Mark Scheme

1. (a) (3^4 = 81) B1 (2^5 = 32) B1 (may be implied) (81 - 32 = 49) A1

Answer: 49

(2 marks)

(b) Division method shown or factor tree M1 (504 = 2 \times 2 \times 2 \times 3 \times 3 \times 7) A1 (= 2^3 \times 3^2 \times 7) A1

Accept: any correct working showing systematic factorisation

(3 marks)

(c) Prime factors of 378: (2 \times 3^3 \times 7) M1 (may be implied) HCF = (2 \times 3^2 \times 7 = 126) A1

Accept: correct answer from any valid method (e.g. listing factors)

(2 marks)

2. (a) (42 + 3x + 2x + 18 = 180) M1 (5x + 60 = 180) A1

Accept: any correct equation showing angles in a triangle sum to 180°

(2 marks)

(b) (5x = 120) M1 (x = 24) A1

(2 marks)

(c) (3 \times 24 = 72°) A1ft

Follow through their value of x

(1 mark)

3. (a) Pattern 4 has (4^2 = 16) grey tiles M1 (may be implied) White tiles = 20 A1

Accept: correct answer from extending pattern

(2 marks)

(b) (n^2) B1

Accept: n squared or n²

(1 mark)

(c) Total tiles = grey + white (= n^2 + (4n + 4)) M1 (= n^2 + 4n + 4) A1 (= (n + 2)^2) A1

Accept: any correct simplified form

(3 marks)

(d) ((n + 2)^2 = 80) M1 (n + 2 = \pm\sqrt{80} = \pm 8.944...) M1 (n = 6.944...) or (n = -10.944...) A1 "n must be a positive integer, so Pattern 80 is impossible" E1

Accept: equivalent argument showing 80 is not a perfect square

(3 marks)

4. Multiply first equation by 3: (9x + 6y = 48) M1 Multiply second equation by 2: (10x - 6y = 18) M1 Add equations: (19x = 66), so (x = \frac{66}{19}) A1 Substitute to find (y = \frac{49}{19}) or (y = 2\frac{11}{19}) A1ft

Accept: any valid method (substitution or elimination) Follow through their x value for y

(4 marks)

5. (a) P(green) = (1 - 0.35 - 0.42 = 0.23) M1 Number of green = (0.23 \times 200) M1 (= 46) A1

(3 marks)

(b) Number of red counters = (0.35 \times 200 = 70) M1 (may be implied) P(both red) = (\frac{70}{200} \times \frac{69}{199}) M1 (= \frac{4830}{39800} = \frac{483}{3980}) A1

Accept: any correct equivalent fraction

(3 marks)

6. Gradient of L is 3 B1 (may be implied) Gradient of M is (-\frac{1}{3}) M1 (2 = -\frac{1}{3}(6) + c) M1 (2 = -2 + c), so (c = 4) A1

Answer: (y = -\frac{1}{3}x + 4)

Accept: equivalent forms such as (3y = -x + 12) or (x + 3y = 12) if asked to convert

(4 marks)

7. Expand first two brackets: ((2x + 3)(x - 5) = 2x^2 - 10x + 3x - 15 = 2x^2 - 7x - 15) M1A1 Multiply by third bracket: ((2x^2 - 7x - 15)(3x + 1)) M1 (= 6x^3 + 2x^2 - 21x^2 - 7x - 45x - 15) (= 6x^3 - 19x^2 - 52x - 15) A1

Award M1 for expanding first pair of brackets A1 for correct intermediate result M1 for multiplying by third bracket A1 for correct final answer

(4 marks)

8. (a) (y = (x^2 - 4x) + 1) (= (x - 2)^2 - 4 + 1) M1A1 (= (x - 2)^2 - 3) A1

Award M1 for method of completing the square A1 for correct squared term A1 for correct constant

(3 marks)

(b) Minimum point is (2, -3) B1ft B1ft

Award B1 for x = 2 B1 for y = -3 Follow through from their completed square form

(2 marks)

(c) B1 for correct parabola shape (U-shaped) B1 for minimum point marked at (2, -3) B1 for y-intercept at (0, 1) and attempt at x-intercepts shown (need not be accurately calculated)

Accept: x-intercepts approximately shown even if not calculated

(3 marks)

9. (a) Using Pythagoras: (x^2 + (2x + 3)^2 = 13^2) M1 (x^2 + 4x^2 + 12x + 9 = 169) A1 (5x^2 + 12x + 9 - 169 = 0) (5x^2 + 12x - 160 = 0) A1 QED

(3 marks)

(b) Using quadratic formula: (x = \frac{-12 \pm \sqrt{144 + 3200}}{10}) M1 (= \frac{-12 \pm \sqrt{3344}}{10}) M1 (= \frac{-12 \pm 57.827...}{10}) A1 (x = 4.58) (3 s.f.) A1

Reject negative solution as x > 0

(4 marks)

(c) Width = 4.5827... m Length = 2(4.5827...) + 3 = 12.165... m M1 Area = 4.5827... × 12.165... = 55.74... m² M1 Number of flowers = (\frac{55.74...}{0.4}) = 139.36... M1 Maximum = 139 flowers A1

Award marks for method; must round down for final answer

(4 marks)

(d) Cost of flowers = 139 × £3.50 = £486.50 M1 Cost of path = 13 × £42 = £546 M1 Total = £486.50 + £546 = £1032.50 A1ft

(3 marks)

(e) Total exceeds £1000 B1 Therefore the customer will receive the discount B1

(2 marks)

10. (a) Volume = (\pi r^2 h = 500) M1 Therefore (h = \frac{500}{\pi r^2}) A1 Surface area = (2\pi r^2 + 2\pi rh) M1 (= 2\pi r^2 + 2\pi r \times \frac{500}{\pi r^2}) (= 2\pi r^2 + \frac{1000}{r}) A1 QED

(4 marks)

(b) (r \approx 4.3) cm B1

Accept: values in range 4.2–4.4 cm

(1 mark)

(c) (S \approx 350) cm² B1

Accept: values in range 345–355 cm²

(1 mark)

(d) Using radius of 3 cm would mean the bottle does not have minimum surface area B1 A radius of approximately 4.3 cm gives minimum material/surface area B1

Accept: equivalent explanations showing understanding that 3 cm is not at the minimum point

(2 marks)

(e) Minimum surface area ≈ 350 cm² (from graph) B1ft Manufacturing cost = 350 × £0.02 = £7.00 M1 Profit = £2.50 - £7.00 = -£4.50 M1 Or: "The company makes a loss of £4.50 per bottle" A1

Accept: recognition that profit is negative (company makes a loss)

(4 marks)

(f) Minimum surface area ≈ 350 cm² B1 15% reduction = 0.85 × 350 = 297.5 cm² M1

For a cylinder with volume 500 cm³: Minimum surface area occurs when (\frac{dS}{dr} = 0) This gives (4\pi r - \frac{1000}{r^2} = 0) M1 (r^3 = \frac{1000}{4\pi} = 79.577...) (r = 4.301...) cm A1 Minimum S = (2\pi(4.301...)^2 + \frac{1000}{4.301...}) (= 116.18... + 232.56... = 348.7) cm² M1

A 15% reduction would give 296.4 cm²

Conclusion: The rival's claim is not possible E1 The minimum surface area for a cylinder with this volume is approximately 349 cm². A 15% reduction would require surface area of 296 cm², which is below the theoretical minimum for any cylindrical design with this volume.

Award E1 for clear conclusion with justification

(6 marks)

11. (a) When t = 0: (P = 200 \times 2^0 = 200 \times 1 = 200) B1

(1 mark)

(b) When t = 5: (P = 200 \times 2^5) M1 (= 200 \times 32 = 6400) A1

(2 marks)

(c) (200 \times 2^t > 50000) M1 (2^t > 250) M1 (t > \log_2(250)) or (t > \frac{\log 250}{\log 2}) M1 (t > 7.966...) Therefore 8 hours (or t = 8) A1

Accept: trial and improvement showing (t = 7) gives 25600 and (t = 8) gives 51200

(4 marks)

(d) When t = 4: (P = 200 \times 2^4 = 200 \times 16) M1 (= 3200) A1

(2 marks)

(e) Actual population = 3100 Percentage error = (\frac{|3200 - 3100|}{3100} \times 100%) M1 (= \frac{100}{3100} \times 100% = 3.23%) (or 3.2%) A1

(2 marks)

(f) Level 3 (6–7 marks): • Checks at least one other time point with correct calculation • Correctly determines whether scientist's claim is valid at all measured points • Discusses at least two relevant limitations (e.g., model assumes perfect exponential growth, doesn't account for limiting factors like nutrients/space, actual data shows deviation from pure exponential pattern) • Makes clear judgement about reliability beyond t = 8 with justification • Uses mathematical evidence to support evaluation • Communication is clear and logical

Level 2 (3–5 marks): • Checks one other time point with some calculation shown • Makes some comment on validity of claim • Discusses at least one limitation in general terms • Some comment on reliability beyond t = 8 • Some use of mathematical evidence • Communication may lack clarity in places

Level 1 (1–2 marks): • Limited checking of other time points • General comments about accuracy • No substantial discussion of limitations • Little or no mathematical evidence • Poor communication

Indicative content:

Check t = 2: Model gives 200 × 4 = 800. Actual = 850. Error = (\frac{50}{850} \times 100% = 5.88%) ✓ Less than 10%

Check t = 6: Model gives 200 × 64 = 12800. Actual = 11900. Error = (\frac{900}{11900} \times 100% = 7.56%) ✓ Less than 10%

Check t = 8: Model gives 200 × 256 = 51200. Actual = 48200. Error = (\frac{3000}{48200} \times 100% = 6.22%) ✓ Less than 10%

The claim is valid for all measured time points.

However, limitations include:

• Real bacterial populations don't grow exponentially forever (nutrients become limited, waste builds up)
• The percentage errors are increasing with time (3.2%, 5.9%, 7.6%), suggesting the model is becoming less accurate
• Beyond t = 8, the model would likely significantly overestimate population
• The model doesn't account for the lag phase or stationary phase of bacterial growth

Conclusion: The model is adequate for short-term predictions up to 8 hours but would not be reliable for longer periods.

(7 marks)

Sample Answers with Examiner Commentary

Question 10(f) — Sample Answers

Grade 9 answer

First, I need to verify the minimum surface area for a cylindrical bottle with volume 500 cm³.

For a cylinder: V = πr²h = 500, so h = 500/(πr²)

Surface area S = 2πr² + 2πrh = 2πr² + 2πr × 500/(πr²) = 2πr² + 1000/r

To minimise, I differentiate with respect to r: dS/dr = 4πr - 1000/r²

Setting this equal to zero: 4πr = 1000/r² 4πr³ = 1000 r³ = 1000/(4π) = 79.577 r = 4.301 cm

Substituting back: S = 2π(4.301)² + 1000/4.301 S = 116.18 + 232.56 = 348.74 cm²

This confirms the graph's minimum of approximately 350 cm².

If the rival's bottle uses 15% less material: 0.85 × 348.74 = 296.43 cm²

This is impossible for a cylindrical bottle with volume 500 cm³ because 348.74 cm² is the absolute minimum surface area achievable. The rival company's claim would require a surface area below this theoretical minimum, which violates the mathematical constraints. The only way to use less material would be to use a non-cylindrical shape, such as a sphere, which has the minimum surface area for a given volume.

Mark: 6/6

Examiner commentary: This is an exemplary response demonstrating clear mathematical reasoning. The candidate correctly uses calculus to verify the minimum surface area (though reading from the graph would suffice), accurately calculates the 15% reduction, and provides a definitive conclusion with full justification. The additional insight about alternative shapes shows depth of understanding. All marking points achieved with excellent mathematical communication.

Grade 6 answer

From the graph, minimum surface area = 350 cm²

15% less material = 350 - (0.15 × 350) = 350 - 52.5 = 297.5 cm²

The rival company claims their bottle uses 297.5 cm².

Looking at the graph, 350 cm² is the minimum point. This is the smallest surface area possible for a cylinder with volume 500 cm³. You can't go below the minimum point, so the rival company's claim is not possible.

The claim is false because 297.5 cm² is less than the minimum of 350 cm² shown on the graph. The graph shows that any radius other than about 4.3 cm gives a higher surface area, so you can't get below 350 cm².

Mark: 3/6

Examiner commentary: This response demonstrates sound understanding of the concept that values below a minimum are impossible. The candidate correctly reads from the graph and calculates the 15% reduction (M1, B1, M1). However, the response lacks mathematical rigour—there's no verification of the minimum using the formula or calculus, and the explanation relies entirely on graph-reading. The conclusion is correct but underdeveloped. To reach higher marks, the candidate needed to verify the minimum value mathematically and provide more detailed justification. The response would benefit from explaining why this is the theoretical minimum for any cylinder with this volume.

Grade 3 answer

The graph shows minimum surface area is about 350 cm².

15% of 350 = 52.5

So rival company's bottle uses 52.5 cm² less material.

52.5 cm² is much less than 350 cm² so yes, the rival's claim is possible. They use less material.

15% is not a big reduction so it seems reasonable that they could make the bottle with less material if they are more efficient in their manufacturing process.

Mark: 1/6

Examiner commentary: This response shows a fundamental misunderstanding of the question. The candidate has calculated the reduction amount (52.5 cm²) but failed to subtract it from 350 to find the claimed surface area of 297.5 cm². More critically, the student hasn't recognised that 350 cm² represents a mathematical minimum—the lowest possible value for any cylindrical design with this volume. The suggestion that "efficient manufacturing" could reduce surface area below the mathematical minimum shows confusion between practical manufacturing efficiency and geometric constraints. The candidate needed to understand that the minimum point on the graph represents a theoretical limit that cannot be exceeded regardless of manufacturing methods. This answer demonstrates weak graphical interpretation and limited understanding of optimization concepts.

Question 11(f) — Sample Answers

Grade 9 answer

First, I'll verify the scientist's claim by checking percentage errors at all time points.

t = 0: Model = 200, Actual = 200, Error = 0% ✓

t = 2: Model = 200 × 2² = 800, Actual = 850 Error = |800 - 850|/850 × 100% = 50/850 × 100% = 5.88% ✓

t = 4: Model = 200 × 2⁴ = 3200, Actual = 3100 Error = 100/3100 × 100% = 3.23% ✓

t = 6: Model = 200 × 2⁶ = 12800, Actual = 11900 Error = 900/11900 × 100% = 7.56% ✓

t = 8: Model = 200 × 2⁸ = 51200, Actual = 48200 Error = 3000/48200 × 100% = 6.22% ✓

The scientist's claim is technically correct—all errors are below 10%.

However, this claim is misleading for several reasons:

Limitations of the model:

1. The model assumes unlimited exponential growth, but bacterial populations face limiting factors (nutrients, space, waste accumulation).

2. Real bacterial growth follows distinct phases: lag phase, exponential phase, stationary phase, and death phase. This model only represents the exponential phase.

3. The percentage errors show an increasing trend (0%, 5.88%, 3.23%, 7.56%, 6.22%). Although they fluctuate, the overall pattern suggests the model is diverging from reality.

4. The actual data at t = 8 (48200) is significantly below the model prediction (51200), suggesting the population growth is already slowing.

Reliability beyond t = 8: The model would not be reliable for predictions beyond t = 8 hours because:

• The population is likely entering the stationary phase where growth rate decreases
• At t = 10, the model would predict 200 × 2¹⁰ = 204,800, but the growth rate is clearly decelerating
• The increasing absolute errors (even if percentage errors remain acceptable) indicate systematic deviation

Conclusion: While the model meets the scientist's 10% criterion for the measured period, it is only adequate for short-term predictions during the exponential growth phase (approximately 0-6 hours). For practical purposes requiring prediction beyond 8 hours, a more sophisticated model incorporating carrying capacity would be essential.

Mark: 7/7

Examiner commentary: This is an outstanding response demonstrating the analytical skills expected of a top-grade candidate. The student systematically checks all data points with correct calculations, accurately evaluates the claim's validity, identifies multiple relevant limitations with scientific understanding, and makes a nuanced judgement about reliability. The distinction between technical correctness and practical usefulness shows sophisticated thinking. The response is well-structured, uses precise mathematical and scientific terminology, and supports all statements with evidence. Full marks awarded.

Grade 6 answer

I'll check if the errors are less than 10% at other times.

t = 2: Model = 200 × 4 = 800, Actual = 850 Error = (850 - 800)/850 × 100% = 50/850 × 100% = 5.9% This is less than 10% ✓

t = 6: Model = 200 × 2⁶ = 12800, Actual = 11900 Error = (12800 - 11900)/11900 × 100% = 900/11900 × 100% = 7.6% This is less than 10% ✓

So the scientist's claim is correct because all the errors are less than 10%.

However, there are some problems with the model. Real bacteria don't grow exponentially forever because they run out of food and space. Also, at t = 8, the model predicts 51200 but the actual is only 48200, which is quite a big difference of 3000 bacteria.

The model probably wouldn't work well beyond t = 8 because the bacteria would start to run out of resources and stop growing so fast. The exponential model would keep predicting higher and higher populations but in reality the population would level off.

The model is okay for the times measured but not reliable for longer predictions.

Mark: 4/7

Examiner commentary: This response demonstrates solid understanding and competent mathematical skills. The candidate correctly checks two additional time points with accurate calculations (M1, A1) and reaches a valid conclusion about the claim (E1). They identify relevant limitations including resource depletion and show awareness that exponential models have limitations. However, the response lacks depth in several areas: no discussion of the trend in percentage errors, limited scientific terminology (e.g., doesn't mention growth phases), and the reliability discussion is somewhat superficial. To access the highest marks, the candidate needed to provide more detailed analysis of the limitations, discuss specific aspects of bacterial growth curves, and develop the argument about long-term reliability with more mathematical or scientific reasoning.

Grade 3 answer

Checking t = 6: Model = 200 × 6 = 1200 Actual = 11900 Error = 11900 - 1200 = 10700 Percentage = 10700/11900 = 89.9%

This is much more than 10% so the scientist is wrong.

The model is not accurate because the percentage error is nearly 90%. This is way too high to be useful.

The model wouldn't be reliable for predictions beyond t = 8 because the errors are too big. The scientist's claim is not valid.

Mark: 1/7

Examiner commentary: This response contains a critical mathematical error that undermines the entire answer. The student has misunderstood the model, calculating 200 × 6 instead of 200 × 2⁶, leading to a completely incorrect prediction and error calculation. This demonstrates weak understanding of exponential notation—a fundamental concept for this question. While the candidate attempts to evaluate the claim and comment on reliability, these conclusions are based on faulty mathematics. The response shows limited understanding of percentage error calculation (used raw difference rather than proportional difference) and lacks any discussion of the model's limitations beyond accuracy. To improve, the student must: (1) revise exponential notation and calculations, (2) understand percentage error as a proportional measure, and (3) develop awareness of model limitations beyond just numerical accuracy. Only 1 mark awarded for attempting to check another data point, despite the incorrect method.