Mark Scheme

Section A — Structured Questions (78 marks)

Question 1

(a) $\frac{3x^2 - 12}{x^2 - 4x + 4}$ (3 marks)

• Factorises numerator: $3(x^2 - 4)$ or $3(x + 2)(x - 2)$ M1
• Factorises denominator: $(x - 2)^2$ or $(x - 2)(x - 2)$ M1
• Cancels and simplifies to $\frac{3(x + 2)}{x - 2}$ A1

Accept: $\frac{3x + 6}{x - 2}$

(b) $5(2x - 3) = 3(x + 4) + 7$ (3 marks)

• Expands to $10x - 15 = 3x + 12 + 7$ or equivalent M1
• Collects terms: $7x = 34$ or equivalent M1
• $x = \frac{34}{7}$ or 4.86 (to 3sf) or $4\frac{6}{7}$ A1

(c) (2 marks)

• Subtracts $u$ from both sides: $v - u = at$ M1
• Divides by $a$: $t = \frac{v - u}{a}$ A1

Question 2

(a) (1 mark)

• 1 (goal) B1

(b) (3 marks)

• Attempts $\frac{\Sigma fx}{\Sigma f}$ or calculates $\Sigma fx$ M1
• $\Sigma fx = 0 + 12 + 22 + 18 + 8 + 5 = 65$ A1
• Mean = $\frac{65}{40} = 1.625$ or 1.63 A1

(c) (2 marks)

• Number of matches with more than 2 goals = $6 + 2 + 1 = 9$ M1
• Probability = $\frac{9}{40}$ or 0.225 A1

(d) (2 marks)

• $P(\text{0 goals in first match}) = \frac{8}{40}$ and $P(\text{0 goals in second match}) = \frac{8}{40}$ M1
• $\frac{8}{40} \times \frac{8}{40} = \frac{64}{1600} = \frac{1}{25}$ or 0.04 A1

Question 3

(a) (2 marks)

• Uses Area = $\frac{1}{2}ab\sin C$ M1
• Area = $\frac{1}{2} \times 9 \times 12 \times \sin 75° = 52.2$ cm² (to 3sf) A1

(b) (4 marks)

• Applies cosine rule: $BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(\angle BAC)$ M1
• Substitutes: $BC^2 = 81 + 144 - 2 \times 9 \times 12 \times \cos 75°$ M1
• $BC^2 = 225 - 216\cos 75° = 225 - 55.9... = 169.1...$ A1
• $BC = 13.0$ cm (to 3sf) A1

(c) (3 marks)

• States that Area = $\frac{1}{2} \times \text{base} \times \text{height}$ M1
• Uses $52.2 = \frac{1}{2} \times 12 \times h$ where $h$ is the perpendicular distance M1
• $h = \frac{52.2 \times 2}{12} = 8.70$ cm (to 3sf) A1

Alternative method:

• Uses $\sin 75° = \frac{h}{9}$ M1
• $h = 9\sin 75°$ M1
• $h = 8.70$ cm A1

Question 4

(a) (2 marks)

• First two terms: 5, 6 B1
• All four terms: 5, 6, 7.2, 8.64 B1

Accept: 5, 6, 7.2, 8.6 or 5, 6, 36/5, 216/25

(b) (2 marks)

• Uses $ar^{n-1}$ or $5 \times 1.2^9$ M1
• 10th term = 25.8 or 25.9 (to 3sf) A1

Accept: 25.8, 25.85, 25.852

(c) (2 marks)

• Uses $S_n = \frac{a(r^n - 1)}{r - 1}$ or $S_{15} = \frac{5(1.2^{15} - 1)}{1.2 - 1}$ M1
• $S_{15} = 193$ or 193.1 (to 4sf) A1

(d) (4 marks)

• Finds sum of first 10 terms of geometric sequence: $S_{10} = \frac{5(1.2^{10} - 1)}{0.2}$ M1
• $S_{10} = 129.16...$ or 129.2 A1
• Uses $S_n = \frac{n}{2}(2a + (n-1)d)$ so $129.16 = \frac{10}{2}(10 + 9d)$ M1
• $d = 2.87$ or 2.87 (to 3sf) or $\frac{258.3 - 100}{90}$ A1

Accept: 2.87, 2.870, 2.9

Question 5

(a) (3 marks)

• Major angle = $360° - 130° = 230°$ M1
• Arc length = $\frac{230}{360} \times 2\pi \times 8$ M1
• = 32.0 cm (to 3sf) A1

(b) (2 marks)

• Area of sector = $\frac{130}{360} \times \pi \times 8^2$ M1
• = 72.8 cm² (to 3sf) A1

(c)(i) (2 marks)

• Angle OAT = 90° (tangent perpendicular to radius)
• In quadrilateral OATB: $90° + 130° + 90° + \angle ATB = 360°$ M1
• $\angle ATB = 50°$ A1

(ii) (3 marks)

• In triangle OAT: angle OAT = 90°, angle AOT = 65° (half of 130°) M1
• Uses $\tan 65° = \frac{AT}{8}$ M1
• $AT = 8\tan 65° = 17.15...$ ≠ 14.0 cm

OR uses angle ATO = 25° (since angles in triangle = 180°)

• Uses $\tan 25° = \frac{8}{AT}$ M1
• $AT = \frac{8}{\tan 25°} = 17.2$ cm

Note: The question as set contains an error. The correct answer is approximately 17.2 cm, not 14.0 cm. Award M1 for setting up correct trigonometry, M1 for correct substitution, A0 for obtaining ≈17.2. In an actual paper, this would be caught in pre-vetting.

Corrected part (ii) for marking purposes:

• Recognizes angle in right triangle with angle ATO = 25° M1
• Uses correct trigonometric ratio M1
• Shows working leading to AT = 17.2 cm and notes discrepancy A1ft

Question 6

(a) (2 marks)

• Applies power rule to each term M1
• $\frac{dy}{dx} = 6x^2 - 10x + 3$ A1

(b) (2 marks)

• Substitutes $x = 2$ into derivative M1
• Gradient = $6(4) - 10(2) + 3 = 24 - 20 + 3 = 7$ A1

(c) (5 marks)

• Sets $\frac{dy}{dx} = 0$: $6x^2 - 10x + 3 = 0$ M1
• Uses quadratic formula: $x = \frac{10 \pm \sqrt{100 - 72}}{12}$ M1
• $x = \frac{10 \pm \sqrt{28}}{12} = \frac{10 \pm 5.292...}{12}$ A1
• $x = 1.27$ or $x = 0.394$ (both to 3sf) A1
• Finds both y-coordinates: when $x = 1.27$, $y = -4.07$; when $x = 0.394$, $y = -6.61$ Coordinates: $(1.27, -4.07)$ and $(0.394, -6.61)$ or better accuracy A1

Accept: $(1.274, -4.068)$ and $(0.393, -6.607)$ or exact forms

(d) (3 marks)

• Finds $\frac{d^2y}{dx^2} = 12x - 10$ M1
• At $x = 1.27$: $\frac{d^2y}{dx^2} = 5.24 > 0$, so minimum A1
• At $x = 0.394$: $\frac{d^2y}{dx^2} = -5.27 < 0$, so maximum A1

Accept: testing either side of stationary points with first derivative

Section B — Extended Response (52 marks)

Question 7

(a) (1 mark)

• $V = \pi r^2 h = 50$ so $h = \frac{50}{\pi r^2}$ B1

(b) (3 marks)

• Surface area = curved surface + 2 circular ends = $2\pi rh + 2\pi r^2$ M1
• Substitutes $h = \frac{50}{\pi r^2}$: $A = 2\pi r \times \frac{50}{\pi r^2} + 2\pi r^2$ M1
• Simplifies: $A = \frac{100}{r} + 2\pi r^2$ A1

(c) (2 marks)

• $\frac{dA}{dr} = -\frac{100}{r^2} + 4\pi r$ or $\frac{dA}{dr} = -100r^{-2} + 4\pi r$ M1 A1

(d) (4 marks)

• Sets $\frac{dA}{dr} = 0$: $-\frac{100}{r^2} + 4\pi r = 0$ M1
• Rearranges: $4\pi r = \frac{100}{r^2}$ or $4\pi r^3 = 100$ M1
• $r^3 = \frac{100}{4\pi} = \frac{25}{\pi}$ A1
• $r = 2.00$ m (to 3sf) or $r = 1.997...$ m A1

(e) (2 marks)

• Finds $\frac{d^2A}{dr^2} = \frac{200}{r^3} + 4\pi$ M1
• At $r = 2$: $\frac{d^2A}{dr^2} = 25 + 4\pi = 37.6 > 0$, therefore minimum A1

Accept: testing values either side of $r = 2$

(f) (5 marks)

• When $r = 2$: $h = \frac{50}{\pi \times 4} = 3.98$ m M1
• Checks constraint: $1.5r = 1.5 \times 2 = 3$ and $h = 3.98 > 3$ M1
• States that constraint is satisfied A1
• Therefore use $r = 2$ m B1
• Surface area = $2\pi(2)^2 + \frac{100}{2} = 8\pi + 50 = 75.1$ m² (to 3sf) A1

(g) (5 marks)

• When $r = 2$ and $V = 50$: $h = 3.98$ m, $A = 75.1$ m² (from part f) B1ft
• For 200 tanks: total area = $200 \times 75.1 = 15020$ m² M1
• Cost = $15020 \times 45 = £676,000$ (accept $676000 or $675900$) A1
• When $r = 2$ (non-optimal case for comparison): recalculate if needed, or note this is optimal M1
• Since $r = 2$ m is the optimal solution, there is no saving; monthly cost is £676,000

Alternative interpretation (if student assumes a different non-optimal radius): If considering arbitrary comparison with different radius that doesn't satisfy optimal conditions, award marks for correct method:

• Calculates surface area for $r = 2$: B1ft
• Calculates total surface area and cost for 200 tanks at optimal: M1 A1
• Calculates cost for 200 tanks at $r = 2$ m (given scenario): M1
• Finds difference: A1

Marking note: Award full credit if student recognizes that $r = 2$ m is already optimal and states saving is zero, OR if question interpretation intended different comparison radius (paper ambiguity), award marks for method shown.

Question 8

(a) (3 marks)

• Gradient $m = \frac{4 - 2}{5 - 1} = \frac{2}{4} = \frac{1}{2}$ M1
• Uses $y - y_1 = m(x - x_1)$ or substitutes into $y = mx + c$ M1
• $y = \frac{1}{2}x + \frac{3}{2}$ or $y = 0.5x + 1.5$ A1

(b) (2 marks)

• $AB = \sqrt{(5-1)^2 + (4-2)^2}$ M1
• $AB = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$ or 4.47 (to 3sf) A1

(c) (2 marks)

• $M = \left(\frac{1+5}{2}, \frac{2+4}{2}\right)$ M1
• $M = (3, 3)$ A1

(d) (3 marks)

• Gradient of perpendicular bisector = $-2$ (negative reciprocal of $\frac{1}{2}$) M1
• Uses point M(3, 3): $y - 3 = -2(x - 3)$ M1
• $y = -2x + 9$ A1

(e) (2 marks)

• Substitutes $x = 3$ into equation of perpendicular bisector M1
• $y = -2(3) + 9 = 3$, which gives point (3, 6), but C is (3, 6), so this is an error in the question setup

Corrected: For C(3, 6) to lie on perpendicular bisector $y = -2x + 9$:

• Substitutes $x = 3$: $y = -6 + 9 = 3 \neq 6$

Marking adjustment: C does not lie on the perpendicular bisector with the given coordinates. Award:

• M1 for substitution method
• A1 for showing calculation (even if shows C does not lie on it)

Examiner note: In a real paper, coordinates would be verified. For marking purposes, award marks for correct method.

(f) (6 marks)

For parallelogram ABCD:

• Uses $\vec{AB} = \vec{DC}$ or $\vec{AD} = \vec{BC}$ M1
• Position 1: $\vec{AB} = (4, 2)$, so $D = C - \vec{AB} = (3, 6) - (4, 2) = (-1, 4)$ M1 A1
• Position 2: $\vec{BA} = (-4, -2)$, so $D = C - \vec{BA} = (3, 6) - (-4, -2) = (7, 8)$ A1
• Area = base × perpendicular height = $|AB| \times$ perpendicular distance from C to AB M1
• Area = $2\sqrt{5} \times$ (perpendicular distance) = $2\sqrt{5} \times \frac{|3/2 - 3/2|...}$

Alternative: Area using vectors = $|\vec{AB} \times \vec{AC}|$

• $\vec{AC} = (2, 4)$, Area = $|4 \times 4 - 2 \times 2| = |16 - 4| = 12$ A1

Both positions give same area = 12 square units (parallelogram property)

(g)(i) (2 marks)

• A'(2, 1), B'(4, 5), C'(6, 3) B2 (Award B1 for two correct)

(g)(ii) (2 marks)

• Reflection in the line $y = x$ B1
• (No additional detail needed as single transformation) B1

Question 9

(a) (1 mark)

• Median at 60th value ≈ 63 kg (accept 62-64 kg) B1

(b) (2 marks)

• $Q_1$ at 30th value ≈ 57 kg, $Q_3$ at 90th value ≈ 69 kg M1
• IQR = $69 - 57 = 12$ kg (accept 11-13 kg depending on reading) A1

(c) (2 marks)

• From graph, number with mass > 72 kg ≈ $120 - 100 = 20$ M1
• Probability = $\frac{20}{120} = \frac{1}{6}$ or 0.167 (accept 0.15-0.20 depending on reading) A1ft

(d) (3 marks)

• From graph: number with mass < 52 kg ≈ 12 M1
• Number with mass > 78 kg ≈ $120 - 115 = 5$ M1
• Total requiring monitoring ≈ $12 + 5 = 17$ (accept 15-20) A1

(e)(i) (4 marks)

• Uses midpoints: 45, 52.5, 57.5, 62.5, 67.5, 72.5, 80 M1
• Calculates $\Sigma fx$: $(45 \times 8) + (52.5 \times 14) + (57.5 \times 20) + (62.5 \times 28) + (67.5 \times 25) + (72.5 \times 15) + (80 \times 10)$ M1
• $\Sigma fx = 360 + 735 + 1150 + 1750 + 1687.5 + 1087.5 + 800 = 7570$ A1
• Mean = $\frac{7570}{120} = 63.1$ kg (to 3sf) A1

(e)(ii) (1 mark)

• Midpoints used / exact values within groups unknown / assumes uniform distribution within classes B1

Accept: "We don't know the exact values, only the groups"

(f) (3 marks)

• First school: mean ≈ 63.1 kg, SD = 9.2 kg B1
• Second school: mean = 63.5 kg, SD = 12.8 kg B1
• Comparison: The schools have similar mean masses (within 0.4 kg), but the second school has greater variability/spread in masses (SD 12.8 kg vs 9.2 kg), indicating less consistency / more diverse range of student masses. B1

Accept any valid comparison noting:

• Similar centres/means
• Greater spread/variability in second school
• More consistent data in first school

Question 10

(a) (1 mark)

• $f(7) = 3(7) - 5 = 16$ B1

(b) (3 marks)

• $gf(x) = g(3x - 5)$ M1
• $= (3x - 5)^2 + 2$ M1
• $= 9x^2 - 30x + 25 + 2 = 9x^2 - 30x + 27$ A1

(c) (2 marks)

• $fg(x) = f(x^2 + 2) = 3(x^2 + 2) - 5$ M1
• $= 3x^2 + 6 - 5 = 3x^2 + 1$ A1

(d) (3 marks)

• $gf(x) = 27$ gives $9x^2 - 30x + 27 = 27$ M1
• $9x^2 - 30x = 0$ M1
• $3x(3x - 10) = 0$ so $x = 0$ or $x = \frac{10}{3}$ (or 3.33 to 3sf) A1

(e) (2 marks)

• Let $y = 3x - 5$, then $x = \frac{y + 5}{3}$ M1
• $f^{-1}(x) = \frac{x + 5}{3}$ A1

(f) (4 marks)

• $f^{-1}(x) = g(x)$ gives $\frac{x + 5}{3} = x^2 + 2$ M1
• Multiplies by 3: $x + 5 = 3x^2 + 6$ M1
• Rearranges: $3x^2 - x + 1 = 0$ A1
• Uses discriminant or quadratic formula: $x = \frac{1 \pm \sqrt{1 - 12}}{6} = \frac{1 \pm \sqrt{-11}}{6}$ No real solutions A1

Accept: "No solutions" or "No real solutions" or shows discriminant < 0

(g)(i) (3 marks)

• $hf(x) = h(3x - 5) = \frac{6}{(3x - 5) - 1}$ M1
• $= \frac{6}{3x - 6} = \frac{6}{3(x - 2)} = \frac{2}{x - 2}$ A1
• Not defined when $x = 2$ B1

(g)(ii) (3 marks)

• $\frac{2}{x - 2} = 2$ M1
• $2 = 2(x - 2)$ M1
• $2 = 2x - 4$, $2x = 6$, $x = 3$ A1

Sample Answers with Examiner Commentary

Question 7 — Sample Answers

Grade A (high distinction) answer*

(a) $V = \pi r^2 h = 50$

Therefore $h = \frac{50}{\pi r^2}$ ✓

(b) Total surface area consists of two circular ends and the curved surface.

$A = 2\pi r^2 + 2\pi rh$

Substituting $h = \frac{50}{\pi r^2}$:

$A = 2\pi r^2 + 2\pi r \times \frac{50}{\pi r^2}$

$A = 2\pi r^2 + \frac{100\pi r}{\pi r^2}$

$A = 2\pi r^2 + \frac{100}{r}$ ✓

(c) $\frac{dA}{dr} = 4\pi r - 100r^{-2}$

$= 4\pi r - \frac{100}{r^2}$ ✓

(d) For minimum surface area, $\frac{dA}{dr} = 0$

$4\pi r - \frac{100}{r^2} = 0$

$4\pi r = \frac{100}{r^2}$

$4\pi r^3 = 100$

$r^3 = \frac{100}{4\pi} = \frac{25}{\pi}$

$r = \sqrt[3]{\frac{25}{\pi}} = 2.00$ m (to 3sf) ✓

(e) $\frac{d^2A}{dr^2} = 4\pi + 200r^{-3} = 4\pi + \frac{200}{r^3}$

When $r = 2$: $\frac{d^2A}{dr^2} = 4\pi + \frac{200}{8} = 12.566... + 25 = 37.6 > 0$

Since the second derivative is positive, this confirms a minimum. ✓

(f) When $r = 2$ m: $h = \frac{50}{\pi \times 4} = \frac{50}{4\pi} = 3.98$ m

Stability constraint requires $h \geq 1.5r = 1.5 \times 2 = 3$ m

Since $3.98 > 3$, the constraint is satisfied.

Therefore use $r = 2$ m.

Surface area = $2\pi(2)^2 + \frac{100}{2} = 8\pi + 50 = 75.1$ m² ✓

(g) For optimal design ($r = 2$ m):

• Surface area per tank = 75.1 m²
• For 200 tanks: $200 \times 75.1 = 15,020$ m²
• Cost = $15,020 \times 45 = £676,000$

For $r = 2$ m with $V = 50$ m³:

• $h = \frac{50}{\pi \times 4} = 3.98$ m
• Surface area = $2\pi(4) + 2\pi(2)(3.98) = 25.13 + 50.0 = 75.1$ m²

Since we are already using $r = 2$ m for comparison, I need to compare with a different radius.

For $r = 2$ m (non-optimized baseline from question): Let me recalculate assuming the question wants comparison with literal $r = 2$ m before optimization check:

Actually, $r = 2$ m IS the optimal value from part (d), so if the tanks were previously made with a different radius, I would need that information.

However, the question states "compared to using tanks with radius 2 metres" — this means $r = 2$ m is the comparison baseline, and I've found it's also optimal.

Therefore, there is no saving as $r = 2$ m is already optimal. The monthly cost is £676,000 whether optimized or not.

Mark: 22/22 marks

Examiner commentary: This is an exemplary response demonstrating complete understanding of optimization with constraints. The student correctly differentiates, finds critical points, verifies using the second derivative test, checks the constraint, and performs all calculations accurately. The working is clearly laid out with proper mathematical notation. In part (g), the student recognizes the potential ambiguity in the question and addresses it logically. Full marks awarded. This represents high Grade A* work: comprehensive, accurate, and showing mathematical maturity in handling a multi-stage problem.

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Grade C (pass) answer

(a) $V = \pi r^2 h = 50$

$h = \frac{50}{\pi r^2}$ ✓

(b) Surface area = $2\pi r^2 + 2\pi rh$

$= 2\pi r^2 + 2\pi r \times \frac{50}{\pi r^2}$

$= 2\pi r^2 + \frac{100}{r}$ ✓

(c) $\frac{dA}{dr} = 4\pi r - \frac{100}{r^2}$ ✓

(d) $4\pi r - \frac{100}{r^2} = 0$

$4\pi r = \frac{100}{r^2}$

$r^3 = \frac{100}{4\pi}$

$r = 1.997$ m ✓ (working shown but rounded as 2.00 m)

(e) When I put in $r = 1.5$, I get $\frac{dA}{dr} = -25.6$

When I put in $r = 2.5$, I get $\frac{dA}{dr} = 15.4$

It changes from negative to positive so it's a minimum. ✓

(f) $h = \frac{50}{\pi \times 4} = 3.98$ m ✓

$1.5 \times 2 = 3$ m

$3.98 > 3$ so it's OK ✓

Surface area = $2\pi \times 4 + \frac{100}{2} = 25.1 + 50 = 75.1$ m² ✓

(g) Surface area for 200 tanks = $200 \times 75.1 = 15,020$ m²

Cost = $15,020 \times 45 = 676,000$ pounds ✓

For $r = 2$ m, the surface area is 75.1 m² so cost is the same.

Saving = £0 ✓

Mark: 15/22 marks

Examiner commentary: This answer demonstrates solid understanding of the core calculus techniques and follows the optimization process. The student earns marks in parts (a)-(d) for correct method and answers. In part (e), the student uses a valid but less sophisticated approach (testing values either side) rather than the second derivative test—this earns the marks but shows weaker technique. Part (f) is mostly correct but the surface area calculation uses $2\pi r^2$ incorrectly written as $2\pi \times 4$ without showing $r^2 = 4$; still arriving at the correct answer earns credit. Part (g) loses marks because the student doesn't clearly establish what "comparison" radius should be used or demonstrate the cost calculation for a genuinely different scenario. The student correctly identifies that $r = 2$ is optimal but doesn't fully engage with the comparative aspect the question requires. This represents mid-Grade C work: competent execution of standard techniques but lacking the depth and precision of higher grades.

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Grade E (near miss) answer

(a) $V = \pi r^2 h = 50$

$\pi r^2 = \frac{50}{h}$ ✗

$r^2 = \frac{50}{\pi h}$

(b) Surface area = $\pi r^2 + 2\pi rh$ ✗ (missing one circular end)

$= \pi r^2 + 2\pi r \times \frac{50}{\pi r^2}$

$= \pi r^2 + \frac{100}{r}$

(c) $\frac{dA}{dr} = 2\pi r - \frac{100}{r^2}$ ✗ (wrong coefficient)

(d) $2\pi r - \frac{100}{r^2} = 0$

$2\pi r = \frac{100}{r^2}$

$2\pi r^3 = 100$

$r^3 = \frac{50}{\pi}$

$r = 2.52$ m ✗ (follow-through error from wrong derivative)

(e) It's a minimum because the graph of surface area curves upward. ✓ (basic understanding, no calculation)

(f) $h = \frac{50}{\pi \times 2.52^2} = 2.51$ m ✗ft

$1.5 \times 2.52 = 3.78$ m

$2.51 < 3.78$ so doesn't work ✗ft

Need to use $r$ that makes $h = 1.5r$

$\frac{50}{\pi r^2} = 1.5r$

$50 = 1.5\pi r^3$

$r = 2.12$ m (partial credit for attempting constraint)

(g) Surface area = $2\pi(2)^2 + \frac{100}{2} = 25.1 + 50 = 75$ m²

200 tanks = $200 \times 75 = 15,000$ m²

Cost = $15,000 \times 45 = £675,000$ ✓ (method correct for one scenario)

[Student doesn't calculate comparison case or saving]

Mark: 6/22 marks

Examiner commentary: This response shows some understanding of the optimization process but makes critical errors early that cascade through the question. In part (a), the student rearranges for $r$ instead of $h$, failing to complete the required result—no marks. Part (b) omits one circular end from the surface area formula (a common error); though some method is shown, the final answer is incorrect—no marks. The derivative in (c) has the wrong coefficient for $2\pi r$ instead of $4\pi r$, indicating the student didn't correctly differentiate $2\pi r^2$—this is a fundamental calculus error. The follow-through errors continue through parts (d) and (f). Part (e) earns 1 mark for showing basic conceptual understanding. In part (f), the student does recognize that the constraint isn't satisfied and attempts to find an alternative value, showing some problem-solving awareness—partial credit. Part (g) shows one calculation but doesn't complete the comparison required. To improve, this student needs to: practice differentiation of polynomial terms more thoroughly, be more careful with formula recall (especially for 3D shapes), and fully read question requirements about making comparisons. This represents Grade E work: some relevant knowledge but with fundamental gaps and lack of accuracy.

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Question 9 — Sample Answers

Grade A (high distinction) answer*

(a) Median is at the $\frac{120}{2} = 60$th value.

From the cumulative frequency curve, reading across from 60 on the vertical axis to the curve and down to the horizontal axis gives approximately 63 kg. ✓

(b) Lower quartile $Q_1$ is at the $\frac{120}{4} = 30$th value ≈ 57 kg

Upper quartile $Q_3$ is at the $\frac{3 \times 120}{4} = 90$th value ≈ 69 kg

Interquartile range = $69 - 57 = 12$ kg ✓

(c) Reading from the graph, approximately 100 students have mass ≤ 72 kg.

Therefore