Logic Short Notes for GATE CS

GATE CS Logic: 4 chapters, 21 previous year questions (100% of Analytical Aptitude), 355 practice questions and one solved question from each chapter.

About Logic Short Notes

Quick revision sheets for Logic in GATE CS. Every chapter is condensed into key formulas, shortcuts and common traps so you can revise 4 chapters fast before the exam.

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

One Solved Question from Each Logic Chapter

Question 1 · Deductive Reasoning, Statements and Syllogisms MCQ

Based on the following statements about a software system, determine how many of the listed conclusions MUST be true.

Statement 1: If a module is tested, then it is documented.

Statement 2: Some documented modules are reviewed.

Statement 3: No untested module is documented.

Conclusion I: Some tested modules are reviewed.

Conclusion II: All reviewed modules are documented.

Conclusion III: Some reviewed modules are tested.

A.
0
B.
1
C.
2
D.
3

Correct Answer:

Step-by-Step Solution

Key idea: This is a casework question on categorical and conditional statements, recognisable by the "how many MUST be true" format combined with conditional premises. The critical move is translating Statement 3 via its contrapositive to discover that two sets are actually equal.

Step 1: Translate each statement into set relations.

Statement 1: Tested $\subseteq$ Documented.
Statement 2: Documented $\cap$ Reviewed $\neq = \emptyset$ (at least one module is both documented and reviewed).
Statement 3: "No untested module is documented" means if a module is not tested, then it is not documented. Symbolically: $\neg$ Tested $⟹ \neg$ Documented.

Step 2: Apply the contrapositive to Statement 3.

The contrapositive of $\neg$ Tested $⟹ \neg$ Documented is: Documented $⟹$ Tested.

This means Documented $\subseteq$ Tested.

Step 3: Combine with Statement 1.

We have Tested $\subseteq$ Documented (from S1) and Documented $\subseteq$ Tested (from S3 contrapositive).

Therefore: Tested $=$ Documented. These are exactly the same set.

Step 4: Evaluate each conclusion.

Conclusion I: "Some tested modules are reviewed." Since Tested $=$ Documented, and Statement 2 says some documented modules are reviewed, it follows that some tested modules are reviewed. MUST be true. $✓$
Conclusion II: "All reviewed modules are documented." Statement 2 only says some documented modules are reviewed. It does not say that every reviewed module is documented. There could be reviewed modules outside the documented set. NOT necessarily true. $\times$
Conclusion III: "Some reviewed modules are tested." Since Tested $=$ Documented, and some documented modules are reviewed (S2), those same modules are both reviewed and tested. MUST be true. $✓$

Step 5: Count. Exactly 2 conclusions (I and III) must be true.

Common trap: Misreading Statement 3 as "No tested module is documented" (a sign error flipping "untested" to "tested"). This would make Tested $\cap$ Documented $= \emptyset$ , which combined with S1 forces Tested $= \emptyset$ , making both I and III false and yielding answer 0.

Answer: 2

Question 2 · Relationships, Ordering and Arrangement Puzzles MCQ

Consider a triangle with its 3 corners and the 3 mid-points of its sides, making 6 distinct points in total. What is the maximum number of these points that can be selected such that no three selected points are collinear?

A.
2
B.
3
C.
4
D.
5

Correct Answer:

Step-by-Step Solution

Key idea: This is a maximum selection problem based on geometric collinearity. We must observe that each side of the triangle contains exactly 3 collinear points (2 corners and 1 mid-point).

Step 1: Total points = 6. We want to find the maximum subset with no 3 collinear points.

Step 2: Check if 5 is possible. If we select 5 points, we exclude exactly 1 point.

If we exclude a corner, the side opposite to it still contains its 2 corners and 1 mid-point (3 collinear points).
If we exclude a mid-point, the other two sides still contain their 2 corners and 1 mid-point each (3 collinear points each).

Thus, any selection of 5 points will contain at least one full side, meaning 3 collinear points. So 5 is impossible.

Step 3: Check if 4 is possible. We can select 2 corners and 2 mid-points, for example, ${C_{1}, C_{2}, M_{2}, M_{3}}$ .

Side $C_{1} C_{2}$ contains only $C_{1}, C_{2}$ (since $M_{1}$ is excluded).
Side $C_{2} C_{3}$ contains only $C_{2}, M_{2}$ (since $C_{3}$ is excluded).
Side $C_{3} C_{1}$ contains only $M_{3}, C_{1}$ (since $C_{3}$ is excluded).

No three points are collinear. Thus, 4 is possible.

Answer: 4

Question 3 · Patterns, Series, Coding and Analogies MCQ

In a certain code language, the word $CRANE$ is coded as $FOZSD$ and the word $GLASS$ is coded as $TTZMH$ . Based on this coding scheme, determine which of the following statements is/are TRUE:

I. The coding rule applies a uniform forward shift of $+ 1$ to all letters after reversing the word.

II. The middle letter of the reversed word is shifted backward by $1$ .

III. The code for the word $LIGHT$ under this scheme is $UIFJM$ .

Options:

A. Only I and II

B. Only II and III

C. Only I and III

D. I, II and III

A.
Only I and II
B.
Only II and III
C.
Only I and III
D.
I, II and III

Correct Answer:

Step-by-Step Solution

Key idea: This is a statement-truth verification question on a constrained reverse-shift pattern. Recognise it by checking if a simple reverse-and-shift holds uniformly; if it fails at a specific position, look for an exception constraint.

Step 1: Test Statement I (Uniform Reverse Shift).

Reverse $CRANE$ to get $ENARC$ .

Apply $+ 1$ shift: $E \to F$ , $N \to O$ , $A \to B$ , $R \to S$ , $C \to D$ .

Result: $FOSBD$ .

The given code is $FOZSD$ . The third letter $A$ did not shift to $B$ ; it shifted to $Z$ (which is $A - 1$ ). Thus, Statement I is FALSE.

Step 2: Test Statement II (Middle Letter Exception).

In $ENARC$ , the middle letter is $A$ (position 3). In the code $FOZSD$ , the middle letter is $Z$ .

Verify with $GLASS$ : Reversed is $SSALG$ . Shift $+ 1$ except middle: $S \to T$ , $S \to T$ , $A \to Z$ , $L \to M$ , $G \to H$ .

Result: $TTZMH$ . This matches the given code perfectly. Thus, Statement II is TRUE.

Step 3: Test Statement III (Apply to $LIGHT$ ).

Reverse $LIGHT$ to get $THGIL$ .

Apply $+ 1$ shift except middle ( $G$ ):

$T (20) + 1 = U (21)$

$H (8) + 1 = I (9)$

$G (7) - 1 = F (6)$

$I (9) + 1 = J (10)$

$L (12) + 1 = M (13)$

Result: $UIFJM$ . This matches Statement III. Thus, Statement III is TRUE.

Answer: Only II and III are true.

Question 4 · Graph, Route and Tournament Reasoning MCQ

An archipelago consists of 9 islands, labeled A through I, separated by a network of 14 navigable channels. Four bridges have already been constructed: between A and B, between B and C, between C and A, and between D and E. We need to build additional bridges so that it is possible to travel between any two island using only bridges. What is the minimum number of additional bridges required?

A.
5
B.
4
C.
6
D.
8

Correct Answer:

Step-by-Step Solution

Key idea: This is a graph connectivity question where existing bridges contain a cycle. The trap is blindly subtracting the total bridge count from $V - 1$ without recognising that one bridge in the cycle is redundant.

Step 1: Model the 9 islands as vertices: A, B, C, D, E, F, G, H, I. So $V = 9$ .

Step 2: Trace existing bridges to identify connected components:

Bridges A–B, B–C, C–A form a cycle on {A, B, C} → one component.
Bridge D–E → one component {D, E}.
Islands F, G, H, I have no bridges → four isolated components.

Total components $C = 1 + 1 + 4 = 6$ .

Step 3: To merge $C$ components into one connected network, we need exactly $C - 1$ new bridges.

$Additional bridges = 6 - 1 = 5$

Wrong path (Option B): A student computes $V - 1 = 8$ and subtracts all 4 existing bridges: $8 - 4 = 4$ . This fails because the cycle A–B–C–A uses 3 bridges but only 2 are needed to connect those 3 islands. One bridge is redundant and does not reduce the component count. The correct method is to count components directly and apply $C - 1$ .

Generalization: When existing bridges contain cycles, blindly subtracting the bridge count from $V - 1$ underestimates the answer. Always count components directly.

Verification: Adding 5 bridges to connect 6 components merges them into 1 network. Total edges = 4 existing + 5 new = 9, which is $V - 1 + 1$ (the extra 1 is the redundant cycle edge). All 9 islands are reachable.

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Logic Short Notes for GATE CS

About Logic Short Notes

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