Consider :
Statement − I : (p ∧ ~ q) ∧ (~ p ∧ q) is a fallacy.
Statement − II : (p → q) ↔ (~ q → ~ p) is a tautology.
Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I
Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I
Statement -I is True; Statement -II is False.
Statement -I is False; Statement -II is True
B.
Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I
p | q | ~p | ~q | p^~q | ~p^q | (p^~q)^(~p^q) |
T T F F |
T F T F |
F F T T |
F T F T |
F T F F |
F F T F |
F F F F |
p | q | ~p | ~q | p ⇒ q | ~ q ⇒ ~ p | (p ⇒ q) ⇔ (~ q ⇒ ~ p) |
T T F F |
T F T F |
F F T T |
F T F T |
T F T T |
T F T T |
T T T T |
Consider the following statements:
(a) Mode can be computed from histogram
(b) Median is not independent of change of scale
(c) Variance is independent of the change of origin and scale. Which of these is/are correct?
only (a)
only (b)
only (a) and (b)
(a), (b) and (c)
C.
only (a) and (b)
Consider the following statements
P: Suman is brilliant
Q: Suman is rich
R: Suman is honest. The negation of the statement ì Suman is brilliant and dishonest if and only if Suman is richî can be ex- pressed as
~ P ^ (Q ↔ ~ R)
~ (Q ↔ (P ^ ~R)
~ Q ↔ ~ P ^ R
~ (P ^ ~ R)↔ Q
B.
~ (Q ↔ (P ^ ~R)
Negation of (PΛ~ R) ↔ Q is ~ ↔(PΛ ~ R)↔Q)
It may also be written as ~ (Q ↔ (PΛ ~ R))
Let p be the statement “x is an irrational number”, q be the statement “y is a transcendental number”, and r be the statement “x is a rational number iff y is a transcendental number”.
Statement –1: r is equivalent to either q or p
Statement –2: r is equivalent to ∼ (p ↔ ∼ q).
Statement −1 is false, Statement −2 is true
Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1
Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.
Statement − 1 is true, Statement − 2 is false.
D.
Statement − 1 is true, Statement − 2 is false.
Given statement r = ∼ p ↔ q
Statement −1 : r_{1} = (p ∧ ∼ q) ∨ (∼ p ∧ q)
Statement −2 : r_{2} = ∼ (p ↔ ∼ q) = (p ∧ q) ∨ (∼ q ∧ ∼ p)
From the truth table of r, r1 and r2,
r = r1.
Hence Statement − 1 is true and Statement −2 is false.
Let S be a non empty subset of R. Consider the
following statement:
P: There is a rational number x∈S such that x > 0.
Which of the following statements is the negation of the statement P?
There is a rational number x∈S such that x ≤ 0.
There is no rational number x∈ S such that x≤0.
Every rational number x∈S satisfies x ≤ 0.
x∈S and x ≤ 0 ⇒ x is not rational.
C.
Every rational number x∈S satisfies x ≤ 0.