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) 3.2 litres — The blue paint is 3 of the 8 equal shares in the mix, that is 3/8 of the total. Three shares are 1.2 litres, so one share is 1.2 ÷ 3 = 0.4 litres. The whole mix is 8 shares: 8 × 0.4 = 3.2 litres. 2 litres is the volume of white paint, 1.92 litres divides the blue paint by the white paint's 5 shares instead of its own 3, and 9.6 litres treats the 1.2 litres as a single share.
- (d) 12 — Method: call the younger brother's age x, write the elder brother's age in terms of x, and form an equation from the total. Working: the elder brother is x + 6, so x + (x + 6) = 30; simplifying gives 2x + 6 = 30, subtracting 6 from both sides gives 2x = 24, and dividing by 2 gives x = 12. Checking: 12 and 18 add up to 30 and differ by 6. Answer: 12. The distractors: 18 comes from solving correctly and then giving the elder brother's age, which is not the age asked for; 15 comes from halving 30 and ignoring the 6-year difference altogether; 24 comes from taking 6 off the total, 30 − 6 = 24, and giving that as an age.
- (b) 3 : 5 — The highest common factor of 27 and 45 is 9. Divide both parts by 9: 27 ÷ 9 = 3 and 45 ÷ 9 = 5, giving 3 : 5. Giving 9 : 15 divides by 3, a common factor but not the highest one, so it is not yet in simplest form. Giving 5 : 3 swaps the order of the two parts. Giving 27 : 45 has not been simplified at all.
- (b) 10 — Corresponding angles are equal, so 5x + 4 = 3x + 24. Subtracting 3x from both sides gives 2x + 4 = 24, then subtracting 4 gives 2x = 20, so x = 10. 14 comes from adding the constants, 4 + 24, instead of subtracting them when rearranging. 19 comes from treating the angles as co-interior (summing to 180°): 5x + 4 + 3x + 24 = 180 gives 8x = 152, so x = 19. 20 correctly reaches 2x = 20 but stops without dividing by 2.
- (d) 9 — Method: each of the first spin's 3 outcomes can be paired with each of the second spin's 3 outcomes, so the two counts are multiplied together. Working: 3 × 3 = 9 ordered pairs. Answer: 9. Watch out: writing down 6 comes either from adding the two spins' counts, 3 + 3, instead of multiplying them, or from listing colour pairs without order, treating (red, blue) as the same entry as (blue, red) — both shortcuts miss outcomes that the ordered list contains. Writing down 3 lists only a single spin's outcomes and never pairs the two spins up at all. And writing down 27 treats the spinner as though it had been spun three times, 3 × 3 × 3, one spin too many.
- (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.
- (b) 2/7 — The comparison asked for is part with part, not part with whole, so the number of cars goes on the bottom. The ratio gives 2 shares of vans for every 7 shares of cars, so the vans are 2/7 of the cars. 2/9 uses the 9 shares in the whole car park as the denominator, 7/2 compares the cars with the vans, and 7/9 is the fraction of the vehicles that are cars.
- (b) 4 — Method: form the equation 30 + 25h = 130, where h is the number of hours, then solve for h. Working: subtract the call-out fee from the total bill: 25h = 130 − 30 = 100. Divide by the hourly rate: h = 100 ÷ 25 = 4. Answer: 4 hours. 5.2 comes from dividing the whole bill by the hourly rate without subtracting the fixed fee first, 130 ÷ 25. 3.5 comes from swapping the fee and the rate, subtracting the rate from the bill and dividing by the fee, (130 − 25) ÷ 30. 6.4 comes from adding the call-out fee to the bill instead of subtracting it, before dividing by the rate, (130 + 30) ÷ 25.
- (c) C = 1.5n — Method: a fixed ratio between C and n means C is always the same multiple of n, and that multiple is the cost of one bottle. Working: 3.00 ÷ 2 = 1.5, 7.50 ÷ 5 = 1.5 and 12.00 ÷ 8 = 1.5, so every bottle costs £1.50 and C = 1.5n. Answer: C = 1.5n. The distractors: C = n + 1 comes from subtracting on the first row, 3 − 2 = 1, and adding that difference instead of multiplying; it fits the first row and fails the other two, which is why three rows are given; C = 3n reads the £3.00 as the price of one bottle when it is the price of two; C = n/1.5 divides the number of bottles by the price of one bottle, which works out how many bottles a pound buys instead of what n bottles cost.
- (b) (12, −8) — Method: to multiply a column vector by a number, multiply every part of the vector by that number. Working: top number 4 × 3 = 12; bottom number 4 × (−2) = −8. Answer: 4a = (12, −8). A candidate who adds 4 to each part instead of multiplying gets (7, 2). A candidate who multiplies only the top number by 4 and leaves the bottom number unchanged gets (12, −2). A candidate who multiplies only the bottom number by 4 and leaves the top number unchanged gets (3, −8).
- (a) 0.09 — Method: two independent events that must both happen are combined by multiplying their probabilities. Working: the same probability 0.3 applies to each day, so the calculation is 0.3 × 0.3. Written as fractions this is 3/10 × 3/10 = 9/100. Answer: the probability is 0.09. The distractors: 0.6 comes from adding 0.3 and 0.3 instead of multiplying them; 0.3 comes from quoting the single-day probability, as though the second day added no further condition; 0.9 comes from multiplying 3 by 3 correctly but keeping only one decimal place in the product instead of two.
- (a) 40 — Multiply the number of choices for each item: 4 × 5 × 2 = 40. 11 comes from adding the three numbers instead of multiplying them. 20 comes from multiplying only the crisps and chocolate bars, 4 × 5, and forgetting the drink. 10 comes from multiplying only the chocolate bars and drinks, 5 × 2, and forgetting the crisps.
- (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.
- (b) 8 — The gradient of a line through the origin is the y-value divided by the x-value at any point on the line. Using (10, 80): gradient = 80 ÷ 10 = 8.
- (b) Draw wider arcs from each crossing point, meeting above. — Once the two arcs cross the line at points either side of P, compasses are opened to a radius greater than before and arcs are drawn from each of those two points so that they meet above (or below) the line; joining that meeting point to P gives the perpendicular. (Joining the two crossing points with a straight line only retraces part of the original line, since both points already lie on it; drawing a circle centred at P through both crossing points does not locate any new point needed for the perpendicular; drawing an arc centred at P through only one crossing point repeats the first step instead of moving on to the second pair of arcs.)
- (b) 2.1 m — The three pieces use 3 × 0.9 = 2.7 m of wood. Remaining wood = 4.8 − 2.7 = 2.1 m. A candidate who miscounts and only subtracts 2 pieces instead of 3 gets 4.8 − 1.8 = 3.0 m. A candidate who adds instead of subtracting gets 4.8 + 2.7 = 7.5 m. A candidate who gives the length used instead of the length remaining gets 2.7 m.
- (a) 9 — Method: the gaps in this sequence are not constant, so work out each of the two terms named from the rule and then subtract the earlier from the later. Working: the 5th term is 5² + 1 = 25 + 1 = 26 and the 4th term is 4² + 1 = 16 + 1 = 17, so the difference is 26 − 17. Answer: 9. The distractors: 7 comes from using the 3rd and 4th terms, one position too early, 17 − 10; 11 comes from using the 5th and 6th terms, one position too late, 37 − 26; 1 comes from subtracting the position numbers, 5 − 4, instead of the terms themselves.
- (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.)
- (a) 300 — Method: round each number to 1 significant figure, then divide the rounded values. Working: 8,900 rounds to 9,000 and 29 rounds to 30; cancelling a zero from each gives 900 ÷ 3. Answer: 300. The distractors: 450 comes from rounding 29 down to 20 instead of to the nearest ten, giving 9,000 ÷ 20; 3,000 comes from rounding 29 to 3 rather than to 30, a place-value slip that divides by a number ten times too small; 307 is the exact quotient rounded to the nearest whole number, worked out in full when the question asks for an estimate.
- (c) 36 — Add the parts of the ratio: 3 + 4 = 7. Divide the total by the number of parts: 84 ÷ 7 = 12, so one part is worth 12. The smaller share has 3 parts: 3 × 12 = 36. (48 is the larger share, using 4 parts instead of 3. 42 comes from splitting 84 in half, treating the ratio as if it were 1:1. 28 comes from dividing 84 by 3 — one of the ratio numbers — instead of dividing by the total number of parts, 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.