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Logic Short Notes for GATE DA

GATE DA Logic: 2 chapters, 5 previous year questions (100% of Analytical Aptitude), 278 practice questions and one solved question from each chapter.

A question from this chapter

Question 1

Level 3: Exam Standard

Let p_{1}, p_{2}, \dots, p_{7} be propositions. We are given that p_{i} \to \neg p_{i + 1} is true for i = 1, 2, \dots, 6, and p_{7} \to p_{1} is true. What is the maximum number of these propositions that can be simultaneously true?

Question 2

Level 3: Exam Standard

A company's profit function P (x) has a single valley at x = 5 . It is given that P (0) = 10, P (5) = - 20, and P (10) = 30 . What is the minimum possible number of integer values of x in the interval [0, 10] for which P (x) \leq 0 ?

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Logic Short Notes for GATE DA

GATE DA Logic: 2 chapters, 5 previous year questions (100% of Analytical Aptitude), 278 practice questions and one solved question from each chapter.

About Logic Short Notes

Quick revision sheets for Logic in GATE DA. Every chapter is condensed into key formulas, shortcuts and common traps so you can revise 2 chapters fast before the exam.

Logic Weightage in GATE DA

Logic accounts for 5 of 5 Analytical Aptitude previous year questions in our bank (100%), about 1.7 per paper across 3 papers.

Logic Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Propositional Logic and Logical Equivalence	Conditional Statements, Tautologies and Logical Equivalence	3	60%	167
Logical Reasoning and Constraint Deduction	Monotonicity-Based Logical Deduction, Constraint-Based Sequencing and Elimination	2	40%	111

One Solved Question from Each Logic Chapter

Question 1 · Propositional Logic and Logical Equivalence NAT

Let $p_{1}, p_{2}, \dots, p_{7}$ be propositions. We are given that $p_{i} \to \neg p_{i + 1}$ is true for $i = 1, 2, \dots, 6$ , and $p_{7} \to p_{1}$ is true. What is the maximum number of these propositions that can be simultaneously true?

Correct Answer:

4.00

Step-by-Step Solution

Key idea: The conditional $A \to \neg B$ means we cannot have $A$ True and $B$ True. This translates to an independent set problem on a graph.

Step 1: Translate the conditions. For $i = 1 \dots 6$ , $p_{i} \to \neg p_{i + 1}$ means we cannot have both $p_{i}$ and $p_{i + 1}$ be True. This forms a path graph $P_{7}$ where adjacent vertices cannot both be True.

Step 2: Analyze the boundary condition. $p_{7} \to p_{1}$ means if $p_{7}$ is True, $p_{1}$ MUST be True. It does NOT prevent them from both being True.

Step 3: Maximize the number of True propositions. We want to find the maximum independent set in $P_{7}$ that also satisfies $p_{7} \to p_{1}$ .

The maximum independent set in $P_{7}$ has size $⌈ 7/2 ⌉ = 4$ . The only set of size 4 is ${1, 3, 5, 7}$ .

Step 4: Check the boundary condition for ${1, 3, 5, 7}$ . Here $p_{7}$ is True and $p_{1}$ is True. The condition $p_{7} \to p_{1}$ becomes $T \to T$ , which is True.

Step 5: Thus, the maximum number of propositions that can be simultaneously True is 4.

Answer: 4.00

Question 2 · Logical Reasoning and Constraint Deduction NAT

A company's profit function $P (x)$ has a single valley at $x = 5$ . It is given that $P (0) = 10$ , $P (5) = - 20$ , and $P (10) = 30$ . What is the minimum possible number of integer values of $x$ in the interval $[0, 10]$ for which $P (x) \leq 0$ ?

Correct Answer:

1.00

Step-by-Step Solution

Key idea: This is a single valley pattern question where you must deduce the location of the roots and then minimize the integer points between them by exploiting the lack of linearity constraints.

Step 1: Analyze the monotonicity intervals.

A single valley at $x = 5$ means $P (x)$ strictly decreases for $x < 5$ and strictly increases for $x > 5$ .

Step 2: Locate the roots.

On $[0, 5]$ : $P (0) = 10 > 0$ and $P (5) = - 20 < 0$ . Since $P$ is strictly decreasing here, it crosses 0 exactly once. Let this root be $r_{1} \in (0, 5)$ .
On $[5, 10]$ : $P (5) = - 20 < 0$ and $P (10) = 30 > 0$ . Since $P$ is strictly increasing here, it crosses 0 exactly once. Let this root be $r_{2} \in (5, 10)$ .

Step 3: Determine where $P (x) \leq 0$ .

The function is negative between the roots, so $P (x) \leq 0$ for $x \in [r_{1}, r_{2}]$ .

Step 4: Minimize the number of integers in $[r_{1}, r_{2}]$ .

We know $P (5) = - 20 \leq 0$ , so $x = 5$ is always in the interval.

To minimize the integers, we want $r_{1}$ and $r_{2}$ to be as close to 5 as possible.

Can $r_{1} > 4$ ? Yes, if $P (4) < 0$ . For example, $P (4) = - 10$ .

Can $r_{2} < 6$ ? Yes, if $P (6) < 0$ . For example, $P (6) = - 10$ .

If $r_{1} \in (4, 5)$ and $r_{2} \in (5, 6)$ , the only integer in $[r_{1}, r_{2}]$ is 5.

Step 5: Verify achievability.

A valid function could pass through $(0, 10), (4, - 10), (5, - 20), (6, - 10), (10, 30)$ with strict monotonicity in the respective intervals.

Answer: 1.00

A question from this chapter

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Logic Short Notes for GATE DA

About Logic Short Notes

Logic Weightage in GATE DA

Logic Chapter Matrix

More from Analytical Aptitude

One Solved Question from Each Logic Chapter

Step-by-Step Solution

Step-by-Step Solution