Question

$A_1,A_2,...,A_{30}$are $30$sets, each having $5$elements and $B_1,B_2,...,B_n$are n sets each with $3$elements. If $\underset{i=1}{\overset{30}{\cup }}{A}_i=\underset{j=1}{\overset{n}{\cup }}{B}_j=S$ and each element of $S$ belongs to exactly $10$ of the $A_i$'s and exactly $9$ of the $B_j$'s, then the value of $n$ is

• A. $15$
• B. $3$
• C. $45$
• D. None of these

Answer 1

The correct option is C

$45$

Explanation for the correct option:

Finding the value of $n$:

Given, $A$'s are $30$ sets with five elements each.

$\sum _{n=1}^{30}n\left(A_i\right)=5\times 30$

$\Rightarrow \sum _{n=1}^{30}n\left(A_i\right)=150\dots \dots \left(1\right)$ [given]

Let $m$ distinct elements in $S$ and each element of $S$ belongs to exactly $10$ of the $A_i$'s then,

$\underset{n=1}{\overset{30}{\Rightarrow \sum }}n\left(A_i\right)=10m\dots \dots \left(2\right)$ [given]

From Equations $\mathrm{\left(}\mathrm{1}\mathrm{\right)}$ and $\mathrm{\left(}\mathrm{2}\mathrm{\right)}$ we get,

$$\begin{gathered} \Rightarrow 10m=150 \\ \Rightarrow m=\frac{150}{10} \\ \Rightarrow m=15 \end{gathered}$$

Similarly,

$\Rightarrow \sum _{j=1}^{n}n\left(B_j\right)=3n\dots \dots \left(3\right)$ [given]

$\Rightarrow \sum _{j=1}^{n}n\left(B_j\right)=9m\dots \dots \left(4\right)$ [given]

Comparing equations $\mathrm{\left(}\mathrm{3}\mathrm{\right)}$ and $\mathrm{\left(}\mathrm{4}\mathrm{\right)}$ we get

$$\begin{gathered} \Rightarrow 3n=9m \\ \Rightarrow n=\frac{9m}{3} \\ \Rightarrow n=3m \\ \Rightarrow n=3\times 15\left[\because m=15\right] \\ \Rightarrow n=45 \end{gathered}$$

Hence, the value of $n$ is $45$.

Therefore, the correct answer is option (C).