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) 1,500 ≤ m < 2,500 — Method: a four-digit figure written to 1 significant figure has been rounded to the nearest 1,000, so the mass lies within half of 1,000, that is 500, of the figure given. Working: 2,000 − 500 = 1,500 and 2,000 + 500 = 2,500. The lower limit is included, because 1,500 kg rounds up to 2,000 kg to 1 significant figure, while 2,500 kg rounds up to 3,000 kg, so the upper limit is not. Answer: 1,500 ≤ m < 2,500. The distractors: 1,950 ≤ m < 2,050 comes from rounding to the nearest 100 instead of to 1 significant figure; 1,000 ≤ m < 3,000 goes a whole 1,000 either side instead of half of it; 1,500 < m ≤ 2,500 has the two limits the wrong way round.
- (c) 2x + 12 — Multiply every term inside the bracket by 2: 2 × x = 2x and 2 × 6 = 12, so 2(x + 6) = 2x + 12. Writing 2x + 6 comes from multiplying only the x-term and leaving the 6 unchanged. Writing x + 12 comes from multiplying only the 6 and leaving the x unchanged. Writing 2x + 8 comes from adding 2 and 6 instead of multiplying them.
- (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.)
- (d) Wrong - the given angle is not the included angle — Sides AB and BC meet at vertex B, so the included angle needed for SAS is angle B, not angle A — the information given is SSA. SSA does not prove congruence: with AB = 10 cm, BC = 7 cm and angle A = 40° there are two different triangles that fit, one with angle C ≈ 74.6° and one with angle C ≈ 105.4°, so Sam's triangles need not be the same shape and size at all. Sam is not correct just because two sides and an angle are equal, since the angle must be the one INCLUDED between those two sides. The condition is not ASA either, because ASA needs two angles, and only one angle is given here. It is also not true that nothing matches — the stated lengths and angle DO match between the two triangles; the problem is which angle was given, not whether the values agree.
- (d) 12 — Method: each result of the first experiment can be paired with every result of the second, so the two counts are combined by multiplying. Working: the coin lands in 2 ways and the dice lands in 6 ways. Each of the 6 dice scores can appear with a head or with a tail, so the table has 6 × 2 rows. Answer: there are 12 different results. The distractors: 8 comes from adding the two counts, 6 + 2, instead of multiplying them; 6 comes from listing the dice scores only and treating the coin as making no difference to the table; 36 comes from working out 6 × 6, counting the coin as though it too had six equally likely results.
- (b) On average a plant grew 1.5 cm taller for each extra day — Method: in the equation of a line, the number multiplying x is the gradient, and a gradient states the change in y produced by an increase of 1 in x, read in the units of the two axes. Working: here x is measured in days and y in centimetres, so the gradient 1.5 carries the units centimetres per day. Testing it on the line, 5 days gives 1.5 × 5 + 4 = 11.5 cm and 6 days gives 1.5 × 6 + 4 = 13 cm, a rise of 1.5 cm for the one extra day. Answer: on average a plant grew 1.5 cm taller for each extra day of watering. The distractors: 1.5 cm as the height before any watering is the value of y when x is 0, which is the other number in the equation, 4 cm, so this swaps the gradient and the intercept; 1.5 cm as the gap between the tallest and the shortest plant reads the gradient as a range, when a range is a difference between two of the 16 plants and a gradient is a rate; 1.5 days for each extra centimetre inverts the rate, dividing days by centimetres instead of centimetres by days, and the line gives 1 cm of growth in two thirds of a day.
- (c) 5 × 10⁻³ — Method: in standard form the coefficient must be at least 1 and less than 10, and for a number smaller than 1 the index is negative and counts the places the decimal point moves to the right. Working: the only significant digit is 5, so the coefficient is 5; moving the decimal point in 0.005 three places to the right gives 5, so the index is −3. Answer: 5 × 10⁻³. The distractors: 0.5 × 10⁻² is the same value written the wrong way, because 0.5 is smaller than 1 and so is not an allowed coefficient; 5 × 10⁻² comes from counting the two zeros after the decimal point instead of the three places the point moves; 5 × 10³ comes from taking the index as positive, which describes a number in the thousands rather than one smaller than 1.
- (c) s = P / 4 — Method: undo the multiplication by 4 by dividing both sides by 4. Working: P = 4s, so dividing both sides by 4 gives s = P / 4. The value s = 4P comes from multiplying by 4 instead of dividing — using the wrong inverse operation. The value s = P + 4 comes from adding 4 instead of dividing. The value s = P − 4 comes from subtracting 4 instead of dividing.
- (a) 1/9 — The ratio copper : tin is 9:1, so write tin over copper: 1/9. (9/1 comes from writing the ratio the wrong way round, copper over tin. 1/10 comes from comparing the tin to the total mass of the alloy, 1 part out of 10. 9/10 comes from comparing the copper to the total mass of the alloy, 9 parts out of 10.)
- (b) second quadrant — The point (−3, 5) has a negative x-coordinate and a positive y-coordinate, and this combination lies in the second quadrant. The first quadrant needs both coordinates positive. The third quadrant needs both coordinates negative. The fourth quadrant needs a positive x-coordinate and a negative y-coordinate.
- (c) 1/36 — Method: the two dice do not affect each other, so the probability that both results happen is the product of the two separate probabilities. Working: a 1 on the red dice has probability 1/6 and a 2 on the blue dice has probability 1/6. Multiplying gives 1 × 1 = 1 on the top and 6 × 6 = 36 on the bottom. Answer: the probability is 1/36. The distractors: 1/18 comes from allowing either order, counting a 2 on the red dice with a 1 on the blue dice as well, although the stem names which dice shows which score; 1/3 comes from adding 1/6 and 1/6 instead of multiplying; 1/6 comes from giving the probability for one dice only and treating the other as making no difference.
- (b) £8 — Rounding to 1 significant figure: £1.85 rounds to £2, and 3.6 kg rounds to 4 kg. The estimate is £2 × 4 = £8. A candidate who used the unrounded values instead of estimating worked out 1.85 × 3.6 = £6.66. A candidate who rounded only the mass and used the exact price worked out 1.85 × 4 = £7.40. A candidate who rounded the price to the nearest 10p instead of 1 significant figure worked out 1.9 × 4 = £7.60.
- (a) x + 3 — Method: "years older" is an addition, and the number of years is added to the age that is known in terms of the letter. Working: Jack's age is x, and Kwame is 3 years older, so Kwame's age is x + 3. Answer: x + 3. The distractors: x − 3 comes from reading "older" as "younger" and subtracting instead of adding; 3x comes from reading "3 years older" as "3 times as old" and multiplying; 3 − x comes from writing the calculation the wrong way round, taking Jack's age from the 3 rather than adding.
- (c) 40 cm — Method: split the total length into the number of parts shown by the ratio, then find the value of the shorter share. Working: the ratio 4:5 has 4 + 5 = 9 parts, so one part is 90 ÷ 9 = 10 cm, and the shorter piece is 4 × 10 = 40 cm. So the shorter piece is 40 cm. Distractor 50 cm is the length of the LONGER piece, not the shorter one. Distractor 45 cm comes from splitting the ribbon into two equal halves, ignoring the ratio. Distractor 10 cm is the value of one part, found correctly but never multiplied by 4.
- (c) −1.5 — Since n = k × m, dividing a number in n by the matching number in m gives k: k = −6 ÷ 4 = −1.5 (check with the bottom numbers: −9 ÷ 6 = −1.5, the same value, confirming n is a scalar multiple of m). 1.5 has the correct size but is missing the negative sign. −10 comes from subtracting the top numbers, −6 − 4, instead of dividing them. −24 comes from multiplying the top numbers, −6 × 4, instead of dividing them.
- (c) 6000 — Method: round each number to 1 significant figure, then multiply the rounded values. Working: 312 rounds to 300 (1 s.f.) and 19 rounds to 20 (1 s.f.). 300 × 20 = 6000. Answer: 6000. 5928 is the exact value of 312 × 19, found by multiplying without rounding first, which is not an estimate. 600 comes from rounding 19 down to 2 instead of to 20, losing a zero from its place value. 6200 comes from rounding 312 to the nearest 10, 310, instead of to 1 significant figure, 300, then multiplying by the correctly rounded 20.
- (d) 5x² — 5x² is a single term: a product of a number and a power of a letter, with no + or − sign attached to anything else. A candidate who picks 5x² + 3x is choosing the expression that contains that term, rather than the term on its own. A candidate who picks 5x² = 20 is drawn in by the equals sign and misses that this makes it an equation. A candidate who picks A = 5x² thinks any statement with a letter and a number combined must be a formula.
- (b) £50 — Method: add the parts of the ratio, divide the amount by the number of parts to find the value of one part, then multiply by the parts in the largest share. Working: 3 + 4 + 5 = 12 parts, £120 ÷ 12 = £10 for one part, and the largest share is 5 parts, so 5 × £10 = £50. Answer: £50. The distractors: £10 is the value of one part only; £30 is the 3-part share, which is the smallest one; £40 is the 4-part share, the middle one.
- (d) 360 km — Method: first find the kilometres per litre by dividing distance by fuel used, then multiply this rate by the new tank size. Working: 180 ÷ 6 = 30 km per litre; 30 × 12 = 360 km. Answer: 360 km. 30 km comes from finding the correct fuel consumption but stopping there, without scaling it up to the full tank. 2160 km comes from multiplying the original distance (180) by the tank size (12) directly, skipping the unit rate. 90 km comes from pairing the numbers the wrong way round: dividing the distance by the new tank size, 180 ÷ 12 = 15, and then multiplying by the original 6 litres, 15 × 6 = 90.
- (a) 42 — Find the value of one part: 18 ÷ 3 = 6. Find the number of dogs: 4 × 6 = 24. Add the cats and the dogs to find the total: 18 + 24 = 42. (24 is the number of dogs only, not the total number of animals. 126 comes from multiplying the number of cats by the total number of parts, 18 × 7, instead of finding one part first. 25 comes from adding the ratio numbers 3 and 4 directly to the number of cats.)
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.