🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (d) 45 — Method: a cumulative frequency is a running total — it counts everybody in every class up to and including the one that ends at the value given. Working: the classes that lie wholly below 30 minutes are 0 ≤ t < 10, 10 ≤ t < 20 and 20 ≤ t < 30, with frequencies 6, 14 and 25, so the running total is 6 + 14 = 20 and then 20 + 25 = 45. Answer: 45 people took less than 30 minutes. The distractors: 25 comes from quoting the frequency of the class 20 ≤ t < 30 on its own instead of the running total; 65 comes from accumulating one class too many and including 30 ≤ t < 40, which is 45 + 20; 35 comes from accumulating from the top downwards, 15 + 20, which counts the people who took 30 minutes or more rather than fewer.
2. (c) 42 — Method: find the target cumulative frequency, 90% of the total, locate the class it falls in from the plotted points, then interpolate: lower boundary, plus the extra distance needed into the class divided by the class's frequency, times its width. Working: 90% of 320 is 0.9 × 320 = 288. The plotted points show a cumulative frequency of 280 at d = 40 and 320 at d = 50, so the class 40 ≤ d < 50 has frequency 320 − 280 = 40 and width 50 − 40 = 10, and 288 falls inside it. The extra distance needed into the class is 288 − 280 = 8, and 8 ÷ 40 × 10 = 2, so the diameter is 40 + 2 = 42. Answer: the estimated diameter is 42 mm. Watch which point and which class the interpolation actually uses: reading off d = 40, the plotted point just below the target, instead of interpolating the extra 8 ball bearings into the next 10 mm, stops one step short of the true answer; finding the diameter below which only 10% lie instead of 90% gives a target of 0.1 × 320 = 32, which falls in the class 10 ≤ d < 20 — the extra distance into that class is 32 − 30 = 2, and 2 ÷ 60 × 10 = 0.3, so this route gives 10 + 0.3 = 10.3, the bottom decile rather than the top 90%; and interpolating within the class 30 ≤ d < 40 instead of 40 ≤ d < 50, as though 288 had not yet reached a cumulative frequency of 280, treats the extra distance as 288 − 190 = 98, and 98 ÷ 90 × 10 = 10.9, giving 30 + 10.9 = 40.9, one class too early.
3. (a) 1/3 — Method: since OABC is a parallelogram, B = OA + OC = a + c. M = (1/2)c, since M is the midpoint of OC. Since AN is twice NC, N is 2/3 of the way along AC from A, so N = a + 2/3(c − a) = (1/3)a + (2/3)c. Working: MN = N − M = (1/3)a + (1/6)c, and MB = B − M = a + (1/2)c. Comparing term by term, 1/3 × (a + (1/2)c) = (1/3)a + (1/6)c, which matches MN exactly. Answer: k = 1/3, so M, N and B lie on a straight line. Giving 2/3 instead is the scalar linking N to B (NB = (2/3)MB), not M to N; giving 1/6 is just MN's c-coefficient read off on its own, without comparing it to MB's c-coefficient at all; and giving 3 is the scalar the wrong way up — it is MB that equals 3 × MN, not the other way round, since MN = k × MB was what was asked for. Always match the direction of the scalar to the vectors exactly as the question states them.
4. (c) n = 4 — Testing n = 4: 4² = 16, and 16 + 4 + 1 = 21, and 21 = 3 × 7 is not a prime number, so this value disproves the claim. Testing n = 1: 1² = 1, and 1 + 1 + 1 = 3, which is prime, so it does not disprove the claim. Testing n = 2: 2² = 4, and 4 + 2 + 1 = 7, which is prime. Testing n = 3: 3² = 9, and 9 + 3 + 1 = 13, which is also prime — a counterexample has to give a result that isn't prime, and only n = 4 does that.
5. (b) x² + y² = 81 — The equation of a circle with centre the origin and radius r is x² + y² = r². Here r = 9, so r² = 9 × 9 = 81, giving x² + y² = 81. Choosing x² + y² = 9 uses the radius itself instead of squaring it. Choosing x² + y² = 18 doubles the radius (9 × 2 = 18) instead of squaring it. Choosing x² − y² = 81 keeps the correct 81 but writes a minus instead of a plus, which is not the equation of a circle.
6. (b) 8 — Method: a circle centred on the origin has equation x² + y² = r², where r is the radius, so the number on the right-hand side is the square of the radius and not the radius itself. Working: comparing x² + y² = 64 with x² + y² = r² gives r² = 64, so r = √64 = 8. Answer: the radius is 8. The distractors: 64 is r² read straight off the equation as though the right-hand side were the radius, which is the commonest error on this form; 32 comes from halving 64, treating the right-hand side as a diameter that has to be halved; 16 is the diameter, 2 × 8, quoted in place of the radius.
7. (c) (1/3)b − (1/3)a — Method: PQ runs from P to Q, so PQ = OQ − OP. Working: PQ = (1/3)b − (1/3)a. Answer: PQ = (1/3)b − (1/3)a. Subtracting the other way round gives (1/3)a − (1/3)b, the same vector pointing back from Q to P instead of P to Q; using 2/3 instead of the 1/3 that OP and OQ were actually given as gives (2/3)b − (2/3)a; and using the full vectors a and b with no scaling at all gives b − a, which is AB, not PQ. Always subtract START from END, OQ − OP, and carry the fraction given in the question through to your final vector.
8. (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.
9. (c) 27 — Method: a frequency is the area of the part of the bar being asked about, so frequency = frequency density × the width of that part. Working: the part asked about runs from 20 to 35, so its width is 35 − 20 = 15 metres; the frequency density there is 1.8 per metre, so the estimate is 1.8 × 15 = 27. Answer: about 27 throws. The distractors: 72 comes from taking the whole bar, 1.8 × 40, and so counting every throw from 20 up to 60; 1.8 comes from reading the height of the bar as a frequency, when a height is a density and only an area is a count; 63 comes from using the upper value 35 as the width, 1.8 × 35, instead of the width 35 − 20.
10. (c) 2.61 — The limit L satisfies L = 12 ÷ (L + 2), so L(L + 2) = 12, giving L² + 2L − 12 = 0 and L = (−2 + √52) ÷ 2 = 2.6056, which is 2.61 to 2 decimal places (the early iterates 4, 2, 3, 2.4, 2.73, ... oscillate around this value before settling). Choosing 3.00 reads off x₃, one of the early oscillating values, before the sequence has settled close to the limit. Choosing 4.00 reads off x₁, the very first iterate, not the value the sequence approaches. Choosing 6.00 comes from writing the limit equation as L = 12 ÷ 2, leaving L itself out of the denominator.
11. (a) (8, 2) — Method: translating a point by a column vector means adding the vector's top number to the x-coordinate and its bottom number to the y-coordinate. Working: x-coordinate 2 + 6 = 8; y-coordinate 5 + (−3) = 2. Answer: B = (8, 2). A candidate who ignores the negative sign and adds 3 instead of −3 gets (8, 8). A candidate who translates in the reverse direction, subtracting the vector from A instead of adding it, gets (−4, 8). A candidate who swaps the vector's top and bottom numbers before adding gets (−1, 11).
12. (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.
13. (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.
14. (a) 2(3x + 1) = 6x + 2 — Expanding 2(3x + 1) = 6x + 2 gives an expression that matches the right-hand side exactly for every value of x — it is an identity. 4x − 3 = 3x + 5 is an ordinary equation with one solution, x = 8. 7 − x = x − 7 is also an ordinary equation with one solution, x = 7. 5x + 1 = 5(x + 1) never holds for any value of x at all, since expanding the right-hand side gives 5x + 5, and 5x + 1 = 5x + 5 would require 1 = 5, which is impossible.
15. (c) Only x = 5 satisfies 3x + 5 = 20, not every value of x. — 3x + 5 = 20 is only true when x = 5, since 3 × 5 + 5 = 20; for any other value of x the two sides are not equal, so it is an equation, not an identity. Saying it cannot be simplified confuses simplifying with the equation/identity distinction, which is about how many values of x make it true. Saying it has an = sign is not a valid test, since identities are also written with an = or ≡ sign. A number on the right-hand side does not decide it either — what matters is whether both sides match for every value of x, not the form of the right-hand side.
16. (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.
17. (d) 4x + 5 — A minus sign directly before a bracket changes the sign of both terms inside it: 6x − (2x − 5) = 6x − 2x + 5 = 4x + 5. The option 4x − 5 comes from only changing the sign of the 2x term and not the −5, giving 6x − 2x − 5. The option 8x − 5 comes from adding 2x instead of subtracting it, as if the minus sign did not apply to the bracket, giving 6x + 2x − 5. The option 8x + 5 repeats that same addition mistake and also changes the sign of the −5 term.
18. (a) (6, 8) — Method: a point lies on the circle x² + y² = 100 exactly when the squares of its two coordinates add to 100, so square both coordinates of each point and add them. Working: for (6, 8), 6² + 8² = 36 + 64 = 100, which matches the right-hand side of the equation. Answer: (6, 8) lies on the circle. The distractors: (3, 4) is the 3, 4, 5 right-angled triangle recalled but never scaled up to a radius of 10, and 3² + 4² = 25, so it lies on the far smaller circle x² + y² = 25; (5, 5) has coordinates adding to 10, which compares the sum of the coordinates with the radius instead of the sum of their squares with r², and 5² + 5² = 50; (10, 10) takes each coordinate separately to equal the radius, and 10² + 10² = 200, which is twice too big.