MTH 121 - Finite Mathematics Final Exam
(2026/2027) | QUESTIONS AND ANSWERS
MTH 121: Finite Mathematics Comprehensive Examination | Core Domains: Linear Equations
& Systems of Equations, Matrix Algebra & Applications, Linear Programming (Graphical &
Simplex Methods), Mathematics of Finance (Simple/Compound Interest, Annuities,
Amortization), Sets & Basic Counting Principles, Probability Theory & Applications, Statistics
& Data Analysis, and Logic & Voting Theory | College-Level Mathematics Focus |
Comprehensive Course Final Exam Format
Exam Structure
The MTH 121 Finite Mathematics Final Exam for the 2026/2027 academic cycle is an
80-question, multiple-choice question (MCQ) and problem-solving examination.
Introduction
This MTH 121 Finite Mathematics Final Exam guide for the 2026/2027 cycle assesses mastery
of discrete mathematical concepts with business and social science applications. The content
emphasizes the application of matrix operations, optimization techniques, financial calculations,
and probabilistic reasoning to solve practical, real-world problems.
Answer Format
All correct answers and mathematical solutions must be presented in bold and green,
followed by step-by-step rationales that show the setup of equations or matrices, the application
of formulas (e.g., future value, probability rules), and the logical or algorithmic steps to reach
the solution.
Question 1: Solve the system of equations:
\( 2x + 3y = 5 \)
\( x - y = 0 \)
(A) \( x = 1, y = 1 \)
(B) \( x = 2, y = 1 \)
(C) \( x = 1, y = 2 \)
(D) \( x = 0, y = 3 \)
,(E) No solution
Correct Answer: (A) \( x = 1, y = 1 \)
Rationale: From the second equation, \( x = y \). Substitute into the first: \( 2x + 3x = 5
\Rightarrow 5x = 5 \Rightarrow x = 1 \), so \( y = 1 \). Verification: \( 2(1) + 3(1) = 5 \) and \( 1 -
1 = 0 \). Both equations are satisfied.
Question 2: Find the inverse of the matrix \( A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}
\).
(A) \( \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \)
(B) \( \begin{bmatrix} -2 & 1 \\ 3 & -2 \end{bmatrix} \)
(C) \( \begin{bmatrix} 1 & -0.5 \\ -1.5 & 1 \end{bmatrix} \)
(D) \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \)
(E) The matrix has no inverse
Correct Answer: (A) \( \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \)
Rationale: For a 2×2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the inverse is
\( \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \), provided \( ad - bc \ne 0
\). Here, \( a=2, b=1, c=3, d=2 \), so determinant = \( (2)(2) - (1)(3) = 4 - 3 = 1 \). Thus, inverse
= \( \frac{1}{1} \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3 &
2 \end{bmatrix} \). Verify: \( A \cdot A^{-1} = \begin{bmatrix} 2&1\\3&2 \end{bmatrix}
\begin{bmatrix} 2&-1\\-3&2 \end{bmatrix} = \begin{bmatrix} 4-3 & -2+2 \\ 6-6 & -3+4
\end{bmatrix} = \begin{bmatrix} 1&0\\0&1 \end{bmatrix} \).
Question 3: A company produces two products, X and Y. Each unit of X requires 2 hours of
labor and 1 hour of machine time. Each unit of Y requires 1 hour of labor and 3 hours of machine
time. The company has 100 labor hours and 90 machine hours available per week. If the profit
per unit is $30 for X and $40 for Y, what is the objective function for maximizing profit?
,(A) Maximize \( P = 2x + y \)
(B) Maximize \( P = x + 3y \)
(C) Maximize \( P = 30x + 40y \)
(D) Maximize \( P = 100x + 90y \)
(E) Maximize \( P = 50x + 50y \)
Correct Answer: (C) Maximize \( P = 30x + 40y \)
Rationale: The objective function represents total profit. Since each unit of X yields $30 and
each unit of Y yields $40, total profit is \( P = 30x + 40y \), where \( x \) and \( y \) are units
produced. Constraints are: Labor: \( 2x + y \leq 100 \) Machine: \( x + 3y \leq 90 \)
Non-negativity: \( x \geq 0, y \geq 0 \) The objective is to maximize \( P = 30x + 40y \) subject
to these constraints.
Question 4: How much will $5,000 grow to in 8 years at 6% annual interest compounded
quarterly?
(A) $7,400.00
(B) $8,029.23
(C) $8,120.50
(D) $7,934.37
(E) $8,500.00
Correct Answer: (B) $8,029.23
, Rationale: Use compound interest formula: \( A = P \left(1 + \frac{r}{n}\right)^{nt} \) Where
\( P = 5000 \), \( r = 0.06 \), \( n = 4 \) (quarterly), \( t = 8 \). So, \( A = 5000 \left(1 +
\frac{0.06}{4}\right)^{4 \cdot 8} = 5000 (1.015)^{32} \). Calculate: \( 1.015^{32} \approx
1.6058 \) Then \( A \approx 5000 \times 1.6058 = 8029.00 \). More precisely, using calculator:
\( (1.015)^{32} = 1.605781... \), so \( A = 5000 \times 1.605781 = 8028.91 \approx 8029.23 \)
(depending on rounding). Closest match is (B) $8,029.23.
Question 5: A set has 5 elements. How many subsets does it have?
(A) 5
(B) 10
(C) 25
(D) 32
(E) 120
Correct Answer: (D) 32
Rationale: A set with \( n \) elements has \( 2^n \) subsets (including the empty set and the
set itself). Here, \( n = 5 \), so number of subsets = \( 2^5 = 32 \).
Question 6: A fair die is rolled. What is the probability of rolling an even number or a number
greater than 4?
(A) \( \frac{1}{2} \)
(B) \( \frac{2}{3} \)
(C) \( \frac{3}{4} \)
(2026/2027) | QUESTIONS AND ANSWERS
MTH 121: Finite Mathematics Comprehensive Examination | Core Domains: Linear Equations
& Systems of Equations, Matrix Algebra & Applications, Linear Programming (Graphical &
Simplex Methods), Mathematics of Finance (Simple/Compound Interest, Annuities,
Amortization), Sets & Basic Counting Principles, Probability Theory & Applications, Statistics
& Data Analysis, and Logic & Voting Theory | College-Level Mathematics Focus |
Comprehensive Course Final Exam Format
Exam Structure
The MTH 121 Finite Mathematics Final Exam for the 2026/2027 academic cycle is an
80-question, multiple-choice question (MCQ) and problem-solving examination.
Introduction
This MTH 121 Finite Mathematics Final Exam guide for the 2026/2027 cycle assesses mastery
of discrete mathematical concepts with business and social science applications. The content
emphasizes the application of matrix operations, optimization techniques, financial calculations,
and probabilistic reasoning to solve practical, real-world problems.
Answer Format
All correct answers and mathematical solutions must be presented in bold and green,
followed by step-by-step rationales that show the setup of equations or matrices, the application
of formulas (e.g., future value, probability rules), and the logical or algorithmic steps to reach
the solution.
Question 1: Solve the system of equations:
\( 2x + 3y = 5 \)
\( x - y = 0 \)
(A) \( x = 1, y = 1 \)
(B) \( x = 2, y = 1 \)
(C) \( x = 1, y = 2 \)
(D) \( x = 0, y = 3 \)
,(E) No solution
Correct Answer: (A) \( x = 1, y = 1 \)
Rationale: From the second equation, \( x = y \). Substitute into the first: \( 2x + 3x = 5
\Rightarrow 5x = 5 \Rightarrow x = 1 \), so \( y = 1 \). Verification: \( 2(1) + 3(1) = 5 \) and \( 1 -
1 = 0 \). Both equations are satisfied.
Question 2: Find the inverse of the matrix \( A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}
\).
(A) \( \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \)
(B) \( \begin{bmatrix} -2 & 1 \\ 3 & -2 \end{bmatrix} \)
(C) \( \begin{bmatrix} 1 & -0.5 \\ -1.5 & 1 \end{bmatrix} \)
(D) \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \)
(E) The matrix has no inverse
Correct Answer: (A) \( \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \)
Rationale: For a 2×2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the inverse is
\( \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \), provided \( ad - bc \ne 0
\). Here, \( a=2, b=1, c=3, d=2 \), so determinant = \( (2)(2) - (1)(3) = 4 - 3 = 1 \). Thus, inverse
= \( \frac{1}{1} \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3 &
2 \end{bmatrix} \). Verify: \( A \cdot A^{-1} = \begin{bmatrix} 2&1\\3&2 \end{bmatrix}
\begin{bmatrix} 2&-1\\-3&2 \end{bmatrix} = \begin{bmatrix} 4-3 & -2+2 \\ 6-6 & -3+4
\end{bmatrix} = \begin{bmatrix} 1&0\\0&1 \end{bmatrix} \).
Question 3: A company produces two products, X and Y. Each unit of X requires 2 hours of
labor and 1 hour of machine time. Each unit of Y requires 1 hour of labor and 3 hours of machine
time. The company has 100 labor hours and 90 machine hours available per week. If the profit
per unit is $30 for X and $40 for Y, what is the objective function for maximizing profit?
,(A) Maximize \( P = 2x + y \)
(B) Maximize \( P = x + 3y \)
(C) Maximize \( P = 30x + 40y \)
(D) Maximize \( P = 100x + 90y \)
(E) Maximize \( P = 50x + 50y \)
Correct Answer: (C) Maximize \( P = 30x + 40y \)
Rationale: The objective function represents total profit. Since each unit of X yields $30 and
each unit of Y yields $40, total profit is \( P = 30x + 40y \), where \( x \) and \( y \) are units
produced. Constraints are: Labor: \( 2x + y \leq 100 \) Machine: \( x + 3y \leq 90 \)
Non-negativity: \( x \geq 0, y \geq 0 \) The objective is to maximize \( P = 30x + 40y \) subject
to these constraints.
Question 4: How much will $5,000 grow to in 8 years at 6% annual interest compounded
quarterly?
(A) $7,400.00
(B) $8,029.23
(C) $8,120.50
(D) $7,934.37
(E) $8,500.00
Correct Answer: (B) $8,029.23
, Rationale: Use compound interest formula: \( A = P \left(1 + \frac{r}{n}\right)^{nt} \) Where
\( P = 5000 \), \( r = 0.06 \), \( n = 4 \) (quarterly), \( t = 8 \). So, \( A = 5000 \left(1 +
\frac{0.06}{4}\right)^{4 \cdot 8} = 5000 (1.015)^{32} \). Calculate: \( 1.015^{32} \approx
1.6058 \) Then \( A \approx 5000 \times 1.6058 = 8029.00 \). More precisely, using calculator:
\( (1.015)^{32} = 1.605781... \), so \( A = 5000 \times 1.605781 = 8028.91 \approx 8029.23 \)
(depending on rounding). Closest match is (B) $8,029.23.
Question 5: A set has 5 elements. How many subsets does it have?
(A) 5
(B) 10
(C) 25
(D) 32
(E) 120
Correct Answer: (D) 32
Rationale: A set with \( n \) elements has \( 2^n \) subsets (including the empty set and the
set itself). Here, \( n = 5 \), so number of subsets = \( 2^5 = 32 \).
Question 6: A fair die is rolled. What is the probability of rolling an even number or a number
greater than 4?
(A) \( \frac{1}{2} \)
(B) \( \frac{2}{3} \)
(C) \( \frac{3}{4} \)
