Question

Suppose $A_{1}, A_{2}, \ldots A_{30}$ are thirty sets, each with five elements and $B_{1}, B_{2}, \ldots, B_{n}$ are $n$ sets each with three elements. Let $\displaystyle\bigcup_{i=1}^{30} A_{i}=\displaystyle\bigcup_{j=1}^{n} B_{j}= S$ If each element of $S$ belongs to exactly ten of the $A_{i}' s$ and exactly nine of the $B_{j}' s$ 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
$\displaystyle\sum_{i=1}^{30} n\left(A_{i}\right)=5 \times 30=150\,\,\,\, (1)$
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
$\displaystyle\sum_{i=1}^{30} n\left(A_{i}\right)=10 m\,\,\,\, (2)$
$\therefore $ From Eq. (1) and (2), we get
$10 \,m = 150$
$\therefore m=15\,\,\, (3)$
Similarly $\displaystyle\sum_{j=1}^{30} n\left(B_{j}\right)=3 n$ and $\displaystyle\sum_{j=1}^{30} n\left(B_{j}\right)=9 \,m$
$\therefore 3 n=9 m$
$\Rightarrow n=\frac{9 m}{3}=3 m$
$=3 \times 15=45\,\,\, [$ from $(3)]$
Hence, $n=45$