answer: the smallest numerator is 7, so the green bag, with 7/24, is the least likely and sits furthest to the left on the 0 to 1 scale. The distractors: the red bag (5/12) comes from finding the largest of the four probabilities instead of the smallest; the blue bag (3/8) comes from scaling 3/8 by changing only the denominator to 24, which turns it into 3/24 and makes it look the smallest; the yellow bag (1/3) comes from comparing numerators alone and assuming the fraction with the numerator 1 must be the smallest.
- (d) 60 — Method: a percentage acts as an operator, so finding 20% of an amount means multiplying it by 20/100, which cancels to 1/5. Working: 20% = 20/100 = 1/5, and 300 ÷ 5 = 60. Answer: 60 seats. The distractors: 15 comes from reading 20% as one twentieth and working out 300 ÷ 20 = 15; 30 comes from finding 10% of 300 and stopping there instead of doubling it; 6 comes from converting 20% to 0.02 rather than 0.2, giving 0.02 × 300 = 6.
- (a) x/5 — Read the decimal as a fraction first: 0.2 is 2 tenths, so the coefficient is 2/10. Dividing the numerator and the denominator by 2 cancels this to 1/5, and a coefficient of 1/5 in front of a letter is written x/5. Reading the digit 2 as 'a half' gives x/2; reading 0.2 as 1/20 gives x/20; dividing only the denominator by 2 leaves 2/5, which is the coefficient 0.4 and twice as large as it should be.
- (a) 5:3 — Divide both prices by their highest common factor, 3: 15 ÷ 3 = 5 and 9 ÷ 3 = 3, giving the ratio 5:3. Choosing 3:5 comes from writing the ratio the wrong way round, as child price to adult price. Choosing 2:3 comes from using the difference between the two prices (15 − 9 = 6) as the first part of the ratio instead of the adult price, then simplifying 6:9 by dividing by 3. Choosing 5:8 comes from comparing the adult price with the total cost of both tickets (£15 out of £24) instead of comparing it with the child price.
- (c) 38° — The interior angles of a hexagon sum to (6 − 2) × 180° = 720°. 130° + 142° + 125° + 150° + 135° = 682°. x = 720° − 682° = 38°. A student who uses 6 × 180° instead of (6 − 2) × 180°, forgetting to subtract the 2, gets 1080° − 682° = 398°. A student who mis-adds the five given angles as 680° instead of 682° gets x = 720° − 680° = 40°. A student who answers with one of the given angles instead of solving for x might write 142°.
- (a) 7/8, 17/20, 0.8, 78% — Method: convert every value to a decimal, then order the decimals from largest to smallest. Working: 7/8 = 0.875, 17/20 = 0.85, 0.8 = 0.8, 78% = 0.78. Ordering from largest to smallest gives 7/8, 17/20, 0.8, 78%. Answer: 7/8, 17/20, 0.8, 78%. '78%, 0.8, 17/20, 7/8' comes from ordering the converted decimals from smallest to largest instead of largest to smallest. '17/20, 7/8, 0.8, 78%' comes from converting 17/20 incorrectly as larger than 7/8, for example treating 17/20 as 0.87 instead of 0.85, swapping the top two. '7/8, 0.8, 17/20, 78%' comes from converting 17/20 incorrectly as equal to 0.8, swapping the middle two.
- (a) 4x + 2 — Expand the bracket: 4(x − 2) = 4x − 8. Then add the 10 that follows the bracket: 4x − 8 + 10 = 4x + 2. The option 4x − 8 comes from expanding the bracket correctly but then forgetting to add the 10. The option 4x + 8 comes from forgetting to multiply the 2 inside the bracket by 4 (treating it as 4x − 2 instead of 4x − 8), then adding 10. The option 4x + 18 comes from a sign error when expanding — treating 4 × (−2) as +8 instead of −8, giving 4x + 8, then adding 10.
- (c) m = 6c — Method: the whole is the sum of the parts in the ratio, and the cement is 1 part, so one part weighs c kg. Working: the mix has 5 + 1 = 6 parts, each of mass c kg, so the total mass is 6 × c, giving m = 6c. Answer: m = 6c. The distractors: m = 5c uses the 5 gravel parts as the multiplier and forgets that the cement is in the mix too, so it gives the mass of the gravel and not the total; m = c + 5 comes from reading the ratio as '5 more than' and adding, which treats a number of parts as a mass in kilograms; m = c/6 turns the relationship upside down, as though the total were shared into the cement rather than the cement multiplied up to the total.
- (c) Ben: 2 × 3 = 6, then 6 + 6 = 12 — Multiplication has priority over addition, so 2 × 3 = 6 is worked out first, then 6 + 6 = 12 — this is Ben's method. Amy adds 6 and 2 before multiplying: 6 + 2 = 8, then 8 × 3 = 24, breaking the priority rule. Chen multiplies the wrong pair of numbers, 6 and 2, instead of 2 and 3: 6 × 2 = 12, then 12 + 3 = 15. Dev applies the right order but slips when multiplying, using 5 instead of 6 for 2 × 3, so the final step becomes 6 + 5 = 11.
- (d) £30.00 — The gradient is (130 − 70) ÷ (5 − 2) = 60 ÷ 3 = £20 per hour. Using the point (2, 70): the cost for 2 hours at £20 per hour is 20 × 2 = £40, so the call-out fee is 70 − 40 = £30. Taking the C-value of the first point as the fee without subtracting the hourly cost gives £70.00 — but that point already includes 2 hours of the hourly rate. Using the gradient itself as the fee, £20.00, confuses the rate per hour with the fixed charge. Subtracting 20 × 3 = 60 instead of 20 × 2 = 40 (using the wrong h-value) gives 70 − 60 = £10.00.
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.