Suppose A_1, A_2, ..., A_ 30 are thirty sets each having 5 elemen… - Vidyadip

MCQ

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 $\bigcup\limits^{30}_\text{i = 1}\text{A}_\text{i}=\bigcup\limits^{\text{n}}_\text{j = 1}\text{B}_\text{j}=\text{S}$ and each element of $S$ belong to exactly $10$ of the ${A_i}^{'s}$ and exactly $9$ of the ${B_j}^{'s},$ then $n$ is equal to:

A
$15$
B
$3$
✓
$45$
D
$35.$

✓

Answer

Correct option: C.

$45$

It is given that each set $\text{A}_\text{j}(1\leq\text{i}\leq30)$ contains $5$ elements and $\bigcup\limits^{30}_\text{i = 1}\text{A}_\text{i}=\text{S}.$
$\therefore\text{n(S)}=30\times5=150$
But, it is given that each element of $S$ belong to exactly $10$ of the ${A_i}^{'s}.$
$\therefore$ Number of distinct elements in $\text{S}=\frac{150}{10}=15......(1)$
It is also given that each set $\text{B}_\text{j}(1\leq\text{j}\leq\text{n})$ contains $3$ elements and $\bigcup\limits^{\text{n}}_\text{j = 1}\text{B}_\text{j}=\text{S}.$
$\therefore\text{ n(S)}=\text{n}\times3=\text{3n}$
Also, each element of $S$ belong to eactly $9$ of ${B_j}^{'s}.$
$\therefore$ Number of distinct elements in $\text{S}=\frac{\text{3n}}{9}......(2)$
From $(1)$ and $(2)$, we have
$\frac{\text{3n}}{9}=15$
$\Rightarrow\text{n} = 45.$
Hence, the correct answer is option $(c).$

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