MTH 201 Business Mathematics I Practice Sets & Model Questions — TU BBM Semester 1

PAPER KHOJ

Practice Examination — Set 1

For BBM · Semester 1

Based on the prescribed TU Faculty of Management syllabus and examination structure

Independently prepared by Paper Khoj. This is not an official Tribhuvan University or Dean's Office question paper.

Course: MTH 201 · Business Mathematics I

Level: Bachelor of Business Management (BBM) · 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

Practice sets and model questions

Practice Set 1

Practice Examination — Set 1

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

Practice sets and model questions

Practice Set 1

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