Question

Let
$p:57$ is an odd prime number
$q:4$ is a divisor of $12$
$r:15$ is the LCM of $3$ and $5$
be three simple logical statements. Which one of the following is true?
(a) $p\vee \left( \sim q\wedge r \right)$
(b) $\sim p\vee \left( q\wedge r \right)$
(c) $\left( p\wedge q \right)\vee \sim r$
(d) $p\vee \left( q\wedge r \right)$
(e) $\left( p\vee q \right)\wedge r$

Answer 1

Hint: Check if the given three statements are true or false and then check which of the statements among the options is true based on the rules of intersection, union and complementation.
Complete step-by-step answer:
We have three statements $p,q,r$. We have to check if the given statements are true or false. We will begin by examining each of the statements.
We have the statement $p$ which states that $57$ is an odd prime number. $57$ is an odd number as it’s not divisible by $2$. However, we can write $57$ as $57=3\times 19$, thus showing that it’s a composite number.
Hence, the statement $p$ is false.
We will now check the statement $q$ which states that $4$ is a divisor of $12$. We know that $\dfrac{12}{4}=3$.
Hence, the statement $q$ is true.
We will now check the statement $r$ which states that $15$ is LCM of $5$ and $3$. We know that LCM of any two prime numbers is the product of numbers itself. As $5$ and $3$ are prime, their LCM is $15$.
Hence, the statement $r$ is true.
We will now check if the given options are true or false.
We know that union (denoted by $\vee$) of any two true (or false) statements is true (or false). Union of a true statement and a false statement is true as well.
Also, intersection (denoted by $\wedge$) of any two true (or false) statements is true (or false). While, intersection of a true and a false statement is false.
The complement (denoted by $\sim$) of a true statement is false and a false statement is true.
We will now use these rules to check all the options.
We will consider option (a) which is $p\vee \left( \sim q\wedge r \right)$. This is the union of a false statement with the intersection of two false statements, which is union of a false statement with a false statement. Hence, this is false.
We will consider option (b) which is $\sim p\vee \left( q\wedge r \right)$. This is the union of a true statement with the intersection of two true statements, which is the union of a true statement with a true statement. Hence, this is true.
We will consider option (c) which is $\left( p\wedge q \right)\vee \sim r$. This is the intersection of a false and a true statement with the union of a false statement, which is union of a false statement with a false statement. Hence, this is false.
We will consider option (d) which is $p\vee \left( q\wedge r \right)$. This is the union of a false statement with the intersection of two true statements, which is the union of a false statement with a true statement. Hence, this is true.
We will consider option (e) which is $\left( p\vee q \right)\wedge r$. This is the union of a true statement and a false statement with the intersection of a true statement, which is the intersection of a true statement with a true statement. Hence, this is true.
Hence, the true statements are $\sim p\vee \left( q\wedge r \right),p\vee \left( q\wedge r \right),\left( p\vee q \right)\wedge r$ which are options (b), (d), (e).
Note: It’s necessary to keep in mind the rules of union, intersection and complementation of logical statements to check if the given statements are true or false.