🔺 Higher stretch — grade 8 and 9 topics

Answer key

1. (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.
2. (c) 17.7 cm — Method: the iteration converges on x, the depth of the box, which is also the side of each square cut away; a square is removed from both ends of the 20 cm side, so the longer side of the base is 20 − 2x. Run the formula three times, then carry out that subtraction. Working: x₁ = (16 × 1² − 1³ + 50) ÷ 60 = 65 ÷ 60 = 1.08333…; x₂ = 67.50636… ÷ 60 = 1.12510…; x₃ = 68.82959… ÷ 60 = 1.14715…. The longer side of the base is 20 − 2 × 1.14715… = 17.70568…, which is 17.7 cm correct to 1 decimal place. Answer: 17.7 cm. The distractors: 1.1 cm is x₃ itself rounded, the depth of the box, given by a candidate who runs the iteration correctly and then stops before the step the question actually asks for; 18.9 cm is 20 − 1.14715…, taking away one square instead of two and forgetting that a corner is cut from each end of that side; 9.7 cm is 12 − 2 × 1.14715…, the shorter side of the base, which measures the wrong edge of the sheet.
3. (b) 0.0309 — Method: P(both defective | at least one defective) = P(both defective) ÷ P(at least one defective). Find each using independence: P(both) = 0.06², P(at least one) = 1 − P(neither) = 1 − 0.94². Working: P(both) = 0.06² = 0.0036. P(neither) = 0.94² = 0.8836, so P(at least one) = 1 − 0.8836 = 0.1164. P(both | at least one) = 0.0036 ÷ 0.1164 = 0.0309 (3 s.f.). Answer: 0.0309. Watch out: leaving the answer as 0.0036 gives P(both defective) itself, not the probability once you already know at least one is defective — you still need to divide by P(at least one defective). Giving 0.0600 answers with the single-component defect rate, ignoring the condition altogether. And 0.5000 assumes that 'at least one' makes the outcomes 'exactly one defective' and 'both defective' equally likely, which is not how these probabilities combine.
4. (b) (12, −9) — Two enlargements one after the other combine into a single scale factor: 1.5 × 2 = 3. Multiplying a vector by a scalar means multiplying both the top number and the bottom number by it: top = 4 × 3 = 12, bottom = −3 × 3 = −9, giving (12, −9). A candidate who adds the scale factor to each number instead of multiplying gets (4 + 3, −3 + 3) = (7, 0). A candidate who multiplies the top number but leaves the bottom number unchanged gets (12, −3). A candidate who multiplies the bottom number but leaves the top number unchanged gets (4, −9). The correct column vector for the poster is (12, −9).
5. (a) (3, 7) — y = f(x) + 5 is a vertical translation of y = f(x) by 5 units up — the translation vector is (0, 5) — so only the y-coordinate of any point changes. Turning point (3, 2) → (3, 2 + 5) = (3, 7). Adding the 5 to the x-coordinate, or treating it as a horizontal shift like y = f(x + 5), moves the wrong coordinate — check first whether the number sits inside or outside the brackets.
6. (a) 19 — Each time x increases by 1, y increases by 3 (4, 7, 10, 13 — a constant difference of 3). So at x = 4, y = 13 + 3 = 16, and at x = 5, y = 16 + 3 = 19. A candidate who stops one step early, giving the value for x = 4 instead of x = 5, answers 16. A candidate who overcounts and adds three steps of 3 instead of two from x = 3 gets 13 + 9 = 22. A candidate who mistakes the y-intercept (4) for the common difference and adds 4 twice from x = 3 gets 13 + 8 = 21.
7. (b) y = −3x/4 + 25 — Method: a straight line that touches a circle at one point is a tangent there, so it is perpendicular to the radius drawn to that point; find the gradient of the radius, take its negative reciprocal, then substitute the point of contact into y − y₁ = m(x − x₁). Working: the radius from (0, 0) to (12, 16) has gradient 16 ÷ 12, which cancels to 4/3, so the tangent has gradient −3/4. Substituting gives y − 16 = −3/4(x − 12), so y − 16 = −3x/4 + 9 and y = −3x/4 + 25. Answer: y = −3x/4 + 25. The distractors: y = 3x/4 + 7 turns the gradient of the radius upside down but leaves it positive, so the perpendicular step is only half done; y = −4x/3 + 32 changes the sign of the radius gradient without turning it upside down, which is the other half left undone; y = −3x/4 − 25 uses the correct gradient but substitutes the point of contact with both signs reversed, writing y + 16 = −3/4(x + 12).
8. (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.
9. (d) 5√2 — For a circle x² + y² = r², the 50 on the right-hand side is r², not r, so the radius is √50. Writing 50 as 25 × 2, the largest square factor times what remains, gives √50 = √25 × √2 = 5√2. Forgetting to square-root 50 at all and giving the value of r² instead gives 50. Halving 50 instead of taking its square root gives 25. Using 25 as the number left outside the square root sign, instead of as the number under it, gives the wrongly simplified 25√2.
10. (d) x = −2, x = 6; y-intercept (0, 12) — Reflecting y = f(x) in the x-axis, to get y = −f(x), negates every y-value but leaves every x-value fixed. The x-intercepts happen where y = 0, and −0 = 0, so they are unaffected: y = −f(x) still crosses the x-axis at x = −2 and x = 6. The y-intercept is the value at x = 0: f(0) = −12, so −f(0) = 12, giving the point (0, 12) — the sign flips because the y-intercept is a nonzero y-value, unlike the roots. Writing 'x = 2, x = −6; y-intercept (0, −12)' comes from confusing −f(x) with f(−x) — reflecting in the y-axis instead of the x-axis, which negates the x-values of the intercepts (turning −2 into 2 and 6 into −6) but leaves f(0) unchanged, since f(−0) = f(0) = −12. Writing 'x = −2, x = 6; y-intercept (0, −12)' comes from forgetting that −f(x) is a reflection at all, and assumes both intercepts stay exactly as they were. Writing 'x = 2, x = −6; y-intercept (0, 12)' correctly negates the y-intercept but wrongly negates the x-intercepts too, as if a reflection in the x-axis also flipped the sign of every x-value.
11. (c) 37.5 — The tangent at (18, 24) is 18x + 24y = 900 (using ax + by = r² with a = 18, b = 24, r² = 900). Setting x = 0 to find the y-intercept: 24y = 900, so y = 37.5. Choosing 900 skips the division by 24 and just repeats the constant. Choosing 50 divides the constant by the x-coefficient 18 instead of the y-coefficient 24. Choosing 1.25 uses the radius 30 instead of r² = 900 as the constant before dividing.
12. (d) 3 and 4 — f(3) = 3³ − 3 × 3² − 4 = 27 − 27 − 4 = −4, and f(4) = 4³ − 3 × 4² − 4 = 64 − 48 − 4 = 12. Since f(3) is negative and f(4) is positive, there is a change of sign, so the solution lies between 3 and 4. A sign slip when expanding −3x², treating it as +3x², gives f(3) = 27 + 27 − 4 = 50, a positive value; taken with the correctly negative f(2) = 8 − 12 − 4 = −8, this reads as a change of sign between 2 and 3. The same sign slip applied at x = 1 gives f(1) = 1 + 3 − 4 = 0, read as the point the sign changes, against f(0) = 0 + 0 − 4 = −4, giving 0 and 1. Shifting the correctly found interval up by one integer, an indexing slip, reports the change as lying between 4 and 5 instead of 3 and 4.
13. (c) 90 — Method: the height of a bar is its frequency density, so twice as tall means twice the frequency density — not twice the frequency, because the two classes have different widths. Then frequency = frequency density × class width. Working: the first bar has frequency density 3 per cm, so the second has frequency density 2 × 3 = 6 per cm; the class 30 ≤ x < 45 is 45 − 30 = 15 cm wide, so its frequency is 6 × 15 = 90. Answer: 90 rods. The distractors: 120 comes from doubling the first bar's frequency instead of its height — the first class holds 3 × 20 = 60 rods, and doubling that ignores the fact that the second class is narrower; 45 comes from using the first bar's frequency density, 3, for the second bar, 3 × 15, and so never using the information that it is twice as tall; 6 comes from stopping at the frequency density of the taller bar and quoting a height as though it were a count.
14. (a) 2.55 — x₁ = 20 ÷ (2 + 5) = 20 ÷ 7 = 2.857142857. x₂ = 20 ÷ (2.857142857 + 5) = 20 ÷ 7.857142857 = 2.545454545, which rounds to 2.55. Reporting x₁ instead of x₂ gives 2.857142857, which rounds to 2.86. Dropping the +5 in the denominator, using xₙ₊₁ = 20 ÷ xₙ, gives x₁ = 20 ÷ 2 = 10 and x₂ = 20 ÷ 10 = 2, which is 2.00. A sign error in the denominator, using xₙ₊₁ = 20 ÷ (xₙ − 5), gives x₁ = 20 ÷ (2 − 5) = −6.666666667 and x₂ = 20 ÷ (−6.666666667 − 5) = −1.714285714, which rounds to −1.71.
15. (c) Translate +1 in y; new range 0 ≤ y ≤ 2 — Adding 1 to sin x translates the graph 1 unit in the positive y-direction, and every y-value on the range increases by 1: −1 + 1 = 0 and 1 + 1 = 2, giving a new range of 0 ≤ y ≤ 2. A translation in the x-direction would not add anything to the y-values, so 'Translate +1 in x; range unchanged' correctly leaves the range at −1 ≤ y ≤ 1 for that (wrong) transformation, but the transformation itself is not what y = sin x + 1 shows. 'Translate +1 in y; new range −2 ≤ y ≤ 0' correctly spots the translation but subtracts 1 from each bound instead of adding it. 'Reflect in the x-axis' mistakes the transformation for a reflection rather than a translation; reflecting sin x in the x-axis does leave the range at −1 ≤ y ≤ 1 unchanged in size, but that is not the transformation y = sin x + 1 actually applies.
16. (c) 9/28 — Method: two linked steps. Total everyone whose test is positive, since the person picked is known to be one of them, then divide the positive tests that belong to people with the condition by that total. Working: 45 positive tests come from people who have the condition and 95 come from people who do not, so 140 tests are positive. The people with the condition give 45/140, and dividing the numerator and the denominator by 5 gives 9/28. Answer: the probability is 9/28. The distractors: 9/10 is 45/50, the probability of a positive test given that the person has the condition, which is the condition and the event the wrong way round and is the figure a candidate quotes when the two are confused; 9/200 is 45/1000, dividing by everyone tested rather than by the 140 who tested positive; 1/20 is 50/1000, the probability that a person has the condition before the test result is used at all.
17. (d) (−2, 6) — Method: add the top numbers of both vectors to the starting x-coordinate, and the bottom numbers of both vectors to the starting y-coordinate. Working: x-coordinate 2 + 3 + (−7) = −2; y-coordinate −1 + 5 + 2 = 6. Answer: (−2, 6). A candidate who only applies vector u and forgets v gets (5, 4). A candidate who only applies vector v and forgets u gets (−5, 1). A candidate who works out the combined vector u + v but forgets to add it to the starting point gets (−4, 7).
18. (c) (−35, −55) — Add the three stages component by component to find the drone's position relative to base: (30+(−10)+15, 40+20+(−5)) = (35, 55). The flight back to base is the negative of this vector, reversing both numbers: (−35, −55). (35, 55) is the vector from base to the drone's position — it forgets to reverse direction for the return flight. (−35, 55) only reverses the top number. (35, −55) only reverses the bottom number.