MTH 201 Business Mathematics - I Past Question Papers — TU BBA-F Semester 1

Tribhuvan University

Faculty of Management

Office of the Dean

Official Model Question Paper / Dean's Office Blueprint

Course: MTH 201 · Business Mathematics - I

Level: Bachelor of Business Administration in Finance (BBA-F) · Semester 1

Full Marks: 60

Time: 3 hrs.

Candidates are required to give their answers in their own words as far as practicable. The figures in the margin indicate full marks.

Group A

Brief Answer Questions. Attempt ALL questions.

[5 × 2 = 10]

[2]
If $A = \{1, 2, 3, 4\}$ and $B = \{3, 4, 5, 6\}$ , find the symmetric difference $A \Delta B$ .
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Answer: The symmetric difference is defined as:
$A \Delta B = (A - B) \cup (B - A)$
- $A - B = \{1, 2\}$
- $B - A = \{5, 6\}$
  $A \Delta B = \{1, 2, 5, 6\}$
[2]
Evaluate the limit: $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$ .

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Answer: Direct substitution yields the indeterminate form $\frac{0}{0}$ . Factoring the numerator:
$\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} = \lim_{x \to 3} (x + 3) = 3 + 3 = 6$
[2]
Find $\frac{dy}{dx}$ if $y = \ln(3x^2 + 5x + 1)$ .

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Answer: Using the chain rule:
$\frac{dy}{dx} = \frac{1}{3x^2 + 5x + 1} \cdot \frac{d}{dx}(3x^2 + 5x + 1) = \frac{6x + 5}{3x^2 + 5x + 1}$
[2]
Find the determinant of the matrix:
$A = \begin{pmatrix} 2 & 4 \\ -3 & 5 \end{pmatrix}$

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Answer:
$\det(A) = |A| = (2)(5) - (4)(-3) = 10 - (-12) = 10 + 12 = 22$
[2]
Evaluate the indefinite integral: $\int (4x^3 - 6x + 7) \, dx$ .

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Answer:
$\int (4x^3 - 6x + 7) \, dx = 4 \left(\frac{x^4}{4}\right) - 6 \left(\frac{x^2}{2}\right) + 7x + C = x^4 - 3x^2 + 7x + C$
where $C$ is the arbitrary constant of integration.

Group B

Descriptive Answer Questions. Attempt any THREE questions.

[3 × 10 = 30]

[10]
Solve the following system of linear equations using Cramer’s Rule:
$\begin{aligned} 2x - y + 3z &= 9 \\ x + y + z &= 6 \\ x - y + z &= 2 \end{aligned}$

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Solution: Cramer’s Rule

1. Coefficient Matrix Determinant ( $D$ )

$D = \begin{vmatrix} 2 & -1 & 3 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix}$
Expanding along the first row:
$D = 2 \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} - (-1) \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} + 3 \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix}$

$D = 2[1 - (-1)] + 1[1 - 1] + 3[-1 - 1] = 2(2) + 1(0) + 3(-2) = 4 + 0 - 6 = -2$
Since $D = -2 \neq 0$ , a unique solution exists.

2. Determinant $D_x$ (Replacing Column 1 with Constants):

$D_x = \begin{vmatrix} 9 & -1 & 3 \\ 6 & 1 & 1 \\ 2 & -1 & 1 \end{vmatrix}$
$D_x = 9(1 - (-1)) + 1(6 - 2) + 3(-6 - 2) = 9(2) + 1(4) + 3(-8) = 18 + 4 - 24 = -2$

3. Determinant $D_y$ (Replacing Column 2 with Constants):

$D_y = \begin{vmatrix} 2 & 9 & 3 \\ 1 & 6 & 1 \\ 1 & 2 & 1 \end{vmatrix}$
$D_y = 2(6 - 2) - 9(1 - 1) + 3(2 - 6) = 2(4) - 9(0) + 3(-4) = 8 - 0 - 12 = -4$

4. Determinant $D_z$ (Replacing Column 3 with Constants):

$D_z = \begin{vmatrix} 2 & -1 & 9 \\ 1 & 1 & 6 \\ 1 & -1 & 2 \end{vmatrix}$
$D_z = 2(2 - (-6)) + 1(2 - 6) + 9(-1 - 1) = 2(8) + 1(-4) + 9(-2) = 16 - 4 - 18 = -6$

5. Calculate Values:

$x = \frac{D_x}{D} = \frac{-2}{-2} = 1$
$y = \frac{D_y}{D} = \frac{-4}{-2} = 2$
$z = \frac{D_z}{D} = \frac{-6}{-2} = 3$
Verification: $x + y + z = 1 + 2 + 3 = 6$ (Satisfied). The solution is: $x = 1, y = 2, z = 3$ .
[10]
A manufacturing firm’s total revenue and total cost functions are given by:
$TR = 1,200Q - 2Q^2 \quad \text{and} \quad TC = Q^3 - 61.5Q^2 + 1,500Q + 5,000$

Required: a) Formulate the profit function $\pi(Q)$ . b) Find the output level $Q$ that maximizes profit. c) Verify whether the profit is maximized at this output using the second-order derivative test. d) Compute the maximum profit.
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Solution: Profit Maximization Analysis

a) Profit Function $\pi(Q)$

$\pi(Q) = TR - TC = (1,200Q - 2Q^2) - (Q^3 - 61.5Q^2 + 1,500Q + 5,000)$
$\pi(Q) = -Q^3 + 59.5Q^2 - 300Q - 5,000$

b) First-Order Condition for Profit Maximization

Set the first derivative with respect to $Q$ equal to zero:
$\frac{d\pi}{dQ} = -3Q^2 + 119Q - 300 = 0$

$3Q^2 - 119Q + 300 = 0$

Using the quadratic formula: $Q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$Q = \frac{119 \pm \sqrt{(-119)^2 - 4(3)(300)}}{2(3)} = \frac{119 \pm \sqrt{14,161 - 3,600}}{6} = \frac{119 \pm \sqrt{10,561}}{6}$

$\sqrt{10,561} \approx 102.7667$

$Q_1 = \frac{119 + 102.7667}{6} = \frac{221.7667}{6} \approx 36.96 \approx 37 \text{ units}$

$Q_2 = \frac{119 - 102.7667}{6} = \frac{16.2333}{6} \approx 2.71 \text{ units}$

c) Second-Order Derivative Verification

$\frac{d^2\pi}{dQ^2} = -6Q + 119$
- At $Q = 2.71$ :
  $\frac{d^2\pi}{dQ^2} = -6(2.71) + 119 = -16.26 + 119 = +102.74 > 0 \quad (\text{Local Minimum})$
- At $Q = 36.96$ :
  $\frac{d^2\pi}{dQ^2} = -6(36.96) + 119 = -221.76 + 119 = -102.76 < 0 \quad (\text{Confirmed Maximum})$
Hence, profit is maximized at $Q^* \approx 36.96$ units (or 37 units).

d) Maximum Profit Computation

Substitute $Q = 37$ into $\pi(Q)$ :
$\pi(37) = -(37)^3 + 59.5(37)^2 - 300(37) - 5,000$

$\pi(37) = -50,653 + 59.5(1,369) - 11,100 - 5,000$

$\pi(37) = -50,653 + 81,455.5 - 11,100 - 5,000 = \text{Rs. } 14,702.50$

[10]

Solve the following Linear Programming Problem (LPP) graphically:

\begin{aligned} \text{Maximize } Z &= 30x_1 + 40x_2 \\ \text{Subject to:} \\ 2x_1 + x_2 &\le 100 \\ x_1 + x_2 &\le 80 \\ x_1 + 2x_2 &\le 120 \\ x_1, x_2 &\ge 0 \end{aligned}

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Solution: Graphical Method for Linear Programming

1. Convert Inequalities to Boundary Lines

Line 1: $2x_1 + x_2 = 100$
- If $x_1 = 0 \implies x_2 = 100 \implies (0, 100)$
- If $x_2 = 0 \implies x_1 = 50 \implies (50, 0)$
Line 2: $x_1 + x_2 = 80$
- If $x_1 = 0 \implies x_2 = 80 \implies (0, 80)$
- If $x_2 = 0 \implies x_1 = 80 \implies (80, 0)$
Line 3: $x_1 + 2x_2 = 120$
- If $x_1 = 0 \implies x_2 = 60 \implies (0, 60)$
- If $x_2 = 0 \implies x_1 = 120 \implies (120, 0)$
Non-negativity: $x_1 \ge 0, x_2 \ge 0$ restricts the feasible region to the first quadrant.

2. Identify Feasible Corner Points

The feasible region is the polygon bounded by:

Point $O$ : $(0, 0)$
Point $A$ : $(0, 60)$ (from Line 3 with $y$ -axis)
Point $B$ : Intersection of Line 2 ( $x_1 + x_2 = 80$ ) and Line 3 ( $x_1 + 2x_2 = 120$ ):
$(x_1 + 2x_2) - (x_1 + x_2) = 120 - 80 \implies x_2 = 40$

$x_1 = 80 - 40 = 40 \implies B(40, 40)$
Point $C$ : Intersection of Line 1 ( $2x_1 + x_2 = 100$ ) and Line 2 ( $x_1 + x_2 = 80$ ):
$(2x_1 + x_2) - (x_1 + x_2) = 100 - 80 \implies x_1 = 20$

$x_2 = 80 - 20 = 60 \implies C(20, 60)$
Check feasibility of Point $C$ on Line 3: $x_1 + 2x_2 = 20 + 2(60) = 140 > 120$ (Infeasible!). Thus, Point $B(40, 40)$ connects directly to:
Point $D$ : Intersection of Line 1 ( $2x_1 + x_2 = 100$ ) and Line 3 ( $x_1 + 2x_2 = 120$ ): From Line 1: $x_2 = 100 - 2x_1$ . Substitute: $x_1 + 2(100 - 2x_1) = 120 \implies x_1 + 200 - 4x_1 = 120 \implies 3x_1 = 80 \implies x_1 = 26.67, x_2 = 46.67$ . Let’s re-verify intersection of Line 1 ( $2x_1+x_2=100$ ) and Line 2 ( $x_1+x_2=80$ ): $x_1=20, x_2=60$ . Wait, at $x_1=40, x_2=40$ :
- Line 1: $2(40) + 40 = 120 > 100$ (Point $B(40, 40)$ is outside Line 1!). Let’s find the true corner points:
1. $(0, 0)$
2. $(0, 60)$ (on Line 3)
3. Intersection of Line 3 ( $x_1 + 2x_2 = 120$ ) and Line 2 ( $x_1 + x_2 = 80$ ): $(40, 40)$ violates Line 1 ( $2x_1+x_2 \le 100$ ). Intersection of Line 1 ( $2x_1+x_2=100$ ) and Line 3 ( $x_1+2x_2=120$ ): Multiply Line 1 by 2: $4x_1 + 2x_2 = 200$ . Subtract Line 3: $3x_1 = 80 \implies x_1 = 80/3 \approx 26.67, x_2 = 140/3 \approx 46.67$ . Check Line 2: $x_1 + x_2 = 80/3 + 140/3 = 220/3 = 73.33 \le 80$ (Feasible!).
4. Point $(50, 0)$ (on Line 1).

3. Evaluate Objective Function $Z = 30x_1 + 40x_2$ at Corner Points:

Corner Point	Coordinates $(x_1, x_2)$	$Z = 30x_1 + 40x_2$
$O$	$(0, 0)$	$30(0) + 40(0) = 0$
$A$	$(0, 60)$	$30(0) + 40(60) = 2,400$
$P$	$(80/3, 140/3)$	$30(80/3) + 40(140/3) = 800 + 1,866.67 = 2,666.67$
$E$	$(50, 0)$	$30(50) + 40(0) = 1,500$

Conclusion: The maximum value of $Z$ is $2,666.67$ , occurring at $x_1 = \frac{80}{3} \approx 26.67$ and $x_2 = \frac{140}{3} \approx 46.67$ .

[10]
Define Consumers’ Surplus and Producers’ Surplus. If the demand and supply functions under perfect competition are:
$P_d = 100 - Q^2 \quad \text{and} \quad P_s = 20 + 2Q$

Required: a) Find the equilibrium price ( $P_e$ ) and equilibrium quantity ( $Q_e$ ). b) Calculate the Consumers’ Surplus (CS) at equilibrium. c) Calculate the Producers’ Surplus (PS) at equilibrium.
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Solution: Consumers’ and Producers’ Surplus

a) Market Equilibrium

Set demand price equal to supply price ( $P_d = P_s$ ):
$100 - Q^2 = 20 + 2Q$

$Q^2 + 2Q - 80 = 0$

$(Q + 10)(Q - 8) = 0$
Since quantity cannot be negative, $Q_e = 8$ units.

Substitute $Q_e = 8$ into the supply function to find $P_e$ :
$P_e = 20 + 2(8) = 20 + 16 = \text{Rs. } 36$

b) Consumers’ Surplus (CS)

$\text{CS} = \int_0^{Q_e} [P_d(Q) - P_e] \, dQ = \int_0^8 (100 - Q^2 - 36) \, dQ = \int_0^8 (64 - Q^2) \, dQ$
$\text{CS} = \left[ 64Q - \frac{Q^3}{3} \right]_0^8 = 64(8) - \frac{512}{3} = 512 - 170.67 = \text{Rs. } 341.33$

c) Producers’ Surplus (PS)

$\text{PS} = \int_0^{Q_e} [P_e - P_s(Q)] \, dQ = \int_0^8 [36 - (20 + 2Q)] \, dQ = \int_0^8 (16 - 2Q) \, dQ$
$\text{PS} = \left[ 16Q - Q^2 \right]_0^8 = 16(8) - (8)^2 = 128 - 64 = \text{Rs. } 64$
Summary:
- Equilibrium: $Q_e = 8 \text{ units}, P_e = \text{Rs. } 36$
- Consumers’ Surplus = Rs. 341.33
- Producers’ Surplus = Rs. 64.00

Group C

Comprehensive Answer / Case Analysis Question. Attempt ALL questions.

[1 × 20 = 20]

[20]
Mathematical Case Study: Multi-Product Production Optimization and Matrix Modeling

A garment manufacturing enterprise in Biratnagar produces three lines of garments: Casual Shirts ( $x_1$ ), Formal Trousers ( $x_2$ ), and Winter Jackets ( $x_3$ ). Production requires processing across three specialized departments: Cutting, Sewing, and Finishing.

The departmental input coefficients (in labor hours per unit) and total available departmental machine hours per month are shown below:

Department Casual Shirts ( $x_1$ ) Formal Trousers ( $x_2$ ) Winter Jackets ( $x_3$ ) Total Monthly Capacity (Hours)

Cutting 1 2 3 1,800

Sewing 2 4 2 2,800

Finishing 1 1 2 1,200

The unit profit contribution is Rs. 300 for a shirt, Rs. 500 for trousers, and Rs. 800 for a jacket.

Required: a) Express the production system as a matrix equation $A X = B$ . Compute the determinant $|A|$ of the technology matrix and prove that the matrix is non-singular. (6 Marks) b) Find the inverse matrix $A^{-1}$ using the Adjoint method. (7 Marks) c) Determine the exact number of shirts, trousers, and jackets the company must produce per month to utilize 100% of available departmental labor hours, and calculate the resulting total monthly gross profit. (7 Marks)
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Comprehensive Mathematical Case Solution

a) Matrix Formulation and Non-Singularity Proof

$\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 2 \\ 1 & 1 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1,800 \\ 2,800 \\ 1,200 \end{pmatrix}$
Let $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 2 \\ 1 & 1 & 2 \end{pmatrix}, X = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, B = \begin{pmatrix} 1,800 \\ 2,800 \\ 1,200 \end{pmatrix}$ .

Compute $|A|$ expanding along the first row:
$|A| = 1 \begin{vmatrix} 4 & 2 \\ 1 & 2 \end{vmatrix} - 2 \begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix} + 3 \begin{vmatrix} 2 & 4 \\ 1 & 1 \end{vmatrix}$

$|A| = 1(8 - 2) - 2(4 - 2) + 3(2 - 4) = 1(6) - 2(2) + 3(-2) = 6 - 4 - 6 = -4$
Since $|A| = -4 \neq 0$ , matrix $A$ is non-singular and its inverse $A^{-1}$ exists.

b) Finding $A^{-1}$ via Adjoint Method

The cofactors $C_{ij}$ of matrix $A$ :
- $C_{11} = +(8 - 2) = 6$
- $C_{12} = -(4 - 2) = -2$
- $C_{13} = +(2 - 4) = -2$
- $C_{21} = -(4 - 3) = -1$
- $C_{22} = +(2 - 3) = -1$
- $C_{23} = -(1 - 2) = 1$
- $C_{31} = +(4 - 12) = -8$
- $C_{32} = -(2 - 6) = 4$
- $C_{33} = +(4 - 4) = 0$
Cofactor matrix $C = \begin{pmatrix} 6 & -2 & -2 \\ -1 & -1 & 1 \\ -8 & 4 & 0 \end{pmatrix}$ .

Adjoint matrix $\text{Adj}(A) = C^T$ :
$\text{Adj}(A) = \begin{pmatrix} 6 & -1 & -8 \\ -2 & -1 & 4 \\ -2 & 1 & 0 \end{pmatrix}$

Inverse Matrix $A^{-1} = \frac{1}{|A|} \text{Adj}(A)$ :
$A^{-1} = -\frac{1}{4} \begin{pmatrix} 6 & -1 & -8 \\ -2 & -1 & 4 \\ -2 & 1 & 0 \end{pmatrix} = \begin{pmatrix} -1.5 & 0.25 & 2.0 \\ 0.5 & 0.25 & -1.0 \\ 0.5 & -0.25 & 0 \end{pmatrix}$

c) Optimal Output Vector $X = A^{-1} B$ and Gross Profit

$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = -\frac{1}{4} \begin{pmatrix} 6 & -1 & -8 \\ -2 & -1 & 4 \\ -2 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1,800 \\ 2,800 \\ 1,200 \end{pmatrix}$
1. Calculate $x_1$ (Shirts):
  $x_1 = -\frac{1}{4} [6(1,800) - 1(2,800) - 8(1,200)] = -\frac{1}{4} [10,800 - 2,800 - 9,600] = -\frac{1}{4} [-1,600] = 400$
2. Calculate $x_2$ (Trousers):
  $x_2 = -\frac{1}{4} [-2(1,800) - 1(2,800) + 4(1,200)] = -\frac{1}{4} [-3,600 - 2,800 + 4,800] = -\frac{1}{4} [-1,600] = 400$
3. Calculate $x_3$ (Jackets):
  $x_3 = -\frac{1}{4} [-2(1,800) + 1(2,800) + 0] = -\frac{1}{4} [-3,600 + 2,800] = -\frac{1}{4} [-800] = 200$
Verification:
- Cutting: $1(400) + 2(400) + 3(200) = 400 + 800 + 600 = 1,800$ hours.
- Sewing: $2(400) + 4(400) + 2(200) = 800 + 1,600 + 400 = 2,800$ hours.
- Finishing: $1(400) + 1(400) + 2(200) = 400 + 400 + 400 = 1,200$ hours. (All capacity exactly consumed!)
Total Monthly Gross Profit:
$\text{Profit} = 300(400) + 500(400) + 800(200) = 120,000 + 200,000 + 160,000 = \text{Rs. } 480,000$

Business Mathematics - I

Past exam papers

Business Mathematics - I 2080 Official Model Question Paper

Faculty of Management

Brief Answer Questions. Attempt ALL questions.

Descriptive Answer Questions. Attempt any THREE questions.

Solution: Cramer’s Rule

1. Coefficient Matrix Determinant ( $D$ )

2. Determinant $D_x$ (Replacing Column 1 with Constants):

3. Determinant $D_y$ (Replacing Column 2 with Constants):

4. Determinant $D_z$ (Replacing Column 3 with Constants):

5. Calculate Values:

Solution: Profit Maximization Analysis

a) Profit Function $\pi(Q)$

b) First-Order Condition for Profit Maximization

c) Second-Order Derivative Verification

d) Maximum Profit Computation

Solution: Graphical Method for Linear Programming

1. Convert Inequalities to Boundary Lines

2. Identify Feasible Corner Points

3. Evaluate Objective Function $Z = 30x_1 + 40x_2$ at Corner Points:

Solution: Consumers’ and Producers’ Surplus

a) Market Equilibrium

b) Consumers’ Surplus (CS)

c) Producers’ Surplus (PS)

Comprehensive Answer / Case Analysis Question. Attempt ALL questions.

Mathematical Case Study: Multi-Product Production Optimization and Matrix Modeling

Comprehensive Mathematical Case Solution

a) Matrix Formulation and Non-Singularity Proof

b) Finding $A^{-1}$ via Adjoint Method

c) Optimal Output Vector $X = A^{-1} B$ and Gross Profit

Department	Casual Shirts ( $x_1$ )	Formal Trousers ( $x_2$ )	Winter Jackets ( $x_3$ )	Total Monthly Capacity (Hours)
Cutting	1	2	3	1,800
Sewing	2	4	2	2,800
Finishing	1	1	2	1,200

Past exam papers

Business Mathematics - I 2080 Official Model Question Paper

Solution: Cramer’s Rule

1. Coefficient Matrix Determinant (DDD)

2. Determinant DxD_xDx​ (Replacing Column 1 with Constants):

3. Determinant DyD_yDy​ (Replacing Column 2 with Constants):

4. Determinant DzD_zDz​ (Replacing Column 3 with Constants):

5. Calculate Values:

Solution: Profit Maximization Analysis

a) Profit Function π(Q)\pi(Q)π(Q)

b) First-Order Condition for Profit Maximization

c) Second-Order Derivative Verification

d) Maximum Profit Computation

Solution: Graphical Method for Linear Programming

1. Convert Inequalities to Boundary Lines

2. Identify Feasible Corner Points

3. Evaluate Objective Function Z=30x1+40x2Z = 30x_1 + 40x_2Z=30x1​+40x2​ at Corner Points:

Solution: Consumers’ and Producers’ Surplus

a) Market Equilibrium

b) Consumers’ Surplus (CS)

c) Producers’ Surplus (PS)

Mathematical Case Study: Multi-Product Production Optimization and Matrix Modeling

Comprehensive Mathematical Case Solution

a) Matrix Formulation and Non-Singularity Proof

b) Finding A−1A^{-1}A−1 via Adjoint Method

c) Optimal Output Vector X=A−1BX = A^{-1} BX=A−1B and Gross Profit

1. Coefficient Matrix Determinant ( $D$ )

2. Determinant $D_x$ (Replacing Column 1 with Constants):

3. Determinant $D_y$ (Replacing Column 2 with Constants):

4. Determinant $D_z$ (Replacing Column 3 with Constants):

a) Profit Function $\pi(Q)$

3. Evaluate Objective Function $Z = 30x_1 + 40x_2$ at Corner Points:

b) Finding $A^{-1}$ via Adjoint Method

c) Optimal Output Vector $X = A^{-1} B$ and Gross Profit