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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- (d) 12 — Method: if the bracelets are identical and no beads are left over, the number of bracelets must divide exactly into both totals, so it is the highest common factor of 24 and 36. Working: 24 = 2³ × 3 and 36 = 2² × 3²; taking the lower index of each shared prime gives 2² × 3 = 4 × 3 = 12. Each bracelet then has 2 red beads and 3 blue beads. Answer: 12. The distractors: 6 comes from taking each shared prime once rather than at its lower index, giving 2 × 3, which is a common factor but not the highest; 72 is the lowest common multiple of 24 and 36, from taking the higher index of each prime instead of the lower; 60 comes from adding the two bead totals instead of looking for a common factor.
- (b) (0, −2) — Method: a translation acts on the two coordinates separately: moving right or left changes the x-coordinate only, moving up or down changes the y-coordinate only, and right and up add while left and down subtract. Working: the point starts at (−4, 7); moving 4 units to the right gives an x-coordinate of −4 + 4 = 0; moving 9 units down gives a y-coordinate of 7 − 9 = −2. Answer: (0, −2). The distractors: (5, 3) comes from pairing each number with the wrong coordinate, adding 9 to −4 and taking 4 from 7; (0, 16) comes from treating 'down' as an addition, giving 7 + 9 = 16 for the second coordinate; (−8, −2) comes from treating 'to the right' as a subtraction, giving −4 − 4 = −8 for the first coordinate.
- (a) 600 cm — Method: a scale of 1 : n means each 1 cm on the plan stands for n cm in real life, so multiply the plan length by the scale. Working: 3 × 200 = 600. Answer: 600 cm. The distractors: 60 cm comes from multiplying by 20 instead of 200; 203 cm comes from adding the scale to the plan length instead of multiplying; 6000 cm comes from reading the scale as 1 : 2000.
- (d) 600 cm² — Enlarging by scale factor 2 makes the new dimensions 10 × 2 = 20 cm and 15 × 2 = 30 cm, so the poster's area = 20 × 30 = 600 cm². A pupil who scales the original area, 150 cm², by the scale factor itself instead of by its square gets 150 × 2 = 300 cm². A pupil who adds the scale factor to each dimension instead of multiplying gets (10 + 2) × (15 + 2) = 204 cm². A pupil who forgets to enlarge the postcard at all just uses the original area, 150 cm². The correct area of the poster is 600 cm².
- (b) 1/2 — Method: for equally likely outcomes the probability is the number of favourable outcomes divided by the total number of outcomes, written in its simplest form. Working: the dice can show 1, 2, 3, 4, 5 or 6, so there are 6 equally likely scores; the even scores are 2, 4 and 6, which is 3 of them, giving 3/6. Answer: 3/6 cancels to 1/2, which sits halfway along the 0 to 1 probability scale. The distractors: 1/3 comes from listing only 2 and 4 as the even scores and writing 2/6; 1/6 comes from giving the probability of one particular even score, such as the 2, rather than any even score; 2/3 comes from listing the even numbers as 0, 2, 4 and 6 and writing 4/6, forgetting that a dice has no 0 on it.
- (b) 20 — Method: each sector of a pie chart stands for a percentage of the whole set of data, so the number of pupils in a sector is that percentage of the total number of pupils. Working: the football sector is 50% and the whole pie chart stands for all 40 pupils, so the number of pupils is 50% of 40 = 0.5 × 40 = 20. Answer: 20 pupils, which is a count of pupils rather than a percentage. The distractors: 10 comes from taking the netball sector, 25% of 40, instead of the football sector; 80 comes from doubling the 40 pupils instead of halving them, which is what multiplying by 50% would do the wrong way round; 40 comes from writing down the total number of pupils in the class and never applying the percentage at all.
- (d) 3/5 — Method: a decimal with one digit after the point is a number of tenths, so it is written over 10 and then cancelled. Working: 0.6 is six tenths, so 0.6 = 6/10; the highest common factor of 6 and 10 is 2, and 6 ÷ 2 = 3 with 10 ÷ 2 = 5. Answer: 3/5. The distractors: 2/3 comes from confusing 0.6 with the recurring decimal 0.666..., which is the one that equals 2/3; 1/6 comes from putting 1 over the single digit after the point; 3/50 comes from using hundredths for a one-place decimal, giving 6/100, which then cancels by 2 to 3/50.
- (d) −2 — Method: run the machine backwards, undoing the operations in the opposite order and swapping each one for its inverse. Working: the machine added 7 last, so take 7 off the output: 1 − 7 = −6; before that the machine had multiplied by 3, so divide: −6 ÷ 3, and a negative divided by a positive stays negative. Answer: −2, which checks because 3 × (−2) + 7 = −6 + 7 = 1. The distractors: 2 comes from dividing 6 by 3 and losing the minus sign; −6 comes from taking the 7 off and stopping there, never undoing the multiplication; −18 comes from multiplying −6 by 3 instead of dividing by 3.
- (d) 40 cm — Method: convert the real length to centimetres, then divide by the scale factor. Working: 10 m = 1000 cm. 1000 ÷ 25 = 40 cm. Wrong options: 0.4 cm comes from dividing 10 by 25 without converting metres to centimetres first; 25000 cm comes from multiplying instead of dividing (1000 × 25); 4 cm comes from converting 10 m to 100 cm (using ×10 instead of ×100) before dividing.
- (d) (6, 2) — Applying the first vector: (2, 1) + (5, −3) = (7, −2), which is the warehouse. Applying the second vector: (7, −2) + (−1, 4) = (6, 2), the delivery address. '(7, −2)' stops at the warehouse and forgets the second flight. '(8, −6)' comes from adding (1, −4) instead of (−1, 4) for the second vector, getting both signs wrong. '(11, −3)' comes from swapping the components of the second vector to (4, −1) before adding.
- (c) 3/8 — Method: write out every result of the three coins as a string of three letters, H for heads and T for tails, count the results that match the description and divide by how many results the list holds. Working: each coin lands two ways and no coin affects another, so the list holds 2 × 2 × 2 = 8 equally likely results. Exactly two heads means one coin lands on tails and the other two on heads, so the results are HHT, HTH and THH — 3 of the 8. Answer: the probability is 3/8. The distractors: 4/8 comes from reading 'exactly two heads' as 'at least two heads' and counting HHH as well; 2/8 comes from a list made without a system, in which HHT and THH are written down and HTH, the result with the tail between the two heads, is missed; 6/8 comes from counting 3 × 2 = 6 ways of picking which two of the three coins show heads, which counts every pair of coins twice, once in each order.
- (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.
- (a) 5p + 7q — Tom's total spend combines like terms: (2p + 6q) + (3p + q) = 5p + 7q, adding the notebook terms (2p + 3p = 5p) and the pencil terms (6q + q = 7q) separately. Answering 12pq adds every coefficient together (2 + 6 + 3 + 1 = 12) and multiplies the letters, combining unlike terms as though notebooks and pencils were the same item. Answering 5p + 6q correctly combines the notebook terms but forgets to add Tuesday's extra pencil to the 6q. Answering 2p + 7q correctly combines the pencil terms but forgets to add Tuesday's 3 extra notebooks to the 2p. The total amount Tom has spent is 5p + 7q pence.
- (c) 16 cm — Corresponding sides of similar triangles are all in the same ratio. Use the pair whose lengths are both known: the scale factor from triangle ABC to triangle PQR is 12 ÷ 6 = 2. Since QR corresponds to BC, multiply BC by that scale factor: 8 × 2 = 16, so QR = 16 cm.
- (b) 31.4 cm — Circumference = π × diameter, so 3.14 × 10 = 31.4 cm. 15.7 cm comes from using the radius, 5 cm, in place of the diameter: 3.14 × 5 = 15.7, which is only half the circumference. 78.5 cm comes from using the area formula π × radius² instead of the circumference formula: 3.14 × 5² = 3.14 × 25 = 78.5. 62.8 cm comes from keeping the 2 from the radius form of the formula, C = 2 × π × radius, but putting the full diameter into it: 2 × 3.14 × 10 = 62.8.
- (a) 2, 3, 4, 5 — Method: work out which whole numbers satisfy both parts of the inequality. Working: n ≥ 2 means n can be 2 or more; n < 6 means n must be less than 6, so 6 itself is not included. The whole numbers that fit both conditions are 2, 3, 4 and 5. Answer: 2, 3, 4, 5. 2, 3, 4, 5, 6 treats < 6 as ≤ 6 and wrongly includes 6. 3, 4, 5 treats ≥ 2 as > 2 and wrongly leaves out 2. 1, 2, 3, 4, 5 wrongly includes 1, which does not satisfy n ≥ 2.
- (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.
- (d) 1200 cm — Method: multiply the drawing length by the scale factor. Working: 4 × 300 = 1200 cm. Wrong options: 0.013 cm comes from dividing instead of multiplying (4 ÷ 300); 304 cm comes from adding the scale factor to the drawing length instead of multiplying; 600 cm comes from using half the scale factor (150) instead of 300.
- (b) 12 — Each of the 3 soups can be paired with each of the 4 bread rolls, so multiply: 3 × 4 = 12. 7 comes from adding the two numbers instead of multiplying them. 3 comes from using only the number of soups and ignoring the bread rolls. 4 comes from using only the number of bread rolls and ignoring the soups.
- (b) 4 : 25 — For similar shapes, the ratio of areas is the ratio of lengths squared: 2² : 5² = 4 : 25. 2 : 5 comes from using the perimeter ratio itself as the area ratio, without squaring it at all. 8 : 125 comes from cubing each part instead of squaring (2³ : 5³) — cubing is the rule for volume, not area. 4 : 5 comes from squaring only the first part of the ratio (2² = 4), and leaving the second part unsquared.
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.