✏️ Paper 1 non-calculator warm-up — Foundation

Answer key

1. (c) £3,200 — Method: find the value of one part of the ratio from the first investor's amount, then work out the second investor's share before adding both together. Working: £1,200 is 3 parts, so one part is £1,200 ÷ 3 = £400. The second investor's share is 5 × £400 = £2,000, and the total is £1,200 + £2,000 = £3,200. So the total invested is £3,200. Distractor £2,000 is only the second investor's share, without adding the first investor's £1,200. Distractor £2,400 comes from doubling the first investor's amount instead of using the ratio. Distractor £6,000 comes from multiplying £1,200 by 5 directly instead of first finding the value of one part.
2. (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.
3. (b) 1 : 1.875 — To write a ratio in the form 1 : n, divide both parts by the first part, 8: 8 ÷ 8 = 1 and 15 ÷ 8 = 1.875, giving 1 : 1.875. Giving 1 : 0.53 divides the wrong way round, computing 8 ÷ 15 instead of 15 ÷ 8. Giving 1.875 : 1 has the two parts of the answer swapped, which is the form n : 1, not 1 : n. Giving 8 : 1.875 divides only the second part by 8, so the first part is still 8, not 1.
4. (b) 30 — Method: the smallest matching total is the lowest common multiple of the two pack sizes. Working: multiples of 6 are 6, 12, 18, 24, 30 …; multiples of 10 are 10, 20, 30 …. The lowest common multiple is 30. 60 comes from working out 6 × 10 = 60, the product of the pack sizes rather than their lowest common multiple. 16 comes from working out 6 + 10 = 16, which is not a common multiple at all. 2 is the highest common factor of 6 and 10, not a total of tickets. Answer: 30.
5. (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.
6. (d) −1.7 — Since the numbers have different signs, find the difference between their sizes: 4.5 − 2.8 = 1.7, then keep the sign of the number further from zero. So −4.5 + 2.8 = −1.7. A candidate who drops the negative sign gets 1.7. A candidate who adds the magnitudes instead of finding the difference gets −(4.5 + 2.8) = −7.3. A candidate who takes the smaller digit from the larger in the tenths column, doing 8 − 5 = 3 instead of borrowing to make 15 − 8 = 7, gets 2.3 and so −2.3.
7. (b) £1,000, so £900 is not enough — Method: round each number to 1 significant figure, multiply to estimate the total cost, then compare the estimate with the money available. Working: 187 rounds to 200 and £4.85 rounds to £5, so the estimate is 200 × 5 = 1,000, and £1,000 is more than the £900 the school has. Answer: £1,000, so £900 is not enough. The distractors: £800 comes from cutting £4.85 down to £4 instead of rounding it up to £5, giving 200 × 4 = 800, and that estimate wrongly suggests the money stretches; £935 comes from rounding the price only and keeping 187 lunches, giving 187 × 5 = 935; £950 comes from rounding 187 to the nearest 10 rather than to 1 significant figure, giving 190 × 5 = 950.
8. (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.
9. (a) 20 — Method: round each number to 1 significant figure, then divide. Working: 397 rounds to 400 and 21 rounds to 20, so the estimate is 400 ÷ 20 = 20. Answer: 20. The distractors: 19 comes from not estimating at all, since 397 ÷ 21 = 18.9 to 1 decimal place, which rounds to 19, while the question asks for an estimate from rounded numbers; 200 comes from dividing 400 by 2 instead of by 20, losing the place value of the rounded divisor; 2 comes from dividing the leading digits only and losing the place value of both numbers.
10. (b) −5 — Using the order of operations, work out the multiplication first: 4 × (−2) = −8. Then 3 + (−8) = −5. A candidate who adds before multiplying gets (3 + 4) × (−2) = −14. A candidate who drops the negative sign on the multiplication gets 3 + 4 × 2 = 11. A candidate who works out the multiplication correctly but gives that as the final answer, forgetting to combine it with the 3, gets −8.
11. (d) 50 g — Method: the milk is 4 parts of the ratio, so use the milk to find the value of one part, then read off the chocolate, which is 1 part. Working: one part = 200 ÷ 4 = 50, and the chocolate is one part. Answer: 50 g. The distractors: 40 g comes from treating the 200 g as the total mass of the mixture and splitting it into 1 + 4 = 5 parts; 250 g is the total mass of the finished mixture, the 200 g of milk plus the chocolate, rather than the chocolate on its own; 800 g comes from multiplying 200 by 4 instead of dividing, which scales the milk up rather than down to the chocolate.
12. (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.
13. (b) (2 + 3) × 5 − 1 — 2 + 3 = 5, then 5 × 5 = 25, then 25 − 1 = 24, so the brackets belong around 2 + 3. Placing them around 5 − 1 instead gives 5 − 1 = 4, then 3 × 4 = 12, then 2 + 12 = 14. Leaving the multiplication bracketed instead changes nothing, because it already had priority: 3 × 5 = 15, then 2 + 15 = 17, then 17 − 1 = 16. Bracketing both 2 + 3 and 5 − 1 uses two pairs instead of the one asked for: 2 + 3 = 5, 5 − 1 = 4, then 5 × 4 = 20.
14. (c) 693 — Method: multiplying a bracket by a number multiplies every term inside it, so 7 × (100 − 1) = 7 × 100 − 7 × 1. Working: 7 × 100 = 700 and 7 × 1 = 7, so the calculation becomes 700 − 7 = 693. Answer: 693. The distractors: 699 comes from subtracting the 1 itself rather than 7 lots of it, giving 700 − 1 = 699; 707 comes from adding the second product instead of subtracting it, giving 700 + 7 = 707; 700 comes from rounding 99 up to 100 and then offering the estimate 7 × 100 as an exact value.
15. (c) 0.75 — Method: a fraction means the numerator divided by the denominator, and carrying out that division gives the decimal. Working: 3 ÷ 4 is worked out by writing 3 as 3.00; 4 divides into 30 tenths 7 times with 2 tenths left over, and 4 divides into the remaining 20 hundredths exactly 5 times, giving 7 tenths and 5 hundredths. Answer: 0.75. The distractors: 0.34 comes from writing the numerator and the denominator as the two digits after the point instead of dividing; 0.7 comes from adding the numerator and the denominator to get 7 and reading that as seven tenths; 0.25 comes from converting the quarter that is left over rather than the three quarters the question asks about.
16. (c) £672 — Simple interest per year = 3% of £600 = £18. Over 4 years the interest is 18 × 4 = £72. Total in the account = £600 + £72 = £672. A student who gives just the interest, without adding it to the principal, writes £72. A student who adds only one year's interest instead of four gets £600 + £18 = £618. A student who wrongly compounds the interest each year gets 600 × 1.03⁴ = £675.31.
17. (c) 36 — Add the parts of the ratio: 3 + 4 = 7. Divide the total by the number of parts: 84 ÷ 7 = 12, so one part is worth 12. The smaller share has 3 parts: 3 × 12 = 36. (48 is the larger share, using 4 parts instead of 3. 42 comes from splitting 84 in half, treating the ratio as if it were 1:1. 28 comes from dividing 84 by 3 — one of the ratio numbers — instead of dividing by the total number of parts, 7.)
18. (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.
19. (a) 36 — Method: find the lowest common multiple of 6 and 9, then move up the list of common multiples until one is greater than 20. Working: the common multiples of 6 and 9 are 18, 36, 54 …. 18 is not greater than 20, so the next one, 36, is the smallest value of n that is greater than 20. 18 is the lowest common multiple itself, but it fails the 'greater than 20' condition. 54 is the common multiple after 36, one step too far. 27 is a multiple of 9 but not of 6, since 27 ÷ 6 is not a whole number. Answer: 36.
20. (c) 2:3 — The white paint is 5 − 2 = 3 litres. The ratio of blue paint to white paint is 2 : 3, which has no common factor, so it is already in simplest form. Getting 2 : 5 compares the blue paint to the total amount of shade instead of to the white paint. Getting 3 : 2 has the two parts the wrong way round. Getting 5 : 3 uses the total amount of shade instead of the blue paint as the first part.