HSC Arts (English Medium) १२ वीं कक्षा - Maharashtra State Board Important Questions

Write the converse and contrapositive of the statement -
“If two triangles are congruent, then their areas are equal.”

Examine whether each of the following statement patterns is a tautology or a contradiction or a contingency.

[~(~p ∧ ~q)] v q

Construct the switching circuit for the following statement : [p v (~ p ∧ q)] v [(- q ∧ r) v ~ p]

Express the following statement in symbolic form and write its truth value.

"If 4 is an odd number, then 6 is divisible by 3 "

Prove that the following statement pattern is equivalent :

(p ∨ q) →  r and (p → r) ∧ (q → r)

If p ˄ q = F, p → q = F, then the truth value of p and q is ______.

Using truth table prove that p ↔ q = (p ∧ q) ∨ (~p ∧ ~q).

Construct the simplified circuit for the following circuit:

Write truth values of the following statements :`sqrt5` is an irrational number but 3 +`sqrt 5` is a complex number.

Using truth table, prove the following logical equivalence:

(p ∧ q) → r ≡ p → (q → r)

Write down the following statements in symbolic form :

(A) A triangle is equilateral if and only if it is equiangular.
(B) Price increases and demand falls

Without using the truth table show that P ↔ q ≡ (p ∧ q) ∨ (~ p ∧ ~ q)

Using truth table examine whether the following statement pattern is tautology, contradiction or contingency `(p^^~q) harr (p->q)`

Write the dual of the following statements: (p ∨ q) ∧ T

Using truth table prove that ∼p ˄ q ≡ (p ˅ q) ˄ ∼p

Write truth values of the following statements: ∃ n ∈ N such that n + 5 > 10.

Write the following statement in symbolic form and find its truth value:
∀ n ∈ N, n² + n is an even number and n² - n is an odd number.

Using truth tables, examine whether the statement pattern (p ∧ q) ∨ (p ∧ r) is a tautology, contradiction or contingency.

Construct the switching circuit for the statement (p ∧ q) ∨ (~ p) ∨ (p ∧ ~ q).

If p, q, r are the statements with truth values T, F, T, respectively then find the truth value of (r ∧ q) ↔ ∼ p