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 true; Statement −2 is not a correct explanation for Statement −1.

The statement p → (q → p) is equivalent to

• p → (p → q) 

• p → (p ∨ q)

• p → (p ∧ q)

• p → (p ∧ q)

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)

• only (a) and (b)

The Boolean expression ~ (p v q )v (-p ∧q ) is equivalent to

• ~q

• ~ p

• p

• q

If p, q, r are simple propositions with truth values T, F, T, then the truth value of $\left(~ \mathrm{p} \vee \mathrm{q}\right) \wedge ~ \mathrm{r} \Rightarrow \mathrm{p}$ is

• true

• false
  
  

• true, if r is false

• true, if q is true

If p : It rains today, q : I go to school, r : I shall meet any friends ands : I shall go for a movie, then which of the following is the proportion? If it does not rain or if I do not go to school, then I shall meet my friend and go for a movie.

• $\left(~ \mathrm{p} \wedge ~ \mathrm{q}\right) \Rightarrow \left(\mathrm{r} \wedge \mathrm{s}\right)$

• $~ \left(\mathrm{p} \wedge \mathrm{q}\right) \Rightarrow \left(\mathrm{r} \wedge \mathrm{s}\right)$

• $~ \left(\mathrm{p} \vee \mathrm{q}\right) \Rightarrow \left(\mathrm{r} \wedge \mathrm{s}\right)$

• None of these

The statement $\left(\mathrm{p} \Rightarrow \mathrm{q}\right) \iff \left(~ \mathrm{p} \wedge \mathrm{q}\right)$ is

• tautology

• contradiction

• Neither (a) nor (b)

• None of these

The negation of $\left(~ \mathrm{p} \wedge \mathrm{q}\right) \vee \left(\mathrm{p} \wedge ~ \mathrm{q}\right)$ is

• $\left(\mathrm{p} \vee ~ \mathrm{q}\right) \vee \left(~ \mathrm{p} \vee \mathrm{q}\right)$

• $\left(~ \mathrm{p} \vee \mathrm{q}\right) \vee \left(~ \mathrm{p} \vee \mathrm{q}\right)$

• $\left(\mathrm{p} \wedge ~ \mathrm{q}\right) \wedge \left(~ \mathrm{p} \vee \mathrm{q}\right)$

• $\left(\mathrm{p} \wedge ~ \mathrm{q}\right) \wedge \left(\mathrm{p} \vee ~ \mathrm{q}\right)$

If a₁, a₂ and a₃ be any positive real numbers, then which of the following statement is not true?

• 3a₁a₂a₃ $\le {{\mathrm{a}}_1}^3 + {{\mathrm{a}}_2}^3 + {{\mathrm{a}}_3}^3$

• $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2} + \frac{{\mathrm{a}}_2}{{\mathrm{a}}_3} + \frac{{\mathrm{a}}_3}{{\mathrm{a}}_1} \ge 3$

• (a₁ + a₂ + a₃)$\left(\frac{1}{{\mathrm{a}}_1} + \frac{1}{{\mathrm{a}}_2} + \frac{1}{{\mathrm{a}}_3}\right) \ge 9$

• (a₁ . a₂ . a₃)$\left(\frac{1}{{\mathrm{a}}_1} + \frac{1}{{\mathrm{a}}_2} + \frac{1}{{\mathrm{a}}_3}\right) \ge 27$

The proposition $~ \left(\mathrm{p} \iff \mathrm{q}\right)$ is equivalent to

• $\left(\mathrm{p} \vee ~ \mathrm{q}\right) \wedge \left(\mathrm{q} \wedge ~ \mathrm{p}\right)$

• $\left(\mathrm{p} \wedge ~ \mathrm{q}\right) \wedge \left(\mathrm{q} \wedge ~ \mathrm{p}\right)$

• $\left(\mathrm{p} \wedge ~ \mathrm{q}\right) \wedge \left(\mathrm{q} \wedge ~ \mathrm{p}\right)$

• None of the above