Sample paper · GCSE Foundation · grades 1–5
GCSE Foundation sample Paper 1 (non-calculator)
The real Paper 1 is 1.5 hour 30 minutes and 80 marks, non-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 no calculator-only items. AO1 / AO2 / AO3 at this tier: 50% / 25% / 25%.
Non-calculatorGCSE Foundation
Answer key: GCSE Foundation sample Paper 1 (non-calculator)
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- (b) 5/9 — Add the parts of the ratio to find how many equal shares make up the whole: 4 + 5 = 9 shares. The blue counters take 5 of those 9 shares, so 5/9 of the counters are blue. 4/9 is the fraction that are red, 4/5 is the ratio copied straight down as a fraction, and 5/4 compares the blue counters with the red counters instead of with the whole box.
- (b) Week 7 — The height after n weeks is 4 + 5(n − 1) = 5n − 1. The plant is taller than 30 cm when its height, 5n − 1, is more than 30, written 5n − 1 > 30; add 1 to both sides (5n > 31) and divide by 5, which gives n > 6.2, so the first whole week is week 7, where the height is 5 × 7 − 1 = 34 cm. Rounding 6.2 down to week 6 without checking gives a height of only 5 × 6 − 1 = 29 cm, which is not yet taller than 30 cm. Counting one week too many gives week 8. Working out 30 ÷ 5 = 6 and then subtracting 1 from the week number instead of from the height gives week 5.
- (b) 5 litres — Method: adding water changes the total volume but adds no salt, so work out the volume of salt, then the total volume that makes that salt 20% of the mixture, then the extra water. Working: 30% of 10 litres is 0.3 × 10 = 3 litres of salt. For the same 3 litres to be 20% of the new mixture, the new total volume is 3 ÷ 0.2 = 15 litres. The water added is the extra volume, 15 − 10 = 5 litres. Answer: 5 litres. The distractors: 3 litres is the volume of salt in the solution, which is the first step and not what the question asks for; 15 litres is the total volume of the new mixture, which counts the 10 litres already in the container as water that was poured in; 2 litres comes from taking 20% of the original 10 litres, applying the new percentage to the old volume instead of to the new one.
- (b) (3, −1) — Method: find the distance from the point to the mirror line, then place the image the same distance on the other side of the line. Working: B is 5 − 2 = 3 units above the line y = 2, so its image is 3 units below the line, at y = 2 − 3 = −1, giving (3, −1); the x-coordinate does not change, since the mirror line is horizontal. Options: (3, −5) comes from reflecting in the x-axis (y = 0) instead of the line y = 2; (−3, 5) comes from reflecting in the y-axis instead of a horizontal line; (3, −4) comes from doubling the distance from the line instead of reflecting it, giving 2 − 2×3 = −4. Answer: (3, −1).
- (a) 80 — Method: list the outcomes that count as a success, write the probability from them, then multiply by the number of rolls. Working: the scores of 3 or more are 3, 4, 5 and 6, which is 4 of the 6 equally likely scores, so the probability is 4/6, which cancels to 2/3. Over 120 rolls the expected number is 120 × 2 ÷ 3 = 80. Answer: about 80 of the rolls would be expected to give 3 or more. The distractors: 60 comes from reading a score of 3 or more as a score above 3 and counting only 4, 5 and 6, giving 120 × 3 ÷ 6 = 60; 40 is the expected number of rolls that are not 3 or more, 120 × 2 ÷ 6 = 40; 20 is 120 ÷ 6 and is the expected count for one single score.
- (d) Only internet users reach the website; others are excluded. — Method: a sample is biased when it systematically leaves out part of the population, or systematically over-represents another part. Working: anyone without internet access, or who does not visit the council's website, has NO chance of being included — the sample is drawn only from internet-using residents, which is not the whole town. Saying too many people might respond because the survey is free confuses bias with sample size — bias is about who CAN be reached, not how many respond. Saying people might lie describes a different problem, response honesty, not who was sampled in the first place. Saying online surveys cannot be anonymous is not a reason connected to bias at all. A sample is biased when part of the population has no chance of being included, whatever the reason for that.
- (d) 28 — Method: round each number to the nearest whole number, then multiply the rounded numbers to get an estimate that can be compared with the assistant's answer. Working: 7.2 rounds to 7, and 3.9 rounds to 4, so the estimate is 7 × 4 = 28. Since 28 is much smaller than 56.16, the assistant's answer cannot be correct. 56 comes from rounding the assistant's answer to the nearest whole number, instead of rounding the two numbers being multiplied and then multiplying them. 35 comes from rounding both numbers correctly but then slipping in the seven times table, writing 7 × 5 = 35 in place of 7 × 4 = 28. 21 comes from rounding 3.9 down to 3 instead of 4, giving 7 × 3 = 21. Answer: 28.
- (c) −8 — The nth term is the first term plus (n − 1) lots of the common difference: 40 + 8 × (−6) = 40 − 48 = −8. A candidate who uses 9 lots of the common difference instead of 8 gets 40 + 9 × (−6) = −14. A candidate who treats the common difference as +6 instead of −6 gets 40 + 8 × 6 = 88. A candidate who uses only 7 lots of the common difference gets 40 + 7 × (−6) = −2.
- (d) 5 — The gradient equals the amount gained divided by the time taken: 15 ÷ 3 = 5 litres per minute.
- (a) the lens-shaped region where the two circles overlap — The region within 30 km of the ship is a circle of radius 30 km around the ship; the region within 20 km of the lighthouse is a circle of radius 20 km around the lighthouse. Since 30 + 20 = 50 km is more than the 40 km between them, and 30 − 20 = 10 km is less than 40 km, the two circles genuinely overlap, in a lens-shaped region where both conditions hold at once. "the whole smaller circle, since it fits inside the larger one" would only be true if the 40 km distance between the centres were small enough for one circle to swallow the other, which it is not here. "no overlap — 30 km and 20 km don't add to 40 km" wrongly assumes the two radii must add up to exactly the distance between the centres for an overlap to exist; any total greater than that distance is enough. "a straight strip, exactly halfway between the two" confuses this with an equidistant locus, which is a different problem from the one asked here.
- (c) 9/25 — The frequency tree already shows the swim-and-cycle branch directly: 54 out of 150, which simplifies to 9/25. Adding both cycling branches together, 54 + 18 = 72, gives the total number of cyclists, so 72/150 = 12/25, not just those who also swim. Using the non-swimmers' cycling figure, 18, gives 18/150 = 3/25, the probability of cycling WITHOUT swimming. Using the swimmers' total of 90, before splitting by cycling, gives 90/150 = 3/5, the probability of swimming on its own.
- (d) 10 — Method: list the pairs systematically, taking each sweet in turn and pairing it only with the sweets that come after it, so that no pair is written down twice. Working: numbering the sweets 1 to 5, the first sweet pairs with 4 others, the second pairs with 3 sweets that come after it, the third with 2 and the fourth with 1, so the total is 4 + 3 + 2 + 1 = 10. Answer: 10. The distractors: 20 comes from working out 5 × 4 and never halving, which counts each pair twice, once in each order; 25 comes from working out 5 × 5, which allows the same sweet to be chosen twice; 9 comes from adding the 5 choices and the 4 remaining choices instead of combining them as a selection of two.
- (c) The 5 must multiply both terms inside the bracket, so 5(x + 2) expands to 5x + 10, which is never equal to 5x + 2 for any value of x. — Expanding the bracket correctly, 5(x + 2) = 5x + 10, since the 5 multiplies both the x and the 2. This is never equal to 5x + 2, since that would require 10 = 2. The option claiming 5(x + 2) means 5 × x + 2 ignores that the 5 must multiply the whole bracket, not just the x-term. The option about working out the bracket first with a value of x misunderstands algebraic expansion, which holds for every x, not just specific ones. The option about addition before multiplication misapplies the order of operations to bracket expansion, which always distributes the outer factor over every term inside, whatever x is.
- (a) 0.6 litres — The ratio 1 : 9 means the solution has 1 + 9 = 10 equal parts in total. Each part is 6 ÷ 10 = 0.6 litres, and disinfectant is 1 part, so Priti needs 0.6 litres of disinfectant. Giving 0.667 litres divides by 9, the number of parts of water, instead of the total number of parts, 10 (6 ÷ 9 ≈ 0.667). Giving 6 litres is the total amount of solution, not just the disinfectant's share of it. Giving 5.4 litres works out the water's share (6 × 9 ÷ 10 = 5.4), not the disinfectant's.
- (a) (5, 2) — Reflecting in the x-axis keeps the x-coordinate the same and changes the sign of the y-coordinate, so (5, −2) maps to (5, 2). (−5, −2) changes the sign of the x-coordinate instead, which is what happens when reflecting in the y-axis. (−5, 2) changes the sign of both coordinates, which is the result of a rotation of 180° about the origin, not a reflection in the x-axis. (2, 5) comes from dropping the minus sign and then swapping the two numbers round, so neither coordinate has actually been reflected.
- (a) 12 — Compare the powers of each prime that appears in both factorisations. In 2² × 3² and 2² × 3 × 7, the prime 2 appears with power 2 in both, and the prime 3 appears with power 2 in one and only power 1 in the other — take the lower power, 3¹. Multiplying the shared primes at their lower powers, 2² × 3, gives 12. Using power 1 for both primes instead of comparing the powers properly, 2 × 3, gives 6, which misses that 2 is common at power 2, not power 1. Multiplying the primes at their higher powers and including 7, which only appears in 84, gives 2² × 3² × 7, which comes to 252 — this is the lowest common multiple, not the highest common factor. Only spotting that 3 is a common prime and overlooking that 2 is common as well gives 3. So the highest common factor of 36 and 84 is 12.
- (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.
- (c) 120 km/h — Method: for a fixed distance the average speed multiplied by the time is constant, and that constant is the distance, so divide the distance by the new time. Working: speed × time = 240, so in 2 hours the speed needed is 240 ÷ 2 = 120 km/h. Answer: 120 km/h. The distractors: 80 km/h is the average speed of the original journey, 240 ÷ 3, which answers for the 3-hour timing rather than the 2-hour one; 160 km/h comes from halving the 3 hours to 1.5 hours and working out 240 ÷ 1.5, instead of using the 2 hours the question gives; 480 km/h comes from multiplying the distance by the 2 hours rather than dividing by it.
- (c) 3:4 — The jacket costs 3/7 of £84, which is 3 × (84 ÷ 7) = 3 × 12 = £36. The bag then costs the rest of the money, £84 − £36 = £48. The ratio of the jacket to the bag is 36:48, which simplifies to 3:4. Writing the fraction spent on the jacket, 3/7, directly as the ratio, without working out that the bag's share is the remaining 4/7, gives 3:7. Giving the ratio the wrong way round, bag to jacket instead of jacket to bag, gives 4:3. Assuming the jacket and bag cost the same, ignoring the fraction given, gives 1:1.
- (c) 9:15:21 — 6 : 10 : 14 simplifies to 3 : 5 : 7 (divide every part by 2). Multiplying every part of 3 : 5 : 7 by 3 gives 9 : 15 : 21, so 9 : 15 : 21 is equivalent to 6 : 10 : 14. Adding 2 to every part of 6 : 10 : 14 gives 8 : 12 : 16, which is not equivalent — ratios are equivalent when every part is multiplied by the same number, not when the same number is added to every part. Doubling only the first two parts, 6 × 2 = 12 and 10 × 2 = 20, but leaving the third part unchanged at 14, gives 12 : 20 : 14 — a scaling applied to two parts and not the third. Cancelling the first two parts correctly, 6 ÷ 2 = 3 and 10 ÷ 2 = 5, then treating the three numbers as a sequence and making the third part the sum of the first two, 3 + 5 = 8, gives 3 : 5 : 8 — the third part was never divided by 2 at all.
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.