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Analytical Aptitude PYQs for GATE DA

GATE DA Analytical Aptitude: 1 units and 2 chapters, weightage from 5 previous year questions across 3 papers, a study order by exam weight and 278 practice q

A question from this chapter

Question 1

2026 PYQ

Level 3: Exam Standard

Assume that a Creative (C) person will Succeed (S) if the person is also Disciplined (D), but will not succeed otherwise. Now, consider the following statements: (i) C \land S \Leftrightarrow D (ii) C \Rightarrow (S ⟺ D) (iii) C \Leftrightarrow ((D \Rightarrow S) \lor \neg S) Which of the following options is correct?

Question 2

2026 PYQ

Level 3: Exam Standard

Rishi and Swathi are students of Class 5. Pavan and Tanvi are students of Class 4. Rishi and Pavan are boys. Swathi and Tanvi are girls. The four students played a total of three games of chess. The games were played one after another. A player who lost a game did not participate in any more games. It was observed that: (i) the first game was the only game where two students of the same class played against each other, (ii) the students of Class 5 won more games than the students of Class 4, and (iii) the boys won two games and the girls won one game. The student who did not lose any game is __________.

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Analytical Aptitude PYQs for GATE DA

GATE DA Analytical Aptitude: 1 units and 2 chapters, weightage from 5 previous year questions across 3 papers, a study order by exam weight and 278 practice questions.

About Analytical Aptitude Previous Year Questions (PYQs)

5 previous year questions from Analytical Aptitude in GATE DA, grouped by chapter with the exam year, answer key and step-by-step solution for each.

GATE DA Analytical Aptitude Unit-wise Weightage from Past Papers

We counted every GATE DA Analytical Aptitude previous year question in our bank (5 questions from 3 papers) and grouped them by unit.

Unit	Chapters	PYQs	Share of section	Avg per paper
Logic	2	5	100%	1.7

Suggested Analytical Aptitude Study Order for GATE DA

Logic: 100% of past Analytical Aptitude questions, about 1.7 per paper.

Start where the marks are. Units at the top of this list have appeared most often in past GATE DA papers.

Units in GATE DA Analytical Aptitude

Logic Previous Year Questions (PYQs) (2 chapters)

All Analytical Aptitude chapters

One Solved Question from Each Analytical Aptitude Chapter

Question 1 · Propositional Logic and Logical Equivalence · 2026 MCQ

Assume that a Creative

(C)

person will Succeed

(S)

if the person is also Disciplined

(D)

, but will not succeed otherwise. Now, consider the following statements:

(i)

C \land S \Leftrightarrow D

(ii)

C \Rightarrow (S ⟺ D)

(iii)

C \Leftrightarrow ((D \Rightarrow S) \lor \neg S)

Which of the following options is correct?

A.
Both (i) and (ii) are TRUE
B.
Only (ii) is TRUE
C.
Both (ii) and (iii) are TRUE
D.
Only (iii) is TRUE

Correct Answer:

Step-by-Step Solution

Insight: "A creative person will succeed if disciplined, but not otherwise" means: if creative, then (succeed iff disciplined).

Exam route: Translate the English sentence step by step. "Creative person will succeed if disciplined" gives $C \to (D \to S)$ . "But not otherwise" adds $C \to (\neg D \to \neg S)$ . Together: $C \to (S ⟺ D)$ . Check which option matches.

Learning route:

Step 1: Parse the sentence structure.

"A Creative (C) person will Succeed (S) if the person is also Disciplined (D), but will not succeed otherwise."

The phrase "but not otherwise" is the key. It means the condition is both necessary and sufficient.

Step 2: Translate the "if" part.

"will succeed if disciplined" $⟹ D \to S$ (within the context of being creative).

Step 3: Translate the "but not otherwise" part.

"will not succeed otherwise" $⟹ \neg D \to \neg S$ (within the context of being creative).

Step 4: Combine 2 and 3.

$(D \to S) \land (\neg D \to \neg S) \equiv S ⟺ D$ .

Step 5: Apply the context.

The whole rule applies to creative people, so: $C \to (S ⟺ D)$ .

Step 6: Evaluate the given statements.

(i) $C \land S ⟺ D$ : This means $(C \land S) ⟺ D$ , which is not equivalent. For example, if $C = F, D = F$ , the original gives $T$ (vacuously), but (i) gives $F ⟺ F = T$ . Wait, let me check $C = F, S = T, D = F$ : original gives $T$ , (i) gives $(F \land T) ⟺ F = F ⟺ F = T$ . Let me check $C = F, S = F, D = T$ : original gives $T$ , (i) gives $(F \land F) ⟺ T = F ⟺ T = F$ . Not equivalent. So (i) is FALSE.

(ii) $C \to (S ⟺ D)$ : This is exactly what we derived. TRUE.

(iii) $C ⟺ ((D \to S) \lor \neg S)$ : Simplify the right side. $(D \to S) \lor \neg S = (\neg D \lor S) \lor \neg S = \neg D \lor (S \lor \neg S) = \neg D \lor T = T$ . So (iii) becomes $C ⟺ T$ , which means $C$ must be true. This is not equivalent to $C \to (S ⟺ D)$ . FALSE.

Answer: Only (ii) is TRUE, which is option B.

Question 2 · Logical Reasoning and Constraint Deduction · 2026 MCQ

Rishi and Swathi are students of Class 5. Pavan and Tanvi are students of Class 4. Rishi and Pavan are boys. Swathi and Tanvi are girls. The four students played a total of three games of chess. The games were played one after another. A player who lost a game did not participate in any more games. It was observed that:
(i) the first game was the only game where two students of the same class played against each other,
(ii) the students of Class 5 won more games than the students of Class 4, and
(iii) the boys won two games and the girls won one game.
The student who did not lose any game is __________.

A.
Pavan
B.
Rishi
C.
Swathi
D.
Tanvi

Correct Answer:

Step-by-Step Solution

Key idea: This is a constraint-based sequencing and elimination question, recognizable by the sequential games, elimination rules, and conditions on wins/classes.

Step 1: Understand the game format. There are 4 students and 3 games. A loser is eliminated. For all 4 students to play, the format must be: Game 1 (A vs B), Game 2 (Winner of G1 vs C), Game 3 (Winner of G2 vs D).

Step 2: Use condition (i). Game 1 is the ONLY game where two students of the same class played. The same-class pairs are (Rishi, Swathi) in Class 5 and (Pavan, Tanvi) in Class 4. Thus, Game 1 must be either (R, S) or (P, T).

Step 3: Test if Game 1 is (P, T). If Class 4 plays each other in G1, Class 4 wins G1. Condition (ii) states Class 5 won more games than Class 4. Since there are 3 games total, Class 5 must win 2 and Class 4 must win exactly 1. This means Class 5 must win G2 and G3. If Class 5 wins G2, the winner is R or S. Then Game 3 would be (R or S) vs the remaining (S or R), which is a same-class game. This violates condition (i). Thus, Game 1 cannot be (P, T).

Step 4: Test if Game 1 is (R, S). Game 1 is R vs S. Class 5 wins G1. To satisfy the quotas (Class 5 wins 2, Class 4 wins 1; Boys win 2, Girls win 1), Rishi (Boy) must win G1. If Swathi won, we would need 2 Boy wins from the remaining games, forcing Class 4 to win 2 games, violating (ii). So Rishi wins G1.

Step 5: Determine Game 2 and 3. Rishi plays Game 2 against Pavan or Tanvi. If Rishi loses G2, the Class 4 student wins, and Class 4 would end up winning 2 games total (violating ii). Thus, Rishi must win G2.

Step 6: Now Rishi has won G1 and G2 (2 Boy wins, 2 Class 5 wins). The quotas for Boys and Class 5 are fully met. Therefore, Game 3 must be won by a Girl from Class 4 to satisfy the remaining quotas (1 Girl win, 1 Class 4 win).

Step 7: Game 3 is Rishi vs Tanvi. Tanvi must win. Since Tanvi won her only game and didn't play before, she never lost any game.

Answer: D