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
GCSE Foundation sample Paper 1 (non-calculator)
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- 1.In a school long jump competition Priya's jump is recorded as 3.8 m and Nadia's jump is recorded as 3.9 m, each correct to the nearest 0.1 m. Priya says she may have jumped further than Nadia. Decide whether Priya is right, and give a reason for your answer.
- 2.A cinema sells adult tickets and child tickets. On one evening, 40 tickets are sold in total. There are 8 more adult tickets sold than child tickets. Work out the number of child tickets sold.
- 3.Which of these ratios is equivalent to 5 : 4?
- 4.A roof has two triangular sections, P and Q. Section P has a sloping edge of 3.6 m and a base edge of 2.4 m, with an angle of 70° between them. Section Q has a sloping edge of 3.6 m and a base edge of 2.4 m, with an angle of 70° between them, arranged the same way round. A roofer wants to cut identical triangular felt panels for both sections without measuring Section Q separately. Which condition confirms that the two sections are congruent?
- 5.A box contains 1 white bead, 4 purple beads and 4 red beads. One bead is taken at random from the box. Work out the probability that the bead is white or purple.
- 6.On a scatter graph of the arm span and the height of some pupils, the points rise from left to right. Write down the type of correlation this shows.
- 7.A fabric is dyed blue and yellow in the ratio 3 : 5. A tailor uses 1.5 m of blue fabric and the matching amount of yellow fabric needed for the ratio. Work out the total length of fabric used.
- 8.Look at the statement 2x + 5 = 17. Considering whether it is an expression, an equation, a formula, or an identity, which classification is correct?
- 9.The table shows values of x and y. x: 3, 5, 8. y: 12, 20, 32. Which statement correctly describes the relationship between x and y?
- 10.The column vector r has top number 2 and bottom number 6. Which of these column vectors is parallel to r?
- 11.The probability that Ben catches his bus on time is 0.8. The probability that Freya catches her bus on time is 0.5. The two events are independent. Work out the probability that both Ben and Freya catch their bus on time.
- 12.Simplify 2³ × 2⁴, giving your answer as a single power of 2.
- 13.The first five terms of a quadratic sequence are 5, 8, 13, 20, 29. Work out the next term in the sequence.
- 14.A blue ribbon is 45 cm long. A red ribbon is 30 cm long. Write the length of the blue ribbon as a fraction of the length of the red ribbon. Give your answer in its simplest form.
- 15.A hexagon has interior angles of 130°, 142°, 125°, 150°, 135° and x°. Work out the value of x.
- 16.Work out √25 + 4² − 12 ÷ 3
- 17.Make x the subject of the formula y = 4x − 3.y = 4x − 3
- 18.Noah runs 2 km in 10 minutes. Work out how long he takes to run 5 km at the same speed.
- 19.Write 0.0038 in standard form.
- 20.Work out 25% of 200.
Answer key
- (b) No — their possible jump lengths do not overlap — Method: each recorded jump stands for the lengths within half of 0.1 m, that is 0.05 m, of the figure recorded, and Priya is right only if the two ranges overlap. Working: Priya's jump is at least 3.8 − 0.05 = 3.75 m and below 3.85 m, because a jump of 3.85 m would have been recorded as 3.9 m; Nadia's jump is at least 3.85 m and below 3.9 + 0.05 = 3.95 m. Every length Priya could have jumped is below 3.85 m and every length Nadia could have jumped is at least 3.85 m, so Nadia jumped further whatever the exact lengths were. Answer: No — their possible jump lengths do not overlap. The distractors: the reason that a recorded jump is exactly the length jumped reaches the same verdict by treating a rounded record as exact, which is the idea this question tests; both jumps being 3.85 m would put 3.85 m inside Priya's range, when a jump of that length is recorded as 3.9 m; Priya jumping up to 3.9 m goes a whole 0.1 m above her record instead of half of it.
- (c) 16 — Let a be the number of adult tickets and c the number of child tickets: a + c = 40 and a = c + 8. Substituting the second equation into the first: (c + 8) + c = 40, so 2c = 32, and c = 16. A candidate who reports the number of adult tickets instead of child tickets would get a = 24. A candidate who splits the 40 tickets evenly, ignoring the difference of 8, would get 20. A candidate who forgets to divide 2c = 32 by 2 would write c = 32.
- (a) 15 : 12 — Multiply both parts of the ratio 5 : 4 by the same number, 3, to get an equivalent ratio: 5 × 3 = 15 and 4 × 3 = 12, giving 15 : 12. Giving 15 : 16 multiplies the two parts by different scale factors (×3 and ×4), which changes the ratio. Giving 9 : 8 adds 4 to each part instead of multiplying, which also changes the ratio. Giving 4 : 5 swaps the order of the two parts.
- (d) SAS — two sides, included angle — Each section has two known sides, 3.6 m and 2.4 m, with the 70° angle between them, matching in both sections; this is exactly the SAS condition, so 'SAS — two sides, included angle' is correct. 'SSS — but only two sides given' is wrong because SSS requires three pairs of equal sides, but only two sides are given for each triangle here. 'ASA — angle between two sides' is wrong because ASA requires two angles with a side between them, but only one angle, 70°, is given, not two. 'Cannot prove — only one angle' is wrong because SAS is specifically designed to prove congruence from exactly two sides and the one angle between them, so no further angle is needed.
- (a) 5/9 — Method: a bead cannot be two colours at once, so white and purple are mutually exclusive and their probabilities are added over the common total. Working: there are 1 + 4 + 4 = 9 beads, so P(white) = 1/9 and P(purple) = 4/9; adding gives 1/9 + 4/9, and 1 + 4 = 5 ninths. Answer: 5/9. The distractors: 4/9 comes from giving the probability of a purple bead alone and forgetting to include the white one; 5/8 comes from counting 5 favourable beads but using 8 as the total, leaving the single white bead out of the count of the box; 5/18 comes from adding 1/9 and 4/9 by adding the denominators as well as the numerators.
- (b) Positive correlation — Method: the type of correlation is named from the direction the points take as the scatter graph is read from left to right. Working: the points rise from left to right, so as the arm span read on the horizontal axis increases, the height read on the vertical axis increases as well; two quantities that increase together show positive correlation. Answer: positive correlation. The distractors: negative correlation comes from naming the direction the wrong way round, since a negative correlation needs the points to fall as the graph is read from left to right; no correlation comes from treating points that are spread out rather than sitting exactly on a line as though they showed no relationship; direct proportion comes from confusing a rising trend with proportion, which would additionally need the line through the points to pass through the origin and would mean doubling one quantity doubles the other.
- (b) 4 m — Blue fabric is 3 parts and this equals 1.5 m, so one part is 1.5 ÷ 3 = 0.5 m. Yellow fabric is 5 parts, so it is 5 × 0.5 = 2.5 m. The total length is 1.5 + 2.5 = 4 m. 2.5 m is the length of yellow fabric only, without adding the blue fabric back in. 1.5 m is just the given length of blue fabric, with the yellow fabric never worked out. 2.4 m comes from swapping the ratio, treating blue as 5 parts and yellow as 3 parts, giving one part as 1.5 ÷ 5 = 0.3 m and yellow as 3 × 0.3 = 0.9 m, then adding 1.5 + 0.9 = 2.4.
- (d) An equation, true for one value of x — 2x + 5 = 17 is only true when x = 6, so it is an equation. A candidate who ignores the equals sign and focuses only on the x terms and numbers present picks the expression option. A candidate who sees an equals sign and assumes it must relate two different quantities, as a formula does, picks that option, missing that there is only one quantity, x, involved here. A candidate who wrongly checks whether both sides look similar in structure rather than testing specific values picks the identity option, missing that the two sides are not equal for every value of x.
- (a) y is always 4 times x. — Check the multiplier for each pair: 12 ÷ 3 = 4, 20 ÷ 5 = 4, 32 ÷ 8 = 4 — the same multiplier every time, so y is always 4 times x. 'x is always 4 times y' comes from writing the multiplier the wrong way round. 'y is always x plus 9' only works for the first pair (3 + 9 = 12); it gives 14 for x = 5 and 17 for x = 8, not 20 and 32. 'y is always double x, plus 6' also only works for the first pair (3 × 2 + 6 = 12); it gives 16 for x = 5, not 20.
- (a) (3, 9) — Two column vectors are parallel only when one is the other multiplied by a single scalar — both numbers scaled by the same factor with the same sign. r has top:bottom ratio 2:6, which simplifies to 1:3. (3, 9) has ratio 3:9, also 1:3, so (3, 9) = 1.5 × r: it is parallel to r. (6, 2) swaps the top and bottom numbers of r, giving ratio 6:2 (3:1), which is not the same ratio. (2, 3) halves only the bottom number of r, giving ratio 2:3, not 1:3. (−2, 6) changes the sign of only the top number of r, giving ratio −2:6, which no longer matches 2:6 throughout.
- (b) 0.4 — The two events are independent, so multiply the two probabilities: 0.8 × 0.5 = 0.4. Choosing 1.3 comes from adding the two probabilities instead of multiplying, 0.8 + 0.5 = 1.3. Choosing 0.3 comes from subtracting the two probabilities, 0.8 − 0.5 = 0.3, instead of multiplying them. Choosing 0.1 comes from using the probability that Ben is NOT on time, 1 − 0.8 = 0.2, and multiplying that by Freya's probability instead, 0.2 × 0.5 = 0.1.
- (d) 2⁷ — When multiplying powers of the same base, the indices add: 3 + 4 = 7, so 2³ × 2⁴ = 2⁷. Multiplying the indices instead of adding them gives 3 × 4 = 12, so 2¹². Subtracting the indices instead of adding them gives 4 − 3 = 1, so 2¹. Multiplying the bases together as well as adding the indices gives 2 × 2 = 4, so 4⁷.
- (b) 40 — The differences between consecutive terms are 3, 5, 7, 9, increasing by 2 each time, so the next difference is 11, giving 29 + 11 = 40. A candidate who reuses the last difference of 9 without increasing it gets 29 + 9 = 38. A candidate who increases the difference by only 1 instead of 2 gets 29 + 10 = 39. A candidate who reuses the first difference of 3 instead of the pattern gets 29 + 3 = 32.
- (d) 3/2 — The length of the blue ribbon goes on the top and the length of the red ribbon goes on the bottom, giving 45/30. Both parts divide by 15: 45 ÷ 15 = 3 on the top, and 30 ÷ 15 = 2 on the bottom. So the fraction is 3/2. It is greater than 1 because the blue ribbon is longer than the red one, and a fraction of one quantity compared with another is allowed to be top-heavy.
- (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°.
- (b) 17 — Roots and powers are worked out first: √25 = 5 and 4² = 16. Division comes next: 12 ÷ 3 = 4. Then addition and subtraction, left to right: 5 + 16 − 4 = 17. A candidate who treated 4² as 4 × 2 = 8, multiplying the base by the exponent instead of squaring it, worked out 5 + 8 − 4 = 9. A candidate who did not evaluate the root and used 25 itself worked out 25 + 16 − 4 = 37. A candidate who ignored the priority of division and worked through 5 + 16 − 12 ÷ 3 strictly left to right got 5 + 16 = 21, then 21 − 12 = 9, then 9 ÷ 3 = 3.
- (d) x = (y + 3)/4 — Two operations have been applied to x: it has been multiplied by 4 and then 3 has been subtracted. Undo them in the opposite order. Adding 3 to both sides gives y + 3 = 4x. Dividing both sides by 4 then gives (y + 3)/4 = x, so x = (y + 3)/4. The bracket matters: writing y/4 + 3 divides only the y by 4 and leaves the 3 untouched. Writing (y − 3)/4 subtracts the 3 instead of adding it, and writing 4(y + 3) multiplies by 4 rather than dividing.
- (d) 25 minutes — Method: find the time for one kilometre, then multiply by the number of kilometres — the unitary method with a rate. Working: 10 ÷ 2 = 5 minutes per km, and 5 × 5 = 25. Answer: 25 minutes. The distractors: 20 minutes comes from multiplying the 10 minutes by 2, the distance in the given rate, instead of by the scale factor 2.5; 50 minutes comes from multiplying 10 by 5, treating the 10 minutes as the time for a single kilometre; 15 minutes comes from adding the 5 km on to the 10 minutes, adding quantities that are not the same kind.
- (a) 3.8 × 10⁻³ — 0.0038 is less than 1, so the power of 10 is negative. Moving the decimal point 3 places gives A = 3.8, so 0.0038 = 3.8 × 10⁻³. A candidate who wrote 3.8 × 10³ used a positive power, which is only correct for numbers of 10 or more. A candidate who wrote 38 × 10⁻⁴ used a value of A outside the required range. A candidate who wrote 3.8 × 10⁻⁴ counted one place too many when moving the decimal point.
- (a) 50 — Method: 25% is one quarter, so 25% of a quantity is the quantity divided by 4. Working: 200 ÷ 4 = 50. Answer: 50. The distractors: 25 comes from writing the percentage itself as the answer; 100 comes from halving, which is 50% not 25%; 800 comes from multiplying by 4 instead of dividing.
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.