GATE DA Logic Previous Year Questions

GATE DA 2025 | Question 15

MCQ · 1 mark · −0.33

Let \(p\) and \(q\) be any two propositions. Consider the following propositional statements.
\(S_{1} : p \to q\), \(S_{2} : \neg p \wedge q\), \(S_{3} : \neg p \vee q\), \(S_{4} : \neg p \vee \neg q\),
where \(\wedge\) denotes conjunction (AND operation), \(\vee\) denotes disjunction (OR operation), and \(\neg\) denotes negation (NOT operation). Which one of the following options is correct? (Note: \(\equiv\) denotes logical equivalence)

\(S_{1} \equiv S_{3}\) \(S_{2} \equiv S_{3}\) \(S_{2} \equiv S_{4}\) \(S_{1} \equiv S_{4}\)

Explanation

Answer: (A)

\(p \to q\) is false only when \(p\) is true and \(q\) is false, and \(\neg p \vee q\) is false in exactly the same case, so \(S_{1} \equiv S_{3}\).

Every other pair differs on at least one row of the truth table:

\(S_{2}\) and \(S_{3}\): with \(p\) and \(q\) both true, \(S_{2}\) is false but \(S_{3}\) is true.
\(S_{2}\) and \(S_{4}\): with both false, \(S_{2}\) is false but \(S_{4}\) is true.
\(S_{1}\) and \(S_{4}\): with both true, \(S_{1}\) is true but \(S_{4}\) is false.

Answer: A

GATE DA 2024 | Question 29

MSQ · 1 mark

Let \(x\) and \(y\) be two propositions. Which of the following statements is a tautology /are tautologies?

\((\neg x \land y) \Longrightarrow (y \Longrightarrow x)\) \((x \land \neg y) \Longrightarrow (\neg x \Longrightarrow y)\) \((\neg x \land y) \Longrightarrow (\neg x \Longrightarrow y)\) \((x \land \neg y) \Longrightarrow (y \Longrightarrow x)\)

Explanation

Answer: (B), (C), (D)

An implication is false only when its premise is true and its conclusion is false. Each premise here is true in exactly one row of the truth table, so check the conclusion in that row.

A: \(\neg x \land y\) needs \(x\) false and \(y\) true. Then \(y \Longrightarrow x\) is true \(\Longrightarrow\) false, which is false. Not a tautology.
B: \(x \land \neg y\) needs \(x\) true and \(y\) false. Then \(\neg x\) is false, so \(\neg x \Longrightarrow y\) is true. Tautology.
C: \(x\) false and \(y\) true make \(\neg x \Longrightarrow y\) true \(\Longrightarrow\) true, which is true. Tautology.
D: \(x\) true and \(y\) false make \(y \Longrightarrow x\) false \(\Longrightarrow\) true, which is true. Tautology.

Answer: B, C, D

GATE DA 2024 | Question 54

MSQ · 2 marks

Let game(ball, rugby) be true if the ball is used in rugby and false otherwise.

Let shape(ball, round) be true if the ball is round and false otherwise.

Consider the following logical sentences:

s1: \(\forall \textit{ball}\ \neg\, \textit{game(ball, rugby)} \Longrightarrow \textit{shape(ball, round)}\)

s2: \(\forall \textit{ball}\ \neg\, \textit{shape(ball, round)} \Longrightarrow \textit{game(ball, rugby)}\)

s3: \(\forall \textit{ball}\ \textit{game(ball, rugby)} \Longrightarrow \neg\, \textit{shape(ball, round)}\)

s4: \(\forall \textit{ball}\ \textit{shape(ball, round)} \Longrightarrow \neg\, \textit{game(ball, rugby)}\)

Which of the following choices is/are logical representations of the assertion,
“All balls are round except balls used in rugby”?

\(s1 \wedge s3\) \(s1 \wedge s2\) \(s2 \wedge s3\) \(s3 \wedge s4\)

Explanation

Answer: (A), (C)

The assertion says two things about every ball: a ball not used in rugby is round, and a ball used in rugby is not round.

s1 states the first part, and s2 is its contrapositive (a ball that is not round is used in rugby), so s1 and s2 say the same thing.
s3 states the second part, and s4 is its contrapositive, so s3 and s4 say the same thing.

A correct representation needs one sentence from each pair, so \(s1 \wedge s3\) (A) and \(s2 \wedge s3\) (C) both work. Option B repeats the first part and would still allow a round rugby ball; option D repeats the second part and would allow a non-rugby ball that is not round.

Answer: A, C