Question

Suppose $A_1,A_2,...,A_{30}$ are thirty sets, each with five elements and $B_1,B_2,...,B_{30}$ are $n$ sets ecah with three elements. Let $\bigcup _{i=1}^{30}{A}_i=\bigcup _{j=1}^n{B}_j=S$

If each element of $S$ belongs to exactly ten of the $A_i^{\prime }s$ and exactly none of the $B_j^{\prime }s$ then $n=$

• A. 45
• B. 35
• C. 40
• D. none of these

Answer 1

The correct option is A 45

Given $A_i$'s are thirty sets with five elements each, so $\sum _{i=1}^{30}n\left(A_i\right)=5\times 30=150$ ...(1)

If there are $m$ distinct elements in $S$ and each element of $S$ belongs to exactly $10$ of the $A_i$'s, we have

$\sum _{i=1}^{30}n\left(A_i\right)=10m$ ...(2)

$\therefore$ from (1) and (2), we get $10m=150\Rightarrow m=15$ ...(3)

Similarly $\sum _{j=1}^{30}n\left(B_j\right)=3n$ and $\sum _{j=1}^{30}n\left(B_j\right)=9m$

$\therefore 3n=9m\Rightarrow \frac{9m}{3}=3m=3\times 15=45$ (from (3))

Hence, $n=45$