⚖️ Foundation to Higher crossover check

Answer key

1. (a) £102 — Method: find the value of one part of the ratio, then work out each group's share before comparing them. Working: the ratio 5:2 has 5 + 2 = 7 parts, so one part is £238 ÷ 7 = £34. Adults donate 5 × £34 = £170 and children donate 2 × £34 = £68, so adults donate £170 − £68 = £102 more than children. So the difference is £102. Distractor £68 is only the children's donation, without finding the difference. Distractor £170 is only the adults' donation, without finding the difference. Distractor £136 comes from doubling the children's donation instead of subtracting it from the adults' donation.
2. (a) 7 — Method: multiply both sides by 3 to clear the fraction, then solve the resulting equation. Working: 2x + 1 = 5 × 3 = 15. Subtract 1: 2x = 14. Divide by 2: x = 7. Answer: 7. 2 comes from ignoring the denominator altogether, treating the equation as 2x + 1 = 5 without multiplying by 3 first. 8 comes from a sign error, adding 1 to 15 instead of subtracting it, giving 2x = 16. 14 comes from correctly reaching 2x = 14 but stopping there, without dividing by 2 to find x.
3. (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.
4. (d) a² + b² = c² — Pythagoras' theorem states that the square of the hypotenuse equals the sum of the squares of the other two sides, so a² + b² = c². "a + b = c" adds the sides directly without squaring them at all. "a² − b² = c²" subtracts the squares instead of adding them. "a² + b² = c" adds the squares correctly but forgets to square the hypotenuse on the other side of the equation.
5. (b) £76.00 — One part of the ratio is £47.50 ÷ 5 = £9.50. The school receives 8 parts, so its share is 9.50 × 8 = £76.00. Dividing £47.50 by 8 instead of 5, treating the charity's amount as if it were 8 parts, gives 47.50 ÷ 8 = 5.9375, then × 5 = £29.69. Adding the charity's amount to the school's amount instead of stopping at the school's own share gives the total collected, 9.50 × 13 = £123.50. Adding one part to the charity's amount instead of multiplying one part by 8 gives 47.50 + 9.50 = £57.00.
6. (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.
7. (a) 44.7 m — Method: the line joining the two tops is the hypotenuse of a right-angled triangle whose horizontal side is the gap between the masts and whose vertical side is the difference in their heights, so Pythagoras' theorem applies. Working: the difference in heights is 50 − 30 = 20 m, so d² = 40² + 20² = 1600 + 400 = 2000 and d = √2000 = 44.721…, which is 44.7 m to 1 decimal place. Answer: 44.7 m. The distractors: 34.6 m comes from subtracting the squares, √(40² − 20²), instead of adding them; 60.0 m comes from adding the two sides of the triangle, 40 + 20, rather than using Pythagoras' theorem; 50.0 m is the height of the taller mast, copied from the question in place of the distance asked for.
8. (c) y = 2x + 1 — Method: the gradient of the line through two points is the change in y divided by the change in x, and the constant is then found by substituting one of the points into y = mx + c. Working: m = (5 − 3) ÷ (2 − 1) = 2 ÷ 1 = 2, so the line is y = 2x + c; substituting x = 1 and y = 3 gives 3 = 2 × 1 + c, so c = 3 − 2 = 1. Answer: y = 2x + 1. The distractors: y = x + 2 comes from taking the gradient as the change in x, 2 − 1 = 1, and then substituting (1, 3) to reach a constant of 2; y = 2x − 1 comes from working out the constant as mx − y, 2 × 1 − 3 = −1, instead of y − mx; y = 2x + 3 comes from using the y-coordinate of (1, 3) as the constant without substituting at all.
9. (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.
10. (c) 1.2 — Add the parts of the ratio: 6 + 1 = 7. Divide the total amount by the number of parts: 8.4 ÷ 7 = 1.2 litres, which is the value of one part and also the amount of syrup, since syrup is 1 part. (7.2 litres is the amount of water, using 6 parts instead of 1. 1.4 comes from dividing 8.4 by 6 — the water's part of the ratio — instead of dividing by the total number of parts, 7. 0.84 comes from dividing 8.4 by 10 instead of by 7.)
11. (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.
12. (a) 225 — Method: find the interior angle of the regular octagon, then subtract it from 360° to find the reflex angle at the same vertex, since the interior angle and the reflex angle together make a full turn. Working: there are 8 − 2 = 6 triangles' worth of angle in the octagon, so the interior angle = 6 × 180 ÷ 8 = 135; reflex angle = 360 − 135 = 225. Answer: 225°. A candidate who stops after finding the interior angle gives 135. A candidate who works out the exterior angle instead, 360 ÷ 8 = 45, gives 45. A candidate who subtracts the exterior angle from 360° instead of the interior angle, working out 360 − 45, gets 315.
13. (a) £1,000 — Method: find the increase, then add it to the original price; the multiplier 1.25 does both steps at once. Working: 25% is one quarter, so 25% of £800 = £800 ÷ 4 = £200, and £800 + £200 = £1,000. Answer: £1,000. The distractors: £200 is the increase on its own, not the new price; £825 comes from adding £25 to £800, treating the 25% as £25; £600 comes from taking the 25% off the price instead of adding it on.
14. (b) x ≤ 4 — Method: undo the number subtracted from 3x first, then divide by 3; the direction of the sign changes only for division by a negative number. Working: adding 1 to both sides of 3x − 1 ≤ 11 gives 3x ≤ 12; dividing both sides by 3, which is positive, gives x ≤ 4. Answer: x ≤ 4. The distractors: x ≥ 4 comes from turning the sign round while dividing by 3; x ≤ 10/3 comes from subtracting 1 from both sides instead of adding it, giving 3x ≤ 10; x ≤ 12 comes from stopping at 3x ≤ 12 and reading off the 12 without dividing by 3.
15. (a) 1.5 — Method: with an even number of values the median is the mean of the two middle values, which for 12 values are the 6th and the 7th once the data are in order. Working: the results are already in order, and 12 ÷ 2 = 6, so the middle pair are the 6th value, 1, and the 7th value, 2; the median is (1 + 2) ÷ 2 = 1.5. Answer: 1.5 brothers and sisters. The distractors: 1 comes from reading the 6th value and stopping there instead of averaging the middle pair; 2 comes from working out the mean, 24 ÷ 12, instead of the median; 6 comes from working out the range, 6 − 0, which measures spread rather than centre.
16. (b) 47 — Method: angles on a straight line add up to 180°. Working: 180 − 38 − 95 = 47. Answer: x = 47. A candidate who uses 360° instead of 180° gets 227. A candidate who subtracts only one of the two given angles from 180° gets 142. A candidate who adds the two given angles instead of subtracting them from 180° gets 133.
17. (d) 60° — Method: the six angles at the centre together make one complete turn of 360°, and because the hexagon is regular they are all equal, so divide 360° by 6. Working: 360 ÷ 6 = 60. Answer: 60°. The distractors: 120° is the interior angle of a regular hexagon, 720 ÷ 6, which is the angle at a vertex and not the angle at the centre; 45° comes from dividing 360 by 8, treating the hexagon as though it had eight sides; 30° comes from halving the angle at the centre, as though each of the six triangles were split again by a line of symmetry.
18. (b) An angle between 90° and 180° — Method: angle names are fixed by two markers, a right angle at 90° and a straight line at 180°. An obtuse angle is one that has opened past the right angle but has not reached the straight line. Working: an obtuse angle is greater than 90° and smaller than 180°. Answer: an angle between 90° and 180°. The distractors: an angle between 0° and 90° is the definition of an acute angle, chosen by a candidate who remembers the wrong side of the right-angle marker; an angle between 180° and 360° is the definition of a reflex angle, chosen by a candidate who knows the angle is past 90° and then takes the largest category; an angle of exactly 90° is the right angle itself, which is the marker an obtuse angle has passed rather than the obtuse angle.
19. (c) 67 — Method: the mean is the total of the values divided by how many values there are, so add first and divide second. Working: the total is 50 + 83 + 68 = 201 points and three matches were played, so the mean is 201 ÷ 3 = 67 points. Answer: 67. The distractors: 68 comes from writing down the median, the middle value of 50, 68, 83, instead of the mean; 33 comes from working out the range, 83 − 50, which measures spread and not centre; 100.5 comes from dividing the total by 2 instead of by the 3 matches played.
20. (b) 5 litres — Method: adding water changes the total volume but adds no salt, so work out the volume of salt, then the total volume that makes that salt 20% of the mixture, then the extra water. Working: 30% of 10 litres is 0.3 × 10 = 3 litres of salt. For the same 3 litres to be 20% of the new mixture, the new total volume is 3 ÷ 0.2 = 15 litres. The water added is the extra volume, 15 − 10 = 5 litres. Answer: 5 litres. The distractors: 3 litres is the volume of salt in the solution, which is the first step and not what the question asks for; 15 litres is the total volume of the new mixture, which counts the 10 litres already in the container as water that was poured in; 2 litres comes from taking 20% of the original 10 litres, applying the new percentage to the old volume instead of to the new one.