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.Write 200 as a product of its prime factors, using index notation.
- 2.A car park charges by the hour. Parking for 1 hour costs £5, 2 hours costs £9, 3 hours costs £13 and 4 hours costs £17, with the cost increasing by the same amount for each extra hour. Work out an expression, in terms of n, for the cost, in pounds, of parking for n hours.
- 3.Two mathematically similar cylinders have heights in the ratio 3 : 4. Write the ratio of their volumes in its simplest form.
- 4.Work out the distance between the point (0, 0) and the point (5, 12).
- 5.In a survey of 45 students, 22 said they like reading, 18 said they like gaming, and 6 said they like both. Work out how many students like exactly one of reading or gaming.
- 6.Five pupils spent these numbers of minutes on their homework: 50, 65, 55, 90, 60. Work out the median time.
- 7.Work out the value of 5³.
- 8.A number, n, is doubled and then 7 is subtracted. The result is 15. Form an equation and solve it to find n.
- 9.It takes 8 painters 6 days to paint a fence. Working at the same rate, work out how many days 4 painters would take to paint the same fence.
- 10.A quadrilateral has angles of 100°, 85° and 95° at three of its vertices. Work out the fourth angle.
- 11.Priya spins a fair spinner with P(red) = 0.3, and separately flips a fair coin with P(heads) = 0.5. Using a tree diagram, work out the probability that she gets red AND heads.
- 12.On one night in Manchester the temperature at 3 am was 8 °C below zero and at 9 am it was 3 °C below zero. Write down the warmer of the two readings.
- 13.A phone plan costs a fixed £15 per month plus 8p per minute of calls. In one month, Jamal's bill was £23.80. Form an equation using m for the number of minutes, and solve it to find how many minutes Jamal used that month.
- 14.Two mathematically similar jugs have heights 8 cm and 12 cm. The smaller jug holds 200 ml when it is full. Work out how much the larger jug holds when it is full.
- 15.A warehouse stores identical cube-shaped crates. Its plan view is a 2 by 4 rectangle of crate positions, and every position is filled to a height of 3 crates, except one corner position, which has only 2 crates stacked on it because a delivery was incomplete. How many crates are there in total?
- 16.A garden has flowers, herbs and vegetables in the ratio 2 : 3 : 4. What fraction of the garden is not vegetables?
- 17.Solve the inequality x − 10 < −3.
- 18.It takes 2 identical pumps 10 hours to empty a flooded basement. Working at the same rate, work out how many hours 5 of these pumps would take to empty the same basement.
- 19.A digital timer truncates every time to 1 decimal place. It shows a swimmer's time for one length as 12.3 seconds. Using t for the swimmer's actual time in seconds, write down the error interval for t.
- 20.A cyclist rides 30 km in 1 hour 30 minutes. Work out the average speed of the cyclist in km/h.
Answer key
- (d) 2³ × 5² — Method: divide repeatedly by the smallest prime number, then write any repeated prime using a power. Working: 200 ÷ 2 = 100, 100 ÷ 2 = 50, 50 ÷ 2 = 25, 25 ÷ 5 = 5, and 5 is prime, so 200 = 2 × 2 × 2 × 5 × 5, written as 2³ × 5². 2² × 5³ swaps the two powers, giving 4 × 125 = 500, not 200. 2³ × 5 leaves out one of the two 5s, giving 8 × 5 = 40, not 200. 2 × 5³ leaves out two of the three 2s, giving 2 × 125 = 250, not 200. Answer: 2³ × 5².
- (a) 4n + 1 — Method: find the increase in cost per hour, then find the constant by adjusting the 1-hour cost. Working: the cost goes up by £4 for each extra hour (9 − 5 = 4), so the coefficient of n is 4. The constant is the 1-hour cost minus the common difference: 5 − 4 = 1. Answer: the nth term is 4n + 1. 4n + 5 comes from using the 1-hour cost, 5, as the constant without subtracting the common difference. 4n − 3 comes from a slip in working out the constant, subtracting the common difference twice (5 − 4 − 4 = −3) instead of once. n + 4 comes from swapping the hourly increase and the constant.
- (c) 27 : 64 — For similar solids, the ratio of volumes is the ratio of lengths cubed: 3³ : 4³ = 27 : 64. 3 : 4 comes from using the height ratio itself as the volume ratio, without cubing it at all. 9 : 16 comes from squaring each part instead of cubing (3² : 4²) — squaring is the rule for area, not volume. 27 : 4 comes from cubing only the first part of the ratio (3³ = 27), and leaving the second part uncubed.
- (b) 13 — Method: the straight-line distance between two points on a grid is the hypotenuse of a right-angled triangle whose shorter sides are the horizontal gap and the vertical gap between them, so Pythagoras' theorem applies. Working: one point is the origin, so the horizontal gap is 5 and the vertical gap is 12. Then d² = 5² + 12² = 25 + 144 = 169, so d = √169 = 13. Answer: 13. The distractors: 17 comes from adding the two gaps, 5 + 12, instead of adding their squares and taking the root; 12 comes from reading off the vertical gap alone and offering that as the whole distance; 7 comes from subtracting the gaps, 12 − 5, as though a distance were a difference.
- (c) 28 — Reading only is 22 − 6 = 16, and gaming only is 18 − 6 = 12, so exactly one of the two is 16 + 12 = 28. Adding 22 and 18 without removing the 6 who like both, 22 + 18 = 40, counts those 6 students twice. Giving 6 mistakes the number who like both for the number who like exactly one. Finding 22 + 18 − 6 = 34 gives the number who like at least one of reading or gaming, but stops there instead of also removing the 6 who like both to leave only those who like exactly one.
- (c) 60 minutes — Method: the median is the middle value once the data have been put in order of size, so the list must be sorted before any position is read. Working: in order the times are 50, 55, 60, 65, 90 minutes; there are 5 values, so the middle position is the third and the time sitting there is 60 minutes. Answer: 60 minutes. The distractors: 64 minutes comes from working out the mean, 320 ÷ 5, instead of the median; 70 minutes comes from taking the time halfway between the shortest and the longest, (50 + 90) ÷ 2; 40 minutes comes from working out the range, 90 − 50, which measures spread rather than centre.
- (c) 125 — Method: a number cubed means multiplying the number by itself three times. Working: 5 × 5 × 5 = 125. 15 comes from working out 5 × 3 = 15, multiplying by the index instead of cubing. 8 comes from working out 5 + 3 = 8 instead of cubing. 53 comes from writing the base and the index next to each other instead of carrying out the calculation. Answer: 125.
- (a) 11 — 2n − 7 = 15, so adding 7 to both sides gives 2n = 22, and dividing by 2 gives n = 11. A candidate who makes a sign error and forms the equation 2n + 7 = 15 gets 2n = 8 and n = 4. A candidate who correctly adds 7 to get 2n = 22 but forgets to divide by 2 gets n = 22. A candidate who ignores the −7 altogether and solves 2n = 15 gets n = 7.5.
- (b) 12 days — This is inverse proportion: fewer painters take longer. Multiply the original numbers to find the total painter-days needed: 8 × 6 = 48 painter-days. Divide by the new number of painters: 48 ÷ 4 = 12 days. Working out 6 × 4 ÷ 8 = 3 days treats it as direct proportion, as if fewer painters needed less time. Stopping at 48 gives the total painter-days, not the number of days. Working out 6 + (8 − 4) = 10 days adds the change in the number of painters straight onto the number of days, treating painters and days as the same kind of quantity. 4 painters take 12 days.
- (d) 80° — The four angles of any quadrilateral add up to 360°. Three of the angles add up to 100 + 85 + 95 = 280°, so the fourth angle is 360 − 280 = 80°. "70°" comes from using 350° as the total instead of 360°, an easy slip on the quadrilateral angle sum. "280°" comes from stopping at the sum of the three given angles and writing that total down as the answer, instead of subtracting it from 360°. "260°" comes from subtracting only the one angle 100° from 360°, instead of subtracting all three given angles.
- (c) 0.15 — Method: for two independent events, multiply along the branches of the tree to find the probability of both outcomes happening together. Working: P(red and heads) = P(red) × P(heads) = 0.3 × 0.5 = 0.15. Answer: 0.15. Watch out: adding the two probabilities, 0.3 + 0.5 = 0.8, does not give the probability of both — probabilities along one path of a tree are multiplied, not added. Writing down 0.5 ignores the spinner altogether and gives only the coin's probability. And writing down 0.65 is the probability of red OR heads, which is 0.3 + 0.5 − 0.15 = 0.65, a different question from the one asked here.
- (a) −3 °C — Method: write each reading as a signed temperature, then choose the one further to the right on a number line. Working: 8 °C below zero is −8 °C and 3 °C below zero is −3 °C. On a number line −3 lies to the right of −8, so it is the warmer reading. Answer: −3 °C. The distractors: −8 °C comes from ordering negatives by the size of their digits, treating −8 as the larger number; 3 °C has the right size but the sign dropped, and a reading of 3 °C is above zero rather than below it; 5 °C comes from working out the difference between the two readings instead of choosing one of them.
- (d) 110 — Method: form the equation 15 + 0.08m = 23.80, where m is the number of minutes, then solve for m. Working: subtract the fixed fee: 0.08m = 23.80 − 15 = 8.80. Divide by the cost per minute: m = 8.80 ÷ 0.08 = 110. Answer: 110 minutes. 1.1 comes from using 8 instead of 0.08 as the cost per minute, forgetting to convert pence to pounds. 297.5 comes from dividing the whole bill by the cost per minute without subtracting the fixed fee first, 23.80 ÷ 0.08. 485 comes from adding the fixed fee to the bill instead of subtracting it, before dividing by the cost per minute, (23.80 + 15) ÷ 0.08.
- (a) 675 ml — How much a jug holds is a volume, and volumes of similar solids scale with the cube of the length scale factor. The length scale factor is 12 ÷ 8 = 1.5, so the volume scale factor is 1.5 × 1.5 × 1.5 = 3.375. The larger jug holds 200 × 3.375 = 675 ml. Multiplying the scale factor by 3 instead of raising it to the power 3 is the mistake to guard against here.
- (b) 23 — If every position were filled to the full height of 3, the total would be 2 × 4 × 3 = 24 crates. One corner position has only 2 crates instead of 3, one crate short of full height there, so the actual total is 24 − 1 = 23. "24" comes from using the full height everywhere and forgetting the one incomplete corner. "22" comes from removing 2 crates for the incomplete corner instead of the 1 that is actually missing (3 − 2 = 1, not 2). "21" comes from removing all 3 crates at that corner, as though the position were completely empty rather than 2 crates short.
- (d) 5/9 — Total parts = 2 + 3 + 4 = 9. Flowers and herbs together make up 2 + 3 = 5 parts, so the fraction that is not vegetables is 5/9. 4/9 comes from finding the fraction of vegetables instead of the fraction that is not vegetables. 5/7 comes from leaving flowers out of the total and using only herbs and vegetables, 3 + 4 = 7, as the total. 3/4 comes from writing the ratio of herbs to vegetables directly as a fraction.
- (d) x < 7 — Method: add 10 to both sides to leave x on its own; adding the same number to both sides never changes the direction of an inequality. Working: adding 10 to both sides of x − 10 < −3 leaves x on the left and −3 + 10 on the right, and −3 + 10 = 7, so x < 7. Answer: x < 7. The distractors: x > 7 comes from turning the sign round while adding, as though every move flipped it; x < −13 comes from subtracting 10 from both sides instead of adding it, giving −3 take away 10; x < 13 comes from ignoring the minus sign on −3 and working out 3 + 10 instead.
- (a) 4 hours — This is inverse proportion: more pumps take less time. Multiply the original numbers to find the total pump-hours needed: 2 × 10 = 20 pump-hours. Divide by the new number of pumps: 20 ÷ 5 = 4 hours. Working out 10 × 5 ÷ 2 = 25 hours treats it as direct proportion, as if more pumps needed more time. Stopping at 20 gives the total pump-hours, not the number of hours. Working out 10 − (5 − 2) = 7 hours subtracts the extra number of pumps straight from the number of hours, treating pumps and hours as the same kind of quantity. 5 pumps take 4 hours.
- (c) 12.3 ≤ t < 12.4 — Method: truncating cuts the later digits off instead of rounding them, so nothing is ever pushed upwards. The displayed value is therefore the smallest the time can be, and the time can run up to, but not reach, the next value the display can show. Working: the display reads 12.3, so the actual time is at least 12.3 seconds; as soon as the time reaches 12.3 + 0.1 = 12.4 seconds the display would read 12.4, so 12.4 is not included. Answer: 12.3 ≤ t < 12.4. The distractors: 12.25 ≤ t < 12.35 is the interval for a time rounded to 1 decimal place, and this display does not round; 12.3 < t ≤ 12.4 excludes the one value the display certainly allows and includes the one it rules out; 12.3 ≤ t ≤ 12.4 treats 12.4 seconds as possible, but at 12.4 seconds the display would no longer read 12.3.
- (d) 20 km/h — Method: average speed = total distance ÷ total time, with the time written in hours. Working: 1 hour 30 minutes = 1.5 hours, and 30 ÷ 1.5 = 20. Answer: 20 km/h. The distractors: 45 km/h comes from multiplying 30 by 1.5 instead of dividing; 15 km/h comes from dividing by 2, as if the ride had taken 2 hours; 30 km/h comes from dividing by the whole hour only and ignoring the extra 30 minutes.
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.