Question

Suppose $A_1,A_2,,A_{30}$ are thirty sets each having $5$ elements and $B_1,B_2,..,B_n$ are $n$ sets each with $3$ elements, let $\bigcup _{i=1}^{30}{A}_i=\bigcup _{j=1}^n{B}_j=S$ and each element of $S$ belongs to exactly $10$ of the $A_{i}s$ and exactly $9$ of the $B_{j}s.$ Then $n$ is equal to

• A. $15$
• B. $3$
• C. $45$
• D. $Noneofthese$

Answer 1

The correct option is C $45$

$\left(c\right)$.
Since each $A_i$ has $5$ elements, we have
$\stackrel{30}{\underset{i=1}{\Sigma }}n\left(A_i\right)=5\times 30=150$. . . $\left(1\right)$
Let $S$ consist of $m$ distinct elements. Since each elements of $S$ belongs to exactly $12$ of the $A_i$ $s$ we also have
$\stackrel{30}{\underset{i=1}{\Sigma }}n\left(A_i\right)=10m$. . . $\left(2\right)$
Hence from $\left(1\right)$ and $\left(2\right)$, $10m=150$ or $m=15$. Again since each $B_i$ has $3$ elements and each element of $S$ belongs to exactly $9$ of the $B_j$ $s$ we have
$\stackrel{30}{\underset{j=1}{\Sigma }}n\left(B_j\right)=3n$ and $\stackrel{30}{\underset{j=1}{\Sigma }}n\left(B_j\right)=9m$
It follows that $3n=9m=9\times 15$
. . . $[\therefore m=15]$
This gives $n=45$.