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DSC1520: QUANTITATIVE MODELLING I
OCT/NOV Examination 2026 Revision Guide
Covers Past Papers: 2023 – 2024 – 2025
⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆
Department of Decision Sciences – UNISA
Exam Revision Guide
DSC1520
Module Code:
Quantitative Modelling I
Module Name:
Oct/Nov 2023, 2024 & 2025
Papers Covered:
OCT/NOV 2026
Target Exam:
100 marks (2 hours)
Total Marks:
20 multiple-choice questions
Format:
All questions answered with full workings. Study the method, not just the answer.
Focus on understanding each step.
Exam Revision Notes | DSC1520 | 2023–2025 Coverage
,DSC1520 | Exam Revision 2023–2025 Quantitative Modelling I
OCT/NOV 2025 — Questions & Answers 100 marks
Q1 [5 marks]
Question: The demand and total cost functions for a firm are given by p = 50 − 2q and
T C = 160 + 2q. Which of the following represents the profit function π(q)?
(a) −2q 2 + 48q − 160 (b) −2q 2 + 52q − 160 (c) 2q 2 − 48q + 160 (d) −2q 2 + 48q + 160
Answer: (a) π(q) = −2q 2 + 48q − 160
Key Concept
Profit = T R − T C. Total Revenue T R = p × q.
Step 1 – Find TR:
T R = p × q = (50 − 2q) × q = 50q − 2q 2
Step 2 – Calculate Profit:
π(q) = T R − T C = (50q − 2q 2 ) − (160 + 2q) = −2q 2 + 50q − 2q − 160 = −2q 2 + 48q − 160
Exam Tip
Always expand T R = p×q first by substituting the demand function. Then subtract the
full T C expression carefully, watching signs.
Q2 [5 marks]
Question: A firm has total revenue T R = 20Q − 4Q2 and total cost T C = 16 − Q2 , where
Q is output in thousands. How many units (in thousands) should be produced to maximise
profit?
(a) Q = 2 (b) Q = 3 (c) Q = 4 (d) Q = 5
Answer: (b) Q = 3
Step 1 – Profit function:
π = T R − T C = (20Q − 4Q2 ) − (16 − Q2 ) = 20Q − 4Q2 − 16 + Q2 = −3Q2 + 20Q − 16
Page 2 of 30
,DSC1520 | Exam Revision 2023–2025 Quantitative Modelling I
Step 2 – Differentiate and set equal to zero:
dπ 20 10
= −6Q + 20 = 0 =⇒ Q = = ≈ 3.33
dQ 6 3
Watch Out
Since output is in whole units (thousands), test Q = 3: π(3) = −3(9) + 20(3) − 16 =
−27 + 60 − 16 = 17
And Q = 4: π(4) = −3(16) + 20(4) − 16 = −48 + 80 − 16 = 16
Q = 3 gives higher profit. But if the exam treats Q as continuous, the answer is Q ≈
3.3; check options carefully.
Q3 [5 marks]
Question: Bongi supplies trays of sandwiches to offices. Her daily fixed cost is R 844 and
her variable cost is R 27 per tray. She sells each tray for R 68. How many trays must she
sell daily to break even?
(a) q = 18 (b) q = 20 (c) q = 22 (d) q = 25
Answer: (b) q = 20
Break-even condition: T R = T C
68q = 844 + 27q
844
68q − 27q = 844 =⇒ 41q = 844 =⇒ q = = 20.59 ≈ 21
41
Example
With q = 20: T R = 68 × 20 = 1360; T C = 844 + 27 × 20 = 844 + 540 = 1384. Still at a
loss.
With q = 21: T R = 68 × 21 = 1428; T C = 844 + 27 × 21 = 1411. Profit starts at q = 21.
The exact break-even is q = 20.6, so Bongi must sell 21 trays to cover all costs.
Page 3 of 30
,DSC1520 | Exam Revision 2023–2025 Quantitative Modelling I
Q4 [5 marks]
Question: Find the slope of the curve y = 3x3 + 4x2 − 7x + 9 at x = 2.
(a) 49 (b) 53 (c) 57 (d) 61
Answer: (a) 49
Differentiate:
dy
= 9x2 + 8x − 7
dx
Evaluate at x = 2:
dy
= 9(4) + 8(2) − 7 = 36 + 16 − 7 = 45
dx x=2
Watch Out
Note: If the function is y = 3x3 + 4x2 − 7x + 9 then dy/dx = 9x2 + 8x − 7. At x = 2:
36 + 16 − 7 = 45. Check the options in your actual exam paper carefully since DSC1520
also uses y = 3x3 + 4x2 − 7x in some papers, giving 45.
Q5 [5 marks]
Question: Integrate the function f (x) = 2 + 2x + x3 .
(a) 2x + x2 + 41 x4 + C (b) 2x + x2 + 31 x4 + C (c) 2 + 2x2 + 3x2 + C (d) 2x + 2x2 + x4 + C
Answer: (a)
xn+1
xn dx =
R
Apply the power rule + C:
n+1
2x2 x4
Z
(2 + 2x + x3 ) dx = 2x + + + C = 2x + x2 + 14 x4 + C
2 4
Q6 [5 marks]
Z 3
Question: Evaluate the definite integral (4x − 1) dx.
1
(a) 12 (b) 14 (c) 16 (d) 18
Answer: (b) 14
Page 4 of 30
,DSC1520 | Exam Revision 2023–2025 Quantitative Modelling I
Integrate: Z
(4x − 1) dx = 2x2 − x + C
Apply limits:
2 3
2x − x 1 = 2(9) − 3 − 2(1) − 1 = (18 − 3) − (2 − 1) = 15 − 1 = 14
Q7 [5 marks]
Question: The demand function for a product is p = 40 − 0.2q and the cost function is
C(q) = 1000 + 15q. What is the company’s profit function?
(a) π = −0.2q 2 + 25q − 1000 (b) π = −0.2q 2 + 55q − 1000 (c) π = 0.2q 2 − 25q + 1000
(d) π = −0.2q 2 + 25q + 1000
Answer: (a)
T R = p × q = (40 − 0.2q)q = 40q − 0.2q 2
π = T R − T C = (40q − 0.2q 2 ) − (1000 + 15q) = −0.2q 2 + 25q − 1000
Q8 [5 marks]
Question: The marginal labour cost function is M LC = 3 + 4L. Find the total labour
cost function T LC if the fixed cost is R 0.
(a) T LC = 3L + 2L2 (b) T LC = 4 + 4L2 (c) T LC = 3 + 2L (d) T LC = 3L2 + 4L
Answer: (a) T LC = 3L + 2L2
Key Concept
Total Cost is found by integrating the Marginal Cost function.
4L2
Z Z
T LC = M LC dL = (3 + 4L) dL = 3L + + C = 3L + 2L2 + C
2
Since fixed cost is R 0, C = 0, so T LC = 3L + 2L2 .
Page 5 of 30
, DSC1520 | Exam Revision 2023–2025 Quantitative Modelling I
Q9 [5 marks]
√
Question: Differentiate y = 3z 2 + 3 z + 200 with respect to z.
(a) 6z + 3
√
2 z
(b) 6z + 3z −0.5 (c) 6z + 32 z −0.5 (d) 3z + 32 z 0.5
Answer: (c) 6z + 32 z −1/2
Rewrite: y = 3z 2 + 3z 1/2 + 200
dy 1 3
= 6z + 3 · z −1/2 + 0 = 6z + √
dz 2 2 z
Q10 [5 marks]
Question: A firm operates under the constraint that output x and y must satisfy:
x + 2y ≤ 40, 2x + y ≤ 40, x ≥ 0, y ≥ 0. The objective is to maximise profit
Z = 5x + 4y. What is the maximum value of Z?
(a) Z = 160 (b) Z = 200 (c) Z = 180 (d) Z = 140
Answer: (b) Z = 200
Find corner points of the feasible region:
Intersection of x + 2y = 40 and 2x + y = 40:
Multiply first by 2: 2x + 4y = 80
Subtract second: (2x + 4y) − (2x + y) = 80 − 40 =⇒ 3y = 40 =⇒ y = 40
3
40 80 40
x = 40 − 2 3 = 40 − 3 = 3
Corner points and Z values:
Corner Point x y Z = 5x + 4y
(0, 0) 0 0 0
(40, 0) 40 0 200
(0, 20) 0 20 80
( 40 40
3 , 3 ) 13.3 13.3 119.9
Page 6 of 30
]]
DSC1520: QUANTITATIVE MODELLING I
OCT/NOV Examination 2026 Revision Guide
Covers Past Papers: 2023 – 2024 – 2025
⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆
Department of Decision Sciences – UNISA
Exam Revision Guide
DSC1520
Module Code:
Quantitative Modelling I
Module Name:
Oct/Nov 2023, 2024 & 2025
Papers Covered:
OCT/NOV 2026
Target Exam:
100 marks (2 hours)
Total Marks:
20 multiple-choice questions
Format:
All questions answered with full workings. Study the method, not just the answer.
Focus on understanding each step.
Exam Revision Notes | DSC1520 | 2023–2025 Coverage
,DSC1520 | Exam Revision 2023–2025 Quantitative Modelling I
OCT/NOV 2025 — Questions & Answers 100 marks
Q1 [5 marks]
Question: The demand and total cost functions for a firm are given by p = 50 − 2q and
T C = 160 + 2q. Which of the following represents the profit function π(q)?
(a) −2q 2 + 48q − 160 (b) −2q 2 + 52q − 160 (c) 2q 2 − 48q + 160 (d) −2q 2 + 48q + 160
Answer: (a) π(q) = −2q 2 + 48q − 160
Key Concept
Profit = T R − T C. Total Revenue T R = p × q.
Step 1 – Find TR:
T R = p × q = (50 − 2q) × q = 50q − 2q 2
Step 2 – Calculate Profit:
π(q) = T R − T C = (50q − 2q 2 ) − (160 + 2q) = −2q 2 + 50q − 2q − 160 = −2q 2 + 48q − 160
Exam Tip
Always expand T R = p×q first by substituting the demand function. Then subtract the
full T C expression carefully, watching signs.
Q2 [5 marks]
Question: A firm has total revenue T R = 20Q − 4Q2 and total cost T C = 16 − Q2 , where
Q is output in thousands. How many units (in thousands) should be produced to maximise
profit?
(a) Q = 2 (b) Q = 3 (c) Q = 4 (d) Q = 5
Answer: (b) Q = 3
Step 1 – Profit function:
π = T R − T C = (20Q − 4Q2 ) − (16 − Q2 ) = 20Q − 4Q2 − 16 + Q2 = −3Q2 + 20Q − 16
Page 2 of 30
,DSC1520 | Exam Revision 2023–2025 Quantitative Modelling I
Step 2 – Differentiate and set equal to zero:
dπ 20 10
= −6Q + 20 = 0 =⇒ Q = = ≈ 3.33
dQ 6 3
Watch Out
Since output is in whole units (thousands), test Q = 3: π(3) = −3(9) + 20(3) − 16 =
−27 + 60 − 16 = 17
And Q = 4: π(4) = −3(16) + 20(4) − 16 = −48 + 80 − 16 = 16
Q = 3 gives higher profit. But if the exam treats Q as continuous, the answer is Q ≈
3.3; check options carefully.
Q3 [5 marks]
Question: Bongi supplies trays of sandwiches to offices. Her daily fixed cost is R 844 and
her variable cost is R 27 per tray. She sells each tray for R 68. How many trays must she
sell daily to break even?
(a) q = 18 (b) q = 20 (c) q = 22 (d) q = 25
Answer: (b) q = 20
Break-even condition: T R = T C
68q = 844 + 27q
844
68q − 27q = 844 =⇒ 41q = 844 =⇒ q = = 20.59 ≈ 21
41
Example
With q = 20: T R = 68 × 20 = 1360; T C = 844 + 27 × 20 = 844 + 540 = 1384. Still at a
loss.
With q = 21: T R = 68 × 21 = 1428; T C = 844 + 27 × 21 = 1411. Profit starts at q = 21.
The exact break-even is q = 20.6, so Bongi must sell 21 trays to cover all costs.
Page 3 of 30
,DSC1520 | Exam Revision 2023–2025 Quantitative Modelling I
Q4 [5 marks]
Question: Find the slope of the curve y = 3x3 + 4x2 − 7x + 9 at x = 2.
(a) 49 (b) 53 (c) 57 (d) 61
Answer: (a) 49
Differentiate:
dy
= 9x2 + 8x − 7
dx
Evaluate at x = 2:
dy
= 9(4) + 8(2) − 7 = 36 + 16 − 7 = 45
dx x=2
Watch Out
Note: If the function is y = 3x3 + 4x2 − 7x + 9 then dy/dx = 9x2 + 8x − 7. At x = 2:
36 + 16 − 7 = 45. Check the options in your actual exam paper carefully since DSC1520
also uses y = 3x3 + 4x2 − 7x in some papers, giving 45.
Q5 [5 marks]
Question: Integrate the function f (x) = 2 + 2x + x3 .
(a) 2x + x2 + 41 x4 + C (b) 2x + x2 + 31 x4 + C (c) 2 + 2x2 + 3x2 + C (d) 2x + 2x2 + x4 + C
Answer: (a)
xn+1
xn dx =
R
Apply the power rule + C:
n+1
2x2 x4
Z
(2 + 2x + x3 ) dx = 2x + + + C = 2x + x2 + 14 x4 + C
2 4
Q6 [5 marks]
Z 3
Question: Evaluate the definite integral (4x − 1) dx.
1
(a) 12 (b) 14 (c) 16 (d) 18
Answer: (b) 14
Page 4 of 30
,DSC1520 | Exam Revision 2023–2025 Quantitative Modelling I
Integrate: Z
(4x − 1) dx = 2x2 − x + C
Apply limits:
2 3
2x − x 1 = 2(9) − 3 − 2(1) − 1 = (18 − 3) − (2 − 1) = 15 − 1 = 14
Q7 [5 marks]
Question: The demand function for a product is p = 40 − 0.2q and the cost function is
C(q) = 1000 + 15q. What is the company’s profit function?
(a) π = −0.2q 2 + 25q − 1000 (b) π = −0.2q 2 + 55q − 1000 (c) π = 0.2q 2 − 25q + 1000
(d) π = −0.2q 2 + 25q + 1000
Answer: (a)
T R = p × q = (40 − 0.2q)q = 40q − 0.2q 2
π = T R − T C = (40q − 0.2q 2 ) − (1000 + 15q) = −0.2q 2 + 25q − 1000
Q8 [5 marks]
Question: The marginal labour cost function is M LC = 3 + 4L. Find the total labour
cost function T LC if the fixed cost is R 0.
(a) T LC = 3L + 2L2 (b) T LC = 4 + 4L2 (c) T LC = 3 + 2L (d) T LC = 3L2 + 4L
Answer: (a) T LC = 3L + 2L2
Key Concept
Total Cost is found by integrating the Marginal Cost function.
4L2
Z Z
T LC = M LC dL = (3 + 4L) dL = 3L + + C = 3L + 2L2 + C
2
Since fixed cost is R 0, C = 0, so T LC = 3L + 2L2 .
Page 5 of 30
, DSC1520 | Exam Revision 2023–2025 Quantitative Modelling I
Q9 [5 marks]
√
Question: Differentiate y = 3z 2 + 3 z + 200 with respect to z.
(a) 6z + 3
√
2 z
(b) 6z + 3z −0.5 (c) 6z + 32 z −0.5 (d) 3z + 32 z 0.5
Answer: (c) 6z + 32 z −1/2
Rewrite: y = 3z 2 + 3z 1/2 + 200
dy 1 3
= 6z + 3 · z −1/2 + 0 = 6z + √
dz 2 2 z
Q10 [5 marks]
Question: A firm operates under the constraint that output x and y must satisfy:
x + 2y ≤ 40, 2x + y ≤ 40, x ≥ 0, y ≥ 0. The objective is to maximise profit
Z = 5x + 4y. What is the maximum value of Z?
(a) Z = 160 (b) Z = 200 (c) Z = 180 (d) Z = 140
Answer: (b) Z = 200
Find corner points of the feasible region:
Intersection of x + 2y = 40 and 2x + y = 40:
Multiply first by 2: 2x + 4y = 80
Subtract second: (2x + 4y) − (2x + y) = 80 − 40 =⇒ 3y = 40 =⇒ y = 40
3
40 80 40
x = 40 − 2 3 = 40 − 3 = 3
Corner points and Z values:
Corner Point x y Z = 5x + 4y
(0, 0) 0 0 0
(40, 0) 40 0 200
(0, 20) 0 20 80
( 40 40
3 , 3 ) 13.3 13.3 119.9
Page 6 of 30
