UGC NET Computer Science Unit 1 & 2 PYQs with Answers

UGC NET Computer Science Unit 1 and Unit 2 form the foundation of the syllabus and frequently feature both direct and concept-based questions in the examination. Unit 1 covers Discrete Structures and Mathematical Logic, while Unit 2 focuses on Digital Logic and Boolean Algebra. 

Practising Previous Year Questions (PYQs) from these units helps candidates identify important topics, understand the difficulty level, and improve problem-solving speed. For aspirants aiming to maximise their scores, solving UGC NET Computer Science Unit 1 & 2 PYQs is one of the most effective preparation strategies. 

UGC NET Computer Science Unit 1 & 2 PYQs

UGC NET Computer Science Unit 1 & 2 Previous Year Questions (PYQs) are an essential resource for candidates preparing for the examination. Practicing PYQs from these units helps candidates understand frequently asked concepts, identify important topics, and become familiar with the exam pattern. 

1. Which of the following is NOT logically equivalent to P → Q?

• A) ~P ∨ Q

• B) ~Q → ~P

• C) ~(P ∧ ~Q)

• D) ~Q → P

Answer: D) ~Q → P

Explanation: P → Q is logically equivalent to its contrapositive (~Q → ~P), to ~P ∨ Q, and to ~(P ∧ ~Q) — all three say "if P then Q." However, ~Q → P says "if not Q then P," which is actually the inverse's converse and carries a completely different meaning, making it not equivalent to P → Q.

2. Consider the Boolean expression given by F = (X + Y + Z)(X' + Y)(Y' + Z). Which of the following Boolean expressions is/are equivalent to F' (Complement of F)?
   A. (X' + Y' + Z')(X + Y')(Y + Z')
   B. (XY' + Z')
   C. (X + Z')(Y' + Z')
   D. (XY' + YZ' + X'Y'Z')
   
   Choose the correct answer from the options given below:

• A) A, B and C Only

• B) B, C and D Only

• C) A, C and D Only

• D) A, B and D Only

Answer: B) B, C and D Only
Explanation: F' is found by applying De Morgan's law to F = (X+Y+Z)(X'+Y)(Y'+Z), giving F' = X'Y'Z' + XY' + YZ'. Simplifying, B, C, and D are all equivalent forms of F', while A is actually the complement of the complement (i.e., F itself), not F'.

3. Match List-I with List-II.

• A) A-III, B-II, C-I, D-IV

• B) A-II, B-III, C-IV, D-I

• C) A-IV, B-III, C-I, D-II

• D) A-IV, B-I, C-III, D-II

Answer: A) A-III, B-II, C-I, D-IV
Explanation: AΔB = (A−B)∪(B−A) → III; A−(B∪C) = (A−B)∩(A−C) → II; A−(B∩C) = (A−B)∪(A−C) → I; A∩(B−C) = (A∩B)−(A∩C) → IV.

4. Let 'G' be a simple, undirected graph with n (n > 10) vertices and k connected components. What is the minimum possible number of edges of G?

• A) k(n−1)

• B) n(k−1)

• C) n+k−2

• D) n−k

Answer: D) n − k
Explanation: To minimize edges, each connected component should be a tree. A tree with m vertices has m−1 edges, so k components together have n−k edges total (since the sum of vertices across all components is n).

5. Consider the following statements:
   P: There exists no simple, undirected and connected graph with 80 vertices and 77 edges.
   Q: All vertices of Euler graph are of even degree.
   R: Every simple, undirected, connected and acyclic graph with 50 vertices has at least two vertices of degree one.
   S: There exits a bipartite graph with more than ten vertices which are 2-2-colorable.

What is the number of correct statements among the above statements?

• A) 1

• B) 2

• C) 3

• D) 4

Answer: C) 3
Explanation: P is false (a tree with 80 vertices has 79 edges, so 77 edges with 80 vertices is possible as a forest). Q, R, and S are all true — Euler graphs require all even degrees, a tree always has ≥2 leaves, and any bipartite graph is 2-colorable.

6. Match List-I with List-II.

Choose the correct answer from the options given below:

• A) A-I, B-II, C-III, D-IV

• B) A-II, B-I, C-III, D-IV

• C) A-III, B-I, C-II, D-IV

• D) A-IV, B-III, C-II, D-I

Answer: B) A-II, B-I, C-III, D-IV
Explanation: P∧(P→Q) = P∧(¬P∨Q) = (P∧¬P)∨(P∧Q) → II; ¬(P∨Q)→(P∧Q) simplifies to a tautology equivalent to P∨Q → I; P→Q = ¬P∨Q → III; P∨(Q∧R) distributes to match → IV.