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

MCQ

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

A
15
B
3
✓
45
D
35

✓

Answer

Correct option: C.

45

45.

Solution:
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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