Question

Suppose $A_1,A_2,.....................,A_{30}$ are thirty sets each having 5 elements each and $B_1,B_2,................B_n$ are $n$ sets each with 3 elements each. Let $\bigcup _{i=1}^{30}{A}_i=\bigcup _{j=1}^n{B}_j=S$ and each element of $S$ belongs to exactly 10 of $A_i$'s and exactly 9 of the $B_j$'s. Then $n$ is equal to

• A. $15$
• B. $3$
• C. $45$
• D. $50$

Answer 1

The correct option is C

$45$

We have $S=\bigcup _{i=1}^{30}$$A_i$

Total number of elements in $A_1,A_2,.....................,A_{30}=30\times 5$

Now, each element in the union $S$ belongs to exactly $10$ of the $A_i$'s
So,
$n\left(S\right)=\frac{1}{10}(30\times 5)=15$

Also total number of elements in $B_1,B_2,................B_n=n\times 3$
Again, $S=\bigcup _{j=1}^n{B}_j$

$\Rightarrow n\left(S\right)=\frac{1}{9}(n\times 3)=\frac{n}{3}$

$\Rightarrow \frac{n}{3}=15$

$\therefore n=45$