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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- (a) 11.5 ≤ L < 12.5 — Rounding to the nearest centimetre means L can be up to half a centimetre below or above 12 before it would round to a different whole number. Half of 1 cm is 0.5 cm, so the lower bound is 12 − 0.5 = 11.5 and the upper bound is 12 + 0.5 = 12.5. A value exactly at the upper bound, 12.5, would round up to 13, not 12, so 12.5 itself is excluded, giving 11.5 ≤ L < 12.5. Writing 11.5 ≤ L ≤ 12.5 wrongly includes 12.5 on both ends. Writing 11 ≤ L < 13 uses a whole centimetre either side instead of half a centimetre. Writing 11.5 < L < 12.5 wrongly excludes the lower bound, which is a value that does round to 12.
- (d) The graph never crosses either axis — Since x ≠ 0, there is no point on the graph where x = 0, so it cannot cross the y-axis; likewise 1/x is never equal to 0 for any x, so it cannot cross the x-axis either — the graph never touches either axis. A candidate who forgets the restriction x ≠ 0 might think the graph behaves like other graphs and passes through the origin, (0, 0). A candidate who correctly rules out the x-axis but forgets that x = 0 is also excluded might say the graph crosses the y-axis but never the x-axis. A candidate who only pictures the branch where x and y are both positive might say the graph has only one branch, in quadrant 1, forgetting the second branch where x and y are both negative.
- (a) The hourly rate is £5 and the fixed fee is £7 — The gradient is (27 − 12) ÷ (4 − 1) = 15 ÷ 3 = £5, the hourly rate. Using the point (1, 12): 12 = 5 × 1 + fee, so the fee is 12 − 5 = £7. That gives 'The hourly rate is £5 and the fixed fee is £7'. Swapping the two figures gives the statement with £7 as the rate and £5 as the fee, which has them the wrong way round. Taking the C-value of the first point, £12, as the fixed fee ignores that 1 hour of hire is already included in that £12. Using 15, the change in C, as the hourly rate without dividing by the change in h (3 hours) gives the statement claiming a £15 hourly rate.
- (d) Yes, by the AA condition — Method: similarity is decided by the angles, and because the three angles of a triangle add up to 180°, two matching pairs force the third pair to match as well. Working: the angles at A and D are both 40° and the angles at B and E are both 70°, so the third angles are both 180° − 40° − 70° = 70° and all three pairs are equal. Two pairs were enough, and that is the AA condition. Answer: yes, by the AA condition. The distractors: 'Yes, by the SSS condition' names a condition about three pairs of sides in proportion, and the question gives no side lengths at all; 'No, because no side lengths are given' treats side information as necessary, which it is for congruence but not for similarity; 'No, because the triangles may be different sizes' turns the definition of similarity into an objection, since similar figures are allowed to differ in size and only their shape must match.
- (d) 1/15 — Method: a probability read from a frequency tree is the count at the end of the branch you want, divided by the total number in the whole experiment. Working: the branch for walking followed by the branch for being late ends with 6 pupils, and the experiment covers all 90 pupils, so the probability is 6/90. Dividing the top and the bottom by 6 gives 1/15. Answer: the probability is 1/15. The distractors: 3/25 is 6/50 and comes from dividing the 6 by the 50 walkers rather than by the whole group, which answers a different question about walkers only; 8/45 is 16/90 and comes from counting every late pupil, the 6 walkers and the 10 others together, instead of only the late walkers; 1/9 is 10/90 and comes from reading the late count on the branch for pupils who do not walk.
- (d) 6 — Method: the mode, or modal value, is the value that occurs most often in the data set, and it is a value from the data rather than a count. Working: size 4 occurs twice, size 5 occurs once, size 6 occurs three times and size 9 occurs once, so the highest frequency is three and the size it belongs to is 6. Answer: 6. The distractors: 3 comes from writing down the frequency of the most common size instead of the size itself; 9 comes from picking the largest size in the list, which confuses the mode with the maximum; 4 comes from stopping at the first size that repeats rather than checking which size repeats most often.
- (c) −5 — Method: the bracket is worked out first, then the multiplication and the division, which stand as separate parts, and the subtraction that joins them is carried out last; subtracting a negative is the same as adding. Working: (−3) + 5 = 2, so the product is 2 × (−4) = −8; the division gives (−6) ÷ 2 = −3; joining them gives −8 − (−3) = −8 + 3 = −5. Answer: −5. The distractors: −11 comes from taking away 3 instead of taking away −3, giving −8 − 3 = −11; −1 comes from working from left to right once the bracket is done, giving −8 − (−6) = −2 and then −2 ÷ 2 = −1; 11 comes from treating the first product as positive because it was worked out from a bracket, giving 8 − (−3) = 11.
- (c) 4 — Method: subtract 3 from both sides, then divide by 4. Working: 4x = 19 − 3 = 16, so x = 16 ÷ 4 = 4. Answer: 4. 4.75 comes from dividing 19 by 4 directly, without subtracting 3 first. 3.75 comes from subtracting 4 instead of 3 before dividing. 5.5 comes from a sign error, adding 3 to 19 instead of subtracting it.
- (d) 1 : 1 — Sugar and butter together make 3 + 5 = 8 parts of the mixture. Comparing flour to this, 8 : 8, simplifies to 1 : 1. Giving 1 : 2 compares flour with the whole mixture (8 + 3 + 5 = 16 parts, giving 8 : 16 = 1 : 2) instead of with the rest of the mixture. Giving 3 : 5 is the ratio of sugar to butter, not of flour to the rest of the mixture. Giving 8 : 3 compares flour only with sugar, leaving butter out altogether.
- (b) 28 — Method: total bricks = (number of squares in the footprint) × (the height of the wall in bricks). Working: the footprint has the row of 5 squares plus the 2 squares in the arm that stands out from the middle of that row, and they do not overlap, giving 5 + 2 = 7 squares; multiplying by the height of 4 bricks gives 7 × 4 = 28. Answer: 28. The distractors: 20 comes from using only the row of 5 and ignoring the arm (5 × 4). 24 comes from treating the bottom square of the arm as if it were one of the row's own squares, so the arm is counted as adding only 1 new square instead of 2 (5 + 1 = 6, then 6 × 4). 32 comes from counting the square of the row directly below the arm a second time as part of the arm (5 + 3 = 8, then 8 × 4).
- (c) 0.15 — Method: for two independent events, multiply along the branches of the tree to find the probability of both outcomes happening together. Working: P(red and heads) = P(red) × P(heads) = 0.3 × 0.5 = 0.15. Answer: 0.15. Watch out: adding the two probabilities, 0.3 + 0.5 = 0.8, does not give the probability of both — probabilities along one path of a tree are multiplied, not added. Writing down 0.5 ignores the spinner altogether and gives only the coin's probability. And writing down 0.65 is the probability of red OR heads, which is 0.3 + 0.5 − 0.15 = 0.65, a different question from the one asked here.
- (c) 32 — Method: a power tells you how many times to multiply the base by itself. Working: 2⁵ = 2 × 2 × 2 × 2 × 2 = 32. Answer: 32. (10 comes from multiplying the base by the power, 2 × 5, instead of using repeated multiplication. 25 comes from swapping the base and the power and working out 5² instead. 16 comes from using one fewer 2, effectively working out 2⁴.)
- (d) x = 2 — Method: expand the bracket by multiplying both terms inside it by 2, then undo the addition and the multiplication in turn. Working: expanding gives 2x + 6 = 10; subtracting 6 from both sides gives 2x = 4; dividing both sides by 2 gives x = 2. Answer: x = 2. The distractors: x = 5 comes from dividing both sides by 2 first, reaching x + 3 = 5 and writing 5 without taking the 3 away; x = 8 comes from adding 6 to both sides instead of subtracting it, giving 2x = 16; x = 3.5 comes from expanding 2(x + 3) as 2x + 3, multiplying only the x by the 2, which leads to 2x = 7.
- (d) 20 km/h — Method: average speed = total distance ÷ total time, with the time written in hours. Working: 1 hour 30 minutes = 1.5 hours, and 30 ÷ 1.5 = 20. Answer: 20 km/h. The distractors: 45 km/h comes from multiplying 30 by 1.5 instead of dividing; 15 km/h comes from dividing by 2, as if the ride had taken 2 hours; 30 km/h comes from dividing by the whole hour only and ignoring the extra 30 minutes.
- (c) AB ⊥ CD — Method: match the description of the lines to the correct symbol. Working: the symbol ⊥ means 'is perpendicular to', used when two lines meet at right angles, which is exactly what is described. Options: ∥ means 'is parallel to', for lines that never meet, so it does not apply here; = compares two lengths as equal, but no lengths are mentioned; ≅ means 'is congruent to', used for whole shapes, not for describing how two lines cross. Answer: AB ⊥ CD.
- (c) 2.00 kg — Convert the tin's mass to kilograms first: 650 g = 0.65 kg. Adding this to the bag's mass gives 0.65 + 1.35 = 2.00 kg. Converting 650 g to kilograms by dividing by 100 instead of 1000 gives 6.5 kg, and adding this to 1.35 kg gives 7.85 kg. Adding the two masses without converting grams to kilograms at all — treating 650 as if it were already measured in kilograms — gives 651.35 kg. Subtracting the tin's mass from the bag's mass instead of adding the two together, 1.35 − 0.65, gives 0.70 kg.
- (d) The roots of x² − 6x + 5 = 0 are x = 1 and x = 5 (since it factorises to (x − 1)(x − 5)), so by symmetry the turning point has x-coordinate 3, not 6. — Factorising, x² − 6x + 5 = (x − 1)(x − 5), so the roots are x = 1 and x = 5. The turning point lies midway between the roots by symmetry: (1 + 5) ÷ 2 = 3. The coefficient of x has no direct role in locating the turning point this way. The option giving −6 makes an arbitrary sign change with no mathematical basis. The option giving 5 wrongly takes just one of the two roots instead of their midpoint.
- (b) 5/7 — The enlargement multiplier is 1.4, which as a fraction is 7/5. To reverse an enlargement, use the reciprocal of the multiplier: flip 7/5 to get 5/7. 7/5 comes from using the enlargement multiplier again, instead of reversing it. 3/5 comes from treating the reverse as 'give back the extra amount', working out 1 − (1.4 − 1) = 0.6, instead of using the reciprocal. 5/2 comes from ignoring the whole number in 1.4 and inverting only the decimal part, 0.4, as if it were the whole multiplier.
- (a) 0.35, 3/8, 0.4 — Method: convert the fraction to a decimal so all three values can be compared directly. Working: 3/8 = 0.375, so the three values are 0.35, 0.375 and 0.4. In order from smallest to largest: 0.35, 3/8 (0.375), 0.4. Answer: 0.35, 3/8, 0.4. "3/8, 0.35, 0.4" comes from assuming a fraction must be smaller than any decimal, without converting it first. "0.4, 3/8, 0.35" comes from writing the numbers in reverse order. "0.35, 0.4, 3/8" comes from converting 3/8 by flipping it to 8/3, making it seem larger than both decimals.
- (c) 1.2 — Add the parts of the ratio: 6 + 1 = 7. Divide the total amount by the number of parts: 8.4 ÷ 7 = 1.2 litres, which is the value of one part and also the amount of syrup, since syrup is 1 part. (7.2 litres is the amount of water, using 6 parts instead of 1. 1.4 comes from dividing 8.4 by 6 — the water's part of the ratio — instead of dividing by the total number of parts, 7. 0.84 comes from dividing 8.4 by 10 instead of by 7.)
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.