✏️ Paper 1 non-calculator warm-up — Foundation

Answer key

1. (a) 9/20 — Method: write the decimal over 100 using its two decimal places, then simplify. Working: 0.45 = 45/100 = 9/20 (dividing both numerator and denominator by 5). Answer: 9/20. 9/100 comes from dividing only the numerator by 5 and leaving the denominator as 100. 9/2 comes from writing the decimal over 10 instead of 100, as if there were only one decimal place, then simplifying 45/10. 9/200 comes from writing the decimal over 1000 instead of 100, as if there were three decimal places, then simplifying 45/1000.
2. (c) 21% — Method: an increase of 10% is a multiplier of 1.1, and two successive increases are found by multiplying the multipliers. Working: 1.1 × 1.1 = 1.21, so the rent is 121% of the original, which is an increase of 21%. Answer: 21%. The distractors: 20% comes from adding the two percentages, which ignores that the second 10% is taken of a larger amount; 121% is the multiplier written as the change rather than the change itself; 11% comes from slipping in the multiplication and getting 1.11 instead of 1.21.
3. (c) 1.2 — Add the parts of the ratio: 6 + 1 = 7. Divide the total amount by the number of parts: 8.4 ÷ 7 = 1.2 litres, which is the value of one part and also the amount of syrup, since syrup is 1 part. (7.2 litres is the amount of water, using 6 parts instead of 1. 1.4 comes from dividing 8.4 by 6 — the water's part of the ratio — instead of dividing by the total number of parts, 7. 0.84 comes from dividing 8.4 by 10 instead of by 7.)
4. (a) 3:2 — There are 24 green sweets and 16 orange sweets. The highest common factor of 24 and 16 is 8. Divide both numbers by 8: 24 ÷ 8 = 3 and 16 ÷ 8 = 2, so the ratio is 3 : 2. Dividing by 4 instead of 8 gives 6 : 4, which still has a common factor of 2, so it is not fully simplified. Writing green sweets to the total number of sweets, 24 : 40, simplifies to 3 : 5 — that compares green to everything, not green to orange, so it answers a different question. Swapping the order gives 2 : 3, green and orange the wrong way round.
5. (a) 36 — Method: for the highest common factor, take the LOWER power of each prime that appears in both numbers. Working: for 2, the lower power is 2² (from 108); for 3, the lower power is 3² (from 72), so the highest common factor is 2² × 3² = 4 × 9 = 36. 216 comes from taking the higher power of each prime instead, 2³ × 3³ = 8 × 27 = 216, which gives the lowest common multiple, not the highest common factor. 108 is simply one of the two numbers, not their highest common factor. 6 comes from multiplying the primes without any powers at all, 2 × 3 = 6. Answer: 36.
6. (d) 8 — Method: whatever is inside the brackets is worked out first, and the multiplication is carried out afterwards. Working: inside the brackets, −3 + 7 = 4, and then 2 × 4 = 8. Answer: 8. The distractors: 1 comes from ignoring the brackets and multiplying first, giving 2 × (−3) = −6 and then −6 + 7 = 1; −20 comes from reading −3 + 7 as −(3 + 7) = −10, so that 2 × (−10) = −20; 6 comes from working the brackets out correctly and then adding the 2 instead of multiplying by it, giving 2 + 4 = 6.
7. (b) 24 — List multiples of 8 and of 12: multiples of 8 are 8, 16, 24, 32; multiples of 12 are 12, 24, 36. The lowest number in both lists is 24, so the lighthouses next flash together after 24 minutes. Multiplying the two numbers together, 8 × 12, gives 96, which double-counts the common factor of 4 shared by 8 and 12. Working out the highest common factor instead of the lowest common multiple gives 4, far too soon a time for both lighthouses to line up again. Adding the two numbers, 8 + 12, gives 20, which is not even a multiple of either 8 or 12. So the lighthouses next flash together after 24 minutes.
8. (a) 300 — Method: round each number to 1 significant figure, then divide the rounded values. Working: 8,900 rounds to 9,000 and 29 rounds to 30; cancelling a zero from each gives 900 ÷ 3. Answer: 300. The distractors: 450 comes from rounding 29 down to 20 instead of to the nearest ten, giving 9,000 ÷ 20; 3,000 comes from rounding 29 to 3 rather than to 30, a place-value slip that divides by a number ten times too small; 307 is the exact quotient rounded to the nearest whole number, worked out in full when the question asks for an estimate.
9. (c) 12 m — Method: the perimeter of a rectangle is twice the sum of the length and the width, so half the perimeter is one length plus one width; split that half using the ratio. Working: half of 64 is 64 ÷ 2 = 32 m, the ratio 5:3 has 5 + 3 = 8 parts, so one part is 32 ÷ 8 = 4 m, and the width is 3 × 4 = 12 m. So the width is 12 m. Distractor 20 m is the length, 5 parts, not the width. Distractor 24 m comes from splitting the whole perimeter, 64 m, into 8 parts and multiplying by 3, forgetting to halve the perimeter first. Distractor 8 m comes from the same slip stopped one step earlier: splitting the whole perimeter into 8 parts, 64 ÷ 8 = 8, and giving that instead of the width.
10. (d) 36 — Method: the fraction is acting as an operator on the whole class, so one third of the class equals 12; the operation has to be reversed, and the inverse of dividing by 3 is multiplying by 3. Working: 1/3 × (number of pupils) = 12, so the number of pupils = 12 × 3 = 36. Answer: 36 pupils. The distractors: 4 comes from applying the operator instead of reversing it, working out 12 ÷ 3 = 4; 18 comes from reading the 12 girls as two thirds of the class, giving 12 ÷ 2 × 3 = 18; 24 comes from working out the number of boys, the other two thirds, as 2 × 12 = 24 and giving that instead of the size of the class.
11. (a) −22 — Method: both multiplications are carried out before the addition, and a positive multiplied by a negative is negative. Working: (−3) × 4 = −12 and 2 × (−5) = −10, so the calculation becomes −12 + (−10) = −22. Answer: −22. The distractors: 22 comes from ignoring the minus signs and working out 3 × 4 + 2 × 5 = 22; 50 comes from working from left to right with no priority at all, giving −12 + 2 = −10 and then −10 × (−5) = 50; −2 comes from taking 2 × (−5) as +10, so that −12 + 10 = −2.
12. (c) 40° — Method: the three angles of a triangle always add up to 180°, so split 180° into the number of parts shown by the ratio. Working: the ratio 2:3:4 has 2 + 3 + 4 = 9 parts, so one part is 180 ÷ 9 = 20°, and the smallest angle is 2 × 20 = 40°. So the smallest angle is 40°. Distractor 60° is the middle-sized angle, 3 parts. Distractor 80° is the largest angle, 4 parts. Distractor 20° is the value of one part, found correctly but never multiplied by 2.
13. (d) 60 — Method: round each number to 1 significant figure and multiply; the estimate then shows whether the calculator answer is sensible. Working: 3.1 rounds to 3 and 19.6 rounds to 20, so the estimate is 3 × 20 = 60. Answer: 60. Hannah's 6.076 is about ten times too small, which is what happens when 19.6 is keyed in as 1.96. The distractors: 62 comes from rounding 19.6 only and leaving 3.1 as it stands, giving 3.1 × 20 = 62; 6 comes from trusting the calculator display rather than checking it against an estimate; 600 comes from rounding 19.6 to 200 instead of to 20, a place-value slip, giving 3 × 200 = 600.
14. (b) £76.00 — One part of the ratio is £47.50 ÷ 5 = £9.50. The school receives 8 parts, so its share is 9.50 × 8 = £76.00. Dividing £47.50 by 8 instead of 5, treating the charity's amount as if it were 8 parts, gives 47.50 ÷ 8 = 5.9375, then × 5 = £29.69. Adding the charity's amount to the school's amount instead of stopping at the school's own share gives the total collected, 9.50 × 13 = £123.50. Adding one part to the charity's amount instead of multiplying one part by 8 gives 47.50 + 9.50 = £57.00.
15. (d) 3,200,000 — 1 million = 1,000,000, so 3.2 million = 3.2 × 1,000,000 = 3,200,000. A candidate who moves the decimal point one place too many gets 32,000,000. A candidate who moves it one place too few gets 320,000. A candidate who writes the .2 as extra thousands instead of hundred-thousands gets 3,002,000.
16. (c) Ben: 2 × 3 = 6, then 6 + 6 = 12 — Multiplication has priority over addition, so 2 × 3 = 6 is worked out first, then 6 + 6 = 12 — this is Ben's method. Amy adds 6 and 2 before multiplying: 6 + 2 = 8, then 8 × 3 = 24, breaking the priority rule. Chen multiplies the wrong pair of numbers, 6 and 2, instead of 2 and 3: 6 × 2 = 12, then 12 + 3 = 15. Dev applies the right order but slips when multiplying, using 5 instead of 6 for 2 × 3, so the final step becomes 6 + 5 = 11.
17. (d) 600 — Method: the number of seats is the number of rows multiplied by the number of seats in each row, so round each number to 1 significant figure and then multiply the rounded values, which is quick because a product of two multiples of ten is found by multiplying the non-zero digits and attaching the zeros. Working: 21 rounds to 20 and 29 rounds to 30; 2 × 3 = 6, and 20 and 30 carry one zero each, so two zeros follow the 6. Answer: about 600 seats. The distractors: 50 comes from adding the two rounded numbers instead of multiplying them, 20 + 30; 60 comes from multiplying 20 by the 3 of 30 and forgetting the zero in 30; 6,000 comes from attaching three zeros to 2 × 3 when 20 and 30 provide only two between them.
18. (c) £22.50 — Method: find the cost of 1 kg, then multiply by the mass wanted. Working: £36 ÷ 2.4 = £15 per kilogram, and 1.5 × £15 = £22.50. Answer: £22.50. The distractors: £15 is the price of 1 kg, which is the first step and not what the question asks for; £54 comes from multiplying the £36 by 1.5 without first reducing it to a price per kilogram; £57.60 comes from using the scale factor upside down, multiplying £36 by 2.4 ÷ 1.5 = 1.6 instead of by 1.5 ÷ 2.4.
19. (c) 600 — Method: round each number to the nearest 100, then add the rounded values. Working: 113 is nearer to 100 than to 200, so it rounds to 100; 491 is nearer to 500 than to 400, so it rounds to 500; adding those gives the estimate. Answer: 600. The distractors: 500 comes from rounding each number down to the hundred below instead of to the nearest hundred, giving 100 + 400; 700 comes from rounding each number up to the hundred above, giving 200 + 500; 604 is the exact total, worked out in full when the question asks for an estimate.
20. (b) 5 — Method: list the factors of each number and pick the largest value that appears in both lists. Working: the factors of 15 are 1, 3, 5 and 15; the factors of 25 are 1, 5 and 25. The values in both lists are 1 and 5, and the larger of those is 5. Answer: 5. The distractors: 3 comes from choosing a factor of 15 without checking that it also divides 25; 15 comes from assuming that the smaller of the two numbers is always a factor of the larger one; 75 is the lowest common multiple of 15 and 25, given by taking the highest power of each prime instead of the lowest.