🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (a) y = −2x + 3 — Method: find the step in the outputs for each step of 1 in the input — falling outputs mean a negative multiplier — then read off the output when the input is 0, because that is the number added on. Working: the outputs 5, 3, 1, −1 fall by 2 each time x rises by 1, so x is multiplied by −2; the output at x = 0 is 3, so 3 is added. Answer: y = −2x + 3, checked at x = 2 by −2 × 2 + 3 = −1. The distractors: y = 2x + 3 comes from taking the size of the step, 2, as the multiplier and ignoring the fact that the outputs are falling; y = −2x − 3 comes from using the correct multiplier but writing the number added on as −3 instead of the output 3 listed at x = 0; y = −x + 4 comes from taking the multiplier as −1, its size read from the step of 1 in the inputs instead of the step of 2 in the outputs and its sign from the fact that the outputs fall, and then fitting the number added on to the pair x = −1, y = 5.
2. (d) P lies outside the circle — The squared distance from the origin to P is 8² + 9² = 64 + 81 = 145, which is greater than 100, so P lies outside the circle. Choosing 'P lies inside the circle' comes from adding the coordinates directly (8 + 9 = 17) instead of squaring and adding them, and comparing 17 with 100. Choosing 'P lies on the circle' comes from squaring 9 incorrectly as 36 (as if it were 6²) instead of 81, giving 64 + 36 = 100, equal to the right-hand side. Choosing 'Cannot be determined' wrongly assumes the equation of the line itself is needed, when the radius and the distance from the origin to P are already enough.
3. (c) 14.1 — Method: in x² + y² = r² the right-hand side is the square of the radius, so take its square root to find the radius, then double the radius because the diameter is twice the radius. Working: r² = 50, so r = √50 = 7.07106…, and the diameter is 2 × 7.07106… = 14.14213…, which is 14.1 correct to 1 decimal place. Answer: 14.1. The distractors: 7.1 is the radius, worked out correctly but never doubled, so it answers a question about the radius rather than the diameter; 100.0 comes from doubling the 50 on the right-hand side, treating r² as though it were already the radius; 25.0 comes from halving the 50, treating r² as though it were already the diameter.
4. (c) Week 16, £37,000 — y = f(x − 6) − 8000 combines a horizontal translation of 6 units RIGHT (subtracting 6 inside the brackets) with a vertical translation of £8000 DOWN (subtracting 8000 outside). Applying both to the maximum (10, 45000): 10 + 6 = 16, so the new maximum is in week 16. And 45000 − 8000 = 37000, so the maximum weekly profit is £37,000.
5. (d) 25 — Method: the median is estimated at position n ÷ 2 in the cumulative frequency table, then interpolated across the class it falls in: lower boundary, plus the fraction of the way through the class, times the class width. Working: there are 80 sacks, so the median sits at position 80 ÷ 2 = 40. Before the class 20 ≤ m < 30 the cumulative frequency is 22, and by the end of it, it is 58, so this class holds the 40th sack; its frequency is 58 − 22 = 36 and its width is 30 − 20 = 10. The extra distance needed into the class is 40 − 22 = 18, and 18 ÷ 36 × 10 = 5, so the median is 20 + 5 = 25. Answer: the estimated median mass is 25 kg. Watch which numbers the interpolation uses: reading off just the lower boundary of the median class, 20, ignores how far into that class the 40th sack actually falls; treating n ÷ 2 = 40 itself as the median mass mistakes a position in the list for a mass in kilograms; and using the target position, 40, as the extra distance into the class instead of subtracting the sacks already counted changes the calculation to 20 + 40 ÷ 36 × 10. That comes to 20 + 11.1 = 31.1, overshooting the class because it never subtracts the 22 sacks already counted before it.
6. (b) 60 — Method: independence means the proportion of part timers among the cyclists is the same as the proportion among all the employees, so find that proportion and apply it to the cyclists. Working: 150 of the 400 employees work part time, which is a proportion of 0.375. Applying it to the 160 cyclists gives 0.375 × 160 = 60 employees. Answer: you would expect 60 employees. The distractors: 310 adds 150 and 160, treating the group who do both as everyone who does one thing or the other; 10 subtracts 150 from 160, reading 'both' as the difference between the two counts; 75 halves the 150 part timers, assuming that independence means they split evenly between cyclists and non cyclists, which would need exactly half the workforce to cycle.
7. (b) 4.39 — x₁ = ∛(150 − 3 × 4²) = ∛(150 − 48) = ∛102 = 4.672 (unrounded). x₂ = ∛(150 − 3 × 4.672²) = ∛(150 − 65.49) = ∛84.51 = 4.39 (2 d.p.). Choosing 4.67 stops after only one iteration, giving x₁ instead of x₂. Choosing 84.51 finds the value inside the cube root for x₂ but never takes the cube root. Choosing 6.32 comes from adding 3xₙ² instead of subtracting it inside the root, which does not match the given formula.
8. (b) (3, −3) — Subtracting column vectors means subtracting the top numbers and subtracting the bottom numbers, in the order given: top = 5 − 2 = 3, bottom = 1 − 4 = −3, giving (3, −3). A candidate who works out v − u instead of u − v, reversing the order, gets (−3, 3). A candidate who makes a sign error on the bottom number, treating 1 − 4 as 3 instead of −3, gets (3, 3). A candidate who adds the vectors instead of subtracting gets (7, 5). Because the question asks for u − v, not v − u, the correct answer is (3, −3).
9. (a) 7 — Multiply the input by 2: 6 × 2 = 12. Then subtract 5: 12 − 5 = 7. A candidate who does the operations in the wrong order, subtracting 5 first and then multiplying by 2, gets (6 − 5) × 2 = 2. A candidate who only carries out the multiplication and forgets to subtract gets 12. A candidate who adds 5 instead of subtracting gets 6 × 2 + 5 = 17.
10. (c) −1.5 — Since n = k × m, dividing a number in n by the matching number in m gives k: k = −6 ÷ 4 = −1.5 (check with the bottom numbers: −9 ÷ 6 = −1.5, the same value, confirming n is a scalar multiple of m). 1.5 has the correct size but is missing the negative sign. −10 comes from subtracting the top numbers, −6 − 4, instead of dividing them. −24 comes from multiplying the top numbers, −6 × 4, instead of dividing them.
11. (a) Positive x-direction, 90°; image is y = sin x. — Writing cos(x − 90°) as cos(x − a) with a = 90 shows this is a horizontal translation, y = f(x − a), which moves the graph 90° in the positive x-direction; the identity cos(x − 90°) = sin x confirms the image is y = sin x. Choosing the negative x-direction reverses the sign inside the bracket — subtracting inside the bracket always translates in the positive x-direction, not the negative one, so that statement is wrong on direction. Getting the direction right but conflating the subtraction inside the bracket with an extra reflection of the output flips the sign of the resulting graph, wrongly giving y = −sin x. Treating the subtraction as if it changed the output directly, rather than the input, wrongly calls this a vertical translation even while still correctly recalling that the image simplifies to y = sin x.
12. (d) a = 4 — Expand the left-hand side: (2x + 3)(x + a) = 2x² + 2ax + 3x + 3a = 2x² + (2a + 3)x + 3a. For this to match 2x² + 11x + 12 for every value of x, the x-coefficients must be equal and the constants must be equal: 2a + 3 = 11 and 3a = 12. Both give a = 4, so a = 4. Writing a = 12 comes from the constant-term equation 3a = 12: reading it as saying a itself is 12, rather than dividing both sides by 3. Writing a = 8 comes from the x-coefficient equation 2a + 3 = 11: working out 11 − 3 = 8 correctly but then stopping, without dividing by the 2 in front of a. Writing a = −4 comes from rearranging 2a + 3 = 11 the wrong way round, as 2a = 3 − 11 = −8, which gives a = −4 instead of a = 4.
13. (d) (2/5)a + (3/5)b — Method: OP = OA + AP, and since AP : PB = 3 : 2 splits AB into 5 equal parts, AP is 3/5 of the whole of AB, with AB = b − a. Working: OP = a + 3/5(b − a) = a − (3/5)a + (3/5)b = (2/5)a + (3/5)b. Answer: OP = (2/5)a + (3/5)b. Using the ratio the wrong way round, as though it read AP : PB = 2 : 3, gives (3/5)a + (2/5)b; adding (3/5)b onto the whole of a without subtracting a inside the bracket first gives a + (3/5)b; and treating the ratio as 1 : 1 gives the midpoint, (1/2)a + (1/2)b. Convert the ratio to a fraction of AB measured from A, matching the ORDER the ratio is stated in, and subtract before you scale.
14. (a) 4/9 — Method: two steps. Find how many cars failed altogether, because the car picked is known to be one of them, then divide the diesel failures by that total. Working: 30 petrol cars and 24 diesel cars failed, so 54 cars failed. The diesel failures give 24/54, and dividing the numerator and the denominator by 6 gives 4/9. Answer: the probability is 4/9. The distractors: 3/10 is 24/80, the probability that a car fails given that it is a diesel car, which is the condition and the event swapped; 3/25 is 24/200, dividing by every car serviced that week rather than by the 54 that failed; 2/5 is 80/200, the probability that a car chosen from the whole week is a diesel car, which ignores the fact that the car picked failed.
15. (a) −8 — −f(x) means take the output value from the table and change its sign, without changing which x-value is looked up. From the table, f(1) = 8, so −f(1) = −8. Reading f(1) = 8 from the table but forgetting to apply the negative sign gives 8. Misreading the row and using f(0) = 5 instead of f(1) = 8, then negating it, gives −5. Confusing −f(x) with f(x) − 1 — taking f(1) = 8 and subtracting 1 instead of negating — gives 7.
16. (c) 76 — Method: the frequency of each class is the area of its bar, frequency density × class width, so work out all three frequencies and add them. Working: the widths are 5, 10 and 25 kg, so the frequencies are 4 × 5 = 20, 2.6 × 10 = 26 and 1.2 × 25 = 30; the total is 20 + 26 + 30 = 76. Answer: 76 dogs. The distractors: 7.8 comes from adding the three frequency densities, 4 + 2.6 + 1.2, treating each height as though it were a count; 107 comes from using each upper class boundary as the width, giving 4 × 5, 2.6 × 15 and 1.2 × 40; 39 comes from using the first class width, 5, for every bar, which ignores the unequal intervals and gives 20 + 13 + 6.
17. (c) 20 ≤ h < 40 — Method: with 80 values the lower quartile is the 80 ÷ 4 = 20th value in order, so build a running total until it first reaches 20. Working: the running totals are 14, then 14 + 22 = 36, then 52, then 80; the 20th plant is past 14 but not past 36, so it lies in the second class. Answer: the lower quartile lies in the class 20 ≤ h < 40. The distractors: 0 ≤ h < 20 comes from believing that the bottom quarter of the data must all sit in the first class, when that class holds only 14 of the 80 plants; 40 ≤ h < 50 comes from using the position 80 ÷ 2 = 40 and so locating the median rather than the lower quartile; 50 ≤ h < 80 comes from counting 20 plants down from the tallest instead of up from the shortest, which locates the upper quartile at the 60th plant.
18. (d) x = 4 and x = −4 — Substituting y = 3 gives x² + 9 = 25, which simplifies to x² = 16, so x = 4 or x = −4. Choosing 'x = 3 and x = −3' uses the given value y = 3 as if it were the x-coordinate. Choosing 'x = 4' alone finds the positive square root of 16 but forgets the negative root. Choosing 'x = 5 and x = −5' skips subtracting 3² = 9 from 25 and takes the square root of 25 directly.