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.A shop assistant says that 7.2 × 3.9 = 56.16. Work out an estimate for 7.2 × 3.9, by rounding each number to the nearest whole number, to show that the assistant’s answer cannot be correct.
- 2.Priya has a budget of £50 for a school trip. The coach costs £14 and each student ticket costs £4. Using the inequality 14 + 4s ≤ 50, work out the greatest number of student tickets, s, she can buy.
- 3.A window cleaner charges a call-out fee plus an amount per window. The total charge, C pounds, is shown on a straight-line graph against the number of windows cleaned, w. The line passes through the points (2, 14) and (6, 26). Work out the gradient of the line.
- 4.A cylinder has a volume of 942 cm³ and a height of 12 cm. Using π = 3.14, work out the radius of the cylinder.
- 5.Two ordinary fair dice are rolled, one after the other. Work out the probability that the first dice shows a 6 and the second dice shows an even number.
- 6.A netball team scored 50, 83 and 68 points in three matches. Work out the mean number of points scored.
- 7.Two fair six-sided dice are rolled and their scores are added together. By listing the possible totals systematically, work out how many different totals are possible.
- 8.The point F has x-coordinate −1 and lies 6 units above the x-axis. Write down the coordinates of F.
- 9.3 builders put up a fence in 12 days. All the builders work at the same rate, so the number of days is inversely proportional to the number of builders. Work out how long 9 builders take to put up the same fence.
- 10.Two angles lie on a straight line. One angle is 37°. Work out the size of the other angle.
- 11.The probability that it rains on a given morning is 0.3. If it rains, the probability that Priya's bus is late is 0.8. If it does not rain, the probability that her bus is late is 0.1. Work out the probability that her bus is late.
- 12.A metal cube has a mass of 540 g and a volume of 60 cm³. Work out its density in g/cm³.
- 13.The cost, in pounds, of hiring a van for d days is given by the formula C = 25 + 18d. A customer's hire cost £133. Work out how many days, d, they hired the van for, by first rearranging the formula to make d the subject.
- 14.A straight-line graph shows the number of pages printed, p, plotted against the time in minutes, t, since a printer started a print job. The line passes through the points (2, 30) and (5, 90). Work out the gradient of the line.
- 15.A ramp is built from two straight metal supports that rest on the flat ground and meet each other at the top, forming a triangle with the ground. One support meets the ground at 34°. The angle between the two supports where they meet is 71°. Work out the angle between the second support and the ground.
- 16.Work out the value of √(16 + 9)
- 17.A colony of bacteria doubles in number every hour. At 9am there are 5 bacteria in the colony. Work out how many bacteria there will be at 12 noon.
- 18.A bag contains 7 red counters and 15 blue counters. Write down the ratio of red counters to blue counters in its simplest form.
- 19.Simplify (y³)⁴, giving your answer as a single power of y.
- 20.A force of 20 N acts on an area of 4 m². Using pressure = force ÷ area, work out the pressure, in pascals.
Answer key
- (d) 28 — Method: round each number to the nearest whole number, then multiply the rounded numbers to get an estimate that can be compared with the assistant's answer. Working: 7.2 rounds to 7, and 3.9 rounds to 4, so the estimate is 7 × 4 = 28. Since 28 is much smaller than 56.16, the assistant's answer cannot be correct. 56 comes from rounding the assistant's answer to the nearest whole number, instead of rounding the two numbers being multiplied and then multiplying them. 35 comes from rounding both numbers correctly but then slipping in the seven times table, writing 7 × 5 = 35 in place of 7 × 4 = 28. 21 comes from rounding 3.9 down to 3 instead of 4, giving 7 × 3 = 21. Answer: 28.
- (b) 9 — Subtract 14 from both sides: 4s ≤ 36. Divide both sides by 4: s ≤ 9, so the greatest number of tickets is 9. A candidate who forgets the £14 coach cost solves 4s ≤ 50, getting s ≤ 12.5, rounded down to 12. A candidate who adds the £14 instead of subtracting it solves 4s ≤ 64, getting s = 16. A candidate who miscalculates 50 − 14 as 32 solves 4s ≤ 32, getting s = 8.
- (a) £3.00 per window — The gradient is the change in C divided by the change in w. Change in C = 26 − 14 = 12. Change in w = 6 − 2 = 4. Gradient = 12 ÷ 4 = £3.00 per window. Dividing the change in w by the change in C instead gives 4 ÷ 12 = £0.33 per window. Dividing 12 by the larger w-value only, 12 ÷ 6 = £2.00 per window, comes from not subtracting the smaller w-value first. Adding the w-values instead of subtracting, 12 ÷ (6 + 2) = £1.50 per window, comes from a sign error when finding the change in w.
- (c) 5 cm — Volume of a cylinder = πr²h, so r² = V ÷ (πh) = 942 ÷ (3.14 × 12) = 942 ÷ 37.68 = 25, and r = √25 = 5 cm. A pupil who finds r² = 25 but forgets to take the square root gives 25 cm. A pupil who forgets to divide by π, using r² = 942 ÷ 12 = 78.5, gets r = √78.5 ≈ 8.9 cm. A pupil who forgets to divide by the height, using r² = 942 ÷ 3.14 = 300, gets r = √300 ≈ 17.3 cm. The correct radius is 5 cm.
- (a) 1/12 — The probability that the first dice shows a 6 is 1/6. The probability that the second dice shows an even number, 2, 4 or 6, is 3/6 = 1/2. Since the two dice are independent, multiply the probabilities: 1/6 × 1/2 = 1/12. A candidate who answers 1/6 has considered only the first dice and forgotten the condition on the second dice. A candidate who answers 1/2 has considered only the second dice and forgotten the condition on the first dice. A candidate who answers 1/36 has treated 'an even number' as a single specific value rather than three possible values, using 1/6 × 1/6.
- (c) 67 — Method: the mean is the total of the values divided by how many values there are, so add first and divide second. Working: the total is 50 + 83 + 68 = 201 points and three matches were played, so the mean is 201 ÷ 3 = 67 points. Answer: 67. The distractors: 68 comes from writing down the median, the middle value of 50, 68, 83, instead of the mean; 33 comes from working out the range, 83 − 50, which measures spread and not centre; 100.5 comes from dividing the total by 2 instead of by the 3 matches played.
- (a) 11 — Method: list every possible total from the smallest to the largest, and count how many different values there are. Working: the smallest total is 1+1=2 and the largest is 6+6=12, and every whole number total from 2 to 12 is possible: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 — that is 11 different totals. Answer: 11. 36 comes from counting the number of possible dice outcomes (6×6) instead of the number of different totals. 10 comes from listing the totals but missing one from the ends of the list, for example starting at 3 instead of 2. 6 comes from counting only the number of different scores on one die, not the totals of both dice together.
- (b) (−1, 6) — '6 units above the x-axis' gives a y-coordinate of 6, and the x-coordinate is given as −1, so F = (−1, 6). (6, −1) comes from swapping the x- and y-coordinates. (−1, −6) comes from treating 'above' as a negative direction instead of positive. (1, 6) comes from dropping the negative sign on the given x-coordinate.
- (a) 4 days — Method: the fence is a fixed amount of work, so builders × days is constant; find that product and divide it by the new number of builders. Working: 3 × 12 = 36 builder-days of work, so with 9 builders the time is 36 ÷ 9 = 4 days. Answer: 4 days. The distractors: 6 days comes from halving the 12 days because there are more builders, rather than dividing by the factor of 3 by which the workforce has grown; 36 days is the constant product of builders and days, given as a number of days instead of being shared between the builders; 9 days comes from taking 3 days off the 12, treating three extra builders as three fewer days, which is additive rather than proportional.
- (a) 143° — Angles on a straight line add up to 180°. Set up 37° + x = 180°. Subtract: x = 180° − 37° = 143°. 53° comes from using 90° as the total, as if the two angles made a right angle, instead of the 180° of a straight line.
- (c) 0.31 — Method: a late bus can be reached along two paths of the tree. Multiply along each path, then add the paths that end in a late bus. Working: the rain path gives 0.3 × 0.8 = 0.24. No rain has probability 1 − 0.3 = 0.7, so the second path gives 0.7 × 0.1 = 0.07. Adding the two paths gives 0.24 + 0.07. Answer: the probability is 0.31. The distractors: 0.9 comes from adding the two branch probabilities 0.8 and 0.1 without first weighting them by how often it rains; 0.24 comes from following the rain path only and ignoring that the bus can also be late when it is dry; 0.27 comes from using 0.3 at the start of both paths, so that the no-rain path is given 0.3 × 0.1 instead of 0.7 × 0.1.
- (b) 9 g/cm³ — Method: density = mass ÷ volume. Working: 540 ÷ 60 = 9. Answer: 9 g/cm³. (0.11 g/cm³ comes from dividing the volume by the mass instead of the mass by the volume. 480 g/cm³ comes from subtracting the volume from the mass instead of dividing. 32400 g/cm³ comes from multiplying the mass by the volume instead of dividing.)
- (d) 6 — Method: rearrange the formula to make d the subject, then substitute C = 133. Working: C = 25 + 18d, so subtracting 25 from both sides gives C − 25 = 18d, then dividing by 18 gives d = (C − 25) / 18. Substituting C = 133: d = (133 − 25) / 18 = 108 / 18 = 6. The value 7.4 comes from dividing 133 by 18 without subtracting the fixed £25 first (133 / 18 ≈ 7.4). The value 4.6 comes from pairing the numbers the wrong way round, working out (133 − 18) / 25 = 4.6. The value 90 comes from subtracting both 25 and 18 from 133 instead of dividing by 18.
- (b) 20 — The gradient is the change in p divided by the change in t: (90 − 30) ÷ (5 − 2) = 60 ÷ 3 = 20. Choosing 0.05 comes from dividing the change in t by the change in p, the wrong way round (3 ÷ 60). Choosing −20 comes from subtracting the coordinates in the wrong order for one part of the calculation, for example (30 − 90) ÷ (5 − 2), giving a negative value. Choosing 18 comes from dividing the second p-coordinate by the second t-coordinate directly (90 ÷ 5) instead of using the change between the two points.
- (a) 75° — The three angles in the triangle formed by the two supports and the ground add up to 180°. Two of the angles are 34° and 71°, so the third angle is 180 − 34 − 71 = 75, which is 75°. 105° comes from adding the two given angles together instead of subtracting them from 180° (34 + 71 = 105). 146° comes from computing 180 − 34 = 146 and forgetting to also subtract 71°. 37° comes from subtracting the two given angles from each other instead of using the triangle's angle sum (71 − 34 = 37).
- (a) 5 — 16 + 9 = 25, then √25 = 5. Splitting the root over the addition instead gives √16 = 4 and √9 = 3, then 4 + 3 = 7 — but a root does not split over a sum like this. Multiplying those two roots instead of adding them gives 4 × 3 = 12. Taking the negative square root instead of the positive one gives −5.
- (a) 40 — From 9am to 12 noon is 3 hours, so the population doubles three times: 5 × 2³ = 40. A candidate who counts the elapsed time as 4 hours (an off-by-one counting error) would reach 5 × 2⁴ = 80. A candidate who counts it as only 2 hours would reach 5 × 2² = 20. A candidate who misreads 'doubles' as 'increases by 2' each hour would compute 5 + 3 × 2 = 11.
- (d) 7:15 — Method: write the two parts in the order asked for, red first, then divide both parts by their highest common factor. Working: the factors of 7 are 1 and 7, and the factors of 15 are 1, 3, 5 and 15, so the only common factor is 1; dividing both parts by 1 leaves both counts unchanged. Answer: 7:15, which is already in its simplest form. The distractors: 15:7 is the ratio of blue to red, the reverse of the order the question asks for; 7:5 comes from cancelling the digit 1 out of 15, which is not a division by a common factor; 1:2 comes from dividing 15 by 7, rounding the result to 2 and writing the ratio as 1 to 2, but 7 is not a factor of 15 so that division is not exact.
- (a) y¹² — Method: when a power is raised to another power, multiply the two indices. Working: (y³)⁴ means y³ × y³ × y³ × y³, which is four lots of three y's multiplied together, so the index is 3 × 4 = 12 and (y³)⁴ = y¹². y⁷ comes from adding the indices, 3 + 4 = 7, which is the rule for multiplying two separate powers, not for raising a power to a power. y⁸¹ comes from working out 3⁴ = 81 and using that as the index, raising the inner index to the outer power instead of multiplying the two indices. 12y comes from multiplying the indices to make 12 but then treating y as a coefficient instead of a power. Answer: y¹².
- (a) 5 — Pressure = force ÷ area = 20 ÷ 4 = 5 pascals. Getting 80 comes from multiplying the force and area instead of dividing. Getting 16 comes from subtracting the area from the force (20 − 4 = 16) instead of dividing. Getting 24 comes from adding the force and area (20 + 4 = 24) 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.