(i) Given compound statement is of the form ' $p v q$ '. Since, the statement ' $p v q$ ' has the truth value $T$ whenever either $p$ or $q$ or both have the truth value $T$.

So, it is true statement.

Its component statements are

$p: 57$ is divisible by 2 . [false]

$q: 57$ is divisible by 3 . [true]

(ii) Given compound statement is of the form ' $p \wedge q$ '. Since, the statement ' $p \wedge q$ ' have the truth value $T$ whenever both $p$ and $q$ have the truth value $T$. So, it is a true statement.

Its component statements are

$p: 24$ is multiple of 4 [true]

$q: 24$ is multiple of 6 . [true]

(iii) It is a false statement. Since ' $p \wedge q$ ' has truth value $F$ whenever either $p$ or $q$ or both have the truth value $F$.

Its component statements are

$p$ : All living things have two eyes. [false]

$q$ : All living things have two legs. [false]

(iv) It is a true statement.

Its component statements are

$p: 2$ is an even number. [true]

$q: 2$ is a prime number. [true]

(i) The number 17 is not prime.

(ii) $2+7 \neq 6$.

(iii) Violets are not blue.

(iv) $\sqrt{5}$ is not a rational number.

(v) 2 is a prime number.

(vi) Every real number is not an irrational number.

(vii) Cow has not four legs.

(viii) A leap year has not 366 days.

(ix) There exist similar triangles which are not congruent.

(x) Area of a circle is not same as the perimeter of the circle.

(i) $p$ : Rahul passed in Hindi.

$q$ : Rahul passed in English.

$p \wedge q$ : Rahul passed in Hindi and English.

(ii) $p: x$ is even integers.

$q: y$ is even integers.

$p \cap q: x$ and $y$ are even integers.

(iii) $p: 2$ is factor of 12 .

$q: 3$ is factor of 12 .

$r: 6$ is factor of 12 .

$p \wedge q \wedge r: 2,3$ and 6 are factor of 12 .

(iv) $p: x$ is an odd integer.

$q:(x+1)$ is an odd integer.

$p \vee q$ : Either $x$ or $(x+1)$ is an odd integer.

(v) $p$ : A number is divisible by 2 .

$q$ : A number is divisible by 3.

$p \vee q$ : A number is either divisible by 2 or 3 .

(vi) $p: x=2$ is a root of $3 x^2-x-10=0$.

$q: x=3$ is a root of $3 x^2-x-10=0$.

$p \vee q$ : Either $x=2$ or $x=3$ is a root of $3 x^2-x-10=0$.

(vii) $p$ : Students can take Hindi as an optional paper.

$q$ : Students can take English as an optional subject.

$p \vee q$ : Students can take Hindi or English as an optional paper.

(i) Let $p$ : All rational numbers are real.

$q$ : All rational numbers are complex.

$\sim p$ : All rational number are not real.

$\sim q$ : All rational numbers are not complex.

$\sim(p \wedge q):$ All rational numbers are not real or not complex. $\quad[\because \sim(p \wedge q)=\sim p \vee \sim q]$

(ii) Let $p$ : All real numbers are rationals.

$q$ : All real numbers are irrational.

Then, the negation of the above statement is given by

$\sim(p \vee q)$ : All real numbers are not rational and all real numbers are not irrational.

$$[\because \sim(p \vee q)=\sim p \wedge \sim q]$$

(iii) Let $p$ : $x=2$ is root of quadratic equation $x^2-5 x+6=0$.

$q: x=3$ is root of quadratic equation $x^2-5 x+6=0$.

Then, the negation of conjunction of above statement is given by

$\sim(p \wedge q): x=2$ is not a root of quadratic equation $x^2-5 x+6=0$ or $x=3$ is not a root of the quadratic equation $x^2-5 x+6=0$.

(iv) Let $p$ : A triangle has 3 -sides.

$q$ : A triangle has 4 -sides.

Then, negation of disjunction of the above statement is given by $\sim(p \vee q)$ : A triangle has neither 3-sides nor 4-sides.

(v) Let $p: 35$ is a prime number.

$q: 35$ is a composite number.

Then, negation of disjunction of the above statement is given by $\sim(p \vee q): 35$ is not a prime number and it is not a composite number.

(vi) Let $p$ : All prime integers are even.

$q$ : All prime integers are odd.

Then negation of disjunction of the above statement is given by

$\sim(p \vee q)$ : All prime integers are not even and all prime integers are not odd.

(vii) Let $p:|x|$ is equal to $x$.

$q:|x|$ is equal to $-x$.

Then negation of disjunction of the above statement is given by

$\sim(p \vee q):|x|$ is not equal to $x$ and it is not equal to $-x$.

(viii) Let $p: 6$ is divisible by 2 .

$q: 6$ is divisible by 3 .

Then, negation of conjunction of above statement is given by

$\sim(p \wedge q): 6$ is not divisible by 2 or it is not divisible by 3

We know that, some of the common expressions of conditional statement $p \rightarrow q$ are

(i) if $p$, then $q$

(ii) $q$ if $p$

(iii) ponly if $q$

(iv) $p$ is sufficient for $q$

(v) $q$ is necesary for $p$

(vi) $\sim q$ implies $\sim p$

So, use above information to get the answer

(i) If the number is odd number, then its square is odd number.

(ii) If you take the dinner, then you will get sweet dish.

(iii) If you will not study, then you will fail.

(iv) If an integer is divisible by 5 , then its unit digits are 0 or 5 .

(v) If the number is prime, then its square is not prime.

(vi) If $a, b$ and $c$ are in $A P$, then $2 b=a+c$.