About Logic Previous Year Questions (PYQs)

21 previous year questions from Logic in GATE CS, grouped by chapter with the exam year, answer key and step-by-step solution for each.

Logic Weightage in GATE CS

Logic accounts for 21 of 21 Analytical Aptitude previous year questions in our bank (100%), about 2.3 per paper across 9 papers.

Logic Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Deductive Reasoning, Statements and Syllogisms	Passage-Based Deductive Inference, Categorical Statements and Syllogisms, Conditional Logic and Contraposition	8	38%	125
Relationships, Ordering and Arrangement Puzzles	Family and Blood Relationships, Spatial Placement and Geometric Arrangements, Ordering, Ranking and Sequential Events, Grid Shading and Constraint Puzzles	5	24%	104
Patterns, Series, Coding and Analogies	Letter Coding and Decoding, Number Series and Array Patterns, Word and Functional Analogies	5	24%	78
Graph, Route and Tournament Reasoning	Graph Connectivity and Minimum Bridges, Knockout Tournament Elimination, Hamiltonian Routes and Round Trips	3	14%	48

Deductive Reasoning, Statements and Syllogisms Previous Year Questions (PYQs) (8 questions)
Relationships, Ordering and Arrangement Puzzles Previous Year Questions (PYQs) (5 questions)
Patterns, Series, Coding and Analogies Previous Year Questions (PYQs) (5 questions)
Graph, Route and Tournament Reasoning Previous Year Questions (PYQs) (3 questions)

More from Analytical Aptitude

GATE CS Analytical Aptitude Previous Year Questions (PYQs)

One Solved Question from Each Logic Chapter

Question 1 · Deductive Reasoning, Statements and Syllogisms · 2026_Set2 MCQ

‘When it is raining, peacocks dance.’

Based only on this sentence, which one of the following options is necessarily true?

A.
Peacocks dance only when it is raining.
B.
When peacocks dance, it is raining.
C.
When peacocks are not dancing, it is not raining.
D.
When it is not raining, peacocks do not dance.

Correct Answer:

C

Step-by-Step Solution

Insight: "When P, Q" translates to $P ⟹ Q$ . The contrapositive $\neg Q ⟹ \neg P$ is logically equivalent and must be true.

Exam route: Identify $P$ (raining) and $Q$ (peacocks dance). The contrapositive is "If peacocks are not dancing, it is not raining." Match with Option C.

Learning route:

Translate the statement: "When it is raining ( $P$ ), peacocks dance ( $Q$ )" means $P ⟹ Q$ .
Recall logical equivalences: The contrapositive $\neg Q ⟹ \neg P$ is always true if $P ⟹ Q$ is true.
Form the contrapositive: "When peacocks are not dancing ( $\neg Q$ ), it is not raining ( $\neg P$ )."
Evaluate options:

Option A & B represent the converse ( $Q ⟹ P$ ), which is not necessarily true.
Option D represents the inverse ( $\neg P ⟹ \neg Q$ ), which is not necessarily true.
Option C represents the contrapositive ( $\neg Q ⟹ \neg P$ ), which must be true.

Answer is C.

Question 2 · Relationships, Ordering and Arrangement Puzzles · 2026_Set2 MCQ

The figure in Panel I below is a grid of cells with four rows and four columns. The numbers on the top and on the left represent the number of cells that are to be shaded in that column and row, respectively. Which one of the options shown in Panel II below represents the grid shaded correctly?

A.
(i)
B.
(ii)
C.
(iii)
D.
(iv)

Correct Answer:

B

Step-by-Step Solution

Insight: Verify total row and column sums first, then use forced moves starting with the most constrained row or column.

Exam route: Row sums = 3 + 1 + 2 + 2 = 8. Column sums = 2 + 2 + 2 + 2 = 8. Consistency check passed. Row 2 requires exactly 1 shaded cell. Inspecting the options: Option (i) has 2 shaded in Row 2 (invalid). Option (iii) has 2 shaded in Row 2 (invalid). Option (ii) has 1 shaded in Row 2. Verify Option (ii) columns: Col 1 has 2, Col 2 has 2, Col 3 has 2, Col 4 has 2. All constraints satisfied. Option (ii) is correct.

Learning route:

Step 1: Consistency Check. Sum of row requirements (3+1+2+2 = 8) must equal sum of column requirements (2+2+2+2 = 8). They match.

Step 2: Identify Extremes. Row 2 requires only 1 shaded cell out of 4. This is the tightest constraint.

Step 3: Option Elimination. Instead of solving the grid from scratch (which is time-consuming), evaluate the given options against the tightest constraint.

Step 4: Check Row 2 in all options. Option (i) shows 2 shaded cells. Option (iii) shows 2 shaded cells. Both are immediately discarded.

Step 5: Verify the survivor. Option (ii) has exactly 1 shaded cell in Row 2. Now verify its columns: Col 1 (Rows 1, 4) = 2. Col 2 (Rows 2, 4) = 2. Col 3 (Rows 1, 3) = 2. Col 4 (Rows 1, 3) = 2. All column sums are exactly 2. Option (ii) is the unique valid configuration.

Question 3 · Patterns, Series, Coding and Analogies · 2025_Set2 MCQ

If IMAGE and FIELD are coded as FHBNJ and EMFJG respectively then, which one among the given options is the most appropriate code for BEACH ?

A.
CEADP
B.
IDBFC
C.
JGIBC
D.
IBCEC

Correct Answer:

B

Step-by-Step Solution

Insight: The coding rule involves reversing the original word and then applying a uniform forward shift of +1 to each letter.

Exam route: Reverse BEACH to get HCAEB. Shift each letter forward by 1: H→I, C→D, A→B, E→F, B→C. Result is IDBFC.

Learning route:

Step 1: Test direct left-to-right shift for IMAGE → FHBNJ. I(9) to F(6) is -3, M(13) to H(8) is -5. Inconsistent.

Step 2: Apply the Reverse Test. Reverse IMAGE to get EGAMI.

Step 3: Calculate shift: E(5)→F(6) [+1], G(7)→H(8) [+1], A(1)→B(2) [+1], M(13)→N(14) [+1], I(9)→J(10) [+1]. The rule is confirmed: Reverse +1.

Step 4: Verify with FIELD. Reverse to DLEIF. Shift +1: D→E, L→M, E→F, I→J, F→G. Result EMFJG. Matches perfectly.

Step 5: Apply to BEACH. Reverse to HCAEB. Shift +1: H→I, C→D, A→B, E→F, B→C. Final code is IDBFC.

Question 4 · Graph, Route and Tournament Reasoning · 2026_Set2 MCQ

Figures (i) and (ii) represent intercity highway systems. The black dots represent cities and the line segments between them represent intercity highways.
A salesperson needs to make a trip. She needs to start from a city, visit each of the remaining cities exactly once, and finally return to the same city from which she started.

Which one of the following options is then true?

A.
Such a trip is possible for (i), but not for (ii).
B.
Such a trip is possible for (ii), but not for (i).
C.
Such a trip is possible for both (i) and (ii).
D.
Such a trip is possible neither for (i) nor for (ii).

Correct Answer:

A

Step-by-Step Solution

Insight: A Hamiltonian cycle requires every vertex to have a degree of exactly 2 within the cycle. Graph (ii) has three vertices of degree 2, which forces a contradiction at the central vertex. Graph (i) is a 4x4 grid, which is bipartite with equal partitions, allowing a valid cycle.

Exam route: For (ii), identify vertices with degree 2. Their incident edges must be in the cycle. This forces the central vertex to have degree 3 in the cycle, which is impossible. Thus, (ii) has no Hamiltonian cycle. For (i), a 4x4 grid has a known Hamiltonian cycle (e.g., a snake pattern that closes). Thus, (i) is possible, (ii) is not.

Learning route:

Understand the goal: A trip visiting every city exactly once and returning to the start is a Hamiltonian cycle.
Analyze Graph (ii): It has 5 vertices. The top-left, bottom, and top-right vertices each have exactly 2 connections (degree 2).
Apply the Degree-Two Vertex Rule: In any Hamiltonian cycle, if a vertex has degree 2, both of its edges must be part of the cycle.
Trace the forced edges in (ii): The three degree-2 vertices force 6 edges. However, these edges all converge on the central vertex, giving it a degree of 3 in the supposed cycle. A cycle can only have degree 2 for every vertex. This is a contradiction, so (ii) is impossible.
Analyze Graph (i): It is a 4x4 grid graph. It is bipartite with 8 black and 8 white vertices. Since the partitions are equal, a Hamiltonian cycle is possible. We can explicitly construct one by tracing the perimeter and weaving through the center without repeating vertices.
Conclusion: Possible for (i), not for (ii).

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Logic PYQs for GATE CS

About Logic Previous Year Questions (PYQs)

Logic Weightage in GATE CS

Logic Chapter Matrix

More from Analytical Aptitude

One Solved Question from Each Logic Chapter

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution