✏️ Paper 1 non-calculator warm-up — Foundation

Answer key

1. (a) 5 — Method: each index is worked out first, and subtracting a negative number is the same as adding the positive. Working: (−2)³ = (−2) × (−2) × (−2) = −8 and (−3)² = (−3) × (−3) = 9, while − (−4) becomes + 4, so the calculation becomes −8 + 9 + 4 = 5. Answer: 5. The distractors: −13 comes from taking (−3)² as −9, giving −8 − 9 + 4 = −13; −3 comes from reading − (−4) as − 4, giving −8 + 9 − 4 = −3; 21 comes from treating every power of a negative number as positive, so that (−2)³ is taken as 8 and the calculation becomes 8 + 9 + 4 = 21.
2. (c) 2/3 — Method: multiply the numerators together and the denominators together, then divide both parts of the result by their highest common factor. Working: 3 × 14 = 42 and 7 × 9 = 63, giving 42/63; the highest common factor of 42 and 63 is 21, and 42 ÷ 21 = 2 with 63 ÷ 21 = 3. Answer: 2/3. The distractors: 17/16 comes from adding the numerators and adding the denominators, giving (3 + 14)/(7 + 9); 27/98 comes from turning the second fraction upside down and multiplying, which divides instead of multiplying and gives 3/7 × 9/14; 2/21 comes from cancelling the 7 into the 14 in the numerator but leaving the 7 in the denominator, giving 6/63.
3. (c) 280 — Method: round each number to 1 significant figure, then multiply the rounded numbers. Working: 6.8 rounds to 7 because the next digit is 8, and 41 rounds to 40 because its next digit is 1, so the estimate is 7 × 40 = 280. Answer: 280. The distractors: 240 comes from cutting 6.8 down to 6 instead of rounding it up to 7, giving 6 × 40 = 240; 350 comes from rounding 41 up to 50 when the digit after its first significant figure is less than 5, giving 7 × 50 = 350; 28 comes from multiplying the leading digits only and losing the place value of the 40, which makes the estimate ten times too small.
4. (c) 150 m — Method: a fraction acts as an operator, so finding 3/7 of a length means dividing by the denominator and multiplying by the numerator. Working: 350 ÷ 7 = 50, so one seventh of the route is 50 m, and three sevenths is 50 × 3 = 150 m. Answer: 150 m. The distractors: 50 m comes from finding one seventh and stopping there instead of multiplying by 3; 1050 m comes from multiplying by the numerator without dividing by the denominator, giving 350 × 3 = 1050; 200 m comes from working out the stretch of the route the footpath does not run alongside, which is 4/7 of 350 m, instead of the stretch it does.
5. (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.
6. (a) 1:8 — Convert 2 kg to grams: 2 kg = 2000 g. The ratio is 250 : 2000. The highest common factor of 250 and 2000 is 250. Divide both parts by 250: 250 ÷ 250 = 1 and 2000 ÷ 250 = 8, so the ratio is 1 : 8. Leaving the kilograms unconverted gives 250 : 2, which simplifies to 125 : 1 — the units on each side are different, so this does not compare like with like. Dividing by 50 instead of 250 gives 5 : 40, which still shares a common factor of 5, so it is not fully simplified. Swapping the order gives 8 : 1, grams to kilograms the wrong way round.
7. (a) £18 — Method: the percentage acts as an operator on the bill, and 15% can be built from 10% and 5%, where 5% is half of 10%. Working: 10% of £120 is £120 ÷ 10 = £12, and 5% is half of that, £6, so the charge is £12 + £6 = £18. Answer: £18. The distractors: £8 comes from reading 15% as one fifteenth and working out £120 ÷ 15 = £8; £12 comes from finding 10% of the bill and stopping there; £138 comes from finding the charge correctly and then giving the new total, £120 + £18, rather than the charge the question asks for.
8. (b) 7 — Method: trap the number between the two square numbers on either side of it, then decide which of them it is nearer to. Working: 6² = 36 and 7² = 49, so √45 lies between 6 and 7; 49 − 45 = 4 while 45 − 36 = 9, so 45 is nearer to 49. Answer: 7. The distractors: 6 comes from taking the square number below 45 and stopping there, without checking which of 36 and 49 is nearer; 22.5 comes from halving 45 instead of looking for the number that multiplies by itself to give 45; 2,025 comes from squaring 45 instead of taking its square root.
9. (b) 12 — Method: round the number of pupils to the nearest 10, divide by the number of passengers each minibus can carry, then round up because a part-full minibus still needs a whole vehicle. Working: 179 rounds to 180 (nearest 10); 180 ÷ 16 = 11.25; 11 minibuses only carry 176 passengers, so a 12th minibus is needed for the rest. Answer: 12. 11 comes from rounding 11.25 to the nearest whole number in the usual way, without checking that the leftover pupils still need transporting. 10 comes from rounding 179 down to 170 instead of to the nearest 10, 180. 180 comes from stopping after rounding the number of pupils, without dividing by the number of passengers each minibus carries at all.
10. (d) 17 — 8 ÷ 2 = 4, then 20 − 4 = 16, then 16 + 1 = 17. Stopping after the subtraction and forgetting to add the final 1 leaves 16. Adding the 4 and the 1 together before subtracting gives 4 + 1 = 5, then 20 − 5 = 15 — the subtraction should use the 4 from the division, not a combined total. Working strictly left to right without giving division priority gives 20 − 8 = 12, then 12 ÷ 2 = 6, then 6 + 1 = 7.
11. (c) 40 cm — Method: split the total length into the number of parts shown by the ratio, then find the value of the shorter share. Working: the ratio 4:5 has 4 + 5 = 9 parts, so one part is 90 ÷ 9 = 10 cm, and the shorter piece is 4 × 10 = 40 cm. So the shorter piece is 40 cm. Distractor 50 cm is the length of the LONGER piece, not the shorter one. Distractor 45 cm comes from splitting the ribbon into two equal halves, ignoring the ratio. Distractor 10 cm is the value of one part, found correctly but never multiplied by 4.
12. (c) £840 — Method: find the total amount raised using the reverse percentage, then subtract the entry fees to find the donations. Working: £1,260 is 60% of the total, so the total is £1,260 ÷ 0.6, and subtracting the entry fees from this total leaves £840 raised through donations. Answer: £840. £2,100 comes from correctly finding the total amount raised but then forgetting to subtract the entry fees, giving the total instead of the donations alone. £504 comes from working out 40% of the entry fees themselves, £1,260 × 0.4 = £504, instead of first finding the total amount raised. £1,890 comes from treating £1,260 as 40% of the total instead of 60%, dividing by 0.4 to get a total of £3,150, and then subtracting the entry fees from that incorrect total.
13. (d) 6,000,000 — Method: round each number to 1 significant figure, then subtract. Working: 8,340,000 rounds to 8,000,000 (1 s.f.); 1,950,000 rounds to 2,000,000 (1 s.f.); 8,000,000 − 2,000,000 = 6,000,000. Answer: 6,000,000. 6,390,000 is the exact difference, found without rounding the numbers first. 8,000,000 comes from rounding the population correctly but forgetting to subtract the capital's population at all. 6,300,000 comes from rounding 8,340,000 to the nearest hundred thousand, 8,300,000, instead of to 1 significant figure, then subtracting the correctly rounded 2,000,000.
14. (d) 19% — Method: write each decrease as a multiplier, multiply the multipliers, then compare the result with 100%. Working: a 10% decrease is a multiplier of 0.9, so the two reductions together give 0.9 × 0.9 = 0.81; the final price is 81% of the original, so the price has fallen by 100% − 81% = 19%. Answer: an overall decrease of 19%. The distractors: 20% comes from adding the two reductions, 10% + 10%, which charges the second 10% against the original price instead of against the already reduced price; 21% comes from using the increase multiplier by mistake, since 1.1 × 1.1 = 1.21, and reading that 21% as a decrease; 81% is the percentage of the original price still being paid, not the percentage taken off.
15. (d) £33.00 — The profit is 3/8 of £24 = (£24 ÷ 8) × 3 = £3 × 3 = £9.00. Selling price = £24 + £9.00 = £33.00. A candidate who gives the profit instead of the selling price gets £9.00. A candidate who subtracts the profit instead of adding it gets £24 − £9 = £15.00. A candidate who works out one eighth of £24 and adds that on, forgetting to multiply by the numerator 3, gets £24 + £3 = £27.00.
16. (c) An under-estimate, by 8 — Method: work out the exact product, then compare it with the estimate; an estimate that is smaller than the exact value is an under-estimate, and the difference between them is the size of the error. Working: 48 × 21 = 48 × 20 + 48 = 960 + 48 = 1,008, and 1,008 − 1,000 = 8, so the estimate falls short. Answer: an under-estimate, by 8. The distractors: an over-estimate by 8 has the size of the error right but the direction wrong, and comes from assuming that rounding 48 up to 50 must push the estimate above the exact value, without allowing for 21 being rounded down; an over-estimate by 19 comes from working out 48 × 21 as 48 × 20 + 21 = 981, adding a 21 where another 48 belongs; the claim that the estimate is exactly right comes from arguing that one number was rounded up and the other down, so the two changes must cancel.
17. (d) 3 : 2 : 5 — The highest common factor of 45, 30 and 75 is 15. Divide each part by 15: 45 ÷ 15 = 3, 30 ÷ 15 = 2 and 75 ÷ 15 = 5, giving 3 : 2 : 5. Giving 9 : 6 : 15 divides by 5, a common factor but not the highest one. Giving 15 : 10 : 25 divides by 3 only, even further from simplest form. Giving 2 : 3 : 5 has the first two parts swapped.
18. (d) 1 × 10⁷ seconds — Method: the time for a journey is the distance divided by the speed, so round each number to 1 significant figure and then divide; dividing numbers in standard form means dividing the coefficients and subtracting the indices. Working: 3.1 × 10¹⁵ rounds to 3 × 10¹⁵ and 2.998 × 10⁸ rounds to 3 × 10⁸; 3 ÷ 3 = 1 for the coefficients, and 15 − 8 = 7 for the indices. Answer: about 1 × 10⁷ seconds. The distractors: 3 × 10⁷ seconds comes from subtracting the indices correctly but leaving the coefficient as 3 instead of dividing 3 by 3; 1 × 10⁻⁷ seconds comes from dividing the speed by the distance instead of the distance by the speed; 9 × 10²³ seconds comes from multiplying the two quantities instead of dividing them, since 3 × 3 = 9 and 15 + 8 = 23.
19. (d) 75% — Method: change a fraction to a percentage by dividing the numerator by the denominator and multiplying by 100. Working: 3 ÷ 4 = 0.75, and 0.75 × 100 = 75. Answer: 75%. The distractors: 34% comes from reading the digits 3 and 4 straight off as a percentage; 0.75% comes from dividing but forgetting to multiply by 100; 133% comes from inverting the fraction and working out 4 ÷ 3 instead.
20. (b) 3 : 5 — The highest common factor of 27 and 45 is 9. Divide both parts by 9: 27 ÷ 9 = 3 and 45 ÷ 9 = 5, giving 3 : 5. Giving 9 : 15 divides by 3, a common factor but not the highest one, so it is not yet in simplest form. Giving 5 : 3 swaps the order of the two parts. Giving 27 : 45 has not been simplified at all.