Question

$A_1,A_2,.....A_n$ are thirty sets, each with five elements and $B_1,B_2,....B_n$ 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 $A_i$'s and exactly nine of the $B_j$'s, then $n$ is

• A. $45$
• B. $35$
• C. $40$
• D. $30$

Answer 1

Answer: (A)

$45$

Since each $A_i$ has $5$ elements, we have
${\sum }_{i=1}^{30}n\left(A_i\right)=5\times 30=\mathrm{150...}\left(i\right)$
suppose $S$ has $m$ distinct elements. Since each element of $S$ belongs to exactly $12$ of $A_i$'s we also have
${\sum }_{i=1}^{n}n\left(A_i\right)=10m...\left(2\right)$
From (1) and (2) $10m=150\Rightarrow$ $m=15$
since 3 elements each of $B_i$ has and each element of $S$ belongs to exactly $9$ of the $B_j$'s we have
${\sum }_{j=1}^{n}n\left(B_j\right)=3n{\sum }_{j=1}^{n}n\left(B_j\right)=9m$
$\Rightarrow 3n=9m\Rightarrow n=3m\Rightarrow n=3\times 15=45$