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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- (c) 6 — Listing systematically: red-black, red-grey, blue-black, blue-grey, green-black, green-grey gives 6 different outfits, matching 3 × 2 = 6. 5 comes from adding the number of T-shirts and shorts instead of multiplying them. 3 comes from counting only the T-shirts. 2 comes from counting only the pairs of shorts.
- (c) x = (y + 24)/4 — Method: expand the bracket first, then undo the operations done to x in reverse order. Working: y = 4(x − 6) = 4x − 24, so y + 24 = 4x, so x = (y + 24)/4. Answer: x = (y + 24)/4. x = (y + 6)/4 comes from expanding the bracket incorrectly, treating 4(x − 6) as 4x − 6 instead of 4x − 24. x = 4y + 96 comes from multiplying by 4 instead of dividing by 4 to undo the multiplication, giving 4(y + 24) = 4y + 96. x = y/4 − 6 comes from dividing by 4 first without expanding the bracket, then subtracting 6 as if the bracket's operation still applied afterwards.
- (d) 13/10 — First find the selling price: £150 + £45 = £195. Put the selling price over the cost price: 195/150. Divide both numbers by their highest common factor, 15: 195÷15 = 13, 150÷15 = 10, giving 13/10. (10/13 comes from writing the prices the wrong way round. 3/10 is just the profit written as a fraction of the cost price, 45/150, not the selling price. 13/23 comes from comparing the selling price to the combined total of the cost price and the selling price, 195/345.)
- (a) No — the angle is not between the two sides — Method: check whether the given angle sits between the two given sides. Working: the 35° angle is marked away from the corner where the 40 cm and 65 cm edges meet, so it is not the included angle — this is SSA, which is not one of the four basic congruence conditions, so it does not guarantee congruence. Options: 'Yes — SAS' wrongly treats any two sides plus any angle as SAS, without checking the angle's position; 'Yes — SSS' wrongly counts the angle as if it were a third side; 'No — only two sides measured' is not the real reason, since SAS itself only needs two sides, so this reasoning is beside the point. Answer: no, the angle is not between the two sides.
- (a) No — the three probabilities sum to 1.10, over 1. — Winning, drawing and losing are exhaustive and mutually exclusive, so their probabilities must sum to exactly 1. Adding Freddie's three values gives 0.45 + 0.3 + 0.35 = 1.10, which is more than 1, so his probabilities cannot all be correct: 'No — the three probabilities sum to 1.10, over 1.' Checking only that each value lies between 0 and 1 accepts them as 'Yes — each probability lies between 0 and 1' without ever adding the three together. Judging by which outcome sounds most likely leads to 'Yes — winning has the highest single probability', which never checks the total either. Noting that a runner cannot win, draw and lose at once, and treating that alone as enough, gives 'Yes — the three outcomes are mutually exclusive' — but mutually exclusive outcomes that are also exhaustive must still sum to 1, and 1.10 does not.
- (d) How much the temperatures varied over the seven days — Method: the range of a set of values is the largest value take away the smallest, so it is built from two values only and it measures the gap they leave between them. Working: the largest of the seven readings is 7 °C and the smallest is 2 °C, so the range is 7 − 2 = 5 °C. That figure says the week's readings covered a band 5 °C wide; it names no particular day and no particular reading. Answer: the range describes how much the temperatures varied over the seven days. The distractors: the temperature that occurred most often is the mode, which here is 3 °C, and a mode counts repeats instead of measuring a gap; the temperature typical of the week is an average, and the range is not an average, since it throws away every value lying between the two extremes; the number of different temperatures recorded is 6, a count of how many distinct values appear, while the range is a difference between two of them.
- (c) 8 — Each of the 4 spinner outcomes can be paired with each of the 2 coin outcomes, so multiply: 4 × 2 = 8. 6 comes from adding the two numbers instead of multiplying them. 4 comes from using only the spinner outcomes and ignoring the coin. 2 comes from using only the coin outcomes and ignoring the spinner.
- (a) (40, 0) — Method: a graph crosses the x-axis where y = 0, which in this context is the moment the tank holds no water, so substituting y = 0 and solving gives the time. Working: 0 = −5x + 200 gives 5x = 200, so x = 200 ÷ 5 = 40 and the crossing point is (40, 0); the tank is empty after 40 minutes. Answer: (40, 0). The distractors: (0, 200) is the y-axis crossing, the 200 litres in the tank at the start, found by substituting x = 0 instead of y = 0; (200, 0) comes from reading the constant 200 as the x-coordinate without dividing by the 5 litres lost each minute; (0, 40) has the right number in the wrong place, with the time written as a y-coordinate.
- (b) £2205.00 — With compound interest each year's interest is worked out on the value at the start of that year, so a 5% rise is a multiplier of 1.05 applied once per year. After the first year: 2000 × 1.05 = 2100. After the second year: 2100 × 1.05 = 2205. The question asks for the value of the investment, not for the interest earned, so the answer is £2205.00.
- (c) They must also be equal — Once two triangles are proved congruent by any condition, including ASA, they are identical in every respect: every pair of corresponding sides and every pair of corresponding angles must be equal, not just the ones originally used to prove the congruence. So the two remaining pairs of corresponding sides must also be equal, making 'they must also be equal' correct. 'They might be equal or not' and 'not enough information to say' both wrongly suggest that congruence only guarantees the specific facts used to prove it, when congruence actually guarantees the triangles are identical overall. 'They must be different' is backwards: the triangles being identical is the entire point of proving congruence, not a reason for a side to differ.
- (c) 23 — Multiples of 4 from 1 to 30 are 4, 8, 12, 16, 20, 24 and 28 — 7 numbers, so n(A) = 7. The universal set has 30 elements, so n(A′) = 30 − 7 = 23. Reporting n(A) itself, 7, without subtracting it from the universal set forgets what the complement means. Estimating the count of multiples of 4 as 30 ÷ 4 = 7.5, rounded to 8, instead of listing them exactly, gives 30 − 8 = 22. Miscounting the universal set as having 31 elements instead of 30, an off-by-one slip, gives 31 − 7 = 24.
- (b) 4/9 — Method: double the butter part of the ratio, keeping flour the same, find the new total, then write butter's part over the new total. Working: the new ratio is flour : butter = 5 : 4, since butter doubles from 2 to 4. New total = 5 + 4 = 9. Fraction of butter = 4/9. Answer: 4/9. 2/9 comes from forgetting to double the butter part and using the original value 2 over the new total of 9. 4/7 comes from doubling the butter part correctly to 4 but keeping the old total of 7 instead of working out the new total. 2/5 comes from using the original ratio 5:2 directly as butter over flour without doubling anything.
- (c) (2, 5) — Reflecting in the x-axis keeps the x-coordinate the same and changes the sign of the y-coordinate: Q = (2, 5). (−2, −5) comes from reflecting in the y-axis instead, which changes the sign of the x-coordinate. (−2, 5) comes from reflecting in both axes. (2, −5) comes from not applying the reflection at all.
- (c) 3 : 4 — Method: a fraction compares a part with the whole, while this ratio compares one part with the other part, so find the fraction that is black before writing the ratio. Working: if 3/7 are blue then the black pens make up 7/7 − 3/7 = 4/7 of the box, so out of every 7 pens 3 are blue and 4 are black, and blue : black = 3 : 4. Answer: 3 : 4. The distractors: 3 : 7 comes from reading the numerator and the denominator of 3/7 straight off as the two parts, which compares the blue pens with the whole box rather than with the black pens; 4 : 3 comes from writing the black pens before the blue pens, reversing the order asked for; 4 : 7 is the same numerator-and-denominator reading applied to the black fraction 4/7, again comparing a part with the whole box.
- (c) 21 cm — Method: the perimeter is the distance all the way round the edge, so every side is counted once and the three side lengths are added. Working: the two equal sides give 8 + 8 = 16 cm, and the third side adds 5 cm to that. Answer: the perimeter is 21 cm. The distractors: 16 cm comes from adding the two 8 cm sides and handing that total in before the third side has been included; 13 cm comes from adding one 8 cm side to the 5 cm side, as though the badge carried only the two different lengths printed on it rather than three sides; 24 cm comes from taking all three sides to be 8 cm and working out 3 × 8, which would be the perimeter only if the badge were equilateral.
- (a) 8 × 10² — Method: to multiply numbers written in standard form, multiply the coefficients and add the indices. Working: 4 × 2 = 8 for the coefficients, and −3 + 5 = 2 for the indices; 8 already lies between 1 and 10, so no adjustment is needed. Answer: 8 × 10². The distractors: 6 × 10² comes from adding the coefficients, 4 + 2, instead of multiplying them; 8 × 10⁸ comes from ignoring the minus sign and adding 3 + 5; 8 × 10⁻¹⁵ comes from multiplying the indices, −3 × 5, instead of adding them.
- (d) 21 — The differences between consecutive triangular numbers increase by 1 each time: 3−1=2, 6−3=3, 10−6=4, 15−10=5. So the next difference is 6, giving 15+6=21. A candidate who keeps the difference the same as the previous step (adding 5 again) would reach 20. A candidate who uses a constant difference of 3 throughout would reach 18. A candidate who instead finds the square of the term number (5²=25, since 15 is the 5th term) would reach 25 — that is the rule for square numbers, not triangular numbers.
- (d) 7/4 — A part-to-part ratio a : b gives the fraction a/b when the first quantity is written as a fraction of the second, so 7 : 4 gives 7/4. Writing 4/7 puts the parts the wrong way round — blue as a fraction of red, not red as a fraction of blue. Writing 7/11 uses the total number of counters, 7 + 4 = 11, as the denominator instead of the number of blue counters — that is red as a fraction of the whole bag, not red as a fraction of blue. Writing 11/7 has both the wrong denominator and the parts inverted.
- (b) 63 — 5 + 2 = 7, then 3² = 9, then 7 × 9 = 63. Ignoring the brackets and applying BIDMAS as if the expression were unbracketed gives 3² = 9, then 2 × 9 = 18, then 5 + 18 = 23. Squaring the bracket instead of the 3 gives 7² = 49, then 49 × 3 = 147 — the power belongs to the 3 alone. Multiplying by 3 before squaring the whole product gives 7 × 3 = 21, then 21² = 441.
- (d) 3/2 — Work out the number of altos: 40% of 30 = 12. The rest are sopranos, so there are 30 − 12 = 18 sopranos. Form the fraction 18/12; both numbers share a factor of 6, so 18 ÷ 6 = 3 and 12 ÷ 6 = 2, giving 3/2. 2/3 comes from writing the fraction the wrong way round, as altos over sopranos (12/18). 3/5 comes from comparing the sopranos with the whole choir (18/30), instead of with the altos. 7/3 comes from miscalculating 30 − 12 as 28 instead of 18, then forming 28/12.
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.