Question

Suppose $A_1,A_2,\cdots ,A_{30}$ are thirty sets each having $5$ elements and $B_1,B_2,\cdots ,B_n$ 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 $A_i^{\prime }s$ and exactly $9$ of the $B_j^{\prime }s.$ Then $n$ is equal to

• A. $15$
• B. $135$
• C. $45$
• D. $90$

Answer 1

Answer: (C)

$45$

Since each $A$ $A_i$ has $5$ elements, we have $\sum _{i=1}^{30}n\left(A_i\right)=5\times 30=150\cdots \left(1\right)$
Let $S$ consist of $m$ distinct elements. Since each element of $S$ belongs to exactly $10$ of the $A_i^{\prime }s,$ we also have $\sum _{i=1}^{30}n\left(A_i\right)=10m\cdots \left(2\right)$

Hence from (1) and (2), $10m=150$ or $m=15$.
Again since each $B_i$ has 3 elements and each element of $S$ belongs to exactly $9$ of the $B_i^{\prime }S$, we have $\sum _{j=1}^{n}n\left(B_i\right)=3n$ and $\sum _{j=1}^{n}n\left(B_j\right)=9m$
It follows that $3n=9m=9\times 15$ $[\therefore m=15]$
This gives $n=45$