Question

Suppose $A_1,A_2,A_3,\cdots ,A_{30}$ are thirty sets each having 5 elements and $B_1,B_2,\cdots ,B_n$ are n sets each with 3 elements. Let ${\bigcup }_{i-1}^{30}{A}_i={\bigcup }_{j-1}^{n}Bj=S$ and each element of S belongs to exactly 10 of the $A_i^{\prime }s$ and exactly 9 of the $B_j^{\prime }s$. Then n is equal to

• A. 15
• B. 3
• C. 45
• D. None of these

Answer 1

The correct option is C

45

$O\left(S\right)=O\left({\bigcup }_{i-1}^{30}{A}_i\right)=\frac{1}{10}(5\times 30)=15$
Since, element in the union S belongs to 10 of $A_i^{\prime }s$
Also, $O\left(S\right)=O\left({\bigcup }_{j-1}^{n}Bj\right)=\frac{3n}{9}=\frac{n}{3},\therefore \frac{n}{3}=15\Rightarrow n=45$.