The maximum number of compound propositions, out of p ∨ r ∨ s, p ∨ r ∨ ~s, p ∨ ~q ∨ s, ~p ∨ ~r ∨ s, ~p ∨ ~r ∨ ~s, ~p ∨ q ∨ ~s, q ∨ r ∨ ~s, q ∨ ~r ∨ ~s, ~p ∨ ~q ∨ ~s that can be made simultaneously true by an assignment of the truth values to p, q, r and s, is equal to ____________ .
Correct Answer: 9
Solution and Explanation
The correct answer is 9
There are total 9 compound propositions, out of which 6 contain ~s. So if we assign s as false, these 6 propositions will be true.
In remaining 3 compound propositions, two contain p and the third contains ~r. So if we assign p and r as true and false respectively, these 3 propositions will also be true.
Therefore, maximum number of propositions that can be true are 9.
Top Questions on mathematical reasoning
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If integers 1 through 5 are used to construct such a number, the value of P is: For the clock shown in the figure, if
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A bag contains Violet (V), Yellow (Y), Red (R), and Green (G) balls. On counting them, the following results are obtained:
(i) The sum of Yellow balls and twice the number of Violet balls is 50.
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(iv) The sum of Violet and twice the number of Red balls is 50.
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If the domain of the function \( f(x) = \frac{1}{\sqrt{3x + 10 - x^2}} + \frac{1}{\sqrt{x + |x|}} \) is \( (a, b) \), then \( (1 + a)^2 + b^2 \) is equal to:
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- Let \( A \) be a \( 3 \times 3 \) matrix such that \( X^T AX = 0 \) for all nonzero \( 3 \times 1 \) matrices \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \). \[ A = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad A = \begin{bmatrix} 1 \\ 4 \\ -5 \end{bmatrix}, \quad A = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \quad A = \begin{bmatrix} 0 \\ 4 \\ -8 \end{bmatrix} \] If \( \det(\text{adj}(2(A + I))) = 2^{\alpha} 3^{\beta} 5^{\gamma} \), \( \alpha, \beta, \gamma \in \mathbb{N} \), then \( \alpha^2 + \beta^2 + \gamma^2 \) is:
Concepts Used:
Mathematical Reasoning
Mathematical reasoning or the principle of mathematical reasoning is a part of mathematics where we decide the truth values of the given statements. These reasoning statements are common in most competitive exams like JEE and the questions are extremely easy and fun to solve.
Types of Reasoning in Maths:
Mathematically, reasoning can be of two major types such as:
- Inductive Reasoning - In this, method of mathematical reasoning, the validity of the statement is examined or checked by a certain set of rules, and then it is generalized. The principle of mathematical induction utilizes the concept of inductive reasoning.
- Deductive Reasoning - The principle is the opposite of the principle of induction. Contrary to inductive reasoning, in deductive reasoning, we apply the rules of a general case to a provided statement and make it true for particular statements. The principle of mathematical induction utilizes the concept of deductive reasoning.