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%.
GCSE Foundation sample Paper 1 (non-calculator)
- 1.A carton of orange juice holds 1.35 litres. Ruby pours the juice equally into 4 identical glasses. Work out how much juice is in each glass, giving your answer as a fraction of a litre in its simplest form.
- 2.The first five cube numbers are 1, 8, 27, 64, 125. Write down the next cube number in the sequence.
- 3.Triangle A and triangle B are mathematically similar. A side of triangle A is 4 cm long and the corresponding side of triangle B is 10 cm long. Write the ratio of the length in triangle A to the length in triangle B in its simplest form.
- 4.Work out the gradient of the line segment joining A(2, 3) and B(6, 11).
- 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 scatter graph has 50 points. Most of them lie close to a rising line of best fit, but two of them lie a long way from that line. Write down how those two points should be treated.
- 7.Work out 3/8 of 96.
- 8.Solve 2x − 2 = 2
- 9.A laptop priced at £520 is first increased by 15%, and then the new price is decreased by 20%. Work out the final price of the laptop.
- 10.A shop stacks identical storage crates in a display. The diagram shows the plan of the floor positions used, with one corner position marked. Every floor position is filled with crates stacked 3 high, except the marked corner position, which is left completely empty. Work out how many crates are used in total.
- 11.Four bags each contain counters, some of which are winning counters. The probability of taking a winning counter is 5/12 from the red bag, 3/8 from the blue bag, 7/24 from the green bag and 1/3 from the yellow bag. Work out which bag has the smallest probability of giving a winning counter.
- 12.Work out the value of .
- 13.The formula for converting a distance in miles, m, to kilometres, k, is k = 8m ÷ 5. Work out k when m = 10.
- 14.The ratio of two lengths is 0.6 : 1.5. Write this ratio in its simplest whole-number form.
- 15.A triangular bunting flag has two equal sides. The angle between those two equal sides is 80°, and the other two angles of the flag are equal to each other. Work out the size of each of the two equal angles.
- 16.Work out √144 − 2 × 3 + √25
- 17.The diagram shows a straight line passing through the origin, drawn on a numbered grid. Which of these points lies on the line?
- 18.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.
- 19.A board game has 5 different character pieces and 4 different colour tokens. The dragon piece can only be used with the gold token. Work out how many different combinations of one character piece and one colour token are possible.
- 20.At an animal shelter, the ratio of cats to dogs is 2:5. There are 18 more dogs than cats. Work out the number of cats.
Answer key
- (d) 27/80 — Method: write the total as a fraction of a litre, then divide by the number of glasses. Working: 1.35 = 27/20, so each glass holds 27/20 ÷ 4 = 27/80 of a litre. Answer: 27/80. 27/20 comes from converting the total correctly to a fraction but forgetting to divide by the number of glasses. 27/5 comes from multiplying the total by 4 instead of dividing. 17/50 comes from rounding 1.35 ÷ 4 to 0.34 before converting to a fraction.
- (a) 216 — The cube numbers are formed by cubing 1, 2, 3, 4, 5, ..., so the next one is 6³ = 216. A candidate who squares 6 instead of cubing it gets 6² = 36. A candidate who assumes the sequence doubles each time gets 125 × 2 = 250. A candidate who uses 5 × 6² instead of 6 × 6 × 6 gets 180.
- (c) 2 : 5 — A ratio is written in the order the question names the two shapes, so triangle A's length comes first: 4 : 10. Both parts divide by 2: 4 ÷ 2 = 2 and 10 ÷ 2 = 5, giving 2 : 5. Lengths are compared using the lengths themselves, so nothing is squared here; squaring both parts would give the ratio of the areas instead.
- (a) 2 — Gradient = (change in y) ÷ (change in x) = (11 − 3) ÷ (6 − 2) = 8 ÷ 4 = 2. "0.5" comes from dividing the change in x by the change in y the wrong way round: 4 ÷ 8. "8" is only the change in y, forgetting to divide by the change in x at all. "−2" comes from a sign error, as if the y-coordinate had decreased rather than increased.
- (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.
- (b) Treat them as outliers and check them before deciding — Method: a point lying a long way from the pattern the rest of the data make is called an outlier, and an outlier is investigated before anything is done with it, because it may be an error in the data or it may be a genuine but unusual case. Working: 48 of the 50 points lie close to the rising line of best fit, so the trend is set by those 48; the two remaining points do not follow it, so they are identified as outliers and checked — a mistake in measuring or recording would be corrected, while a genuine reading would be kept and reported. Answer: treat them as outliers and check them before deciding what to do with them. The distractors: deleting them at once assumes that every point far from the line must be an error, which throws away real data; moving the line so that it passes through them assumes a line of best fit must touch particular points, when it is drawn to follow all 50; taking them as proof that there is no correlation lets two points overturn the pattern that the other 48 agree on.
- (d) 36 — To find 3/8 of 96, divide by the denominator and multiply by the numerator: 96 ÷ 8 = 12, and 12 × 3 = 36. Dividing 96 by the numerator instead of the denominator, 96 ÷ 3 = 32, uses the wrong part of the fraction. Using 3/4 instead of 3/8, perhaps misreading the denominator, gives 96 × 3/4 = 72. Finding 96 ÷ 8 = 12 correctly but forgetting to multiply by the numerator 3 leaves just 12 as the final answer.
- (a) x = 2 — Method: undo the subtraction first, then undo the multiplication, doing the same to both sides each time. Working: adding 2 to both sides gives 2x = 4, and dividing both sides by 2 gives x = 2. Answer: x = 2. The distractors: x = 4 comes from stopping at 2x = 4 and writing 4 as the value of x; x = 0 comes from subtracting 2 from both sides instead of adding it, giving 2x = 0; x = 8 comes from multiplying 4 by 2 instead of dividing by 2.
- (c) £478.40 — Method: apply the percentage increase, then apply the percentage decrease to the new price. Working: after the increase, the laptop costs £520 × 1.15. Multiplying this result by 0.80 gives the final price, £478.40. Answer: £478.40. £494 comes from combining the two percentages into a single net change (15% − 20% = −5%) and applying it directly, £520 × 0.95 = £494, instead of applying the two changes one after the other. £416 comes from applying only the 20% decrease to the original price, £520 × 0.80 = £416, forgetting the increase entirely. £598 comes from applying only the 15% increase and stopping there, forgetting to apply the decrease at all.
- (a) 33 — Method: total crates = (number of floor positions that actually have crates on them) × (the stack height). Working: there are 4 × 3 = 12 floor positions in the whole arrangement, but one corner position is left empty, leaving 11 filled positions; each filled position is stacked 3 crates high, so 11 × 3 = 33. Answer: 33. The distractors: 36 comes from forgetting to remove the empty corner and using all 12 positions (12 × 3). 35 comes from removing only one crate for the empty corner instead of the full stack of 3 (36 − 1). 11 comes from counting the filled floor positions and stopping there, forgetting that each one carries a stack 3 crates high.
- (c) The green bag (7/24) — Method: fractions can only be ordered once they share a denominator, so rewrite all four over the lowest common denominator and compare the numerators. Working: the lowest common denominator of 12, 8, 24 and 3 is 24, and scaling gives 5/12 = 10/24, 3/8 = 9/24, 7/24 stays as it is, and 1/3 = 8/24; the numerators are then 10, 9, 7 and 8. Answer: the smallest numerator is 7, so the green bag, with 7/24, is the least likely and sits furthest to the left on the 0 to 1 scale. The distractors: the red bag (5/12) comes from finding the largest of the four probabilities instead of the smallest; the blue bag (3/8) comes from scaling 3/8 by changing only the denominator to 24, which turns it into 3/24 and makes it look the smallest; the yellow bag (1/3) comes from comparing numerators alone and assuming the fraction with the numerator 1 must be the smallest.
- (c) 0.001 — Method: a negative index means 'one over' the positive power, so $10^{-3}$ means one over $10^{3}$. Working: ten cubed is 1000, and one over 1000 is 0.001. 1000 comes from ignoring the negative sign and working out ten cubed instead of its reciprocal. −1000 comes from ignoring what the negative index does to the size, while still writing a negative sign on the large value. −0.001 comes from correctly finding the size, 0.001, but wrongly keeping a negative sign because the index was negative. Answer: 0.001.
- (a) 16 — k = 8m ÷ 5 = (8 × 10) ÷ 5 = 80 ÷ 5 = 16. A candidate who multiplies 8 by 10 but forgets to divide by 5 gets 80. A candidate who divides 10 by 8 and multiplies by 5, using the fraction upside down, gets 10 × 5 ÷ 8 = 6.25. A candidate who adds instead of substituting into the formula gets 8 + 10 = 18.
- (d) 2:5 — Multiply both parts by 10 to clear the decimals: 6 : 15. Divide both parts by their common factor 3: 2 : 5.
- (b) 50° — Method: the three angles of a triangle add up to 180°, and the two angles opposite the equal sides are equal, so subtract the given angle from 180° and halve the remainder. Working: 180° − 80° = 100°, and 100° ÷ 2 = 50°. Answer: 50°. The distractors: 100° comes from subtracting from 180° and forgetting to halve, so it is the two equal angles together; 40° comes from halving the 80° that is given rather than halving what is left of the 180°; 80° comes from assuming that the two base angles must match the angle between the equal sides.
- (b) 11 — Method: roots and the multiplication are worked out before the addition and subtraction, and what is left is then worked through from left to right. Working: √144 = 12, √25 = 5 and 2 × 3 = 6, so the calculation becomes 12 − 6 + 5, which gives 6 + 5 = 11. Answer: 11. The distractors: 1 comes from carrying out the addition before the subtraction, giving 12 − (6 + 5) = 12 − 11 = 1; 35 comes from working from left to right with no priority, giving 12 − 2 = 10, then 10 × 3 = 30 and 30 + 5 = 35; 7 comes from combining the two roots as √(144 + 25) = √169 = 13 and then subtracting the product, giving 13 − 6 = 7.
- (c) (2, 6) — Method: substitute the x-coordinate of each point into the rule for the line (y = 3 × x) and compare it with the point's y-coordinate. Working: the line passes through the origin and rises 3 squares for every 1 square across, so at x = 2 the line's y-value is 3 × 2 = 6, giving the point (2, 6). Answer: (2, 6). Distractor refutation: (2, 3) comes from counting only 3 squares up in total between the origin and x = 2, instead of 3 squares up for every 1 square across, halving the true rise. (3, 2) comes from swapping the x-coordinate and the y-coordinate round. (2, 5) comes from a miscounted gridline, landing one square below the line.
- (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.
- (d) 17 — Without the restriction there would be 5 × 4 = 20 combinations. The dragon piece can only be paired with the gold token, so of the 4 tokens, 3 are not allowed with the dragon piece, giving 20 − 3 = 17 valid combinations. 20 comes from ignoring the restriction completely. 19 comes from subtracting only 1 of the 3 invalid dragon combinations instead of all 3, 20 − 1 = 19. 16 comes from multiplying only the 4 non-dragon pieces by the 4 tokens, 4 × 4 = 16, and forgetting to add back the one valid combination of the dragon piece with the gold token.
- (b) 12 — The difference between the parts of the ratio is 5 − 2 = 3 parts, and this is worth 18. Divide to find one part: 18 ÷ 3 = 6. Cats have 2 parts: 2 × 6 = 12. (30 is the number of dogs, using 5 parts instead of 2. 6 is the value of one part — the number of cats is 2 lots of this, not just one. 9 comes from dividing 18 by 2 and stopping there, instead of dividing by the difference in parts, 3, and then multiplying by 2.)
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.