Question

Suppose $A_1,A_2,.....,A_{30}$ are thirty sets each with five elements and $B_1,B_2,.....,B_n$ are $n$ sets each with three elements such that ${\bigcup }_{i=1}^{30}{A}_i={\bigcup }_{j=1}^n{B}_j=S$ and each element of $S$ belongs to exactly $10$ of the $A_i$'s and exactly $9$ of the $B_j$'s. Then $n=$

• A. $15$
• B. $3$
• C. $45$
• D. $9$

Answer 1

The correct option is C $45$

Since, each $A_i$ has $5$ elements, we have
${\sum }_{i=1}^{30}n\left(A_i\right)=5\times 30=150$ ...(1)
Let set $S$ consists of $k$ elements.
Since, each elements in $S$ belongs to exactly $10$ of $A_i$'s
Therefore,
${\sum }_{i=1}^{30}n\left(A_i\right)=10k$ ...(2)
Hence, from (1) and (2)
$10k=150\Rightarrow k=15$
Now, since each $B_j$ has $3$ elements and each $S$ belongs to exactly $9$ of $B_j$'s
Therefore,
${\sum }_{j=1}^{n}n\left(B_j\right)=3n$ and ${\sum }_{j=1}^{n}n\left(B_j\right)=9k$
$\Rightarrow 3n=9k=9\times 15$
$\Rightarrow n=45$