Mark Scheme

Section A — Structured Questions (60 marks)

Question 1

(a) $4.56 \times 10^{-4}$ (1 mark)

(b)

• $\frac{3.6 \times 10^5}{1.2 \times 10^{-2}} = \frac{3.6}{1.2} \times 10^{5-(-2)}$ (1 mark)
• $= 3 \times 10^7$ (1 mark)

(c) (i)

• $4.5 \times 10^6$ ml $= 4.5 \times 10^6 \div 1000$ litres (1 mark)
• $= 4.5 \times 10^3$ litres (or 4500 litres) (1 mark)

(ii)

• Time $= \frac{4500}{750}$ (1 mark)
• $= 6$ minutes (1 mark)

(d)

• Population density $= \frac{2.87 \times 10^5}{4.30 \times 10^2}$ (1 mark)
• $= \frac{2.87}{4.30} \times 10^{5-2} = 0.6674... \times 10^3$ (1 mark)
• $= 667$ people per km² (to 3 s.f.) (1 mark)

Accept: 667.4 rounded to 667

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Question 2

(a)

• Fraction $= \frac{42}{150}$ (1 mark)
• $= \frac{7}{25}$ (simplest form) (1 mark)

(b)

• Percentage $= \frac{28}{150} \times 100$ (1 mark)
• $= 18.67%$ or $18.7%$ or $\frac{56}{3}%$ (1 mark)

Accept: 18.66%, 18.666...%, 19% (to 2 s.f.)

(c) (i)

• Angle $= \frac{35}{150} \times 360°$ (1 mark)
• $= 84°$ (1 mark)

(ii)

• Angle $= \frac{20}{150} \times 360°$ (1 mark)
• $= 48°$ (1 mark)

(d)

• Number choosing Jerk chicken OR Pelau $= 25 + 42 = 67$ (1 mark)
• Probability $= \frac{67}{150}$ (1 mark)

Accept: 0.447 or 0.45 (to 2 d.p.), 44.7% or 45%

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Question 3

(a)

• $AC^2 = AB^2 + BC^2 = 8^2 + 15^2$ (Pythagoras' theorem) (1 mark)
• $AC^2 = 64 + 225 = 289$, therefore $AC = 17$ cm (1 mark)

(b)

• $\tan(\angle BAC) = \frac{BC}{AB} = \frac{15}{8}$ (1 mark)
• $\angle BAC = \tan^{-1}(1.875) = 61.9°$ (to 1 d.p.) (1 mark)

Accept: 61.93° rounded to 61.9°, or 62.0°

(c)

• Area $= \frac{1}{2} \times AB \times BC$ or $\frac{1}{2} \times 8 \times 15$ (1 mark)
• $= 60$ cm² (1 mark)

(d)

• Area of triangle $= \frac{1}{2} \times AC \times BD = 60$ (from part c) (1 mark)
• $\frac{1}{2} \times 17 \times BD = 60$ (1 mark)
• $BD = \frac{120}{17} = 7.06$ cm (to 3 s.f.) (1 mark)

Accept: 7.059 rounded appropriately

(e)

• Volume $= $ Area of cross-section $\times$ length (1 mark)
• $= 60 \times 20 = 1200$ cm³ (1 mark)

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Question 4

(a) (i)

• $6x^2 - 9x = 3x(2x - 3)$ (2 marks)

Award 1 mark for partially correct factorisation (e.g., extracting $3$ or $x$ only)

(ii)

• $x^2 - 7x + 12 = (x - 3)(x - 4)$ (2 marks)

Award 1 mark for finding factors that multiply to give 12 and add to give -7

(b)

• $10x - 15 = 3x + 8$ (expanding brackets) (1 mark)
• $7x = 23$ (collecting terms) (1 mark)
• $x = \frac{23}{7}$ or $3\frac{2}{7}$ or $3.29$ (to 3 s.f.) (1 mark)

(c)

• $\frac{V}{\pi h} = r^2$ (dividing both sides by $\pi h$) (2 marks)
• $r = \sqrt{\frac{V}{\pi h}}$ or $r = \pm\sqrt{\frac{V}{\pi h}}$ (1 mark)

Award 2 marks if intermediate step shown clearly; 1 mark if jump made directly with minor error

(d) (i)

• $C = 5 + 2.5(12)$ (1 mark)
• $C = 5 + 30 = 35$ dollars (1 mark)

(ii)

• $67.50 = 5 + 2.5m$ (1 mark)
• $62.50 = 2.5m$ (1 mark)
• $m = 25$ miles (1 mark)

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Question 5

(a) Modal class: 31 – 40 (1 mark)

(b)

Mark	Frequency (f)	Midpoint (x)	fx
1–10	3	5.5	16.5
11–20	5	15.5	77.5
21–30	8	25.5	204
31–40	9	35.5	319.5
41–50	5	45.5	227.5
Total	30		845

• Correct midpoints calculated (1 mark)
• $\sum fx$ calculated correctly (1 mark)
• Mean $= \frac{\sum fx}{\sum f} = \frac{845}{30}$ (1 mark)
• $= 28.2$ (to 3 s.f.) or $28.17$ or $28$ (to 2 s.f.) (1 mark)

Accept equivalent working; award marks for correct process even if arithmetic error

(c)

• Number scoring more than 30 $= 9 + 5 = 14$ (1 mark)
• Probability $= \frac{14}{30} = \frac{7}{15}$ (1 mark)

Accept: 0.467, 0.47 (to 2 d.p.), 46.7%, 47%

(d)

• Correct scale and axes labelled (1 mark)
• Points plotted at correct coordinates (5.5, 3), (15.5, 5), (25.5, 8), (35.5, 9), (45.5, 5) (1 mark)
• Points joined with straight lines to form polygon (1 mark)

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Question 6

(a) (i)

• $T_1 = 3(1) - 5 = -2$ (1 mark)

(ii)

• $T_4 = 3(4) - 5 = 7$ (1 mark)

(b)

• $3n - 5 = 40$ (1 mark)
• $3n = 45$ (1 mark)
• $n = 15$ (the 15th term) (1 mark)

(c)

• $T_{n+1} = 3(n+1) - 5$ (1 mark)
• $= 3n + 3 - 5 = 3n - 2$ (1 mark)

Accept: $T_{n+1} = 3n - 2$ without working if correct

(d)

• $T_{n+1} - T_n = (3n - 2) - (3n - 5)$ (1 mark)
• $= 3n - 2 - 3n + 5 = 3$ (constant) (1 mark)

Accept alternative demonstrations using specific terms

(e) (i)

• First four terms: 7, 11, 15, 19 (2 marks)

Award 1 mark for 3 correct terms

(ii)

• $T_{20} = a + 19d = 7 + 19(4)$ (1 mark)
• $= 7 + 76 = 83$ (1 mark)

Accept use of formula $T_n = a + (n-1)d$

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Section B — Extended Response (40 marks)

Question 7

(a) (i)

• Area of plot $= 45 \times 30$ (1 mark)
• $= 1350$ m² (1 mark)

(ii)

• Area of house $= 15 \times 12 = 180$ m² (1 mark)

(iii)

• Area of garden $= \pi r^2 = \pi \times 5^2$ (1 mark)
• $= 25\pi$ m² (1 mark)

(b)

• Area for car park $= 1350 - 180 - 25\pi$ (1 mark)
• $= 1350 - 180 - 25(3.142)$ (1 mark)
• $= 1350 - 180 - 78.55 = 1091$ m² (to nearest m²) (1 mark)

Accept: 1091.45 rounded to 1091

(c) (i)

• Perimeter $= 2(45 + 30) = 2(75)$ (1 mark)
• $= 150$ m (1 mark)

(ii)

• Total cost $= 150 \times 45.50$ (1 mark)
• $= $6825$ or $6825.00$ (1 mark)

(d) (i)

• Length $= 15 + 2(0.5) = 16$ m, Width $= 12 + 2(0.5) = 13$ m (2 marks)

Award 1 mark for one dimension correct

(ii)

• Area $= 16 \times 13$ (1 mark)
• $= 208$ m² (1 mark)

(iii)

• Total cost $= 208 \times 125$ (1 mark)
• $= $26000$ (1 mark)

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Question 8

(a) (i)

• Correct axes drawn with appropriate scales and labels (1 mark)
• All 5 points plotted correctly (within ±1 mm tolerance): A(2.5, 180), B(4.0, 310), C(1.5, 120), D(5.5, 425), E(3.0, 240) (3 marks)

Award 2 marks for 4 points correct, 1 mark for 3 points correct

(ii)

• Positive correlation (1 mark)

Accept: "strong positive correlation" or "as advertising increases, tourist arrivals increase"

(b)

• Line of best fit drawn passing through or near the mean point and showing the trend with approximately equal numbers of points above and below (2 marks)

Award 1 mark for line showing general trend but poorly positioned

(c) (i)

• Reading from graph at $x = 4.5$: approximately 360–380 thousand tourists (2 marks)

Award 1 mark for reading attempt from graph; accept range 350–390 depending on line of best fit

(ii)

• Reading from graph at $y = 350$: approximately $4.2–4.6$ million dollars (2 marks)

Award 1 mark for reading attempt; accept range 4.0–4.8 depending on line of best fit

(d)

• $6.8$ million is outside the range of the data / extrapolation is unreliable (1 mark)
• The relationship may not continue in the same way beyond the data range / other factors may become important (1 mark)

Accept equivalent explanations about extrapolation risks

(e)

• Mean $= \frac{2.5 + 4.0 + 1.5 + 5.5 + 3.0}{5}$ (1 mark)
• $= \frac{16.5}{5} = 3.3$ million dollars (1 mark)

(f) (i)

• Increase $= 1.5 \times 1.6 = 2.4$ million or $1.5 + (1.5 \times 0.6)$ (1 mark)
• New budget $= 2.4$ million dollars (1 mark)

(ii)

• Reading from graph at $x = 2.4$: approximately 165–185 thousand tourists (2 marks)

Award 1 mark for correct method using line of best fit

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Question 9

(a)

• For oranges: $4x + 8y \leq 400$ (2 marks)
• For apples: $6x + 10y \leq 540$ (2 marks)

Award 1 mark each for correct inequality even if slight error in formation

Accept simplified forms: $x + 2y \leq 100$ and $3x + 5y \leq 270$

(b)

• $x \geq 30$ (1 mark)

(c)

• Correct axes with scale 2 cm to 10 baskets, axes labelled (1 mark)
• Line $4x + 8y = 400$ (or simplified form) drawn correctly (1 mark)
• Line $6x + 10y = 540$ (or simplified form) drawn correctly (1 mark)
• Line $x = 30$ drawn correctly (1 mark)
• Feasible region identified correctly (1 mark)
• Unwanted regions shaded appropriately (1 mark)

Key intercepts: For $x + 2y = 100$: $(0, 50)$ and $(100, 0)$; For $3x + 5y = 270$: $(0, 54)$ and $(90, 0)$

(d)

• Reading from graph where $x = 30$ intersects the feasible region boundary (1 mark)
• Maximum Deluxe baskets $= 32$ or $33$ (depending on accuracy of graph) (1 mark)

Accept range 31–34 depending on drawing accuracy

(e) (i)

• $P = 12x + 18y$ (2 marks)

Award 1 mark for correct terms but incorrect formation

(ii)

• Testing corner points of feasible region: (2 marks)
  • $(30, 32)$: $P = 12(30) + 18(32) = 360 + 576 = 936$
  • $(30, 35)$: $P = 12(30) + 18(35) = 360 + 630 = 990$
  • Other relevant corners depending on graph
• Identifying maximum occurs at vertex of feasible region (1 mark)
• $x = 30$, $y = 32$ to $35$ (depending on graph) (1 mark)
• Maximum profit $= $936$ to $990$ (1 mark)

Accept answers consistent with candidate's graph; award marks for method if arithmetic error

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Sample Answers with Examiner Commentary

Question 8 — Sample Answers

Grade I (Distinction) answer

(a) (i) [Candidate draws accurate scatter diagram with properly labelled axes: "Advertising spent ($ millions)" on horizontal axis from 0 to 6, "Tourist arrivals (thousands)" on vertical axis from 0 to 500, using specified scale of 2 cm per 1 million dollars and 2 cm per 100 thousand tourists. All five points plotted accurately at coordinates matching the table.]

(ii) The correlation is positive. As the amount spent on advertising increases, the number of tourist arrivals also increases.

(b) [Candidate draws a line of best fit passing through approximately (3.3, 255) which is near the mean point, with roughly equal numbers of points above and below the line, extending from (1, 100) to (6, 450).]

(c) (i) Reading from my line of best fit at $4.5 million, the number of tourist arrivals is approximately 370 thousand tourists.

(ii) Reading from my line of best fit at 350 thousand tourists, the advertising spent is approximately $4.3 million.

(d) It would not be reliable to use the graph for Territory F because $6.8 million is outside the data range (which only goes up to $5.5 million). This is called extrapolation. The linear relationship shown in the graph may not continue beyond the range of observed data, and other factors might affect tourist arrivals at higher spending levels.

(e) Mean advertising = $\frac{2.5 + 4.0 + 1.5 + 5.5 + 3.0}{5} = \frac{16.5}{5} = 3.3$ million dollars

(f) (i) Increase = 60% of $1.5 million = $0.6 \times 1.5 = 0.9$ million New budget = $1.5 + 0.9 = 2.4$ million dollars

(ii) Reading from my line of best fit at $2.4 million, Territory C can expect approximately 175 thousand tourist arrivals.

Mark: 23/23 marks

Examiner commentary: This is an exemplary response demonstrating all the key skills required at distinction level. The scatter diagram is accurately constructed with correct scale, properly labelled axes, and precisely plotted points. The line of best fit is appropriately positioned. The candidate provides clear readings from the graph with approximate values that are consistent with their drawn line. Part (d) shows sophisticated understanding of extrapolation and its limitations, using correct statistical terminology. All calculations are accurate and working is shown systematically. This candidate demonstrates the mathematical reasoning and communication skills expected at Grade I.

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Grade III (Pass) answer

(a) (i) [Candidate draws scatter diagram with axes labelled, scale mostly correct but one or two points plotted slightly inaccurately, approximately 2-3 mm off correct position.]

(ii) Positive correlation

(b) [Candidate draws a line of best fit but it does not pass through the optimal position—too many points on one side of the line.]

(c) (i) Looking at my graph when x = 4.5, y is about 365 thousand.

(ii) When y = 350, x is about $4.5 million.

(d) Because $6.8 is too high and not on the graph. It is outside the range so we cannot tell what will happen.

(e) Mean = $\frac{2.5 + 4.0 + 1.5 + 5.5 + 3.0}{5} = 3.3$ million dollars

(f) (i) New budget = $1.5 \times 1.6 = 2.4$ million dollars

(ii) From the graph at 2.4 million it is about 180 thousand tourists.

Mark: 14/23 marks

Examiner commentary: This response demonstrates secure understanding of the basic concepts but lacks the precision and depth required for higher marks. The scatter diagram has minor plotting inaccuracies (1 mark deducted). The line of best fit is drawn but poorly positioned (1 mark only awarded). The readings from the graph are reasonable attempts and earn credit. Part (d) identifies that extrapolation is problematic and mentions being outside the range (1 mark awarded) but the explanation lacks detail about why this matters statistically (second mark not awarded). Calculations in parts (e) and (f)(i) are correct. The response would benefit from more detailed explanations, greater accuracy in graph work, and use of precise mathematical terminology.

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Grade V (Near miss) answer

(a) (i) [Candidate draws axes but scale is inconsistent—horizontal axis marked 0, 2, 4, 6 but without proper 2 cm spacing. Three points plotted correctly, two points significantly off position.]

(ii) It is correlation.

(b) [Candidate attempts a line but it connects only two points rather than showing the trend of all data.]

(c) (i) 4.5 million = 350 thousand tourists [candidate appears to have confused axes or misread]

(ii) 350 thousand = about $3.5 million

(d) The number is too big for the graph.

(e) Mean = $2.5 + 4.0 + 1.5 + 5.5 + 3.0 = 16.5$ million dollars

(f) (i) 60% = 0.6 $1.5 + 0.6 = 2.1$ million dollars

(ii) About 170 thousand

Mark: 7/23 marks

Examiner commentary: This response shows some understanding but contains significant errors that prevent the candidate from accessing higher marks. In part (a)(i), the scale is incorrectly applied (0 marks for scaling, 1 mark for attempting a scatter diagram with some correct points). Part (a)(ii) does not specify the type of correlation (0 marks). The line of best fit does not represent the trend appropriately (0 marks). Part (c) shows confusion, possibly mixing up axes (0 marks awarded). Part (d) lacks sufficient explanation of extrapolation (1 mark for recognizing it's problematic). Part (e) shows the critical error of not dividing by 5—the candidate has found the sum, not the mean (0 marks). Part (f)(i) adds the percentage value rather than calculating 60% of the original amount (0 marks). To improve, this candidate needs to: carefully follow scale instructions; understand that correlation must be described (positive/negative); calculate means correctly; and show clear working for percentage increases.

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Question 9 — Sample Answers

Grade I (Distinction) answer

(a) The store has 400 oranges available. Standard baskets use 4 oranges and Deluxe baskets use 8 oranges, so: $4x + 8y \leq 400$

The store has 540 apples available. Standard baskets use 6 apples and Deluxe baskets use 10 apples, so: $6x + 10y \leq 540$

(b) The store wants to make at least 30 Standard baskets: $x \geq 30$

(c) [Candidate draws accurate graph with:

• Both axes labelled "Number of Standard baskets" (x-axis) and "Number of Deluxe baskets" (y-axis)
• Correct scale: 2 cm = 10 baskets on both axes, ranging from 0 to 100 on x-axis and 0 to 60 on y-axis
• First line: $4x + 8y = 400$ simplified to $x + 2y = 100$, plotted through (0, 50) and (100, 0)
• Second line: $6x + 10y = 540$ simplified to $3x + 5y = 270$, plotted through (0, 54) and (90, 0)
• Vertical line at $x = 30$
• Feasible region correctly identified as the area bounded by all three constraints and the axes
• Unwanted regions clearly shaded]

(d) When $x = 30$, I need to find where this intersects the boundary of the feasible region. Substituting $x = 30$ into $x + 2y = 100$: $30 + 2y = 100$ $2y = 70$ $y = 35$

Substituting $x = 30$ into $3x + 5y = 270$: $90 + 5y = 270$ $5y = 180$ $y = 36$

The constraint $x + 2y \leq 100$ is more restrictive, so the maximum number of Deluxe baskets is 35.

(e) (i) Profit = (profit per Standard basket × number of Standard baskets) + (profit per Deluxe basket × number of Deluxe baskets) $P = 12x + 18y$

(ii) To maximize profit, I need to test the corner points of the feasible region:

Corner points:

• $(30, 0)$: $P = 12(30) + 18(0) = 360$
• $(30, 35)$: $P = 12(30) + 18(35) = 360 + 630 = 990$
• $(90, 0)$: $P = 12(90) + 18(0) = 1080$ [but this violates $x \geq 30$ constraint being the main binding one; candidate checks if this point is in feasible region]

Actually, I need to find where the constraints intersect: When $x + 2y = 100$ and $3x + 5y = 270$: From first equation: $x = 100 - 2y$ Substituting: $3(100 - 2y) + 5y = 270$ $300 - 6y + 5y = 270$ $-y = -30$ $y = 30$, so $x = 100 - 60 = 40$

Corner points in feasible region:

• $(30, 0)$: $P = 360$
• $(30, 35)$: $P = 990$
• $(40, 30)$: $P = 12(40) + 18(30) = 480 + 540 = 1020$
• $(90, 0)$: $P = 1080$

The maximum profit is $1080 when $x = 90$ and $y = 0$.

Wait, let me check if $(90, 0)$ satisfies all constraints:

• $x \geq 30$: Yes, $90 \geq 30$ ✓
• $x + 2y \leq 100$: $90 + 0 = 90 \leq 100$ ✓
• $3x + 5y \leq 270$: $270 + 0 = 270 \leq 270$ ✓

Therefore, maximum profit occurs at $x = 90$, $y = 0$, giving $P = $1080$.

Mark: 20/20 marks

Examiner commentary: This is an outstanding response showing complete mastery of linear programming. The candidate correctly formulates both inequalities with clear explanation of the reasoning. The graph is accurately constructed with proper scale, correct lines, and appropriate shading. Part (d) demonstrates sophisticated algebraic verification of the graphical reading. In part (e), the candidate systematically identifies all corner points of the feasible region, tests each one, and correctly identifies the maximum. The self-correction and verification shown (checking that $(90, 0)$ satisfies all constraints) demonstrates the critical thinking expected at distinction level. The working is methodical, clearly communicated, and mathematically precise throughout.

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Grade III (Pass) answer

(a) For oranges: $4x + 8y \leq 400$ For apples: $6x + 10y \leq 540$

(b) $x \geq 30$

(c) [Candidate draws graph with correct axes and scale. The line $x + 2y = 100$ is drawn correctly through (0, 50) and (100, 0). The line $3x + 5y = 270$ is slightly inaccurate, plotted through (0, 54) and (85, 0) instead of (90, 0). The line $x = 30$ is correct. Feasible region is identified but shading is ambiguous—could be interpreted either way.]

(d) From my graph, when $x = 30$, the maximum $y$ is about 34 Deluxe baskets.

(e) (i) $P = 12x + 18y$

(ii) From the graph, the corner points are:

• $(30, 0)$: $P = 12(30) = 360$
• $(30, 34)$: $P = 12(30) + 18(34) = 360 + 612 = 972$
• $(85, 0)$: $P = 12(85) = 1020$

Maximum profit is $1020 when $x = 85$ and $y = 0$.

Mark: 13/20 marks

Examiner commentary: This is a competent response that demonstrates good understanding of the core concepts but contains inaccuracies that reduce the mark. The inequalities in part (a) are correct (4 marks). Part (b) is correct (1 mark). The graph shows understanding of the method but has plotting errors—one line is incorrectly positioned, losing 1 mark (5/6 marks awarded for graphing). Part (d) shows a reasonable reading from the graph (2 marks). The profit expression in part (e)(i) is correct (2 marks). However, in part (e)(ii), the candidate has only tested three corners and missed the actual maximum at $(90, 0)$ due to the graphing error. The method is sound—testing corners and identifying which gives maximum profit—so 3/5 marks are awarded. To improve, this candidate needs greater accuracy in plotting lines (checking intercepts algebraically) and should verify corner points more systematically.

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Grade V (Near miss) answer

(a) Oranges: $4x + 8y = 400$ Apples: $6x + 10y = 540$

(b) $x = 30$

(c) [Candidate draws axes with correct labels but inconsistent scale (spacing not uniform). Attempts to draw lines but $3x + 5y = 270$ is significantly incorrect, appearing to pass through (0, 50) and (60, 0). The line $x = 30$ is drawn. No shading shown, and feasible region not clearly identified.]

(d) Looking at the graph where $x = 30$, $y$ is about 30.

(e) (i) Profit $= 12x + 18y$

(ii) Testing some points:

• When $x = 30$ and $y = 30$: Profit $= 12 \times 30 + 18 \times 30 = 360 + 540 = 900$
• When $x = 50$ and $y = 20$: Profit $= 12 \times 50 + 18 \times 20 = 600 + 360 = 960$

Maximum profit is $960.

Mark: 6/20 marks

Examiner commentary: This response shows partial understanding but contains fundamental errors. In part (a), the candidate writes equalities instead of inequalities (0 marks—this is a critical error showing misunderstanding of the constraint concept). Part (b) incorrectly writes an equality rather than an inequality (0 marks). The graph attempt shows some understanding of the plotting process but is significantly inaccurate and lacks clear identification of the feasible region (2/6 marks—some credit for attempting to plot lines with labelled axes). Part (d) provides a reading, though it's unclear if this is within the actual feasible region (1 mark for attempting to read from graph). Part (e)(i) is correct (2 marks). Part (e)(ii) shows the candidate testing points but these appear to be randomly chosen rather than corner points of the feasible region, and no systematic approach is evident (1 mark for attempting to test some coordinates). To progress, this candidate must: understand the difference between equations and inequalities in constraint problems; improve graphing accuracy substantially; and learn the corner-point method for optimization systematically.