Question

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 $\bigcup\limits_{i = 1}^{30} {{A_i}} = \bigcup\limits_{j = 1}^n {{B_j}} = S$ and each element of $S$ belongs to exactly $10$ of the ${A_i}'s$ and exactly $9$ of the ${B_j}'s$. Then $n$ is equal to:
(A) $15$
(B) $3$
(C) $45$
(D) $35$

Answer 1

Hint: First find the total number of elements in both the sets. Given in the question that the number of elements in both the sets are equal. So equate the total number of elements found for set A to the total number of elements in the set B to get the desired answer.

Complete step-by-step answer:
Given, \[{A_1},{A_2},......,{A_{30}}\] are thirty sets each having $5$ elements.
Therefore, no. of elements in $\bigcup\limits_{i = 1}^{30} {{A_i}} $, i.e., ${A_1} \cup {A_2} \cup {A_3} \cup .......{A_{30}}$
$ = 30 \times 5$
$ = 150$ elements
But each element of ${A_i}$ is used $10$ times,
So, $S = \dfrac{{150}}{{10}} = 15$ …… (1)
Also given , \[{B_1},{B_2},......,{B_n}\] are $n$ sets each with $3$ elements.
Therefore, no. of elements in $\bigcup\limits_{j = 1}^n {{B_j}} $, i.e., ${B_1} \cup {B_2} \cup {B_3} \cup .......{B_n}$
$ = n \times 3$
$ = 3n$ elements
But each element of ${B_j}$ is used $9$ times,
So, $S = \dfrac{{3n}}{9} = \dfrac{n}{3}$ ….. (2)
From (1) and (2), we get-
$\dfrac{n}{3} = 15$
$ \Rightarrow n = 15 \times 3$
$ \Rightarrow n = 45$

Therefore n is equal to 45.

Note: The number of elements in the union of a set $\bigcup\limits_{m = 1}^m {{A_m}} $ with each having $n$ elements is $mn$, if the elements are not repeated. In mathematics, the symbol $\bigcup\limits_{m = 1}^m {{A_m}} $ is used for union summation of sets. The value below the union operator gives us the starting integer, while the top value gives us the upper bound. Therefore, $\bigcup\limits_{m = 1}^m {{A_m}} = {A_1} \cup {A_2} \cup {A_3} \cup ...........{A_m}$.