✏️ Paper 1 non-calculator warm-up — Foundation

Answer key

1. (d) 2³ × 5² — Method: divide repeatedly by the smallest prime number, then write any repeated prime using a power. Working: 200 ÷ 2 = 100, 100 ÷ 2 = 50, 50 ÷ 2 = 25, 25 ÷ 5 = 5, and 5 is prime, so 200 = 2 × 2 × 2 × 5 × 5, written as 2³ × 5². 2² × 5³ swaps the two powers, giving 4 × 125 = 500, not 200. 2³ × 5 leaves out one of the two 5s, giving 8 × 5 = 40, not 200. 2 × 5³ leaves out two of the three 2s, giving 2 × 125 = 250, not 200. Answer: 2³ × 5².
2. (b) 8:12 — Method: two ratios are equivalent when one is obtained from the other by multiplying, or dividing, both parts by the same number. Working: multiplying both parts of 2:3 by 4 gives 2 × 4 = 8 and 3 × 4 = 12, and the check runs the other way too, since the highest common factor of 8 and 12 is 4 and dividing both parts by 4 returns 2:3. Answer: 8:12. The distractors: 8:3 comes from multiplying only the first part by 4 and leaving the second part alone; 12:8 comes from multiplying both parts by 4 correctly but then writing the two parts the wrong way round; 4:5 comes from adding 2 to each part instead of multiplying, and adding the same amount to both parts changes the ratio.
3. (c) 1 : 4 — Convert to the same unit: 1.2 litres = 1200 ml, since 1 litre = 1000 ml. This gives the ratio 300 : 1200. Divide both parts by their highest common factor, 300, to get 1 : 4. Giving 3 : 12 divides by 100 only, which is a common factor but not the highest one. Giving 1 : 1.2 has not converted 1.2 litres into millilitres, so the two parts are not in the same unit. Giving 4 : 1 swaps the order, comparing water to concentrate instead of concentrate to water.
4. (c) 200 students — Method: find the percentage who have a brother or a sister, taking care that the students with both are not counted twice, then take that percentage from 100% and apply the result to the 400 students. Working: 25% + 40% = 65%, but the 15% with both has been counted in each of those figures, so 65% − 15% = 50% have a brother or a sister; that leaves 100% − 50% = 50%, and 50% of 400 = 200. Answer: 200 students. The distractors: 260 students is 65% of 400, the number with a brother or a sister when the 15% overlap is counted twice; 140 students comes from taking that same uncorrected 65% away from the 400; 300 students comes from subtracting only the 25% with a brother and ignoring the sisters altogether.
5. (a) 24 — Method: rearrange the relationship so that the lowest common multiple stands alone; it is the product of the two numbers divided by their highest common factor. Working: 48 = 2 × the lowest common multiple, so the lowest common multiple is 48 ÷ 2 = 24. Checking, 24 is in the 6 times table and in the 8 times table. Answer: 24. The distractors: 48 comes from giving the product of the two numbers and never dividing by the highest common factor; 96 comes from multiplying by the highest common factor instead of dividing by it; 12 comes from dividing by the highest common factor twice, once for each of the two numbers.
6. (a) 4:3:1 — Convert every part to the same unit: 2 m = 200 cm, so the ratio is 200 : 150 : 50. Dividing all three parts by 50 gives 4 : 3 : 1. Writing 2 : 150 : 50 has not converted 2 m into centimetres, so the units do not match. Writing 3 : 4 : 1 has the first two parts the wrong way round. Writing 4 : 3 : 2 comes from an arithmetic slip on the last part: 50 ÷ 50 = 1, not 2.
7. (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.
8. (b) 75 g — Method: split the total mass into the number of parts shown by the ratio, then find the mass of tin. Working: the ratio 7:3 has 7 + 3 = 10 parts, so one part is 250 ÷ 10 = 25 g, and the mass of tin is 3 × 25 = 75 g. So the alloy contains 75 g of tin. Distractor 175 g is the mass of copper, not tin. Distractor 125 g comes from splitting the alloy into two equal halves, ignoring the ratio. Distractor 25 g is the value of one part, found correctly but never multiplied by 3.
9. (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.
10. (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.
11. (a) 13/40 — Method: write the decimal over 1000 using its three decimal places, then simplify. Working: 0.325 = 325/1000 = 13/40 (dividing both numerator and denominator by 25). Answer: 13/40. 13/4 comes from writing the decimal over 100 instead of 1000, as if there were only two decimal places. 8/25 comes from rounding 0.325 down to 0.32 before converting. 3/8 comes from recalling the learned conversion 3/8 = 0.375 and matching it to 0.325 because both are three-place decimals beginning with 3, instead of converting the decimal given.
12. (c) 0.7, 0.72, 3/4 — Method: numbers written in different forms cannot be compared as they stand, so every fraction is turned into a decimal by dividing the numerator by the denominator, and the decimals are then compared place by place from the left. Working: 3/4 means 3 ÷ 4 = 0.75, so the three values to compare are 0.7, 0.75 and 0.72; written to two decimal places they are 0.70, 0.75 and 0.72, and the hundredths digits 0, 5 and 2 put 0.70 first, 0.72 next and 0.75 last; written again in the forms the question used, the order from smallest is 0.7, then 0.72, then 3/4. Answer: 0.7, 0.72, 3/4. The distractors: 3/4, 0.7, 0.72 comes from turning 3/4 into 0.34 by writing the numerator and the denominator as the two digits after the point, which makes the fraction the smallest of the three; 0.72, 3/4, 0.7 comes from the belief that the more digits a decimal has the smaller it must be, which puts both 0.72 and 0.75 below 0.7 and 0.72 below 0.75; 3/4, 0.72, 0.7 comes from comparing the three values correctly but listing them largest first, against an instruction to start with the smallest.
13. (a) 20 — Method: round each number to 1 significant figure, then divide the rounded values. Working: 588 rounds to 600 (1 s.f.) and 31 rounds to 30 (1 s.f.). 600 ÷ 30 = 20. Answer: 20. 17 comes from cutting 588 down to 500, keeping the leading digit as it stands instead of rounding it up to 1 significant figure, 600, then dividing by the correctly rounded 30. 200 comes from misreading the rounded divisor 30 as 3, giving 600 ÷ 3 instead of 600 ÷ 30. 19 is the exact value of 588 ÷ 31 rounded to the nearest whole number, found without rounding the numbers first.
14. (d) 11/12 — Convert both mixed numbers to improper fractions with a common denominator. 2 3/4 = 11/4, which is 33/12, and 1 5/6 = 11/6, which is 22/12. Subtracting, 33/12 − 22/12 gives 11/12, already in its simplest form. Forgetting to borrow, and instead subtracting the fraction parts the other way round to avoid a negative, 10/12 − 9/12 gives 1/12; adding that to the whole-number difference of 1 gives 13/12. Subtracting only the fraction parts, 9/12 − 10/12, and reporting just the size of that difference gives 1/12, which ignores the whole numbers altogether. Adding the two improper fractions instead of subtracting them, 33/12 + 22/12, gives 55/12. So 2 3/4 − 1 5/6 = 11/12.
15. (b) £220 — Method: round each ticket price to the nearest £5, multiply each rounded price by the number of tickets, then add the two totals. Working: the adult price £38.50 rounds to £40, and 4 × £40 = £160; the child price £19.75 rounds to £20, and 3 × £20 = £60; £160 + £60 = £220. Answer: £220. £213.25 is the exact total cost, found without rounding the prices first, so it is not an estimate. £160 comes from including the cost of the adult tickets only and forgetting the three children's tickets. £200 comes from swapping the two ticket quantities, using 3 adults and 4 children instead of 4 adults and 3 children.
16. (d) 2:3:5 — The highest common factor of 12, 18 and 30 is 6. Divide every part by 6: 12 ÷ 6 = 2, 18 ÷ 6 = 3, 30 ÷ 6 = 5, giving 2 : 3 : 5. Dividing by 2 instead of 6 gives 6 : 9 : 15, which still shares a common factor of 3, so it is not fully simplified. Dividing by 3 instead of 6 gives 4 : 6 : 10, which still shares a common factor of 2, so it is not fully simplified either. Swapping the first two parts gives 3 : 2 : 5, the parts in the wrong order.
17. (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.
18. (a) 50 — Method: 25% is one quarter, so 25% of a quantity is the quantity divided by 4. Working: 200 ÷ 4 = 50. Answer: 50. The distractors: 25 comes from writing the percentage itself as the answer; 100 comes from halving, which is 50% not 25%; 800 comes from multiplying by 4 instead of dividing.
19. (c) 150 — Since 90 students represent 3 of the 5 equal parts, one part is 90 ÷ 3 = 30, and the whole year group is five parts: 30 × 5 = 150. Applying the fraction forwards to 90 instead of reversing it, 90 × 3/5 = 54, treats the given number as the whole rather than as three fifths of it. Finding one part correctly as 30 but forgetting to scale up to the whole year group leaves 30 as the final answer. Treating 90 as the whole year group and adding on 2/5 of 90 for the students who do not walk, 90 + (90 × 2/5) = 126, applies the missing fraction to the wrong base amount.
20. (b) 5 litres — Method: adding water changes the total volume but adds no salt, so work out the volume of salt, then the total volume that makes that salt 20% of the mixture, then the extra water. Working: 30% of 10 litres is 0.3 × 10 = 3 litres of salt. For the same 3 litres to be 20% of the new mixture, the new total volume is 3 ÷ 0.2 = 15 litres. The water added is the extra volume, 15 − 10 = 5 litres. Answer: 5 litres. The distractors: 3 litres is the volume of salt in the solution, which is the first step and not what the question asks for; 15 litres is the total volume of the new mixture, which counts the 10 litres already in the container as water that was poured in; 2 litres comes from taking 20% of the original 10 litres, applying the new percentage to the old volume instead of to the new one.