⚖️ Foundation to Higher crossover check

Answer key

1. (d) 2x = 10 — Method: collect the x-terms on one side and the constants on the other, keeping the equals sign balanced. Working: 5x − 3x = 8 + 2, giving 2x = 10. Answer: 2x = 10. 2x = 6 comes from a sign error moving the constant, subtracting instead of adding: 8 − 2 = 6. 8x = 6 comes from adding the x-terms instead of subtracting them (5x + 3x = 8x), alongside the constant sign error. 2x = −10 comes from a sign error moving the constant the other way: −2 − 8 = −10.
2. (b) 24 — Method: let one part of the ratio be worth x, write both groups in terms of x, and use the fact that the two groups end up equal. Working: the boys are 3x and the girls are 4x; after the 6 boys join, 3x + 6 = 4x, so x = 6; the girls are 4 parts, so 4 × 6 = 24. Answer: 24 girls. The distractors: 18 is the number of boys before the 6 join, which is 3 × 6; 30 comes from adding the 6 new members to the girls as well as to the boys; 42 is the total number of members in the choir before the 6 boys join, the 18 boys and the girls together.
3. (c) x = 5 — Method: every point on a vertical line has the same x-coordinate however far up or down the line it lies, so the equation of a vertical line fixes x at a number and does not involve y at all. Working: of the four equations only x = 5 fixes x; it is satisfied by (5, 0), (5, 1), (5, 7) and by every other point whose x-coordinate is 5, and those points form a vertical line. Answer: x = 5. The distractors: y = 5 fixes the y-coordinate instead of the x-coordinate, which gives a horizontal line; y = 5x is a line through the origin with gradient 5, steep but not vertical, and it has a different y-value for every x; x = y fixes neither coordinate and is the line through the origin with gradient 1.
4. (a) y = −2x − 1 — Method: the gradient is the change in y divided by the change in x with both differences taken in the same order, and the constant then comes from substituting either point into y = mx + c. Working: m = (−9 − 3) ÷ (4 − (−2)) = (−12) ÷ 6 = −2, so the line is y = −2x + c; substituting (−2, 3) gives 3 = −2 × (−2) + c = 4 + c, so c = 3 − 4 = −1. Answer: y = −2x − 1. The distractors: y = −2x + 1 comes from rearranging 3 = 4 + c the wrong way round and taking the constant as 4 − 3; y = 2x + 7 comes from losing the minus sign when −12 is divided by 6 and then substituting correctly, 3 = 2 × (−2) + c; y = −(1/2)x + 2 comes from writing the gradient upside down as the change in x over the change in y, 6 ÷ (−12).
5. (a) 3 — Method: subtract the constant term from both sides first, then divide by the coefficient of x. Working: 5x = 17 − 2 = 15; x = 15 ÷ 5 = 3. Answer: x = 3. 3.8 comes from adding 2 instead of subtracting it: (17 + 2) ÷ 5 = 3.8. 1.4 comes from dividing by 5 before subtracting the 2, the wrong order: 17 ÷ 5 = 3.4, then 3.4 − 2 = 1.4. 15 comes from correctly subtracting the 2 but then forgetting to divide by 5.
6. (a) y = −2x + 7 — Using y = −2x + c, and substituting the point (3, 1): 1 = −2(3) + c, so 1 = −6 + c, and c = 7, giving y = −2x + 7. A candidate who uses the y-coordinate of the point directly as the y-intercept, instead of solving for c, would write y = −2x + 1. A candidate who drops the negative sign on the gradient would write y = 2x + 7. A candidate who makes a sign error when isolating c, writing c = 1 − 6 = −5 instead of c = 1 + 6 = 7, would write y = −2x − 5.
7. (a) x > 4 — Subtract 2x from both sides: 3x − 3 > 9. Add 3 to both sides: 3x > 12. Divide both sides by 3: x > 4. A candidate who subtracts 3 from 9 instead of adding gets 3x > 6, so x > 2. A candidate who divides correctly but wrongly flips the inequality (as if dividing by a negative) gets x < 4. A candidate who multiplies by 3 instead of dividing gets x > 36.
8. (a) 40 — Method: the mean is the total divided by how many values there are, so rearranging gives total = mean × number of values. Working: the mean is 8 and there are 5 numbers, so the total is 8 × 5 = 40. Answer: 40, and checking, 40 ÷ 5 = 8, which is the mean given. The distractors: 13 comes from adding the mean and the count, 8 + 5, instead of multiplying them; 1.6 comes from dividing the mean by the count, 8 ÷ 5, which reverses the relationship; 8 comes from quoting the mean itself as the total, which is only true when there is a single number.
9. (a) x = 6 — Method: a minus sign in front of a bracket changes the sign of every term inside it, so expand the bracket first and then simplify. Working: expanding gives 4x − 2x + 6 = 18, which simplifies to 2x + 6 = 18; subtracting 6 from both sides gives 2x = 12, and dividing both sides by 2 gives x = 6. Answer: x = 6. The distractors: x = 12 comes from leaving the −6 unchanged when the bracket is removed, giving 4x − 2x − 6 = 18 and so 2x = 24; x = 24 comes from reaching 2x = 12 correctly and then multiplying by 2 instead of dividing by 2; x = 3 comes from dividing by 2 too early, at 2x + 6 = 18, and dividing only the 2x and the 18 while leaving the 6 untouched, which gives x + 6 = 9.
10. (a) x = 2 — Method: undo the subtraction first, then undo the multiplication, doing the same to both sides each time. Working: adding 2 to both sides gives 2x = 4, and dividing both sides by 2 gives x = 2. Answer: x = 2. The distractors: x = 4 comes from stopping at 2x = 4 and writing 4 as the value of x; x = 0 comes from subtracting 2 from both sides instead of adding it, giving 2x = 0; x = 8 comes from multiplying 4 by 2 instead of dividing by 2.
11. (d) 77 — Method: angles that meet at a point add up to 360°. Working: 82 + 105 + 96 = 283; 360 − 283 = 77. Answer: 77°. A candidate who gives the sum of the three known angles and forgets to subtract it from 360° gets 283. A candidate who leaves out the 105° angle, working out 360 − 82 − 96, gets 182. A candidate who leaves out the 82° angle, working out 360 − 105 − 96, gets 159.
12. (d) 15 — Substituting x = 3 into y = 5x gives y = 5 × 3 = 15. A candidate who adds instead of multiplying would get 5 + 3 = 8. A candidate who ignores the coefficient 5 and just copies the value of x would write y = 3. A candidate who multiplies correctly but treats x as −3 by mistake would get y = 5 × (−3) = −15.
13. (d) x ≤ 30 — Method: add the fixed cost to the cost of x balloons, set that total against the £150 limit with the sign that 'at most' calls for, then solve. Working: the total spend is 60 + 3x pounds, so 60 + 3x ≤ 150; subtracting 60 from both sides gives 3x ≤ 90; dividing both sides by 3, a positive number, gives x ≤ 30. Answer: x ≤ 30. The distractors: x ≤ 90 comes from taking the cake off the budget and stopping at 3x ≤ 90, reading the 90 as a number of balloons when it is the money left for them; x ≤ 50 comes from dividing the whole £150 by 3 and leaving the cake out of the calculation altogether; x ≥ 30 comes from reading 'at most' as 'at least', which reverses the condition.
14. (a) −10 — Method: multiply each bracket out, treating the second bracket as being multiplied by −3 because it is subtracted, then collect like terms. Working: 4(2x − 1) = 8x − 4 and −3(x + 2) = −3x − 6, so the expression becomes 8x − 4 − 3x − 6 − 5x; the x terms give 8x − 3x − 5x = 0, so no term in x survives, and the numbers give −4 − 6 = −10. Answer: −10. The distractors: 2 comes from expanding −3(x + 2) as −3x + 6, leaving the numbers −4 + 6; 5x − 10 comes from forgetting the final −5x, so the x terms give 8x − 3x = 5x; −7 comes from multiplying the 4 over only the first term of its bracket, giving 8x − 1 and so the numbers −1 − 6.
15. (d) x = 6 — Method: with an unknown on both sides, first collect the x terms on one side by subtracting the smaller x term from both sides, then deal with the numbers. Working: subtracting x from both sides gives x + 7 = 13, and subtracting 7 from both sides gives x = 6. Answer: x = 6. The distractors: x = 20 comes from adding 7 to 13 instead of subtracting it once the x terms have been collected; x = 2 comes from collecting the x terms by adding them, giving 3x + 7 = 13 and then 3x = 6; x = −6 comes from subtracting 2x from both sides to get 7 = −x + 13, reaching −6 = −x and then copying the sign straight across instead of dividing by −1.
16. (b) 5 m — Method: the brace, the width and the height form a right-angled triangle in which the brace faces the right angle, so it is the hypotenuse and Pythagoras' theorem applies, a² + b² = c². Working: c² = 3² + 4² = 9 + 16 = 25, so c = √25 = 5. Answer: 5 m. The distractors: 7 m comes from adding the two sides, 3 + 4, instead of adding their squares; 25 m comes from stopping at c² = 25 and forgetting to take the square root; 12 m comes from multiplying 3 × 4, which gives the area of the gate in square metres and not a length across it.
17. (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.
18. (c) 5(x + y − 1) — Method: take out the highest common factor of all three terms and divide every term by it, the number term included. Working: the highest common factor of 5x, 5y and −5 is 5; dividing gives 5x ÷ 5 = x, 5y ÷ 5 = y and −5 ÷ 5 = −1, so the bracket holds x + y − 1. Answer: 5(x + y − 1), which multiplies back out to 5x + 5y − 5. The distractors: 5(x + y + 1) comes from dividing −5 by 5 and losing the minus sign; 5(x + y − 5) comes from dividing only the terms containing a letter by 5 and carrying the −5 into the bracket unchanged; 5(xy − 1) comes from collecting the unlike terms 5x and 5y as 5xy before factorising.
19. (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.
20. (b) £20 — Method: add the parts of the ratio, divide the amount by the number of parts to find the value of one part, then take the parts belonging to the person asked about. Working: 1 + 2 = 3 parts, £60 ÷ 3 = £20 for one part, and Isla has 1 part. Answer: £20. The distractors: £40 is the 2-part share, which belongs to her brother; £30 comes from reading 1:2 as one part out of two and halving the £60; £120 comes from multiplying the £60 by 2 instead of dividing it into 3 parts.