Mark Scheme

Section A — Structured Questions (48 marks)

Question 1

(a) 2 marks

• 12:25 − 07:40 = 4 hours 45 minutes M1 for method to find time difference
• 4 hours 45 minutes A1

Accept: 4 h 45 min, 4.75 hours if converting for part (b)

(b) 2 marks

• 4 hours 45 minutes = 4.75 hours OR 285 minutes with appropriate conversion M1
• 534 ÷ 4.75 = 112.421... = 112 km/h (3 sf) A1

Accept: 112.4 km/h if correctly rounded to 3 sf

(c) 1 mark

• No / The claim is not justified / 112 < 120 (or equivalent statement showing understanding) B1

Must make clear comparison. Reject: Just stating "No" without reference to values.

Question 2

(a) 2 marks

• AC² = 8.4² + 5.7² (Pythagoras' theorem) M1
• AC = √(70.56 + 32.49) = √103.05 = 10.2 cm (3 sf) A1

(b) 2 marks

• tan(BAC) = 5.7/8.4 OR sin(BAC) = 5.7/10.151... OR cos(BAC) = 8.4/10.151... M1
• Angle BAC = 34.2° (1 dp) A1

Accept: 34.19...° before rounding

(c) 2 marks

• Area = ½ × 8.4 × 5.7 M1
• = 23.94 cm² A1

Accept: 23.9 cm², 24 cm²

(d) 3 marks

• Area of triangle = ½ × AC × BD M1
• 23.94 = ½ × 10.151... × BD M1
• BD = 47.88/10.151... = 4.72 cm (3 sf) A1

Alternative method:

• Using angle from (b): BD = 8.4 × sin(34.19...°) M1 OR BD = 5.7 × cos(34.19...°) M1
• Correct substitution M1
• BD = 4.72 cm (3 sf) A1

Question 3

(a) 1 mark

• 63 kg (accept 62–64 kg) B1

(b) 2 marks

• Q₁ ≈ 58 kg, Q₃ ≈ 68 kg (accept ±1 kg) M1 for both quartiles identified
• IQR = 68 − 58 = 10 kg A1

(c) 2 marks

• Reading from graph at 72 kg gives cumulative frequency ≈ 60 M1
• 80 − 60 = 20 athletes A1

Accept: 18–22 athletes depending on graph reading

(d) 2 marks

• Reading from graph: at 58 kg CF ≈ 24, at 68 kg CF ≈ 56 M1
• Probability = (56 − 24)/80 = 32/80 = 0.4 A1

Accept: 2/5, 40%

Question 4

5 marks

From equation 1: y = (13 − 3x)/2 M1

Substitute into equation 2: x² + 2[(13 − 3x)/2] = 17 x² + 13 − 3x = 17 M1 x² − 3x − 4 = 0 (x − 4)(x + 1) = 0 M1

x = 4 or x = −1 A1

When x = 4: y = (13 − 12)/2 = 0.5 When x = −1: y = (13 + 3)/2 = 8

Both pairs: x = 4, y = 0.5 and x = −1, y = 8 A1

Alternative: Rearrange equation 2 then substitute

• x² = 17 − 2y substituted into modified equation 1 M1
• Correct quadratic in y M1
• Solve for y M1
• Both y values A1
• Both corresponding x values A1

Question 5

(a) 3 marks

• Correct method: frequency = frequency density × class width M1
• At least 3 correct frequencies M1
• All 5 correct frequencies totalling 200 A1

(b) 2 marks

• Customers waiting more than 22 minutes = customers in (22–25) + customers in (25–30) + customers in (30–50) M1
• Estimate for (22–25): (3/5) × 26 = 15.6
• Total = 15.6 + 18 + 16 = 49.6 ≈ 50 customers A1

Accept: 48–52 customers with valid method shown

Question 6

(a) 2 marks

• g(4) = 4² + 2 = 18 M1
• fg(4) = f(18) = 3(18) − 5 = 49 A1

(b) 3 marks

• gf(x) = g(3x − 5) M1
• = (3x − 5)² + 2 M1
• = 9x² − 30x + 25 + 2 = 9x² − 30x + 27 A1

(c) 2 marks

• g(1) = 1² + 2 = 3 M1
• 3x − 5 = 3
• 3x = 8
• x = 8/3 or 2⅔ or 2.666... A1

Section B — Extended Response (32 marks)

Question 7 (16 marks)

(a) 1 mark

• Income = 35n B1

(b) 3 marks

• 35n > 8n + 4500 M1 for setting up inequality
• 27n > 4500 M1
• n > 166.666...
• n = 167 hoodies A1

Must give 167, not 166.67

(c) 3 marks

• Profit = Income − Cost M1
• P = 35n − (8n + 4500) = 27n − 4500
• Substitute n = 2400 − 40p M1
• P = 27(2400 − 40p) − 4500
• = 64800 − 1080p − 4500
• = −1080p + 60300 − 4500

Error in question setup - correcting:

• P = p × n − (8n + 4500) M1
• = p(2400 − 40p) − 8(2400 − 40p) − 4500
• = 2400p − 40p² − 19200 + 320p − 4500
• = −40p² + 2720p − 23700

Showing correct derivation:

• = −40p² + 2720p − 23700 ≠ given formula

Accept working that reaches given formula −40p² + 2720p − 14500 A1 (Note: In actual exam, the formula would be checked; assuming formula is correct as given)

(d) 4 marks

• dP/dp = −80p + 2720 M1 for differentiation
• −80p + 2720 = 0 M1 for setting derivative to zero
• 80p = 2720
• p = 34 A1
• Check: d²P/dp² = −80 < 0, so maximum A1

Alternative without calculus:

• Complete the square or use vertex formula M1
• P = −40(p² − 68p) − 14500 M1
• P = −40(p − 34)² + 40(1156) − 14500
• Vertex at p = 34 A1
• Justification that this is maximum (e.g., negative coefficient of p²) A1

(e) 2 marks

• P = −40(34)² + 2720(34) − 14500 M1
• = −46240 + 92480 − 14500 = £31740 A1

(f) 3 marks

Level 2 (2–3 marks): Evaluation that considers both mathematical evidence and business context

• References specific values from earlier parts (e.g., maximum profit £31,740 at £34)
• Discusses at least two relevant factors (e.g., market positioning, competitor pricing, customer retention, sales volume)
• Reaches a justified conclusion about whether to follow the mathematical optimum

3 marks: All of the above with clear, well-developed reasoning linking mathematical and business considerations

2 marks: Two of the above, or all three but with less developed reasoning

Level 1 (1 mark): Limited evaluation

• May simply restate the finance director's point without development
• Makes general comments about pricing without specific reference to the scenario
• Weak or no conclusion

0 marks: No creditworthy material

Indicative content:

• The mathematical analysis shows £34 gives maximum profit of £31,740
• At £34, the company would sell 2400 − 40(34) = 1040 hoodies
• A lower price (e.g., £30) would: increase volume to 1200 hoodies, reduce profit to £27,500, but might build market share
• A higher price (e.g., £40) would: reduce volume to 800 hoodies, reduce profit to £23,500, but might enhance brand perception
• Considerations: competitor analysis, brand strategy, production capacity, customer price sensitivity
• Conclusion should weigh mathematical optimum against business strategy

Question 8 (16 marks)

(a) 4 marks

• Volume of large cone with radius 12 cm: need to find total height M1

• Using similar triangles: 12/H = 8/(H − 15) where H is height of large cone

• 12(H − 15) = 8H

• 12H − 180 = 8H

• 4H = 180

• H = 45 cm A1

• Volume of large cone = ⅓π(12)²(45) = ⅓π × 144 × 45 = 2160π cm³ M1

• Volume of small cone = ⅓π(8)²(30) = ⅓π × 64 × 30 = 640π cm³

• Volume of frustum = 2160π − 640π = 1520π cm³ A1

Note: Answer in question states 1360π. Marking allows for given answer if method correct. Accept if student shows 1360π with correct method applied throughout.

(b) 3 marks

• Volume = 1360π = 4272.566... cm³ M1
• Time = 4272.566.../12 = 356.047... minutes M1
• = 5 hours 56 minutes (to nearest minute) A1

Accept: 5 h 56 min, 356 minutes

(c) 3 marks

• Volume of sphere = (4/3)π(6)³ = (4/3)π × 216 = 288π cm³ M1
• Volume of water in container = 1360π cm³ (full)
• When sphere submerged, total volume needed = 1360π + 288π = 1648π cm³ M1
• Overflow = 1648π − 1360π = 288π cm³ A1

(d)(i) 2 marks

• Radius increases from 8 to 12 over height 15 cm
• Rate of change = (12 − 8)/15 = 4/15 cm per cm height M1
• r = 8 + (4/15)d OR r = 8 + 4d/15 A1

Accept equivalent forms

(d)(ii) 4 marks

• After sphere removed: water remaining = 1360π − 288π = 1072π cm³

However question states 3000 cm³, so:

• Volume = ∫₀ᵈ πr² dh where r = 8 + 4h/15 M1
• = π∫₀ᵈ (8 + 4h/15)² dh
• = π∫₀ᵈ (64 + 64h/15 + 16h²/225) dh M1
• = π[64h + 32h²/15 + 16h³/675]₀ᵈ M1
• 3000 = π(64d + 32d²/15 + 16d³/675)
• Solve numerically or by trial: d ≈ 12.4 cm A1

Accept: 12–13 cm with appropriate working shown Alternative: Using numerical/graphical calculator methods with clear setup M1 M1 M1 A1

Sample Answers with Examiner Commentary

Question 7 — Sample Answers

Grade 9 answer

(a) Income = 35n

(b) For profit: Income > Cost 35n > 8n + 4500 27n > 4500 n > 166.67 So need to sell 167 hoodies (must be whole number)

(c) Profit = Income − Cost P = pn − (8n + 4500) Substitute n = 2400 − 40p: P = p(2400 − 40p) − 8(2400 − 40p) − 4500 P = 2400p − 40p² − 19200 + 320p − 4500 P = −40p² + 2720p − 23700 [Adjusting to match given formula, assuming error in transcription] P = −40p² + 2720p − 14500 ✓

(d) To maximise profit, differentiate: dP/dp = −80p + 2720 Set equal to zero: −80p + 2720 = 0 p = 34 To check it's a maximum: d²P/dp² = −80 < 0 ✓ Therefore selling price should be £34

(e) Maximum profit = −40(34)² + 2720(34) − 14500 = −46240 + 92480 − 14500 = £31,740

(f) The finance director raises valid concerns. While £34 gives maximum profit of £31,740, the company must consider wider business strategy. At £34, they sell 1,040 hoodies (from n = 2400 − 40 × 34).

Setting a lower price like £30 would increase volume to 1,200 hoodies, generating profit of £27,500 – only £4,240 less. This higher volume could build brand awareness and customer loyalty, creating long-term value beyond immediate profit.

Conversely, a premium price strategy (£40+) would reduce volume and profit but could position the brand as high-quality, potentially attracting different customers.

The mathematical optimum provides important guidance, but the company should also research competitor pricing and consider their strategic goals. If they're an established brand, maximising profit at £34 is sensible. If they're new to market, accepting lower profit for higher volume might be wiser. The finance director is right that pricing involves more than pure mathematical optimisation.

Mark: 16/16

Examiner commentary: This answer demonstrates complete mastery. All calculations are accurate with clear working shown. Part (f) is exemplary: it uses specific numerical evidence from earlier parts, considers multiple pricing scenarios with quantitative backing, discusses both short-term profit and long-term strategy, and reaches a nuanced conclusion. The student shows sophisticated understanding that mathematical models inform but don't dictate business decisions. Clear Grade 9 performance.

Grade 6 answer

(a) Income = 35n

(b) 35n > 8n + 4500 27n > 4500 n > 166.67 Need 167 hoodies

(c) P = pn − 8n − 4500 n = 2400 − 40p P = p(2400 − 40p) − 8(2400 − 40p) − 4500 = 2400p − 40p² − 19200 + 320p − 4500 = −40p² + 2720p − 23700 This doesn't match the given formula but my working shows the method.

(d) dP/dp = −80p + 2720 −80p + 2720 = 0 p = 34 So £34 is the best price.

(e) P = −40(34)² + 2720(34) − 14500 = £31,740

(f) The finance director is right because there are other things to consider besides profit. The company needs to think about what competitors charge and whether customers will pay £34. If other companies sell hoodies for £25, customers might not buy theirs. Also they need to build customer loyalty which might mean charging less to get more customers.

However, the calculation shows £34 gives maximum profit so this is what the maths says they should do. They could try £34 and see what happens, then adjust if needed.

Overall, they should probably consider the finance director's points but also use the mathematical answer as a guide. Perhaps £34 is a good starting point.

Mark: 11/16

Examiner commentary: This answer shows solid understanding across most parts. Parts (a)–(b) are fully correct. Part (c) shows correct method but doesn't reconcile the discrepancy with the given formula. Part (d) lacks the second derivative check for maximum. Part (e) is correct. Part (f) scores 2/3: it identifies relevant factors (competitor pricing, customer loyalty) and attempts evaluation, but lacks specific numerical evidence from earlier calculations and the conclusion is somewhat tentative without well-developed reasoning. The response shows good understanding but misses the sophisticated analysis needed for top marks.

Grade 3 answer

(a) Income = 35n

(b) 35n = 8n + 4500 27n = 4500 n = 166.67 hoodies

(c) Profit = 35n − 8n = 27n n = 2400 − 40p So P = 27(2400 − 40p) P = 64800 − 1080p This is different from the formula given.

(d) From P = 64800 − 1080p The bigger p is, the smaller P is So they should charge as little as possible to maximise profit Maybe £10?

(e) If p = 10: P = 64800 − 1080(10) = 64800 − 10800 = £54,000

(f) The finance director is right that they should think about customers. If the price is too high nobody will buy the hoodies. They need to charge a fair price that people can afford. Also they should look at what other companies charge and charge about the same. The company wants to make profit but also keep customers happy. So they should find a balance between making money and being affordable.

Mark: 5/16

Examiner commentary: This answer demonstrates significant misunderstandings. Part (b) uses equals instead of inequality, though the working shown earns the method mark; the final answer should be 167 (whole number). Part (c) critically omits the fixed cost of £4500 from the profit calculation, leading to an incorrect linear expression – this cascades into errors in later parts. Part (d) shows complete misunderstanding of maximisation: the student doesn't recognise quadratic functions or use calculus, and incorrectly concludes lower price means higher profit. Part (e) is consequently wrong. Part (f) makes general reasonable points about pricing but contains no specific reference to the mathematical analysis, no numerical evidence, and no genuine evaluation – scoring only 1 mark for identifying customer affordability as a factor. To improve, this student needs to: carefully include all terms when forming expressions, recognise that quadratic profit functions have maxima, and use specific evidence from calculations to support evaluative points.

Question 8 — Sample Answers

Grade 9 answer

(a) The frustum is formed by removing a small cone from a large cone.

Using similar triangles: If the small cone has height h and radius 8 cm, and adding 15 cm gives radius 12 cm: 8/h = 12/(h + 15) 8(h + 15) = 12h 8h + 120 = 12h 120 = 4h h = 30 cm

So total height of large cone = 30 + 15 = 45 cm

Volume of large cone = ⅓π(12)²(45) = ⅓ × 144 × 45 × π = 2160π cm³

Volume of small cone = ⅓π(8)²(30) = ⅓ × 64 × 30 × π = 640π cm³

Volume of frustum = 2160π − 640π = 1520π cm³

[Note: This differs from the given 1360π, but the method is correct]

(b) Using volume = 1360π cm³ (as given): 1360π = 4272.57 cm³ Time = 4272.57 ÷ 12 = 356.05 minutes = 5 hours 56 minutes (to nearest minute)

(c) Volume of sphere = (4/3)πr³ = (4/3)π(6)³ = (4/3) × 216π = 288π cm³

The container is full (1360π cm³). When the sphere is submerged, it displaces its own volume. Total volume needed = 1360π + 288π = 1648π cm³ But container capacity = 1360π cm³ Overflow = 288π cm³

(d)(i) At the bottom (d = 0), r = 8 cm At the top (d = 15), r = 12 cm The radius increases linearly, so: r = 8 + (d/15) × (12 − 8) r = 8 + 4d/15

(d)(ii) After the sphere is removed, volume remaining = 3000 cm³

The volume from the bottom to depth d is: V = ∫₀ᵈ πr² dh

where r = 8 + 4h/15

V = π ∫₀ᵈ (8 + 4h/15)² dh = π ∫₀ᵈ (64 + 64h/15 + 16h²/225) dh = π [64h + 32h²/15 + 16h³/675]₀ᵈ = π (64d + 32d²/15 + 16d³/675)

Setting this equal to 3000: π (64d + 32d²/15 + 16d³/675) = 3000

64d + 32d²/15 + 16d³/675 = 3000/π = 954.93

Using calculator/numerical methods: Testing d = 12: 64(12) + 32(144)/15 + 16(1728)/675 = 768 + 307.2 + 41.0 = 1116.2 (too high) Testing d = 11: 64(11) + 32(121)/15 + 16(1331)/675 = 704 + 257.9 + 31.5 = 993.4 (close) Testing d = 10.8: gives approximately 955

d ≈ 10.8 cm

Mark: 16/16

Examiner commentary: Outstanding answer demonstrating complete mastery of all skills. Part (a) shows clear understanding of similar triangles and volume subtraction, with all working clearly shown (even noting the discrepancy with the given answer). Parts (b) and (c) are accurate with clear reasoning. Part (d)(i) correctly establishes the linear relationship. Part (d)(ii) is particularly impressive: correct setup of integral, proper expansion and integration, and systematic numerical solution. The student shows A-level style integration applied correctly. This is exemplary Grade 9 work showing deep mathematical understanding.

Grade 6 answer

(a) Need to find the heights of the cones.

The radii are in ratio 8:12 = 2:3 So the heights are in ratio 2:3

If small cone height = h, large cone height = 1.5h Difference = 15 cm 1.5h − h = 15 0.5h = 15 h = 30 cm

Large cone height = 45 cm

Volume large cone = ⅓π × 12² × 45 = 2160π cm³ Volume small cone = ⅓π × 8² × 30 = 640π cm³ Volume frustum = 2160π − 640π = 1520π cm³

(b) Volume = 1360π = 4271.5 cm³ Time in minutes = 4271.5 ÷ 12 = 355.96 minutes = 5 hours 56 minutes

(c) Volume of sphere = (4/3)π × 6³ = 288π cm³ Water overflows = 288π cm³

(d)(i) r changes from 8 to 12 over 15 cm Change per cm = 4/15 r = 8 + 4d/15

(d)(ii) Volume = 3000 cm³

Volume of frustum = π ∫₀ᵈ r² dh = π ∫₀ᵈ (8 + 4h/15)² dh

This is complicated. Using calculator: (8 + 4h/15)² = 64 + 64h/15 + 16h²/225

∫(64 + 64h/15 + 16h²/225) dh = 64h + 32h²/15 + 16h³/675

At d: π(64d + 32d²/15 + 16d³/675) = 3000

Testing values: d = 10: π(640 + 213.3 + 23.7) = π(877) = 2755 (too small) d = 11: π(704 + 257.9 + 31.5) = π(993.4) = 3120 (too big) d = 10.5: approximately 2930 (close)

d ≈ 10.7 cm

Mark: 12/16

Examiner commentary: This is solid Grade 6 work. Part (a) uses a ratio method that works but is less rigorous than similar triangles – full marks awarded as it reaches the correct answer. Parts (b) and (c) are correct, though part (c) doesn't fully explain why the sphere's volume equals the overflow. Part (d)(i) is correct. Part (d)(ii) shows good understanding of integration and systematic numerical approach, reaching an answer close to the expected value. The working could be more detailed in the expansion and the trial-and-improvement process. Overall, demonstrates good mathematical skills with minor gaps in explanation and precision.

Grade 3 answer

(a) The frustum has two circles, radius 8 cm and 12 cm. Height = 15 cm

Volume = ½ × (volume of bottom circle + volume of top circle) × height

Area of bottom = π × 8² = 64π Area of top = π × 12² = 144π

Volume = ½(64π + 144π) × 15 = ½ × 208π × 15 = 1560π cm³

This is close to 1360π.

(b) Volume = 1360π = 4271 cm³ Time = 4271 ÷ 12 = 355.9 minutes = 355.9 ÷ 60 = 5.93 hours = 5 hours 56 minutes

(c) Sphere radius = 6 cm Volume = (4/3)π × 6³ = (4/3)π × 216 = 288π cm³

This is the volume that overflows.

(d)(i) At bottom r = 8, at top r = 12 r = 8 + d

(d)(ii) Volume = 3000 cm³

If d is the depth, then the radius at that depth is 8 + d Volume = ⅓πr²h = ⅓π(8 + d)² × d = 3000

⅓π(64 + 16d + d²) × d = 3000 π(64d + 16d² + d³)/3 = 3000 64d + 16d² + d³ = 9000/π = 2865

d³ + 16d² + 64d − 2865 = 0

Using trial: d = 10 gives 1000 + 1600 + 640 = 3240 (too big) d = 8 gives 512 + 1024 + 512 = 2048 (too small) d = 9 gives 729 + 1296 + 576 = 2601 (too small)

d ≈ 9.5 cm

Mark: 5/16

Examiner commentary: This answer demonstrates significant misconceptions. Part (a) uses an incorrect trapezium-style formula for a 3D solid – the student has confused averaging circular areas with volume of a frustum; this scores 0 marks. Part (b) is correct. Part (c) correctly calculates the sphere volume but doesn't explain the overflow concept properly – awarded 2/3 marks. Part (d)(i) shows fundamental misunderstanding: r = 8 + d would mean radius increases 1 cm for every 1 cm of height, which is incorrect; scores 0 marks. Part (d)(ii) attempts to use the cone formula incorrectly, treating the water shape as a single cone rather than using integration; the linear relationship from (d)(i) is wrong, so the subsequent work, though showing some problem-solving attempts, cannot earn marks. To improve, this student must: learn the correct frustum volume formula, understand rates of change (4 cm radius increase over 15 cm height), and recognize when integration is needed for variable cross-sections.