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 12 as a product of its prime factors.
- 2.The expression 6n + 15 is written as 3(2n + 5). Which statement is correct?
- 3.5 tins of soup cost £8 in total. Work out the multiplier that changes the number of tins into the total cost in pounds.
- 4.Write down which quadrant contains the point (−3, 5) when plotted on a coordinate grid.
- 5.A survey of 160 employees at a company recorded whether they cycle to work, using a frequency tree. The first branch splits them into 90 who work full-time and 70 who work part-time. Of the full-time employees, 27 cycle to work. Of the part-time employees, 14 cycle to work. Work out the probability that an employee, chosen at random from the 160, cycles to work. Give your answer as a fraction in its simplest form.
- 6.Two classes sat the same test. The 30 pupils in Class A had a mean mark of 72. The 20 pupils in Class B had a mean mark of 82. Work out the mean mark of all 50 pupils.
- 7.Oliver drives 95 km at an average speed of 50 km/h. Work out an estimate for the time the journey takes, by rounding the distance to the nearest 100 km.
- 8.Work out the value of 3a − 2b when a = 5 and b = 4
- 9.y is directly proportional to x. When x = 5, y = 18. Work out the value of y when x = 15.
- 10.Each of these has an exact value. Write down the one whose value is the greatest.
- 11.Bag A contains 3 red balls and 2 blue balls. Bag B contains 1 red ball and 3 blue balls. One ball is taken at random from each bag. Work out the probability that both balls are red.
- 12.A space probe is 7.5 × 10⁸ km from Earth. Write this distance as an ordinary number.
- 13.Write down the expression that means the same as 'subtract 5 from n, then divide the result by 2'.
- 14.A car uses 6 litres of petrol to travel 90 km. Write the ratio of the distance travelled in km to the amount of petrol used in litres, in its simplest form.
- 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.Work out 3³ + 2⁴.
- 17.Simplify 2b − 1 + 2b − 4 + 2b
- 18.A bag contains red counters and blue counters in the ratio 5:3. There are 56 counters in the bag altogether. Work out how many counters are blue.
- 19.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.
- 20.A tray holds 8 muffins and 20 cupcakes. Write the ratio of the number of muffins to the number of cupcakes in its simplest form.
Answer key
- (a) 2² × 3 — Method: divide repeatedly by the smallest prime that goes in, until 1 is reached, then write the primes used as a product with indices. Working: 12 ÷ 2 = 6, 6 ÷ 2 = 3 and 3 ÷ 3 = 1, so the primes used are 2, 2 and 3, which is written as 2² × 3. Answer: 2² × 3. The distractors: 2 × 6 comes from stopping at the first factor pair without splitting the 6, which is not prime; 2 × 3 comes from listing each prime once and losing the repeat, and it multiplies to 6 rather than 12; 2 × 3² puts the index on the wrong prime and multiplies to 18.
- (c) 3 and 2n + 5 are both factors of 6n + 15 — Factors are the parts multiplied together to make an expression; terms are the parts added together. In 3(2n + 5) the 3 and the bracket 2n + 5 are multiplied, so both of them are factors of 6n + 15. The parts added together are 6n and 15, and those are its two terms, not its factors. A factor need not be a number: a bracket containing letters is a factor in exactly the same way.
- (c) 1.6 — The multiplier from tins to pounds is found by dividing the cost by the number of tins: £8 ÷ 5 = 1.6. 0.625 comes from dividing the wrong way round, tins ÷ cost (5 ÷ 8), which is the multiplier from pounds back to tins. 3 comes from subtracting the number of tins from the cost (8 − 5), instead of dividing. 40 comes from multiplying the cost by the number of tins (8 × 5), instead of dividing.
- (b) second quadrant — The point (−3, 5) has a negative x-coordinate and a positive y-coordinate, and this combination lies in the second quadrant. The first quadrant needs both coordinates positive. The third quadrant needs both coordinates negative. The fourth quadrant needs a positive x-coordinate and a negative y-coordinate.
- (b) 41/160 — In total, 27 + 14 = 41 of the 160 employees cycle to work, so the probability is 41/160 (41 and 160 share no common factor, so this is already in its simplest form). Writing 27/160 is wrong because it only counts the full-time cyclists and leaves out the 14 part-time cyclists. Writing 41/90 is wrong because it uses the full-time total (90) as the denominator instead of the whole survey (160). Writing 1/5 is wrong because it only uses the part-time branch, simplifying 14/70 to 1/5 and ignoring the full-time cyclists completely. The probability is 41/160.
- (c) 76 marks — Method: a mean of means only works when the groups are the same size, so rebuild each class's total mark, add the totals and divide by the number of pupils altogether. Working: Class A scored 30 × 72 = 2160 marks and Class B scored 20 × 82 = 1640 marks, giving 2160 + 1640 = 3800 marks between 50 pupils, so the overall mean is 3800 ÷ 50 = 76 marks. Answer: 76 marks. The distractors: 77 marks comes from averaging the two class means, (72 + 82) ÷ 2, which ignores the different class sizes; 78 marks comes from attaching each mean to the other class's size, (30 × 82 + 20 × 72) ÷ 50; 3800 marks comes from stopping at the combined total and never dividing by 50.
- (d) 2 hours — Method: the time for a journey is the distance divided by the speed, so round the distance first and then divide by the speed. Working: 95 km rounds to 100 km, and 100 ÷ 50 = 2; the speed is in kilometres per hour, so the answer is a number of hours. Answer: 2 hours. The distractors: 1 hour comes from rounding the distance down to 50 km to match the speed, so that the journey looks like a single hour of driving; 30 minutes comes from dividing the speed by the distance, 50 ÷ 100, instead of the distance by the speed; 1 hour 54 minutes is the exact time, 95 ÷ 50 = 1.9 hours, worked out in full when the question asks for an estimate.
- (d) 7 — Method: substitute both values, work out the two multiplications first, and only then subtract. Working: 3a = 3 × 5 = 15 and 2b = 2 × 4 = 8, so the expression becomes 15 − 8 = 7. Answer: 7. The distractors: 23 comes from adding the two products instead of subtracting, giving 15 + 8; −7 comes from subtracting the wrong way round and working out 8 − 15; 52 comes from working from left to right instead of following the order of operations, giving 3 × 5 = 15, then 15 − 2 = 13, then 13 × 4.
- (c) 54 — Method: y = kx, so k = y ÷ x. Working: k = 18 ÷ 5 = 3.6. At x = 15: y = 3.6 × 15 = 54. Wrong options: 28 comes from adding the change in x (10) onto y instead of scaling; 6 comes from treating the relationship as inverse proportion (k = 5 × 18 = 90, then y = 90 ÷ 15 = 6); 60 comes from rounding the constant up to 4 instead of using 3.6.
- (c) tan 45° — Method: replace each ratio by its exact value, then compare. Working: a right-angled triangle with a 45° angle is isosceles, so its opposite and adjacent sides are equal and the tangent of 45° is exactly 1. The others are cos 30° = √3/2, about 0.87; sin 45° = √2/2, about 0.71; and cos 60° = 1/2. Answer: tan 45°, the only one of the four that reaches 1. Reading √3/2 as though it were √3, about 1.73, makes cos 30° look the largest, but the division by 2 is part of the value. Ranking by the size of the angle also fails here, because the cosine of an angle falls as the angle grows.
- (d) 3/20 — The probability of taking a red ball from Bag A is 3/5, and the probability of taking a red ball from Bag B is 1/4. Since the two picks are independent, the probabilities are multiplied: 3/5 × 1/4 = 3/20. A candidate who answers 4/9 has added the numerators and added the denominators, (3+1)/(5+4), instead of multiplying the two fractions. A candidate who answers 17/20 has added the two fractions, 3/5 + 1/4, instead of multiplying them. A candidate who answers 3/25 has misread Bag B as also containing 5 balls, using 1/5 instead of 1/4.
- (b) 750,000,000 — Method: multiplying by 10⁸ moves the decimal point eight places to the right, and every empty place is filled with a zero. Working: 10⁸ = 100,000,000, and moving the decimal point in 7.5 eight places to the right gives 7.5 × 100,000,000. Answer: 750,000,000. The distractors: 7,500,000,000 comes from removing the decimal point first to make 75 and then writing eight zeros after it, which carries the digits one place too far; 600 comes from reading 10⁸ as 10 × 8 = 80 and working out 7.5 × 80; 0.000000075 comes from moving the decimal point eight places to the left, as though the index were negative.
- (a) (n − 5)/2 — Method: do the subtraction first, keep it together as a single bracket, then divide that whole bracket by 2. Working: 'subtract 5 from n' is (n − 5); 'divide the result by 2' means the whole bracket goes over 2, giving (n − 5)/2. Answer: (n − 5)/2. n/2 − 5 comes from dividing n by 2 first and only then subtracting 5, the wrong order. 2(n − 5) comes from multiplying by 2 instead of dividing. (5 − n)/2 comes from subtracting n from 5 instead of subtracting 5 from n, the wrong way round.
- (a) 15:1 — Write the ratio distance : petrol using the numbers in the question: 90 : 6. Divide both parts by their highest common factor, 6, to give 15 : 1. (1:15 comes from writing the ratio the wrong way round, petrol : distance. 5:2 comes from dividing only the distance by 6 and leaving the 6 litres unchanged, giving 15:6, which does simplify to 5:2 — the error is the step before, because both parts of a ratio must be divided by the same number. 14:1 comes from subtracting at the first step instead of dividing, 90 − 6 = 84, and then writing 84 : 6 and dividing both parts by 6 to get 14 : 1; subtracting has no part in simplifying a ratio.)
- (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.
- (c) 43 — Method: work out each power separately before adding. Working: 3³ = 27 and 2⁴ = 16, so 3³ + 2⁴ = 27 + 16 = 43. Answer: 43. (25 comes from using 3² instead of 3³, giving 9 + 16. 35 comes from working out 2⁴ as 2 × 4 = 8 instead of 2 × 2 × 2 × 2, giving 27 + 8. 432 comes from multiplying the two powers together instead of adding them.)
- (a) 6b − 5 — Method: sort the expression into terms in b and number terms, then collect each kind, keeping the sign in front of every term. Working: the b terms give 2b + 2b + 2b = 6b; the number terms are −1 and −4, and −1 − 4 = −5. Answer: 6b − 5. The distractors: 6b − 3 comes from working out the difference 4 − 1 instead of adding two negative numbers; 6b + 5 comes from adding 1 and 4 while ignoring both minus signs; 8b − 5 comes from multiplying the coefficients 2, 2 and 2 instead of adding them.
- (d) 21 — Add the parts: 5 + 3 = 8. Divide the total by the number of parts: 56 ÷ 8 = 7, so one part is worth 7 counters. Blue has 3 parts: 3 × 7 = 21. (35 is the number of red counters, using 5 parts instead of 3. 28 comes from splitting 56 counters in half instead of in the ratio 5:3. 7 is the value of one part — the number of blue counters is 3 lots of this, not just one.)
- (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.
- (c) 2:5 — Divide both numbers by their highest common factor, 4: 8 ÷ 4 = 2 and 20 ÷ 4 = 5, giving the ratio 2:5. Choosing 5:2 comes from writing the ratio the wrong way round, as cupcakes to muffins. Choosing 2:3 comes from using the difference between the two amounts (20 − 8 = 12) as the second part of the ratio instead of the number of cupcakes, then simplifying 8:12 by dividing by 4. Choosing 2:7 comes from comparing the muffins with the total number of items on the tray (8 out of 28) instead of comparing them with the cupcakes.
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.