Question

Consider the following statements.

I. If $A_n$ is the set of first $n$ prime numbers, Then $\underset{n=2}{\overset{10}{\cup }}{A}_n$ is equal to $\{2,3,5,7,11,13,17,19,23,29\}$

II. If $A$ and $B$ are two sets such that $n(A\cup B)=50,n\left(A\right)=28,n\left(B\right)=32,$ then $n(A\cap B)=10.$

Which of these is correct?

• A. Only I is true.
• B. Only II is true.
• C. Both are true.
• D. Both are false.

Answer 1

The correct option is C

Both are true.

Explanation for the correct option:

Step 1. Check statement I.

If $A$and $B$ are two sets such that $A$ is a subset of $B$, then $A\cup B=B$.

Here $A_2\subset A_3\subset A_4\dots \subset A_{10}$

The superset is $A_{10}$.

We have to find $A_2\cup A_3\cup \dots A_{10}.$

It will be equal $A_{10}$.

$A_{10}$ is the set of the first $10$ prime numbers,

So $A_{10}=\{2,3,5,7,11,13,17,19,23,29\}$

Thus statement I is correct.

Step2. Check statement II.

If $A$ and $B$ are two sets such that $n(A\cup B)=50,n\left(A\right)=28,n\left(B\right)=32,$

We know that

$$\begin{gathered} n(A\cup B)=n\left(A\right)+n\left(B\right)–n(A\cap B) \\ n(A\cup B)=50,n\left(A\right)=28,n\left(B\right)=32 \\ \Rightarrow 50=28+32–n(A\cap B) \\ \Rightarrow n(A\cap B)=60–50 \\ \Rightarrow n(A\cap B)=10 \end{gathered}$$

.Thus statement II is true.

Both I and II are true

Hence ,the correct option is C.