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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- 1.3.4 × 10⁵ is written as an ordinary number. Work out this number.
- 2.Two numbers have a sum of 24. Their difference is 6. Work out the two numbers.
- 3.The ratio of the number of pages Mia has read to the number of pages Leo has read is 7 : 4. Mia has read 84 pages. Work out how many pages they have read altogether.
- 4.An angle is measured with a protractor as 47°. What type of angle is this?
- 5.The probability that a pupil chosen at random has a nut allergy is 1/10. There are 30 pupils in a class. Work out how many of the 30 pupils would be expected to have a nut allergy.
- 6.A vertical line chart shows the number of goals scored by a football team in each of its 20 matches: 0 goals in 4 matches, 1 goal in 7 matches, 2 goals in 6 matches and 3 goals in 3 matches. Write down the modal number of goals.
- 7.Write down the reciprocal of 0.2
- 8.The point A lies on the x-axis. Which statement about A must be true?
- 9.Two cars travel at constant speeds. Car A travels 150 km in 3 hours. Car B travels 180 km in 4 hours. Which car is faster, and what is its speed?
- 10.A square-based pyramid has a square base and four identical triangular sloping faces, with its apex directly above the centre of the base. How many planes of symmetry does it have?
- 11.A frequency tree records how 90 pupils travel to school and whether they were late. The first pair of branches splits the 90 pupils into 50 who walk and 40 who do not walk. On the walking branch, 6 were late and 44 were not. On the other branch, 10 were late and 30 were not. One of the 90 pupils is chosen at random. Work out the probability that the pupil walks to school and was late.
- 12.Work out (−36) ÷ (−6) × 2
- 13.Solve the simultaneous equations x + y = 1 and 2x + y = 5.
- 14.A map has a scale of 1 : 100. A path measures 5 cm on the map. Work out the real length of the path, in metres.
- 15.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.
- 16.A tin of soup has a mass of 450 g. A bag of potatoes has a mass of 2.6 kg. Work out the total mass, in kilograms.
- 17.A box contains 6 chocolates. Write down an expression for the total number of chocolates in c boxes.
- 18.In a mix of mortar the ratio of the mass of sand to the mass of cement is 5 : 2. Write the mass of cement as a fraction of the total mass of the mix.
- 19.Write 0.06 as a fraction in its simplest form.
- 20.1 metre = 100 cm. Work out how many square centimetres there are in 1 square metre.
Answer key
- (d) 340,000 — 3.4 × 10⁵ means moving the decimal point in 3.4 five places to the right, giving 340,000. Moving it only four places gives 34,000, one place short. Moving it six places gives 3,400,000, one place too many. Treating the exponent as negative instead of positive moves the point the wrong way, giving 0.000034.
- (c) 15 and 9 — Method: write the two facts as two equations in the same pair of letters and add them, because the letter with opposite signs cancels. Working: with x the larger number and y the smaller, x + y = 24 and x − y = 6; adding gives 2x = 30, so x = 15, and substituting into x + y = 24 gives y = 9. Answer: 15 and 9, which add to 24 and differ by 6. The distractors: 18 and 6 come from halving 24 to 12 and then adding and subtracting the whole difference of 6 instead of half of it, which leaves a difference of 12; 15 and 21 come from finding the larger number correctly and then adding 6 to it instead of subtracting; 15 and 6 come from finding the larger number and then taking the 6 in the question to be the smaller number itself.
- (d) 132 — Method: find the value of one part of the ratio, use it to find Leo's pages, then add both amounts together. Working: 84 ÷ 7 = 12 (value of one part). Leo's pages = 12 × 4 = 48. Total = 84 + 48 = 132. Wrong options: 48 gives only Leo's pages and forgets to add Mia's; 147 comes from reversing the ratio parts (84 ÷ 4 × 7 = 147) and stopping there; 231 comes from reversing the ratio parts and then adding Mia's pages (84 + 147).
- (b) Acute — An acute angle is any angle less than 90°. Since 47° is less than 90°, it is acute. An obtuse angle is between 90° and 180°, which 47° is not. A reflex angle is greater than 180°, which is far too big for 47°. A right angle is exactly 90°, not 47°. The angle is acute.
- (b) 3 pupils — Method: an expected frequency is the probability multiplied by the number of trials, so multiply the probability by the number of pupils. Working: 30 × 1/10 means finding one tenth of 30, and 30 ÷ 10 = 3. Answer: 3 pupils would be expected to have a nut allergy. The distractors: 27 pupils comes from working out how many are expected NOT to have the allergy, 30 − 3, instead of how many are; 10 pupils comes from reading the 10 in the fraction 1/10 as the number of pupils; 1 pupil comes from reading the numerator of the fraction as the expected number.
- (b) 1 — The four frequencies are 4, 7, 6 and 3 matches, and the largest of these is 7, which corresponds to 1 goal, so the modal number of goals is 1. Choosing 7 confuses the frequency, how many matches, with the number of goals itself. Choosing 2 uses the second-largest frequency, 6 matches, instead of the largest. Choosing 3 uses the smallest frequency, which belongs to the fewest matches, not the most.
- (c) 5 — 0.2 = 1/5, and turning the fraction upside down gives the reciprocal 5/1 = 5. Writing −0.2 mistakes the reciprocal for the negative of the number. Working out 1 − 0.2 = 0.8 mistakes the reciprocal for subtracting the number from 1. Writing 0.5 is the reciprocal of 2, not of 0.2 — a place-value slip that drops the decimal, since 1 ÷ 0.2 = 5 while 1 ÷ 2 = 0.5.
- (a) y = 0 — Method: in the pair (x, y) the first coordinate measures how far left or right of the origin a point is and the second how far above or below the x-axis it is, so a point on an axis has one of those measurements equal to zero. Working: the x-axis is the horizontal line through the origin, so a point sitting on it is neither above nor below that line and its second coordinate is zero, while its first coordinate may be positive, negative or zero. Answer: y = 0. The distractors: x = 0 is the condition for lying on the y-axis, the other axis; x > 0 comes from assuming a point on the x-axis must be to the right of the origin, which is true only of part of that axis; x = y holds only at the origin, which is one point of the x-axis rather than a property shared by all of them.
- (b) Car A, 50 km/h — Method: speed = distance ÷ time for each car, then compare. Working: Car A = 150 ÷ 3 = 50 km/h. Car B = 180 ÷ 4 = 45 km/h. Since 50 > 45, Car A is faster, travelling at 50 km/h. Wrong options: Car B, 45 km/h correctly finds Car B's speed but wrongly names the slower car as faster; Car A, 45 km/h picks the correct car but uses Car B's speed by mistake; Car B, 50 km/h picks the wrong car but uses Car A's correct speed value.
- (d) 4 — Method: a plane of symmetry must pass through the apex and cut the base along one of the base's own lines of symmetry. Working: a square has 4 lines of symmetry (2 through opposite edge midpoints, 2 through opposite corners), and each of these, combined with the apex, gives one plane of symmetry of the pyramid. A student who answers 2 has only found the planes through the edge midpoints, or only the ones through the corners, and missed the other pair. A student who answers 8 has doubled the correct count, perhaps confusing it with a different solid. A student who answers 1 has only spotted the one obvious front-to-back plane. Answer: 4.
- (d) 1/15 — Method: a probability read from a frequency tree is the count at the end of the branch you want, divided by the total number in the whole experiment. Working: the branch for walking followed by the branch for being late ends with 6 pupils, and the experiment covers all 90 pupils, so the probability is 6/90. Dividing the top and the bottom by 6 gives 1/15. Answer: the probability is 1/15. The distractors: 3/25 is 6/50 and comes from dividing the 6 by the 50 walkers rather than by the whole group, which answers a different question about walkers only; 8/45 is 16/90 and comes from counting every late pupil, the 6 walkers and the 10 others together, instead of only the late walkers; 1/9 is 10/90 and comes from reading the late count on the branch for pupils who do not walk.
- (c) 12 — Method: division and multiplication have equal priority, so they are carried out in the order they are written, from left to right; a negative divided by a negative is positive. Working: (−36) ÷ (−6) = 6, and then 6 × 2 = 12. Answer: 12. The distractors: 3 comes from carrying out the multiplication first, (−6) × 2 = −12 followed by (−36) ÷ (−12) = 3; −12 comes from treating a negative divided by a negative as negative, giving −6 and then −6 × 2 = −12; 6 comes from stopping at the division and never carrying out the multiplication by 2.
- (a) x = 4, y = −3 — Method: both equations contain +y with the same coefficient, so subtracting one equation from the other removes y. Working: (2x + y) − (x + y) = 5 − 1 gives x = 4; substituting x = 4 into x + y = 1 gives 4 + y = 1, so y = −3. Answer: x = 4, y = −3, which also satisfies 8 − 3 = 5. The distractors: x = 4, y = 5 comes from rearranging x + y = 1 as y = 1 + x; x = −2, y = 3 comes from eliminating x by doubling the first equation and then reading −y = 3 as y = 3; x = 6, y = −5 comes from adding the constants instead of subtracting them while eliminating y, taking x as 5 + 1.
- (c) 5 m — The scale 1 : 100 means 1 cm on the map represents 100 cm in real life. The path is 5 cm on the map, so the real length is 5 × 100 = 500 cm. Convert to metres: 500 cm = 5 m. Dividing instead of multiplying, 5 ÷ 100 = 0.05, gives 0.05 m — the scale must be used to make the real object bigger than the map, not smaller. Leaving the answer as 500 without converting to metres and calling it 500 m mistakes centimetres for metres. Using a scale of 1 : 1000 instead of the given 1 : 100 gives 5 × 1000 = 5000 cm = 50 m, ten times too large.
- (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.
- (c) 3.05 kg — There are 1000 g in 1 kg, so 450 g = 450 ÷ 1000 = 0.45 kg. The total mass is 0.45 + 2.6 = 3.05 kg. 7.1 kg divides the grams by 100 instead of 1000, 2.645 kg divides them by 10 000, and 452.6 kg adds the two numbers without converting the grams at all.
- (a) 6c — Each box has 6 chocolates, so c boxes have 6 × c = 6c chocolates. A candidate who adds the number of boxes to the number per box instead of multiplying gets 6 + c. A candidate who subtracts instead of multiplying gets c − 6. A candidate who divides instead of multiplying gets c ÷ 6.
- (c) 2/7 — Method: each number in a ratio counts parts, so the whole is all the parts added together, and the fraction wanted is the cement parts over that total. Working: the mix has 5 + 2 = 7 parts altogether, and 2 of those parts are cement, so the cement is 2/7 of the mix. Answer: 2/7. The distractors: 2/5 comes from writing the cement parts over the sand parts, comparing one part with the other part instead of with the whole; 5/7 comes from writing the sand parts over the total, which answers about the wrong material; 7/2 comes from writing the total over the cement parts, turning the fraction upside down.
- (b) 3/50 — Method: write the decimal over the power of ten that matches the number of digits after the point, counting every digit including a zero, then divide the numerator and the denominator by their highest common factor. Working: 0.06 has two digits after the point, so it is 6 hundredths and can be written as 6/100; the highest common factor of 6 and 100 is 2, and 6 ÷ 2 = 3 with 100 ÷ 2 = 50. Answer: 3/50. The distractors: 3/5 comes from ignoring the zero straight after the point and converting 0.6 instead, giving 6/10, which cancels to 3/5; 3/500 comes from counting three decimal places instead of two and writing 6/1000, which cancels to 3/500; 1/6 comes from putting 1 over the digits after the point, as though 0.06 meant one sixth.
- (b) 10000 cm² — Method: an area conversion factor is the square of the length conversion factor, because both sides of the square are scaled. Working: a square metre is a square of side 100 cm, so its area is 100 × 100 = 10000 cm². Answer: 10000 cm². The distractors: 100 cm² comes from using the length factor without squaring it; 200 cm² comes from doubling the length factor instead of squaring it; 1000000 cm² comes from cubing the factor, which is the conversion for a volume, not an area.
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.