Question

Suppose $A_1,A_2,\dots A_{30}$ are thirty sets, each with five elements and $B_1,B_2,\dots ,B_n$ are $n$ sets each 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 nine of the $B_j^{\prime }s$ then $n=$

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

Answer 1

Answer: (A)

Given $A_i^{\prime }$ 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 $A_i^{\prime }s$, we have
$\sum _{i=1}^{30}n\left(A_i\right)=10m(2)$
$\therefore$ From Eq. (1) and (2), we get
$10m=150$
$\therefore m=15(3)$
Similarly $\sum _{j=1}^{30}n\left(B_j\right)=3n$ and $\sum _{j=1}^{30}n\left(B_j\right)=9m$
$\therefore 3n=9m$
$\Rightarrow n=\frac{9m}{3}=3m$
$=3\times 15=45[$ from $(3)]$
Hence, $n=45$