⚖️ Foundation to Higher crossover check

Answer key

1. (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.
2. (d) 4 — Method: collect the x terms on one side and the number terms on the other. Working: subtract 3x from both sides: 4x − 5 = 11. Add 5 to both sides: 4x = 16. Divide by 4: x = 4. Answer: 4. 0.6 comes from adding the x terms instead of subtracting when collecting them, 7x + 3x = 10x, and also subtracting the constants the wrong way round, 11 − 5 = 6, giving 10x = 6. 1.5 comes from correctly collecting the x terms as 4x but subtracting the constants the wrong way round, 11 − 5 instead of 11 + 5. −4 comes from moving the x terms to the wrong side, giving 3x − 7x instead of 7x − 3x, along with a matching sign error on the constants.
3. (d) 23.57 °C — Method: add all seven temperatures, divide by the number of readings and round only at the end. Working: 22 + 24 + 23 + 25 + 26 + 21 + 24 = 165, and 165 ÷ 7 = 23.5714…, which rounds to 23.57 to 2 decimal places. Answer: 23.57 °C. The distractors: 24 °C comes from writing down the mode, the only temperature recorded twice, instead of the mean; 27.5 °C comes from dividing the total by 6 instead of by the 7 days recorded; 5 °C comes from working out the range, 26 − 21, which is a measure of spread and not an average.
4. (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.
5. (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.
6. (c) 5 — Method: the mode is the value that occurs most often, so count how many times each different value appears and compare the counts. Working: 3 appears twice, 5 appears three times, 7 appears once and 8 appears once, so the highest frequency is three and the value carrying it is 5. Answer: 5. The distractors: 3 comes from writing down the frequency of the most common answer instead of the answer itself; 7 comes from taking the middle number of the list as it was written, which applies the median without ordering the data and without answering the question asked; 8 comes from picking the largest value, which confuses the mode with the maximum.
7. (d) An obtuse angle — Method: compare the angle with the two markers that separate the angle names, a right angle at 90° and a straight line at 180°. Working: 108° is greater than 90° and smaller than 180°, so it lies between the right angle and the straight line. Answer: an obtuse angle. The distractors: an acute angle is one below 90°, and is chosen by a candidate who checks only that 108° is less than 180°; a reflex angle is one above 180°, and is chosen by a candidate who checks only that 108° is more than 90° and then takes the largest category; a right angle is exactly 90°, and is chosen by reading 108° as near enough to 90° instead of comparing it properly.
8. (c) 40° — Method: the three angles of a triangle always add up to 180°, so split 180° into the number of parts shown by the ratio. Working: the ratio 2:3:4 has 2 + 3 + 4 = 9 parts, so one part is 180 ÷ 9 = 20°, and the smallest angle is 2 × 20 = 40°. So the smallest angle is 40°. Distractor 60° is the middle-sized angle, 3 parts. Distractor 80° is the largest angle, 4 parts. Distractor 20° is the value of one part, found correctly but never multiplied by 2.
9. (d) (x − 8)(x + 8) — x² − 64 = x² − 8², a difference of two squares, which factorises as (x − 8)(x + 8). A candidate who treats it as a perfect square with a repeated negative factor gets (x − 8)(x − 8), which expands to x² − 16x + 64 — wrong on both the middle and constant terms. A candidate who uses a repeated positive factor gets (x + 8)(x + 8), which expands to x² + 16x + 64. A candidate who picks a different factor pair of 64, such as 4 and 16, without checking that the middle term cancels, gets (x − 4)(x + 16), which expands to x² + 12x − 64 — the wrong middle term.
10. (d) 5x ≤ 30 — Method: write the total cost as the cost of one book times the number of books, then turn the limit into an inequality sign; 'at most' allows the limit to be reached but not passed. Working: x books at £5 each cost 5x pounds; that total must not go above £30, and spending exactly £30 is allowed, so the two sides are joined by ≤ and the inequality is 5x ≤ 30. Answer: 5x ≤ 30. The distractors: 5x ≥ 30 comes from reading 'at most' as 'at least', which reverses the condition; x + 5 ≤ 30 comes from adding the price of a book to the number of books instead of multiplying; 5x < 30 comes from reading 'at most £30' as 'less than £30', which wrongly rules out spending the whole £30.
11. (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.
12. (d) x = 2 — Method: expand the bracket by multiplying both terms inside it by 2, then undo the addition and the multiplication in turn. Working: expanding gives 2x + 6 = 10; subtracting 6 from both sides gives 2x = 4; dividing both sides by 2 gives x = 2. Answer: x = 2. The distractors: x = 5 comes from dividing both sides by 2 first, reaching x + 3 = 5 and writing 5 without taking the 3 away; x = 8 comes from adding 6 to both sides instead of subtracting it, giving 2x = 16; x = 3.5 comes from expanding 2(x + 3) as 2x + 3, multiplying only the x by the 2, which leads to 2x = 7.
13. (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.
14. (a) The means are equal, and Class B's marks are the more consistent because its range is smaller. — Method: comparing two distributions needs two things — a measure of average and a measure of spread — and each must be put into the context of the question. Working: both classes have a mean mark of 70, so on average the two classes scored the same; the range measures spread, and Class A's range of 30 marks is three times Class B's range of 10 marks, so Class B's marks sit closely around the mean while Class A's are far more spread out. Answer: the means are equal, and Class B's marks are the more consistent because its range is smaller. The distractors: the reply crediting Class A with more consistency reverses the meaning of the range, treating a larger range as tighter data when a larger range means more spread; the reply that Class A's mean mark is higher compares the wrong pair of figures, reading the range of 30 as an average; the reply that Class B's mean mark is higher reads the spread correctly but its claim about the means is false, since both means are 70.
15. (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.
16. (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.
17. (c) y = 3x + 5 — Method: find the gradient from the two points, then substitute one point into y = mx + c to find c. Working: gradient = (14 − 2) ÷ (3 − (−1)) = 12 ÷ 4 = 3. Using the point (3, 14): 14 = 3(3) + c, so 14 = 9 + c, giving c = 5. Answer: y = 3x + 5. y = 6x − 4 comes from mishandling the negative x-coordinate, treating 3 − (−1) as 3 − 1 = 2, so the gradient becomes 12 ÷ 2 = 6, and then c = 14 − 18 = −4. y = 3x + 23 comes from a sign error isolating c, adding 9 to 14 instead of subtracting it: c = 14 + 9 = 23. y = x/3 + 13 comes from dividing the change in x by the change in y instead of the other way round, giving a gradient of 4 ÷ 12 = 1/3, and then c = 14 − 1 = 13.
18. (b) 55° — Method: complementary angles are a pair that add up to 90°, so subtract the known angle from 90°. Working: 90 − 35 = 55. Answer: 55°. The distractors: 145° comes from subtracting from 180°, which is the rule for supplementary angles and not for complementary ones; 325° comes from subtracting from 360°, the total of the angles at a point; 90° is the total that the pair must make, written down in place of the missing angle.
19. (d) £195 — Substituting x = 3 into y = 45x + 60 gives y = 45 × 3 + 60 = 135 + 60 = 195. A candidate who forgets to add the call-out fee would get only 45 × 3 = £135. A candidate who adds the hours to the fee and the rate instead of multiplying would get 45 + 60 + 3 = £108. A candidate who multiplies both the hourly rate and the call-out fee by the number of hours would get 45 × 3 + 60 × 3 = £315.
20. (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.