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.Which of these is written correctly in standard form?
- 2.The diagram shows two curves drawn on the same grid. Curve A is the graph of . Which equation could represent curve B?
- 3.In a car park there are 35 cars and 15 vans. Write the ratio of the number of vans to the number of cars in its simplest form.
- 4.A fish tank is a cuboid measuring 40 cm long, 25 cm wide and 30 cm high. Water is poured in to a depth of 20 cm. Work out the volume of water in the tank, in litres.
- 5.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.
- 6.The number of pets owned by each of 19 pupils in a class is recorded: 0 pets — 7 pupils, 1 pet — 3 pupils, 2 pets — 4 pupils, 3 pets — 5 pupils. Work out the median number of pets.
- 7.Which of these decimals lies between 1/4 and 2/5?
- 8.y = 3x − 2. Work out the value of y when x = 4.y = 3x − 2
- 9.A 5 kg bag of potatoes costs £4.50. Work out the cost per kilogram.
- 10.Triangle ABC has AB = 5 cm, BC = 7 cm and CA = 9 cm. Triangle DEF has DE = 5 cm, EF = 7 cm and FD = 9 cm. Which condition proves the two triangles are congruent?
- 11.A bag contains 2 white counters and 8 black counters. One counter is taken at random from the bag. Work out the probability that the counter is not black.
- 12.To estimate the cost of buying 38.7 m of rope at £21.40 per metre, both numbers are first rounded to 1 significant figure. Work out the estimate.
- 13.A tank already contains some water and is then filled at a constant rate. After 2 minutes it contains 50 litres in total; after 7 minutes it contains 150 litres in total. Work out the rate at which water is added, in litres per minute.
- 14.y is directly proportional to x. When x = 7, the value of y is 21. Work out the value of x when y = 12.
- 15.Lines PQ and RS are parallel. A straight line crosses PQ at point X and crosses RS at point Y. Angle PXY = 63°. Angle PXY and angle XYS are co-interior (allied) angles, which add up to 180°. Work out the size of angle XYS.
- 16.Work out the lowest common multiple of 4 and 6.
- 17.The first five terms of a sequence are 2, 5, 8, 11, 14. Work out an expression, in terms of n, for the nth term.
- 18.A map has a scale of 2 cm : 5 km. A footpath measures 8 cm on the map. Work out the real length of the footpath, in kilometres.
- 19.Is 120 divisible by 5? Give a reason for your answer.
- 20.Four graphs are described below. Each shows one quantity plotted against another. Write down the description of the graph that shows two quantities in direct proportion.
Answer key
- (a) 7.5 × 10⁴ — Standard form is A × 10ⁿ, where A is between 1 and 10 (1 ≤ A < 10) and n is an integer — 7.5 × 10⁴ satisfies all of this. 12 × 10³ fails because 12 is not below 10. 0.5 × 10⁴ fails because 0.5 is not at least 1. 6.2 × 4⁵ fails because standard form always uses a power of 10, not a power of 4.
- (d) $y = x^2 + 3$ — Method: a point lies on a curve only if substituting its x-coordinate into the equation gives back its y-coordinate, so read one or two points off curve B and test each equation. Working: curve B crosses the y-axis at (0, 3), and its lowest point is also (0, 3); substituting x = 0 into y = x² + 3 gives 0² + 3 = 3, which matches. Checking a second point: at x = 2 curve B is at y = 7, and 2² + 3 = 4 + 3 = 7, which matches as well. Answer: curve B has equation y = x² + 3. Distractor refutation: y = x² − 3 comes from reading the 3 as a move down instead of a move up; substituting x = 0 gives −3, so that curve would cross the y-axis three squares below the origin, while curve B crosses it three squares above. y = (x − 3)² comes from putting the 3 inside the brackets; substituting x = 0 gives (−3)² = 9, and that curve's lowest point is at (3, 0), three squares to the right along the x-axis, whereas curve B has its lowest point on the y-axis. y = x² + 3x comes from attaching the 3 to the x term instead of writing it on its own; substituting x = 0 gives 0² + 3 × 0 = 0, so that curve passes through the origin, and curve B does not pass through the origin.
- (c) 3 : 7 — Method: a ratio 'of A to B' is written with A first, and it is simplified by dividing both parts by their highest common factor. Working: vans are named first, so the ratio starts as 15 : 35; the highest common factor of 15 and 35 is 5, and 15 ÷ 5 = 3 while 35 ÷ 5 = 7. Answer: 3 : 7. The distractors: 7 : 3 comes from writing the two numbers in the order they appear in the question, cars before vans, instead of the order asked for; 3 : 10 comes from comparing the vans with the total number of vehicles, 15 : 50, a part-to-whole ratio in place of the part-to-part ratio asked for; 15 : 35 is the right comparison left unsimplified, and the question asks for the simplest form.
- (a) 20 litres — Volume of water = length × width × depth of water = 40 × 25 × 20 = 20 000 cm³. Since 1000 cm³ = 1 litre, divide by 1000: 20 000 ÷ 1000 = 20 litres. A pupil who uses the full height of the tank, 30 cm, instead of the water depth, 20 cm, gets 40 × 25 × 30 = 30 000 cm³ = 30 litres. A pupil who forgets to convert cm³ to litres at all gives 20 000 litres. A pupil who divides by 1000 twice by mistake gets 20 000 ÷ 1000 ÷ 1000 = 0.02 litres. The correct volume of water is 20 litres.
- (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.
- (b) 1 — Method: for data in a frequency table, find the position of the median using (n + 1) ÷ 2, then read off the value at that position from the cumulative frequencies. Working: there are 19 pupils, so the median is the 10th value. The cumulative frequencies are 7 (up to 0 pets), 10 (up to 1 pet), 14 (up to 2 pets) and 19 (up to 3 pets). The 10th value falls at the end of the '1 pet' group, so the median is 1 pet. Giving 0 pets is the mode — the category with the highest frequency, 7 — not the median. Giving 3, the highest number of pets minus the lowest, finds the range, a different statistic entirely. Giving 19 states the total number of pupils, not a number of pets at all. Find the middle POSITION first, then read off the value it belongs to — do not confuse it with the mode, the range or the total.
- (d) 0.3 — Converting the fractions to decimals, 1/4 = 0.25 and 2/5 = 0.4, so any decimal between 0.25 and 0.4 is a valid answer, and 0.3 fits. Confusing 1/4 with 1/5 and converting it as 0.2 instead of 0.25 gives a value below the true lower bound. Confusing 2/5 with 1/2 and converting it as 0.5 instead of 0.4 gives a value above the true upper bound. Converting the fractions correctly but choosing a decimal above the true upper bound of 0.4 instead of between the two values gives 0.45.
- (c) 10 — Method: substitute the value of x, then follow the order of operations, carrying out the multiplication before the subtraction. Working: with x = 4 the rule reads y = 3 × 4 − 2; the multiplication gives 3 × 4 = 12, and taking 2 away gives 12 − 2. Answer: y = 10. The distractors: 14 comes from adding the 2 instead of subtracting it; 12 comes from working out 3 × 4 and stopping there, leaving the −2 unused; 6 comes from subtracting first, 4 − 2, and multiplying the result by 3.
- (c) £0.90 — Cost per kilogram = total cost ÷ mass. £4.50 ÷ 5 = £0.90. £1.11 comes from dividing the mass by the cost instead of the cost by the mass (5 ÷ 4.50). £22.50 comes from multiplying the cost by the mass instead of dividing (£4.50 × 5). £0.50 comes from subtracting the two numbers (5 − 4.50) instead of dividing.
- (d) SSS - three sides equal — All three pairs of corresponding sides are equal in length (5 cm, 7 cm and 9 cm in both triangles), so the triangles are congruent by SSS. SAS needs an angle to be given as well as two sides, but no angle is given here. ASA needs two angles and the side between them, but no angles are given at all. RHS needs a right angle and the hypotenuse, but no angle is stated to be a right angle.
- (a) 1/5 — Method: find the probability that the counter is black, then use the fact that an event and its complement add to 1. Working: there are 2 + 8 = 10 counters, so P(black) = 8/10 = 4/5; the complement is 1 − 4/5, and writing 1 as 5/5 gives 5/5 − 4/5. Answer: 1/5. The distractors: 4/5 comes from giving the probability that the counter is black instead of its complement; 1/2 comes from assuming the two colours are equally likely because there are only two of them; 1/4 comes from writing the 2 white counters over the 8 black counters rather than over all 10 counters.
- (d) £800 — Rounding 38.7 to 1 significant figure gives 40, and rounding 21.40 to 1 significant figure gives 20. Multiplying the rounded values gives an estimate of 40 × 20 = £800. Rounding 21.40 to the nearest whole number instead of to 1 significant figure gives 21, and 40 × 21 = £840, one place value too fine for the price. Adding the rounded values instead of multiplying them gives 40 + 20 = £60. Rounding both numbers to 2 significant figures instead of 1, giving 39 and 21, produces 39 × 21 = £819.
- (b) 20 — The rate is the gradient: the change in litres divided by the change in time. From 2 to 7 minutes, the tank gains 150 − 50 = 100 litres over 7 − 2 = 5 minutes, so the rate is 100 ÷ 5 = 20 litres per minute. 100 comes from working out the change in litres but forgetting to divide by the change in time. 25 comes from using only the first reading, 50 ÷ 2, and ignoring the second reading entirely. 0.05 comes from dividing the change in time by the change in litres instead of the other way round, 5 ÷ 100.
- (a) 4 — Method: find the constant of proportionality from the pair given, write the equation, then substitute the new value of y and solve. Working: k = 21 ÷ 7 = 3, so y = 3x; putting y = 12 gives 12 = 3x, and x = 12 ÷ 3 = 4. Answer: 4. The distractors: 36 comes from multiplying by the constant instead of dividing by it, 12 × 3, which is the proportion set up upside down; 84 comes from multiplying 12 by the 7 from the first pair, using a value of x as though it were the constant; 9 comes from working out 12 − 3, treating the equation as y = x + 3 rather than y = 3x.
- (c) 117° — Co-interior (also called allied) angles between parallel lines add up to 180°. Angle PXY and angle XYS are co-interior angles, so angle PXY + angle XYS = 180°. 180° − 63° = 117°, so angle XYS = 117°. A student who mistakes co-interior angles for alternate angles, which are equal rather than supplementary, answers 63° instead. A student who treats the two angles as complementary, subtracting from 90° instead of 180°, gets 90° − 63° = 27°. A student who assumes the transversal meets the parallel lines at right angles answers 90°.
- (a) 12 — List the multiples of each number: multiples of 4 are 4, 8, 12, 16, 20, 24; multiples of 6 are 6, 12, 18, 24. The lowest number that appears in both lists is 12. Picking 24, a common multiple but not the lowest one, gives an answer that is too big. Picking 6, the larger of the two original numbers rather than a common multiple, ignores that the lowest common multiple must appear in both lists. Working out the highest common factor instead of the lowest common multiple gives 2. So the lowest common multiple of 4 and 6 is 12.
- (a) 3n − 1 — Method: find the common difference between consecutive terms, then find the constant by adjusting the first term. Working: 5 − 2 = 3, 8 − 5 = 3, 11 − 8 = 3, so the common difference is 3 and the coefficient of n is 3. The constant is the first term minus the common difference: 2 − 3 = −1. Answer: the nth term is 3n − 1. 3n + 2 comes from using the first term, 2, as the constant without subtracting the common difference. 3n − 2 comes from a slip when working out the constant, treating 2 − 3 as −2 instead of −1. 2n + 3 comes from swapping the common difference and the first term.
- (b) 20 km — The scale 2 cm : 5 km means each 1 cm on the map represents 5 ÷ 2 = 2.5 km in real life. The footpath is 8 cm on the map, so its real length is 8 × 2.5 = 20 km. 40 km comes from multiplying 8 by 5 directly, ignoring that the scale's '2 cm' has to be divided out first: 8 × 5 = 40. 3.2 km comes from dividing 8 by 2.5 instead of multiplying: 8 ÷ 2.5 = 3.2. 5 km comes from multiplying 2.5 by the scale's '2' instead of by the footpath's 8 cm: 2.5 × 2 = 5.
- (a) Yes, because 120 ends in 0 — Method: a whole number divides exactly by 5 when its last digit is 5 or 0, so look at the final digit. Working: the final digit of 120 is 0, so 120 is a multiple of 5; the division confirms it, since 5 × 24 = 120 with nothing left over. Answer: Yes, because 120 ends in 0. The distractors: the option that says yes because 120 is even reaches the right conclusion from the wrong test, since being even is the test for divisibility by 2, and 14 is even but is not a multiple of 5; saying no because 5 does not divide into 12 comes from ignoring the final digit and testing only the leading digits; saying no because the digits add to 3 applies the digit-sum test, which works for 3 and for 9 but not for 5.
- (c) A straight line through the origin, rising from left to right — Two quantities are in direct proportion when one is a constant multiple of the other, so their graph is a straight line through the origin: when one quantity is 0 the other is 0 as well, and doubling one doubles the other. A straight line crossing the vertical axis at 5 has a fixed amount added on, so when the horizontal quantity is 0 the vertical quantity is 5, not 0 — a straight line on its own is not enough for direct proportion. A curve that falls steeply and then levels off without touching either axis shows inverse proportion: one quantity grows as the other shrinks, and their product stays the same. A horizontal line at a height of 3 shows a quantity that does not change at all as the other one grows, so it is not proportional to it.
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.