🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (d) −2 — y = f(x) − 6 is f(x) shifted down by 6, so every y-value on the graph decreases by 6. At x = 0, f(0) = 4, so the new y-value is 4 − 6 = −2. Adding 6 instead of subtracting gives 10; writing down the shift itself, −6, or leaving the original value 4 unchanged both skip the translation altogether.
2. (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.
3. (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.
4. (a) n² + n — n² + n = n(n + 1), the product of two consecutive integers. One of any two consecutive integers is always even, so their product is always even, whatever whole number n is. n² − n + 1 = n(n − 1) + 1 is always ODD, since n(n − 1) is even and adding 1 makes it odd — the opposite of what's asked. 2n + 1 is always odd by definition, not even. n² + 1 is not always even at all: it depends on whether n is odd or even, and testing n = 2 gives 5, which is odd.
5. (c) (2n + 1)² + (2n + 3)² = (4n² + 4n + 1) + (4n² + 12n + 9) = 8n² + 16n + 10 = 8(n² + 2n + 1) + 2, and n² + 2n + 1 is an integer, so the sum is always 2 more than a multiple of 8. — Expand each square carefully: (2n + 1)² = 4n² + 4n + 1 and (2n + 3)² = 4n² + 12n + 9, since the cross term is 2 × 2n × 3 = 12n. Adding gives 8n² + 16n + 10, and factorising out 8 from every term that can hold one gives 8(n² + 2n + 1) + 2; since n² + 2n + 1 is always an integer, the sum is always 2 more than a multiple of 8. The attempt reaching 8(n² + 2n) + 10 has the correct expansion but stops the factorisation one step early — it never pulls a further 8 out of the 10 (10 = 8 + 2), so 'always 10 more than a multiple of 8' should be reduced to 'always 2 more than a multiple of 8'. The attempt reaching 2(4n² + 8n + 5) also has the correct expansion, and the factorisation is true, but 'always even' only shows the sum is a multiple of 2 — being even is necessary but nowhere near sufficient to be a multiple of 8, and the argument never finds the extra factor of 4. The fourth attempt makes an expansion slip, using (2n + 3)² = 4n² + 9 instead of 4n² + 12n + 9 — dropping the 12n cross term entirely — so it works from the wrong expression 8n² + 4n + 10 throughout, and no amount of correct working afterwards can recover the right conclusion.
6. (c) 72 — Method: on a histogram the frequency of a class is the area of its bar, so frequency = frequency density × class width. Working: the class 50 ≤ m < 80 has width 80 − 50 = 30 grams and a frequency density of 2.4 per gram, so the frequency is 2.4 × 30 = 72. Answer: 72 pebbles. The distractors: 192 comes from using the upper class boundary, 80, as the width, giving 2.4 × 80; 12.5 comes from dividing the width by the density, 30 ÷ 2.4, which reverses the area rule; 2.4 comes from reading the height of the bar as the frequency itself, the commonest mistake on histograms, where a height is a density and only an area is a count.
7. (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.
8. (a) 49 — Method: add the frequencies of the classes that lie wholly below 35 seconds, then use linear interpolation for the class that 35 cuts through, assuming the calls in that class are spread evenly across it. Working: below 30 seconds there are 15 + 24 = 39 calls; the value 35 lies in the class 30 ≤ t < 50, which is 20 seconds wide and holds 40 calls, and 35 is 35 − 30 = 5 seconds into it, so the estimated share is (5 ÷ 20) × 40 = 10 calls; the estimate is 39 + 10 = 49. Answer: about 49 calls met the target. The distractors: 79 comes from adding the whole of the class 30 ≤ t < 50, 39 + 40, and so counting calls of up to 50 seconds as being under 35; 39 comes from stopping at the class boundary 30 and ignoring the part class altogether; 69 comes from measuring the part of the class from 35 up to 50 instead of from 30 up to 35, giving (15 ÷ 20) × 40 = 30 and then 39 + 30.
9. (a) (4, −8), a minimum point — The transformation x → x + 1 inside f translates the graph 1 unit to the LEFT, so the x-coordinate becomes 5 − 1 = 4. The minus sign in front of f reflects the graph in the x-axis, so the y-coordinate becomes −8, and a reflection in the x-axis turns every maximum into a minimum, so (4, −8) is a minimum point. Writing '(4, −8), a maximum point' gets the coordinates right but forgets that a reflection in the x-axis swaps maximum and minimum points. Writing '(6, −8), a minimum point' comes from translating 1 unit to the RIGHT instead of the left — f(x + 1) always moves the graph in the negative x-direction. Writing '(4, 8), a minimum point' gets the x-coordinate and the type right, but forgets to actually negate the y-coordinate, even though it does correctly reclassify the point as a minimum.
10. (b) (3, 10) — Method: multiply every part of p by 2, then add the matching parts of q. Working: 2p = (4, 6); adding q gives top 4 + (−1) = 3 and bottom 6 + 4 = 10. Answer: 2p + q = (3, 10). A candidate who forgets to double p first, working out p + q instead, gets (1, 7). A candidate who doubles q instead of p, working out p + 2q, gets (0, 11). A candidate who subtracts q instead of adding it, working out 2p − q, gets (5, 2).
11. (c) The class with times from 10 up to 20 — Method: to find the median class from a histogram, first turn each bar's frequency density into a frequency using density × class width, build up the cumulative frequency, and find the first class whose cumulative frequency reaches or passes n ÷ 2. Working: the four classes have widths 10, 10, 20 and 20, so their frequencies are 5 × 10 = 50, 2 × 10 = 20, 1.5 × 20 = 30 and 1 × 20 = 20, which add to the 120 visitors stated. The median sits at position 120 ÷ 2 = 60. The cumulative frequency is 50 after the first class and 50 + 20 = 70 after the second, so the 60th visitor is reached during the second class. Answer: the median lies in the class 10 ≤ t < 20. Watch which class each shortcut lands on: the tallest bar belongs to the first class, with the highest frequency density, 5 — but the tallest bar shows where visitors are packed most densely, not where the middle visitor falls, and picking it lands one class too early, at 0 ≤ t < 10; taking half of the total TIME span instead of half of the total NUMBER of visitors, 60 minutes ÷ 2 = 30 minutes, lands in the class 20 ≤ t < 40, confusing a value on the horizontal axis with a position in the data; and using the full 120 visitors as the target position, rather than 120 ÷ 2 = 60, reaches all the way to the last class, 40 ≤ t < 60, treating the whole data set's size as though it were the position of a single middle value.
12. (c) a + (1/2)c — Method: OABC is a parallelogram, so OB = OA + AB, and since AB is equal and parallel to OC, AB = c; this gives OB = a + c. M is the midpoint of AB, so AM = (1/2)AB = (1/2)c. Working: OM = OA + AM = a + (1/2)c. Answer: OM = a + (1/2)c. Adding the whole of AB instead of half of it gives a + c, which is OB, not OM; halving the whole diagonal OB instead of just AB gives (1/2)a + (1/2)c, the midpoint of OB rather than of AB; and flipping the sign on the c-term gives a − (1/2)c, which points back the wrong way along AB. Halve only the side you are told to halve, and check the sign before you commit to an answer.
13. (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).
14. (c) y = x² — Method: test a candidate rule against every pair given, not just one — a rule that fits one pair and fails another is not the rule. Working: the outputs 1, 4, 9 rise by 3 and then by 5, so they are not going up in equal steps and the input is not simply multiplied by a fixed number; comparing each output with its own input gives 1 × 1 = 1, 2 × 2 = 4 and 3 × 3 = 9, and all three pairs fit. Answer: y = x². The distractors: y = 3x comes from fitting only the last pair, where 3 × 3 = 9, and reading that 3 as a multiplier; y = 3x − 2 comes from assuming a multiply-then-add rule and using the first step in the outputs, 4 − 1 = 3, as the multiplier — it fits the first two pairs and fails the third; y = 2x comes from fitting only the pair 2 and 4 and reading every output as double its input.
15. (d) 25 ≤ t < 45 — Method: the height of a bar on a histogram is the frequency density, so work out frequency ÷ class width for every class and compare the four heights. Working: 20 ÷ 10 = 2 for the first class; 12 ÷ 15 = 0.8 for the second; 50 ÷ 20 = 2.5 for the third; 60 ÷ 30 = 2 for the fourth. Answer: the largest of 2, 0.8, 2.5 and 2 is 2.5, so the tallest bar is the one for 25 ≤ t < 45. The distractors: 45 ≤ t < 75 comes from picking the class with the greatest frequency, 60, and treating a frequency as a height — but that class is three times as wide, so its 60 competitors are spread thinly; 10 ≤ t < 25 comes from dividing the class width by the frequency, 15 ÷ 12, and picking the largest of those reversed values; 0 ≤ t < 10 comes from assuming the narrowest class must always give the tallest bar, which is only true when the frequencies are equal.
16. (c) 4x − 3 — ff(x) means f(f(x)): substitute f(x) into f in place of x. f(f(x)) = 2 × f(x) − 1 = 2 × (2x − 1) − 1. Expanding the bracket: 2 × (2x − 1) = 4x − 2. Combining the constant terms: −2 − 1 = −3, so f(f(x)) = 4x − 3. Writing 4x − 2 comes from expanding 2(2x − 1) correctly to get 4x − 2, then forgetting to subtract the outer 1 at all. Writing 4x² − 4x + 1 comes from reading ff(x) as f(x) multiplied by itself, (2x − 1)(2x − 1) = 4x² − 4x + 1, instead of substituting f(x) into f. Writing 4x − 1 comes from doubling the coefficient of x in the original rule directly, without actually substituting f(x) into f at all.
17. (d) 4.521 — x₁ = √(3 × 4 + 7) = √19 = 4.358898944. x₂ = √(3 × 4.358898944 + 7) = √20.076696833 = 4.480702716. x₃ = √(3 × 4.480702716 + 7) = √20.442108148 = 4.521294964, which rounds to 4.521. Mislabelling the starting value x₀ as x₁, so that the working stops one iteration too early, reports the true x₂ = 4.480702716, which rounds to 4.481. Working out one iteration too many reports the true x₄ = √(3 × 4.521294964 + 7) = 4.534741987, which rounds to 4.535. Applying the formula in the wrong order, calculating √(3xₙ) + 7 at every step instead of √(3xₙ + 7), gives, from x₀ = 4: √12 + 7 = 10.464101615, then √(3 × 10.464101615) + 7 = 12.602883619, then √(3 × 12.602883619) + 7 = 13.148873950, which rounds to 13.149.
18. (b) 55 — Method: count the classes that lie wholly above 8 minutes, then use linear interpolation for the class that 8 cuts through, assuming the delays in that class are spread evenly. Working: the class 10 ≤ d < 20 lies wholly above 8 and holds 25 buses; the value 8 lies in the class 5 ≤ d < 10, which is 5 minutes wide and holds 75 buses, and the part above 8 runs from 8 to 10, a width of 2, so the estimated share is (2 ÷ 5) × 75 = 30 buses; the estimate is 30 + 25 = 55. Answer: about 55 refunds. The distractors: 100 comes from adding the whole of the class 5 ≤ d < 10, 75 + 25, and so refunding buses only 5 minutes late; 25 comes from using only the class 10 ≤ d < 20 and ignoring the part class that 8 minutes cuts through; 70 comes from taking the part of the class from 5 up to 8 instead of from 8 up to 10, giving (3 ÷ 5) × 75 = 45 and then 45 + 25.