⚖️ Foundation to Higher crossover check

Answer key

1. (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.
2. (d) £40 — Method: add the two prices to make one expression in x, turn 'at least' into ≥, solve the inequality, then read the lowest possible price off the boundary of the solution set. Working: the two items cost x + (2x + 30) = 3x + 30 pounds, so 3x + 30 ≥ 150; subtracting 30 from both sides gives 3x ≥ 120; dividing both sides by 3 gives x ≥ 40, and the smallest value the shirt price is allowed to take is the boundary, £40. Answer: £40. The distractors: £50 comes from leaving the £30 out of the total and solving 3x ≥ 150; £60 comes from using the trousers expression on its own, 2x + 30 ≥ 150; £120 comes from stopping at 3x ≥ 120 and reading the 120 as the price of the shirt without dividing by 3.
3. (a) 36 — Method: add the parts of the ratio, divide the total by the number of parts to find the value of one part, then multiply by the number of parts in the share asked for. Working: 1 + 2 + 3 = 6 parts, 72 ÷ 6 = 12 for one part, and the largest number is 3 parts, so 3 × 12 = 36. Answer: 36. The distractors: 12 is the value of one part, which is the smallest of the three numbers rather than the largest; 24 is 2 parts, the middle number; 216 comes from multiplying 72 by 3 instead of dividing 72 by the 6 parts first.
4. (a) x = 4 — Method: undo the addition of 4 first, then undo the multiplication by 6. Working: subtracting 4 from both sides gives 6x = 24, and dividing both sides by 6 gives x = 4. Checking: 6 × 4 + 4 = 28. Answer: x = 4. The distractors: x = 24 comes from stopping at 6x = 24 and writing 24 as the value of x; x = 18 comes from subtracting 6 from 24 instead of dividing by 6; x = 144 comes from multiplying 24 by 6 instead of dividing by 6.
5. (c) 22 kg — Method: multiply the mean by the number of parcels to rebuild the total mass, then subtract the masses that are known. Working: four parcels with a mean mass of 17 kg have a total mass of 17 × 4 = 68 kg; the three known parcels total 12 + 16 + 18 = 46 kg; so the fourth parcel has mass 68 − 46 = 22 kg. Answer: 22 kg. The distractors: 68 kg comes from stopping at the total mass of all four parcels; 17 kg comes from assuming the missing parcel must have the mean mass; 5 kg comes from multiplying the mean by 3, the number of parcels whose mass is given, leaving 51 − 46 = 5.
6. (c) 15/17 — Method: cos θ = adjacent ÷ hypotenuse, so find the hypotenuse with Pythagoras' theorem first and then decide which short side is next to θ. Working: the hypotenuse is √(8² + 15²) = √(64 + 225) = √289 = 17 cm. The angle θ is opposite the 8 cm side, so the side next to it is the 15 cm side, and cos θ = 15 ÷ 17. Answer: 15/17. The distractors: 8/17 is sin θ, opposite over hypotenuse, used in place of the cosine; 8/15 is tan θ, opposite over adjacent; 17/15 comes from writing the cosine ratio upside down, as hypotenuse over adjacent.
7. (d) x² − 49 — Area = length × width = (x + 7)(x − 7). Expanding: x × x = x², x × (−7) = −7x, 7 × x = 7x, 7 × (−7) = −49. The two middle terms −7x and 7x cancel, leaving x² − 49. A candidate who misremembers the difference-of-two-squares result as a sum gets x² + 49. A candidate who makes a sign error and treats both middle terms as −7x instead of cancelling gets x² − 14x − 49. A candidate who confuses area with perimeter and simply adds the length and width gets 2x.
8. (b) 10 — Corresponding angles are equal, so 5x + 4 = 3x + 24. Subtracting 3x from both sides gives 2x + 4 = 24, then subtracting 4 gives 2x = 20, so x = 10. 14 comes from adding the constants, 4 + 24, instead of subtracting them when rearranging. 19 comes from treating the angles as co-interior (summing to 180°): 5x + 4 + 3x + 24 = 180 gives 8x = 152, so x = 19. 20 correctly reaches 2x = 20 but stops without dividing by 2.
9. (b) (x − 5)(x + 2) — We need two numbers that multiply to −10 and add to −3: these are −5 and 2, since −5 × 2 = −10 and −5 + 2 = −3. So x² − 3x − 10 = (x − 5)(x + 2). A candidate who swaps the signs, using +5 and −2, gets (x + 5)(x − 2), which expands to x² + 3x − 10 — the wrong middle term. A candidate who picks the factor pair 1 and 10 instead of 2 and 5 gets (x − 10)(x + 1), which expands to x² − 9x − 10. A candidate who makes both factors negative gets (x − 5)(x − 2), which expands to x² − 7x + 10 — the wrong sign on the constant term.
10. (b) y = 84 − 14x — Method: the distance still to go is the whole route minus the distance already covered, and at a steady speed the distance covered after x hours is the speed multiplied by x. Working: after x hours Freya has cycled 14x km, so the distance remaining is the 84 km route take away 14x, giving y = 84 − 14x, a straight line with intercept 84 and gradient −14. Answer: y = 84 − 14x. The distractors: y = 84 + 14x comes from adding the distance cycled to the length of the route rather than taking it away, so the ride would grow longer the further she goes; y = 14x − 84 comes from carrying out the subtraction the wrong way round, which is negative for the whole of the ride; y = 14 − 84x comes from swapping the two numbers over, treating 84 km as the hourly speed and 14 km as the length of the route.
11. (a) 49° — Method: an exterior angle of a triangle equals the sum of the two interior angles that are not next to it, which here are the angles at A and at C; since those two are equal, the exterior angle is twice the angle at A. Working: 2 × angle A = 98, so angle A = 98 ÷ 2 = 49. Answer: 49°. The distractors: 82° is the interior angle at B, 180 − 98, given in place of the angle at A; 41° comes from finding that interior angle of 82° and halving it, 82 ÷ 2, instead of halving the exterior angle; 98° comes from taking the exterior angle to be equal to the angle at A on its own, with no halving at all.
12. (a) 5 — The symbol ≤ means n can equal 5 or any number less than 5, so 5 is included and is the largest integer value. A candidate who treats the inequality as strict, as if it were n < 5, answers 4. A candidate who confuses ≤ with ≥ and looks for a value just above the boundary answers 6. A candidate who makes a sign error and reads the inequality as n ≤ −5 answers −5.
13. (d) 29 — Method: with an even number of values there is no single middle value, so the median is the mean of the two values either side of the middle. Working: the six numbers are already in order and 6 ÷ 2 = 3, so the middle pair are the third and fourth values, 22 and 36; their mean is (22 + 36) ÷ 2 = 58 ÷ 2 = 29. Answer: 29, which lies between the two middle values as a median of an even data set must. The distractors: 22 comes from taking the lower of the two middle values and stopping there instead of averaging the pair; 36 comes from taking the larger value of that pair because it sits just past the halfway point of the list; 34 comes from working out the range, 47 − 13, instead of a measure of centre.
14. (c) 15 — The exterior angle is 180° − 156° = 24°, and the number of sides of a regular polygon is 360° divided by the exterior angle, so 360 ÷ 24 = 15. 24° is the exterior angle itself, stopping one step before the final division. 17 comes from finding 15 correctly and then adding 2, muddling the exterior angle rule with the (n − 2) that appears in the interior angle sum formula. 7.5 comes from dividing 180 by the exterior angle instead of 360, using the angles on a straight line rather than the total of the exterior angles of a polygon.
15. (b) 3.5 — Method: work out each percentage of its number separately, then subtract the smaller result from the larger one. Working: 45% of 70 = 31.5, and 35% of 80 = 28, so the difference is 31.5 − 28 = 3.5. Answer: 3.5. 11.5 comes from pairing the percentages with the wrong numbers, working out 35% of 70 = 24.5 and 45% of 80 = 36, and finding their difference. 59.5 comes from adding the two correct results, 31.5 + 28, instead of subtracting them. 10 comes from simply subtracting the two percentages themselves, 45 − 35, without applying them to the numbers at all.
16. (c) 32 — Method: alternate angles between parallel lines are equal, so 2x + 10 = 74. Working: subtracting 10 from both sides gives 2x = 64; dividing by 2 gives x = 32. Answer: x = 32. A candidate who forgets to subtract 10 first and divides 74 by 2 directly gets 37. A candidate who treats the angles as co-interior instead of alternate, so that the two expressions add to 180° rather than being equal, gets 48 after solving. A candidate who makes a sign error and treats the equation as 2x equalling 10 minus 74 instead of 74 minus 10 gets −32.
17. (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.
18. (a) 3/8, 41%, 0.43 — Method: convert every value to a decimal so they can be compared on the same scale, then order them. Working: 3/8 = 0.375, 41% = 0.41, and 0.43 stays as 0.43, so from smallest to largest the decimals are 0.375, 0.41, 0.43, giving the order 3/8, 41%, 0.43. Answer: 3/8, 41%, 0.43. The order 3/8, 0.43, 41% comes from comparing 0.43 and 41% as raw digits (43 versus 41) without converting 41% into the decimal 0.41 first, wrongly placing 0.43 before 41%. The order 0.43, 41%, 3/8 comes from placing the values in completely reversed order, from largest to smallest instead of smallest to largest. The order 41%, 0.43, 3/8 comes from ordering the values by their TYPE (percentage, then decimal, then fraction) rather than by their actual size.
19. (c) 6x(2x + 3) — The highest common factor of 12x² and 18x is 6x. Dividing each term by 6x gives 12x² ÷ 6x = 2x and 18x ÷ 6x = 3, so 12x² + 18x = 6x(2x + 3). A candidate who only takes out the number 6 (missing the x) gets 6(2x² + 3x), which is not fully factorised. A candidate who only takes out 2x (missing the extra factor of 3 in 6) gets 2x(6x + 9), also not fully factorised — the bracket still shares a common factor. A candidate who takes out 3x instead of the full 6x gets 3x(4x + 6), which again is not fully factorised since 4x + 6 shares a common factor of 2.
20. (c) x ≤ 4 — Subtract 9 from both sides: −2x ≥ 1 − 9 = −8. Divide both sides by −2, flipping the inequality: x ≤ 4. A candidate who divides by −2 but forgets to flip the inequality gets x ≥ 4. A candidate who divides −8 by −2 but keeps a negative sign gets x ≤ −4. A candidate who miscalculates 1 − 9 as −10 gets x ≤ 5.