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.A spreadsheet shows that 812 − 397 = 315. Work out an estimate for 812 − 397, by rounding each number to the nearest 100, to check whether the spreadsheet's answer is reasonable.
- 2.Make x the subject of the formula y = 5x + 3.y = 5x + 3
- 3.3 pencils cost £1.20. Work out the cost of 7 pencils.
- 4.Triangle A has a vertex at (1, 2). Triangle A is enlarged, centre (1, 1), to give triangle B, whose corresponding vertex is at (1, 5). Work out the scale factor of the enlargement.
- 5.180 gardeners at an allotment were asked whether they grow vegetables organically. 100 of the gardeners are on the north plots. 64 of the north-plot gardeners grow organically. 30 of the south-plot gardeners do not grow organically. Work out how many south-plot gardeners grow organically.
- 6.A pictogram shows the number of cars sold by a garage in Plymouth each month, where each whole car symbol represents 8 cars sold and a half symbol represents 4 cars. March shows 3 whole symbols and one half symbol. Work out how many cars were sold in March.
- 7.Work out (−36) ÷ (−6) × 2
- 8.Two numbers have a sum of 24. Their difference is 6. Work out the two numbers.
- 9.A blue ribbon is 45 cm long. A red ribbon is 30 cm long. Write the length of the blue ribbon as a fraction of the length of the red ribbon. Give your answer in its simplest form.
- 10.Triangle ABC is extended so that side BC is extended beyond C to a point D, forming exterior angle ACD. Angle BAC = 48° and angle ABC = 67°. Which single angle fact gives angle ACD = 115° in one step, without first working out the interior angle at C?
- 11.A raffle at a school fair sells 500 tickets. Mia buys 25 of the tickets. One ticket is drawn at random to win the prize. Work out the probability that Mia wins the raffle. Give your answer as a fraction in its simplest form.
- 12.Work out 2/5 + 1/4, giving your answer in its simplest form.
- 13.The volume of a cuboid is given by the formula V = lwh. Work out V, in cm³, when l = 2, w = 3 and h = 6.
- 14.A jug holds 3 litres of a drink that is 60% fruit juice. 1 litre of water is added to the jug. Work out the percentage of the new mixture that is fruit juice.
- 15.A shopkeeper builds a display from two cuboid boxes, shown in the diagram. She wants to cover the front of the display with coloured paper. Work out the total area of the front elevation, in square centimetres.
- 16.A recipe uses 0.625 kg of flour. Write this mass as a fraction of a kilogram, in its simplest form.
- 17.Work out the value of 5a + 2 when a = 3
- 18.y is inversely proportional to x, and y = 20 ÷ x. Work out the value of y when x = 4.
- 19.A measuring jug shows a volume of 340 ml, correct to the nearest 20 ml. Work out the smallest possible volume in the jug.
- 20.A block of wood has a mass of 60 g and a volume of 100 cm³. Work out the density of the block, in g/cm³.
Answer key
- (b) 400 — Method: round each number to the nearest 100, then subtract the rounded values. Working: 812 rounds to 800 (nearest 100) and 397 rounds to 400 (nearest 100). 800 − 400 = 400. Answer: 400. 500 comes from rounding 397 down to 300 instead of up to the nearest 100, 400. 300 comes from rounding 812 down to 700 instead of up to the nearest 100, 800. 415 is the exact value of 812 − 397, found without rounding first, so it is not an estimate — the spreadsheet's answer of 315 is too far from the estimate of 400 to be correct.
- (b) x = (y − 3)/5 — To make x the subject of y = 5x + 3, first subtract 3 from both sides to get y − 3 = 5x, then divide both sides by 5: x = (y − 3)/5. Writing x = (y + 3)/5 keeps the division correct but does not change the sign of the 3 when moving it across. Writing x = y/5 − 3 divides only the y term by 5 and leaves the 3 as a separate subtraction, instead of subtracting first and dividing the whole expression. Writing x = 5(y − 3) applies the correct order of subtracting 3 first, but then multiplies by 5 instead of dividing — the inverse of 5x is division, not multiplication. The correct rearrangement is x = (y − 3)/5.
- (a) £2.80 — Method: find the cost of one pencil first, then use it to find the cost of 7 pencils. Working: £1.20 ÷ 3 = £0.40 for one pencil, and £0.40 × 7 = £2.80. So 7 pencils cost £2.80. Distractor £8.40 comes from multiplying £1.20 by 7 without first dividing by 3 to find the cost of one pencil. Distractor £1.20 comes from assuming the price stays the same no matter how many pencils are bought. Distractor £0.40 is the cost of one pencil, found correctly but never multiplied by 7.
- (b) 4 — The scale factor is the distance from the centre to the image, divided by the distance from the centre to the object: (5 − 1) ÷ (2 − 1) = 4 ÷ 1 = 4. (2.5 comes from dividing the raw y-coordinates, 5 ÷ 2, without first subtracting the centre's coordinate; 3 comes from subtracting the two distances instead of dividing them; 0.25 comes from dividing the distances the wrong way round.)
- (b) 50 — There are 180 − 100 = 80 south-plot gardeners. 30 of them do not grow organically, so the rest do: 80 − 30 = 50. Writing 64 is wrong because that is the number of NORTH-plot gardeners who grow organically, not south. Writing 30 again is wrong because that is the number of south-plot gardeners who do NOT grow organically — the question asks for those who do. Writing 80 is wrong because that is the whole south-plot total, without subtracting the 30 who do not grow organically. The answer is 50.
- (c) 28 — Method: multiply the number of whole symbols by the value of one symbol, then add the value of any half symbol shown. Working: 3 whole symbols represent 3 × 8 = 24 cars. The half symbol represents 4 cars. Total cars sold in March = 24 + 4 = 28. Leaving out the half symbol, 3 × 8 = 24, undercounts by exactly the value of that half symbol. Treating the half symbol as if it were a full symbol, 4 × 8 = 32, overcounts because it doubles the value the half symbol is worth. Giving 3.5 reports the number of symbols shown, not the number of cars they represent — the key still needs to be applied. Always apply the key to every symbol shown, including a half symbol, rather than reading off the symbol count itself.
- (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.
- (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) 3/2 — The length of the blue ribbon goes on the top and the length of the red ribbon goes on the bottom, giving 45/30. Both parts divide by 15: 45 ÷ 15 = 3 on the top, and 30 ÷ 15 = 2 on the bottom. So the fraction is 3/2. It is greater than 1 because the blue ribbon is longer than the red one, and a fraction of one quantity compared with another is allowed to be top-heavy.
- (c) exterior angle = sum of the two interior opposite angles — Angle ACD is the exterior angle at C, and the exterior angle of a triangle is always equal to the sum of the two interior angles at the other two vertices — here, angle BAC and angle ABC — so angle ACD = 48° + 67° = 115°. "angles in a triangle add up to 180°" is a true fact, but on its own it only gives the INTERIOR angle at C (180 − 48 − 67 = 65°), not the exterior angle 115° — reaching 115° that way still needs a further step. "angles on a straight line add up to 180°" is the fact that links the interior and exterior angles at C to each other, not the fact that gives 115° directly from the two OTHER angles. "exterior angle is always double the smallest angle" is not a genuine angle fact: double the smallest angle here is 48° × 2 = 96°, not 115°.
- (d) 1/20 — Mia has 25 of the 500 tickets, so the probability she wins is 25/500 = 1/20, dividing the top and the bottom by 25. Writing 19/20 is wrong because that is the probability she does NOT win (1 − 1/20 = 19/20), the opposite of what is asked. Writing 1/25 is wrong because it puts 1 over the number of tickets Mia holds, as though her 25 tickets were the whole raffle — the denominator has to be the 500 tickets sold, not her own share. Writing 1/10 is wrong because it treats the raffle as having only 250 tickets instead of the actual 500: 25/250 = 1/10. The probability that Mia wins is 1/20.
- (d) 13/20 — Method: write both fractions over a common denominator, then add the numerators. Working: 2/5 = 8/20 and 1/4 = 5/20, so 2/5 + 1/4 = 8/20 + 5/20 = 13/20, which is already in its simplest form. Answer: 13/20. 1/3 comes from adding the numerators and the denominators separately: (2+1)/(5+4) = 3/9 = 1/3. 13/40 comes from converting both fractions to twentieths correctly but then adding the denominators as well as the numerators: (8+5)/(20+20) = 13/40. 3/20 comes from adding the original numerators (2+1) but keeping them over the common denominator without converting them first: 3/20.
- (d) 36 cm³ — V = lwh = 2 × 3 × 6 = 36 cm³. A candidate who adds all three numbers instead of multiplying gets 2 + 3 + 6 = 11 cm³. A candidate who multiplies only two of the three numbers, forgetting the length, gets w × h = 3 × 6 = 18 cm³. A candidate who multiplies l × w and w × h separately and adds the two products gets (2 × 3) + (3 × 6) = 6 + 18 = 24 cm³.
- (c) 45% — Method: adding water changes the total volume but not the amount of fruit juice, so find the juice, find the new total volume, and write the first as a percentage of the second. Working: 3 × 0.6 = 1.8 litres of fruit juice; the new volume is 3 + 1 = 4 litres; 1.8 ÷ 4 = 0.45, which is 45%. Answer: 45%. The distractors: 60% is the strength before the water goes in, and assumes that adding water leaves the strength unchanged; 15% comes from dividing the 60% by the 4 litres of mixture instead of dividing the 1.8 litres of juice by the 4 litres; 75% is the fraction of the new mixture that came out of the original jug, 3 litres out of 4, which ignores that only 60% of that 3 litres was juice.
- (a) 4700 cm² — Method: split the T-shaped outline into the two rectangles it is made from, find the area of each, then add them together. Working: the wide base gives a rectangle 90 cm × 30 cm = 2700 cm²; the narrower block on top gives a rectangle 40 cm × 50 cm = 2000 cm²; adding these, 2700 + 2000 = 4700 cm². Answer: 4700 cm². The distractors: 2700 cm² comes from finding only the area of the base rectangle and forgetting to add the block on top. 2000 cm² comes from finding only the area of the top block and forgetting the base. 7200 cm² comes from treating the whole outline as one large rectangle, 90 cm wide by (30 + 50) = 80 cm tall, instead of splitting it into the two separate rectangles that actually make up the shape.
- (c) 5/8 — Method: write the decimal over 1000 using its three decimal places, then simplify. Working: 0.625 = 625/1000 = 5/8 (dividing both numerator and denominator by 125). Answer: 5/8. 25/4 comes from writing the decimal over 100 instead of 1000, as if there were only two decimal places. 31/50 comes from rounding 0.625 to 0.62 before converting. 8/5 comes from simplifying correctly to 5/8 and then writing the fraction upside down.
- (a) 17 — Substitute a = 3 into 5a + 2: 5 × 3 + 2 = 15 + 2 = 17. 10 comes from treating 5a as 5 + a instead of 5 × a, giving 5 + 3 + 2. 25 comes from adding the 2 to a before multiplying by 5, 5 × (3 + 2). 15 comes from working out 5 × 3 correctly but forgetting to add the 2.
- (d) 5 — Method: in an equation of inverse proportion the constant is divided by the value of x. Working: y = 20 ÷ x with x = 4 gives y = 20 ÷ 4 = 5. Answer: 5. The distractors: 80 comes from multiplying 20 by 4 instead of dividing, which is direct proportion applied to an inverse relationship; 16 comes from working out 20 − 4, treating the constant as something to subtract from; 0.2 comes from dividing the wrong way round, 4 ÷ 20.
- (a) 330 ml — Correct to the nearest 20 ml means the true volume could be up to 10 ml (half of 20) either side of 340 ml. The smallest possible volume is 340 − 10 = 330 ml. A candidate who subtracted the full 20 ml instead of half of it worked out 340 − 20 = 320 ml. A candidate who added instead of subtracted, finding the largest possible volume instead of the smallest, worked out 340 + 10 = 350 ml. A candidate who halved the interval again by mistake, using 5 ml instead of 10 ml, worked out 340 − 5 = 335 ml.
- (d) 0.6 g/cm³ — Density = mass ÷ volume. 60 ÷ 100 = 0.6 g/cm³. 1.67 g/cm³ comes from dividing the volume by the mass instead of the mass by the volume (100 ÷ 60). 6000 g/cm³ comes from multiplying the mass by the volume instead of dividing (60 × 100). 40 g/cm³ comes from subtracting the mass from the volume (100 − 60) instead of dividing.
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.