⚖️ Foundation to Higher crossover check

Answer key

1. (c) 33 — Method: multiply the mean by the number of tests to get the total marks, then subtract the marks that are already known. Working: four tests with a mean of 29 give a total of 29 × 4 = 116 marks; the first three marks total 31 + 26 + 26 = 83; so the fourth mark is 116 − 83 = 33. Answer: 33, and checking, (31 + 26 + 26 + 33) ÷ 4 = 116 ÷ 4 = 29. The distractors: 116 comes from stopping at the total for all four tests; 29 comes from assuming the missing mark must be the mean itself; 4 comes from multiplying the mean by 3, the number of marks given, leaving 87 − 83 = 4.
2. (d) 125° — Method: two angles that sit next to each other at a crossing point lie on a straight line, so they add up to 180°; the angle that faces the given one across the point is the equal one, and that is not the angle asked for here. Working: 180° − 55° = 125°. Answer: 125°. The distractors: 55° is the angle vertically opposite the given one, taken by a candidate who reads “next to” as the facing angle and applies the equal-angles rule to the wrong pair; 35° comes from subtracting from 90°, treating the pair as complementary instead of as angles on a straight line; 90° comes from assuming that a line crossing a pair of parallel lines must meet them at right angles, which the question never says.
3. (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.
4. (d) 10.3 m — By Pythagoras' theorem, the hypotenuse = √(5² + 9²) = √(25 + 81) = √106 = 10.29...≈ 10.3 m. "106 m" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "14 m" comes from adding the two shorter sides, 5 + 9, instead of using Pythagoras' theorem at all. "10.2 m" comes from rounding 10.29...m down to 10.2 instead of correctly rounding it up to 10.3.
5. (d) 0 — Method: the range is the largest value minus the smallest value, whatever those two values turn out to be. Working: every value is 10, so the largest value is 10 and the smallest value is 10 as well, and the range is 10 − 10 = 0. Answer: 0 — a range of nothing says the data do not vary at all. The distractors: 10 comes from writing down the repeated value itself instead of the difference between the extremes; 20 comes from adding the largest and the smallest, 10 + 10, instead of subtracting; 40 comes from adding all four values, which gives the total sold and not a measure of spread.
6. (d) 21 — Add the parts: 5 + 3 = 8. Divide the total by the number of parts: 56 ÷ 8 = 7, so one part is worth 7 counters. Blue has 3 parts: 3 × 7 = 21. (35 is the number of red counters, using 5 parts instead of 3. 28 comes from splitting 56 counters in half instead of in the ratio 5:3. 7 is the value of one part — the number of blue counters is 3 lots of this, not just one.)
7. (a) 7 — Subtract 4 from both sides: 3h ≤ 21. Divide both sides by 3: h ≤ 7, so the greatest whole number of hours is 7. A candidate who forgets the £4 fixed fee solves 3h ≤ 25, getting h ≤ 8.33, rounded down to 8. A candidate who adds the £4 instead of subtracting it solves 3h ≤ 29, getting h ≤ 9.67, rounded down to 9. A candidate who reaches h ≤ 7 but wrongly assumes the boundary value cannot be used, thinking that spending the whole £25 is not allowed, answers 6.
8. (c) x = 4 — Method: undo the addition of 1 first, then undo the multiplication by 2. Working: subtracting 1 from both sides gives 2x = 8, and dividing both sides by 2 gives x = 4. Answer: x = 4. The distractors: x = 8 comes from stopping at 2x = 8 and writing 8 as the value of x; x = 5 comes from adding 1 to both sides instead of subtracting it, giving 2x = 10; x = 16 comes from multiplying 8 by 2 instead of dividing by 2.
9. (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.
10. (c) x ≥ 3 — Method: collect the number terms first; the x term is negative, so the final step multiplies both sides by −1, and that is the one step that turns the inequality sign round. Working: subtracting 5 from both sides of 5 − x ≤ 2 gives −x ≤ −3; multiplying both sides by −1 turns −x into x and −3 into 3, and because the multiplier is negative the ≤ becomes ≥, so x ≥ 3. Answer: x ≥ 3. The distractors: x ≤ 3 comes from multiplying by −1 without turning the sign round, the commonest slip on this type; x ≤ −3 comes from reading −x ≤ −3 as though the minus sign could simply be rubbed off the left-hand side; x ≥ −3 comes from turning the sign round correctly but leaving the right-hand side at −3 instead of multiplying it by −1 as well.
11. (d) Line 3 — Method: check each line of Chloe's working against the correct algebraic step. Working: Line 1 to Line 2 is correct, subtracting 2x from both sides gives 3x − 4 = 8. But Line 2 to Line 3 should add 4 to both sides, giving 3x = 12, not 3x = 4 — the constant −4 has been dropped rather than removed correctly. Line 4 follows correctly from Chloe's own, incorrect, Line 3. Answer: Line 3. Line 1 is simply the original equation, copied out correctly. Line 2 correctly subtracts 2x from both sides of Line 1. Line 4 divides Chloe's own Line 3 by 3 correctly — the arithmetic there is fine, the mistake happened one line earlier.
12. (c) 7.1 cm — Method: a shorter side is opposite the 45° angle and the hypotenuse is known, so sin θ = opposite ÷ hypotenuse gives that side directly. Working: sin 45° = x ÷ 10, so x = 10 × sin 45° = 7.071…, which is 7.1 to 1 decimal place. Answer: 7.1 cm. The distractors: 5.0 cm comes from halving the hypotenuse, which is the rule for the side opposite a 30° angle and not a 45° one; 14.1 cm comes from dividing by sin 45° instead of multiplying by it, which makes a shorter side longer than the hypotenuse; 10.0 cm comes from taking tan 45° = 1 and concluding that the shorter side matches the hypotenuse.
13. (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.
14. (d) 30 — Method: 50% is one half, so 50% of a quantity is the quantity divided by 2. Working: 60 ÷ 2 = 30. Answer: 30. The distractors: 120 comes from multiplying by 2 instead of dividing; 3000 comes from multiplying by 50 without dividing by 100; 6 comes from finding 10% instead of 50%.
15. (b) 5 — Method: add 9 to both sides, then divide by 5. Working: 5x = 16 + 9 = 25, so x = 25 ÷ 5 = 5. Answer: 5. 3.2 comes from dividing 16 by 5 directly, without adding 9 first. 1.4 comes from a sign error, subtracting 9 from 16 instead of adding it, then dividing by 5. 25 comes from correctly working out 5x = 25 but stopping there, without dividing by 5 to find x.
16. (c) 80 — Method: the range is a measure of spread and is found by subtracting the smallest value from the largest. Working: the largest number sold is 100 and the smallest is 20, so the range is 100 − 20 = 80. Answer: 80. The distractors: 100 comes from writing down the largest value and never subtracting the smallest; 60 comes from working out the median, the middle value of 20, 40, 60, 70, 100, instead of the range; 58 comes from working out the mean, 290 ÷ 5, which measures centre rather than spread.
17. (b) 21 cm — By Pythagoras' theorem, the other side = √(29² − 20²) = √(841 − 400) = √441 = 21 cm. "9 cm" comes from subtracting the two given lengths directly, 29 − 20 = 9, instead of subtracting their squares. "441 cm" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "35 cm" comes from adding the squares of the two given lengths instead of subtracting them, √(29² + 20²) = √1241 ≈ 35, treating both given lengths as if they were the two shorter sides rather than a shorter side and the hypotenuse.
18. (d) 21.2 m — Method: the cable is the hypotenuse of a right-angled triangle whose vertical side is the drop from the roof to the bracket and whose horizontal side is 15 m, so use Pythagoras' theorem. Working: the drop is 20 − 5 = 15 m, so c² = 15² + 15² = 225 + 225 = 450 and c = √450 = 21.213…, which is 21.2 m to 1 decimal place. Answer: 21.2 m. The distractors: 25.0 m comes from using the whole 20 m height of the roof as the vertical side and forgetting that the bracket is already 5 m up; 30.0 m comes from adding the two sides of the triangle, 15 + 15, instead of using Pythagoras' theorem; 15.0 m is the horizontal distance on its own, which would be the length of the cable only if it ran level.
19. (c) A solid circle at −1 with the arrow pointing right — Method: a number line picture of an inequality carries two decisions: the circle at the boundary says whether the boundary value itself belongs to the solution set, and the arrow says which way the solutions run. Working: the sign is ≥, which includes equality, so x = −1 is itself a solution and the circle drawn at −1 is filled in; testing a value above the boundary, 4 ≥ −1 is true, and testing one below it, −5 ≥ −1 is false, so the solutions lie above −1 and the arrow runs to the right. Answer: a solid circle at −1 with the arrow pointing right. The distractors: an open circle with the arrow pointing right comes from treating ≥ as a strict >, which would shut the boundary value out; a solid circle with the arrow pointing left comes from reading the statement backwards, as if it said −1 ≥ x; an open circle with the arrow pointing left comes from making both of those mistakes at once.
20. (b) £13,860 — Method: apply the first year's percentage decrease, then apply the second year's percentage decrease to the new value. Working: after the first year, the car is worth £17,500 × 0.88. Multiplying this result by 0.90 gives the value at the end of the second year, £13,860. Answer: £13,860. £13,650 comes from adding the two percentages together (12% + 10% = 22%) and applying a single 22% decrease, £17,500 × 0.78 = £13,650, instead of applying the decreases one after the other. £15,750 comes from applying only the second year's 10% decrease to the original price, forgetting the first year's decrease entirely, £17,500 × 0.90 = £15,750. £15,400 comes from applying only the first year's 12% decrease and stopping there, forgetting to apply the second year's decrease at all.