⚖️ Foundation to Higher crossover check

Answer key

1. (c) 12 m — Method: the perimeter of a rectangle is twice the sum of the length and the width, so half the perimeter is one length plus one width; split that half using the ratio. Working: half of 64 is 64 ÷ 2 = 32 m, the ratio 5:3 has 5 + 3 = 8 parts, so one part is 32 ÷ 8 = 4 m, and the width is 3 × 4 = 12 m. So the width is 12 m. Distractor 20 m is the length, 5 parts, not the width. Distractor 24 m comes from splitting the whole perimeter, 64 m, into 8 parts and multiplying by 3, forgetting to halve the perimeter first. Distractor 8 m comes from the same slip stopped one step earlier: splitting the whole perimeter into 8 parts, 64 ÷ 8 = 8, and giving that instead of the width.
2. (c) 8x − 12 — Multiply each term inside the bracket by 4: 4 × 2x = 8x and 4 × (−3) = −12, so 4(2x − 3) = 8x − 12. A candidate who forgets to multiply the second term by 4 gets 8x − 3. A candidate who makes a sign error, treating 4 × (−3) as +12, gets 8x + 12. A candidate who adds 4 to the bracket instead of multiplying gets 2x + 1.
3. (c) y = 200 − 8x — Method: in a linear model the amount present at the start is the constant term and the steady rate of change is the gradient, which is negative because the amount is falling. Working: at x = 0 minutes there are 200 litres, so the constant term is 200; 8 litres are lost every minute, so after x minutes 8x litres have gone and the amount left is y = 200 − 8x. Answer: y = 200 − 8x. The distractors: y = 8x + 200 treats the draining as filling, so the butt would gain 8 litres a minute; y = 200x − 8 swaps the two numbers over, using the starting 200 litres as the rate per minute and the 8 litres per minute as the starting amount; y = −8x − 200 makes the starting amount negative as well as the rate, so the butt would begin 200 litres in deficit.
4. (d) x = 5 — Method: expand the bracket by multiplying both terms inside it by 5, then undo the addition and the multiplication in turn. Working: expanding gives 5x + 15 = 40; subtracting 15 from both sides gives 5x = 25; dividing both sides by 5 gives x = 5. Answer: x = 5. The distractors: x = 8 comes from dividing both sides by 5 first, reaching x + 3 = 8 and writing 8 as the value of x without taking the 3 away; x = 11 comes from adding 15 to both sides instead of subtracting it, giving 5x = 55; x = 7.4 comes from expanding 5(x + 3) as 5x + 3, multiplying only the x by the 5, which leads to 5x = 37.
5. (a) 65° — Method: a line that divides an angle into two equal parts gives each part half of the original angle, so halve 130°. Working: 130 ÷ 2 = 65. Answer: 65°. The distractors: 130° is the whole of angle ABC, written down without halving it; 50° comes from working out 180 − 130, using the angles on a straight line instead of dividing the angle in two; 32.5° comes from dividing by 4 instead of by 2, as though the line split the angle into four equal parts.
6. (b) −5 — Method: a point whose x-coordinate is 0 lies on the y-axis, so its y-coordinate is the constant c; once m and c are both known the equation can be written down and x substituted into it. Working: the line passes through (0, 1), so c = 1 and the equation is y = −3x + 1; substituting x = 2 gives y = −3 × 2 + 1 = −6 + 1 = −5. Answer: −5. The distractors: 7 comes from ignoring the minus sign on the gradient and working out 3 × 2 + 1; 5 comes from working out 3 × 2 = 6 and then using the minus sign to take the constant off the product, 6 − 1 = 5; −7 comes from subtracting the constant instead of adding it, −6 − 1 = −7.
7. (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.)
8. (d) 50 — The angles in any triangle add up to 180°, so x = 180 − 90 − 40 = 50. "90" comes from subtracting only the 90° angle from 180° and forgetting to also subtract the 40°. "130" comes from adding 90° and 40° together instead of subtracting them from 180°. "45" comes from halving the 90° angle instead of using the angle sum of the triangle at all.
9. (b) x ≤ −2 — Method: take the number term off both sides, then divide by the coefficient of x; the sign turns round only when you divide BY a negative number, and here you divide by 3. Working: subtracting 6 from both sides of 3x + 6 ≤ 0 gives 3x ≤ −6; dividing both sides by 3, which is positive, gives x ≤ −2. Answer: x ≤ −2. The distractors: x ≥ −2 comes from turning the sign round because the right-hand side has become negative, which is not the rule; it is the sign of the divisor that matters; x ≤ 2 comes from moving the 6 across without changing its sign, giving 3x ≤ 6; x ≤ −18 comes from multiplying both sides by 3 instead of dividing by it.
10. (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.
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. (d) 3x² + 4x − 7 — Method: the minus sign in front of the second bracket changes the sign of every term inside it; then collect like terms. Working: removing the brackets gives 5x² + 3x − 2 − 2x² + x − 5; the squared terms give 5x² − 2x² = 3x², the x terms give 3x + x = 4x, and the number terms give −2 − 5 = −7. Answer: 3x² + 4x − 7. The distractors: 7x² + 2x + 3 comes from adding the two brackets instead of subtracting, giving 5x² + 2x², 3x − x and −2 + 5; 3x² + 2x + 3 comes from applying the minus sign to 2x² only, leaving −x and +5 unchanged so that 3x − x = 2x and −2 + 5 = 3; 3x² + 4x + 3 comes from changing the signs of the terms with letters but leaving +5 as it stood, so the number terms give −2 + 5 = 3.
13. (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.
14. (c) y = 4x − 5 — Parallel lines have the same gradient, so the new line has gradient 4; since it passes through (0, −5), its y-intercept is −5, giving y = 4x − 5. A candidate who drops the negative sign on the y-intercept would write y = 4x + 5. A candidate who changes the sign of the gradient, instead of keeping it the same for a parallel line, would write y = −4x − 5. A candidate who confuses m and c, using the y-intercept of the first line (3) as the gradient of the second, would write y = 3x − 5.
15. (b) 13 — Using the sum of interior angles formula, (n − 2) × 180° = 1980°, so n − 2 = 1980 ÷ 180 = 11, and n = 11 + 2 = 13. 11 stops after the division, forgetting to add 2 back to find n. 15 adds 2 twice by mistake, giving 11 + 2 + 2. 22 divides 1980 by 90 instead of 180.
16. (b) −1% — Method: write each change as a multiplier and multiply them. A 10% fall is × 0.9 and a 10% rise is × 1.1. Working: 0.9 × 1.1 = 0.99, so the final price is 99% of the original, which is 1% less. Answer: an overall change of −1%. The distractors: 0% comes from assuming a 10% fall and a 10% rise cancel — they do not, because the rise is 10% of a smaller amount; +1% has the size right but the sign wrong, from reading the multiplier 0.99 as 1% above 1 instead of 1% below it; −2% comes from finding the 1% fall and then counting it once for each of the two changes.
17. (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.
18. (d) 6 — Method: a value satisfies x ≥ 7 when it is greater than 7 or exactly equal to 7, so test each value against the boundary. Working: 8 is greater than 7 and 100 is greater than 7, so both satisfy the inequality; 7 is equal to the boundary and ≥ includes equality, so 7 satisfies it as well; 6 is less than 7, so 6 is the one value that fails. Answer: 6. The distractors: 7 is chosen by candidates who read ≥ as a strict 'greater than' and so shut the boundary value out of the solution set; 8 is chosen by reading the question as asking which value DOES satisfy the inequality and taking the smallest such value; 100 is chosen by the same misreading, taking instead the value furthest above the boundary.
19. (a) 1260° — The sum of the interior angles of a polygon with n sides is (n − 2) × 180°. For a nonagon, n = 9, so the sum is (9 − 2) × 180° = 7 × 180° = 1260°. 1620° uses 9 × 180° without subtracting 2 from n first. 140° is the size of a single interior angle of a regular nonagon (1260° ÷ 9), not the sum of all nine. 1440° uses (n − 1) × 180° = 8 × 180° instead of (n − 2) × 180°.
20. (c) 19 — Method: set up the equation 15 + 9n = 186, then subtract the deposit and divide by the cost per student. Working: 9n = 186 − 15 = 171; n = 171 ÷ 9 = 19. Answer: 19 students. 20.67 comes from dividing the whole £186 by £9 without first subtracting the deposit: 186 ÷ 9 ≈ 20.67. 11.8 comes from swapping the two amounts round, subtracting £9 and dividing by £15: (186 − 9) ÷ 15 = 11.8. 22.33 comes from adding the deposit instead of subtracting it: (186 + 15) ÷ 9 ≈ 22.33.