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) 28.8 km/h — A compound unit is converted one part at a time. There are 3600 seconds in an hour, so in one hour the cyclist travels 8 × 3600 = 28 800 metres. There are 1000 metres in a kilometre, so 28 800 m = 28 800 ÷ 1000 = 28.8 km/h. 28 800 km/h leaves the distance in metres, 0.48 km/h converts the seconds to minutes rather than to hours, and 2.22 km/h divides by 3.6 instead of multiplying.
- (d) (3, 0) — Method: every point on the x-axis has y-coordinate 0, so substituting y = 0 into the equation and solving gives the x-coordinate of the crossing point. Working: 0 = 2x − 6 gives 2x = 6, so x = 6 ÷ 2 = 3 and the graph crosses the x-axis at (3, 0). Answer: (3, 0). The distractors: (0, −6) is where the graph crosses the y-axis, found by substituting x = 0 rather than y = 0; (−3, 0) comes from moving the 6 across the equals sign without changing its sign, giving 2x = −6; (6, 0) comes from reading the constant straight off as the crossing point and never dividing by the gradient 2.
- (b) 4/5 — Write the mass of the beans over the mass of the soup: 400/500. Divide the top and bottom by 100 to get 4/5. Choosing 5/4 comes from writing the soup's mass over the beans' mass, the wrong way round. Choosing 1/5 comes from finding the difference in mass (500 − 400 = 100) and writing that over the mass of the soup, instead of using the mass of the beans. Choosing 5/9 comes from writing the mass of the soup over the total mass of both tins (500 out of 900), instead of over the mass of the beans.
- (b) 4 m — The two triangles are similar, so height ÷ shadow is the same for both objects. For the post, height ÷ shadow = 1.2 ÷ 1.8. Flagpole height = 6 × (1.2 ÷ 1.8) = 4 m. A student who multiplies the flagpole's shadow by the post's height without applying the ratio gets 6 × 1.2 = 7.2 m. A student who assumes a constant DIFFERENCE between shadow and height instead of a constant ratio finds the post's shadow minus height, 1.8 − 1.2 = 0.6 m, then subtracts that from the flagpole's shadow to get 6 − 0.6 = 5.4 m.
- (c) 5/8 — Method: the sections are all the same size, so every section is equally likely; count the sections that are not red and write that count over the total number of sections. Working: 8 − 3 = 5 sections are not red, and there are 8 sections altogether. Answer: 5/8, a value between 0 and 1 and a little above the halfway point of the scale. The distractors: 3/8 comes from giving the probability that the spinner does land on red; 5/11 comes from adding the 3 red sections to the 8 sections to make a total of 11 instead of using the 8 sections that exist; 1/2 comes from assuming that 'red' and 'not red' must be equally likely because there are only two possibilities.
- (b) 120, unreliable — x = 25 is outside 1 to 12 — The line of best fit is y = 4x + 20. 4 × 25 = 100, and 100 + 20 = 120, so the estimate is y = 120. But x = 25 lies far outside the plotted range of 1 to 12 years, so this is an extrapolation, and the estimate is not reliable. Reaching 100 instead of 120 comes from 4 × 25 = 100 with the intercept of 20 left out — still correctly flagged as unreliable, but the wrong value. Calling the estimate reliable simply because it was calculated correctly, giving 120, wrongly assumes that a correct calculation is automatically trustworthy, ignoring that x = 25 lies far beyond the data actually collected. Reaching 68, from 4 × 12 = 48 and 48 + 20 = 68, substitutes x = 12, the top of the plotted range, instead of the assistant's actual x = 25, and wrongly calls that reliable because 12 lies inside the range.
- (a) 0 — Method: BIDMAS works through the index first, then the multiplication, then the subtraction. Working: 5² = 25, then 4 × 25 = 100, and finally 100 − 100 = 0. Answer: 0. The distractors: 2400 comes from working from left to right and subtracting first, giving (100 − 4) × 25 = 96 × 25 = 2400; −300 comes from multiplying before applying the index, giving (4 × 5)² = 20² = 400 and then 100 − 400 = −300; 60 comes from reading 5² as 5 × 2 = 10, so that 4 × 10 = 40 and 100 − 40 = 60.
- (b) t − 1 — Method: "fewer than" means a subtraction, and the amount taken away is taken away from the quantity being compared with. Working: Grace has t books and Noah has one book fewer, so Noah has t with 1 taken off, written t − 1. Answer: t − 1. The distractors: t + 1 comes from reading "fewer" as "more" and adding instead of subtracting; 1 − t comes from writing the subtraction the wrong way round, taking Grace's books from the 1 rather than the 1 from Grace's books; t comes from giving the number of books Grace has, which is not the person the question asks about.
- (d) £120 — Method: add the parts of the ratio, divide the prize by the number of parts to find the value of one part, then multiply by the parts in the share asked for. Working: 2 + 3 + 4 = 9 parts, £360 ÷ 9 = £40 for one part, and the share asked for is 3 parts, so 3 × £40 = £120. Answer: £120. The distractors: £40 is the value of one part rather than of a 3-part share; £80 is the 2-part share; £160 is the 4-part share, which is what a candidate reaches by picking the largest share instead of the one the question names.
- (c) Scalene — Method: equal angles in a triangle sit opposite equal sides, so compare the three angles with each other. Working: 10°, 58° and 112° are all different, so no two sides of the triangle are equal either, and a triangle with no equal sides is scalene. Answer: scalene. The distractors: isosceles is chosen by candidates who pair up the two acute angles, 10° and 58°, as base angles without checking that they are actually equal; equilateral is chosen by candidates who check that the three angles add to 180° and take that as meaning the triangle is regular; right-angled is chosen by candidates who see that 112° is larger than 90° and classify the triangle as containing a right angle, when in fact none of the three angles is 90°.
- (a) 2/3 — Method: 'in A or in B' means every score that belongs to at least one of the two sets; a score that belongs to both is still only one outcome, so it is listed once. Working: the even scores are 2, 4 and 6; the scores greater than 4 are 5 and 6. Listing the scores that appear in either set gives 2, 4, 5 and 6, with 6 written once. That is 4 of the 6 faces, or 4/6. Answer: the probability is 2/3. The distractors: 5/6 comes from adding the sizes of the two sets, 3 + 2, so that the score 6 is counted in both and appears twice; 1/2 comes from using the even scores alone; 1/6 comes from giving the probability that the score is in both sets, which is the single score 6, rather than in either of them.
- (d) 10π cm — Method: for a circle, circumference = π × diameter. Working: circumference = π × 10 = 10π cm. Answer: 10π cm. 20π cm comes from using the formula 2 × π × radius but plugging in the diameter as if it were the radius: 2 × π × 10 = 20π. 25π cm comes from confusing circumference with area, using π × radius² with radius 5: π × 5² = 25π. 5π cm comes from halving the diameter and then multiplying by π, as if circumference were π × radius: π × 5 = 5π.
- (d) w ≤ 630 — 'No more than 630 kg' means the weight can be exactly 630 kg or anything less, so the correct inequality is w ≤ 630, using 'less than or equal to' to include the limit itself. Writing w < 630 excludes 630 kg itself, as though the limit could not be reached exactly. Writing w ≥ 630 reverses the direction, describing a minimum weight rather than a maximum. Writing w > 630 both reverses the direction and excludes the boundary value. The inequality describing the lift's weight limit is w ≤ 630.
- (a) 1.5 km — Multiply the map length by the scale: 6 × 25 000 = 150 000 cm. Convert to kilometres: 150 000 cm = 1.5 km. Dividing by only 1000 instead of the full conversion when changing units gives 150 km, a hundred times too large. Misreading the scale as 1 : 2500 instead of 1 : 25 000 gives 6 × 2500 = 15 000 cm = 0.15 km, a hundred times too small. Leaving the answer as 150 000 without converting units at all, and calling it 150 000 km, mistakes centimetres for kilometres completely.
- (a) 750 — Method: to change kilograms to grams, multiply by 1000. Working: 0.75 × 1000 = 750. Answer: 750 g. A candidate who multiplies by 100 instead of 1000 gets 75. A candidate who multiplies by 10000 instead of 1000 gets 7500. A candidate who divides by 1000 instead of multiplying gets 0.00075.
- (d) 15:20 — Method: add all three journey segments to the departure time, converting the total number of minutes into hours and minutes. Working: 55 + 20 + 40 = 115 minutes = 1 hour 55 minutes. 13:25 + 1 hour 55 minutes = 15:20. Answer: 15:20. (15:00 comes from leaving out the 20 minute break and adding only 55 + 40 = 95 minutes. 14:25 comes from leaving out the 55 minute first leg and adding only 20 + 40 = 60 minutes. 14:40 comes from leaving out the 40 minute second leg and adding only 55 + 20 = 75 minutes.)
- (c) (6n + 4)/2 — Six full boxes hold 6 lots of n pencils, which is 6n, and the 4 loose pencils are added on, so the shop has 6n + 4 pencils altogether. Sharing them equally between 2 classes divides that whole total by 2, and brackets are what show that the division applies to all of it: (6n + 4)/2. Without the brackets, 6n + 4/2 halves only the loose pencils; 6(n + 4)/2 adds the loose pencils to every box before the division; 2(6n + 4) doubles the total instead of halving it.
- (b) 8:12 — Method: two ratios are equivalent when one is obtained from the other by multiplying, or dividing, both parts by the same number. Working: multiplying both parts of 2:3 by 4 gives 2 × 4 = 8 and 3 × 4 = 12, and the check runs the other way too, since the highest common factor of 8 and 12 is 4 and dividing both parts by 4 returns 2:3. Answer: 8:12. The distractors: 8:3 comes from multiplying only the first part by 4 and leaving the second part alone; 12:8 comes from multiplying both parts by 4 correctly but then writing the two parts the wrong way round; 4:5 comes from adding 2 to each part instead of multiplying, and adding the same amount to both parts changes the ratio.
- (a) 1/2, 3/5, 2/3, 5/6 — Convert all four fractions to a common denominator of 30: 2/3 is 20/30, 3/5 is 18/30, 5/6 is 25/30, and 1/2 is 15/30. Ordering by these numerators, smallest to largest, gives 15/30, 18/30, 20/30, 25/30, which is 1/2, 3/5, 2/3, 5/6. Ordering by the size of the numerator in the original fractions, 1, 2, 3, 5, rather than converting to a common denominator, gives the wrong order 1/2, 2/3, 3/5, 5/6, because it ignores that the denominators are different. Ordering largest to smallest instead of smallest to largest, as the question asks, gives 5/6, 2/3, 3/5, 1/2. Using the rule "the bigger the denominator, the smaller the fraction" to place the last two, so that 5/6 is put below 2/3 because 6 is bigger than 3, gives 1/2, 3/5, 5/6, 2/3 — that rule only holds when the numerators are the same, and here 20/30 really is smaller than 25/30. So the correct order, smallest to largest, is 1/2, 3/5, 2/3, 5/6.
- (c) They are in direct proportion, because y ÷ x = 2.5 for both pairs. — Testing direct proportion means checking that y ÷ x is the same for every pair: 15 ÷ 6 = 2.5 and 25 ÷ 10 = 2.5, so the quantities are in direct proportion. Saying they are not in proportion because x + y differs uses addition, which is not the correct test for proportion. Saying they are not in proportion because y − x differs also uses the wrong test — subtraction, not division. Saying they are in proportion because x × y is 90 and 250 uses multiplication, which is the test for inverse proportion, and the two products are not even equal to each other, so this option also contradicts itself.
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.