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.Two bags of mixed nuts are combined. Bag A has peanuts and cashews in the ratio 2 : 3. Bag B has peanuts and cashews in the ratio 1 : 4. Both bags contain the same total number of nuts. Work out the fraction of the combined mixture that is peanuts.
- 2.Simplify 5x + 3y − 2x + y.
- 3.A furniture designer makes a scale model of a wardrobe using a scale of 1 : 8. The real wardrobe is 2.4 m tall and 1.2 m wide. Work out the perimeter of the front face of the model wardrobe, in centimetres.
- 4.In triangle ABC, angle A is 2x°, angle B is 3x° and angle C is 4x°. Work out the size of angle B.
- 5.Spinner C has 3 equal sections, numbered 2, 3 and 5. Spinner D has 6 equal sections, numbered 1 to 6. Both spinners are spun once, and the two numbers are multiplied together. Work out the probability that the product is a multiple of 5.
- 6.A scatter graph of the number of ice creams sold at a seaside kiosk in Bournemouth and the number of sunburn cases treated at a nearby pharmacy, recorded on the same 30 days, shows strong positive correlation. Which statement about this correlation is correct?
- 7.At a swimming pool the ratio of adults to children is 3 : 5. Of the children, 2/5 are boys. What fraction of the people at the pool are boys?
- 8.A gym charges a reduced price for the first month of membership and a standard price for every month after that. The total cost is £5 for 1 month, £12 for 2 months, £19 for 3 months and £26 for 4 months. Work out an expression, in terms of n, for the total cost, in pounds, of n months of membership.
- 9.A car travels 90 km using 6 litres of petrol. Work out how far the car can travel, at the same rate, using 9 litres of petrol.
- 10.A solid is built from centimetre cubes standing on a table. Seen from the front, the left-hand column is 2 cubes high, the middle column is 2 cubes high and the right-hand column is 1 cube high. Work out how many squares make up the front elevation.
- 11.A two-way table records how 80 students travel to school. 45 of the students are boys and the rest are girls. 27 of the boys walk to school. Work out the probability, as a fraction, that a student chosen at random from the boys walks to school.
- 12.Work out an estimate for 79.3 − 24.6, by rounding each number to the nearest whole number.
- 13.Solve the simultaneous equations 3x + 2y = 16 and x + y = 7. Work out the value of y.
- 14.A plan is drawn to a scale of 1 : 25. A real fence is 10 m long. Work out the length of the fence on the plan, in centimetres.
- 15.Triangle ABC has AB = 9 cm, BC = 6 cm and angle A = 35°. Triangle DEF has DE = 9 cm, EF = 6 cm and angle D = 35°. Which of the following correctly identifies the condition shown here?
- 16.A roll of ribbon is 8.4 m long. Ribbon is cut into pieces that are each 0.6 m long. Work out how many complete pieces can be cut from the roll.
- 17.A rule turns each input x into an output y. The inputs are x = −1, 0, 1, 2 and the matching outputs are y = 5, 3, 1, −1. Work out the rule.
- 18.In a fruit drink, 3/8 of the total volume is orange juice and the rest is apple juice. Write down the ratio of orange juice to apple juice, in its simplest form.
- 19.A charity raises money from a raffle and a cake sale in the ratio 5 : 3. Altogether the charity raises £320. Work out how much money the cake sale raised.
- 20.A car is bought for £12000. Its value falls by 15% in the first year and by 10% in each year after that. Work out the value of the car 3 years after it was bought.
Answer key
- (b) 3/10 — Since each bag's ratio has 5 parts and both bags contain the same total number of nuts, imagine each bag has 5 nuts: Bag A has 2 peanuts and Bag B has 1 peanut, so together there are 2 + 1 = 3 peanuts out of a combined 5 + 5 = 10 nuts, giving 3/10. 1/5 comes from using only Bag A's peanuts, 2 out of 10, without adding Bag B's peanuts. 1/10 comes from using only Bag B's peanut, without adding Bag A's peanuts. 3/5 comes from writing the combined peanuts over the number of parts in one bag instead of the combined total number of nuts.
- (c) 3x + 4y — Collect the x terms: 5x − 2x = 3x. Collect the y terms: 3y + y = 4y. So 5x + 3y − 2x + y = 3x + 4y. A candidate who subtracts the y terms instead of adding them (3y − y) gets 3x + 2y. A candidate who adds 2x instead of subtracting it (5x + 2x) gets 7x + 4y. A candidate who wrongly combines the x and y terms into a single term gets 6xy.
- (d) 90 cm — Method: scale each dimension by the scale factor, then find the perimeter. Working: model height = 240 ÷ 8 = 30 cm; model width = 120 ÷ 8 = 15 cm. Perimeter = 2 × (30 + 15) = 90 cm. Wrong options: 11.25 cm comes from squaring the scale factor as if finding an area (720 ÷ 64); 510 cm comes from scaling only one dimension and leaving the other at full size; 720 cm comes from finding the real perimeter (2 × (240 + 120)) but forgetting to scale it down at all.
- (a) 60° — Method: the angles of a triangle add up to 180°, so add the three expressions, solve for x and then substitute back into the expression for angle B. Working: 2x + 3x + 4x = 9x, so 9x = 180 and x = 20. Angle B is 3x, so angle B = 3 × 20 = 60. Answer: 60°. The distractors: 20° is the value of x, from stopping as soon as the equation is solved instead of substituting back; 120° comes from using 360° as the angle sum, which gives x = 40 and 3x = 120; 80° is 4x, the angle at C, from substituting into the wrong expression.
- (a) 4/9 — There are 3 × 6 = 18 equally likely outcomes. The product is a multiple of 5 whenever Spinner C shows 5, whatever Spinner D shows, which is 6 outcomes, or whenever Spinner D shows 5 and Spinner C shows 2 or 3, which is 2 more outcomes. This gives 6 + 2 = 8 outcomes, so the probability is 8/18 = 4/9. Choosing 1/3 comes from only counting the 6 outcomes where Spinner C shows 5 and forgetting the 2 outcomes where Spinner D shows 5 instead. Choosing 1/9 comes from only counting the 2 outcomes where Spinner D shows 5 and forgetting the 6 outcomes where Spinner C shows 5. Choosing 1/2 comes from counting 9 outcomes instead of 8, by listing the case where Spinner C shows 5 and Spinner D shows 5 twice over.
- (c) Neither causes the other; sunshine links both. — Both ice cream sales and sunburn cases tend to rise on hot, sunny days, so the amount of sunshine is a third factor linked to both — neither variable causes the other. Saying ice cream sales cause the sunburn assumes a causal link in one direction that the correlation alone cannot establish. Saying sunburn cases cause the ice cream sales assumes the reverse causal link, which is no more justified. Saying a strong correlation always means causation is the general error this question is testing: correlation, however strong, does not by itself prove that one variable causes the other.
- (a) 1/4 — First find the fraction of the people who are children: the pool is 3 + 5 = 8 equal shares and the children take 5, so the children are 5/8 of the people. The boys are 2/5 of that, and 'of' means multiply: 2/5 × 5/8 = 10/40 = 1/4. Answering 2/5 gives the boys as a fraction of the children only, 3/20 multiplies by the adults' fraction 3/8 instead of the children's, and 3/5 is the fraction of the children who are girls.
- (d) 7n − 2 — Method: find how much the total cost rises each month, then find the constant that fits the cost for one month. Working: the cost rises by £7 for each extra month (12 − 5 = 7, 19 − 12 = 7, 26 − 19 = 7), so the cost has the form 7n + c. Substituting n = 1: 7(1) + c = 5, so c = −2. Answer: the total cost in pounds is 7n − 2. The value 7n comes from leaving out the constant. The value 7n + 5 comes from using the cost of one month as the constant directly, without subtracting the monthly rise first. The value 5n + 7 comes from swapping the roles of the cost of one month, £5, and the monthly rise, £7 — using the cost of one month as the coefficient of n and the rise as the constant.
- (a) 135 km — Method: find the distance travelled on one litre, then scale up to 9 litres. Working: 90 ÷ 6 = 15 km per litre, so 15 × 9 = 135 km. Answer: 135 km. 45 km comes from working out the extra distance for the extra 3 litres (15 × 3) but forgetting to add the original 90 km. 60 km comes from using the ratio the wrong way round, 90 × 6 ÷ 9, instead of finding the rate per litre first. 99 km comes from simply adding the number of litres, 9, onto the original distance, 90, instead of scaling the whole journey.
- (c) 5 — The front elevation shows one square for every cube visible from the front, column by column: the left-hand column is 2 cubes high, so it contributes 2 squares; the middle column is 2 cubes high, so it contributes 2 more; the right-hand column is 1 cube high, so it contributes 1. The total is 2 + 2 + 1 = 5 squares. "6" comes from drawing a full 3 by 2 rectangle, treating every column as if it reached the greatest height. "4" comes from losing a square from one of the two tall columns, counting 2 + 1 + 1. "3" comes from counting one square per column — the width of the solid — and ignoring the heights altogether.
- (d) 3/5 — 'Chosen from the boys' restricts the sample space to the 45 boys, of whom 27 walk. So the probability is 27/45 = 3/5. Using all 80 students as the denominator instead of just the boys gives 27/80, which is the probability that a student chosen from everyone is a boy who walks — not what was asked. Using the 18 boys who do NOT walk (45 − 27) as the numerator instead of the 27 who do gives 18/45 = 2/5. Dividing the 27 walking boys by the number of girls (80 − 45 = 35) instead of by the number of boys gives 27/35.
- (b) 54 — Method: round each number to the nearest whole number, then subtract the rounded values. Working: 79.3 rounds to 79 (nearest whole number) and 24.6 rounds to 25 (nearest whole number). 79 − 25 = 54. Answer: 54. 54.7 is the exact value of 79.3 − 24.6, found without rounding first, so it is not an estimate. 55 comes from rounding 24.6 down to 24 instead of up to the nearest whole number, 25, giving 79 − 24. 59 comes from rounding 24.6 to the nearest 10, 20, instead of to the nearest whole number, 25, giving 79 − 20.
- (b) 5 — From x + y = 7, x = 7 − y. Substituting into 3x + 2y = 16: 3(7 − y) + 2y = 16, so 21 − 3y + 2y = 16, giving 21 − y = 16, so y = 5 (then x = 2). A candidate who forgets to multiply the y-term inside the bracket by 3 would write 21 − y + 2y = 16, giving 21 + y = 16, so y = −5. A candidate who subtracts the two equations directly, (3x + 2y) − (x + y) = 16 − 7, gets 2x + y = 9, and if they wrongly treat this as giving y alone, ignoring the x term, they would answer 9. A candidate who reports the value of x instead of y would answer 2.
- (d) 40 cm — Method: convert the real length to centimetres, then divide by the scale factor. Working: 10 m = 1000 cm. 1000 ÷ 25 = 40 cm. Wrong options: 0.4 cm comes from dividing 10 by 25 without converting metres to centimetres first; 25000 cm comes from multiplying instead of dividing (1000 × 25); 4 cm comes from converting 10 m to 100 cm (using ×10 instead of ×100) before dividing.
- (a) SSA - not sufficient to prove congruence — The angle given, angle A, is not the angle between sides AB and BC — it is not the included angle — so this data is SSA (side, side, angle), which is not sufficient to prove congruence on its own; two triangles can share this SSA information without being congruent. SAS is wrong because the given angle is not the one included between the two given sides. ASA is wrong because only one angle (A) is given, not two. AAS is wrong for the same reason — only one angle is given, not two.
- (a) 14 — Multiply both numbers by 10 to clear the decimals: 8.4 becomes 84 and 0.6 becomes 6. Then divide: 84 ÷ 6 = 14, so 14 complete pieces can be cut. Scaling only the divisor by 10 and leaving the dividend as 8.4 gives 8.4 ÷ 6 = 1.4, which rounds down to 1 complete piece — the dividend was never converted. Scaling only the dividend by 10 and leaving the divisor as 0.6 gives 84 ÷ 0.6 = 140. Rounding the divisor from 0.6 to 0.7 before dividing, trading accuracy for a rounder number, gives 8.4 ÷ 0.7 = 12. So 14 complete pieces of ribbon can be cut.
- (a) y = −2x + 3 — Method: find the step in the outputs for each step of 1 in the input — falling outputs mean a negative multiplier — then read off the output when the input is 0, because that is the number added on. Working: the outputs 5, 3, 1, −1 fall by 2 each time x rises by 1, so x is multiplied by −2; the output at x = 0 is 3, so 3 is added. Answer: y = −2x + 3, checked at x = 2 by −2 × 2 + 3 = −1. The distractors: y = 2x + 3 comes from taking the size of the step, 2, as the multiplier and ignoring the fact that the outputs are falling; y = −2x − 3 comes from using the correct multiplier but writing the number added on as −3 instead of the output 3 listed at x = 0; y = −x + 4 comes from taking the multiplier as −1, its size read from the step of 1 in the inputs instead of the step of 2 in the outputs and its sign from the fact that the outputs fall, and then fitting the number added on to the pair x = −1, y = 5.
- (c) 3 : 5 — If orange juice is 3/8 of the total, apple juice is the remaining 1 − 3/8 = 5/8. The ratio of orange to apple is therefore 3 : 5. Inverting gives 5 : 3, apple to orange instead of orange to apple. Using the denominator 8 as the second part of the ratio, 3 : 8, compares orange juice to the whole drink rather than to the apple juice alone. Pairing the total 8 with the apple fraction's numerator 5 gives 8 : 5, which mixes a whole-total figure with a part figure.
- (d) £120 — Total parts = 5 + 3 = 8, so one part is worth 320 ÷ 8 = 40 pounds. The cake sale is 3 parts, so it raised 3 × 40 = 120 pounds. £200 comes from working out the raffle's share, 5 × 40, instead of the cake sale's share. £40 comes from finding the value of one part but forgetting to multiply by 3. £192 comes from dividing the total by 5 instead of 8 to find the value of one part, 320 ÷ 5 = 64, then multiplying by 3, 3 × 64 = 192.
- (d) £8262 — A fall of 15% is a multiplier of 0.85 and a fall of 10% is a multiplier of 0.9, and each multiplier acts on the value at the start of its own year. After year 1: 12000 × 0.85 = 10200. After year 2: 10200 × 0.9 = 9180. After year 3: 9180 × 0.9 = 8262. The value 3 years after the car was bought is £8262. Adding the percentages to make a single fall of 35% would be wrong, because the later falls are taken from smaller values.
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.