Question

Suppose \[{A_1},{A_2},{A_3}........{A_{30}}\] are thirty sets each with five elements and \[{B_1},{B_2},{B_3}........{B_n}\] are n sets each with three elements such that \[\mathop U\limits_{i = 1}^{30} {A_i} = \mathop U\limits_{i = 1}^n {B_j} = S\]and each element of S belongs to exactly 10 of the A and exactly 9 of the B. Then n is equal to
A) 35
B) 45
C) 55
D) 65

Answer 1

Hint: Here first we will calculate the number of elements in \[{A_i}\]’s . Then we will find the elements of A which belong to S. Similarly we will find the number of elements of B which belong to S and then finally find the value of n.

Complete step-by-step answer:
It is given that:
\[{A_1},{A_2},{A_3}........{A_{30}}\]are thirty sets and each set has 5 elements
Therefore, the total number of elements in \[{A_i}\]’s is given by:-
\[ = 5 \times 30\]
\[ = 150\]elements
Therefore, total elements in \[{A_i}\]’s are 150 elements.
Now it is given that:-
Each element of S belongs to exactly 10 of the A.
Therefore, the elements in S is given by:-
\[ = \dfrac{{150}}{{10}}\]
\[ = 15\]
Therefore, number of elements in S are 15 elements…………………………….(1)
Now it is given that:-
 \[{B_1},{B_2},{B_3}........{B_n}\] are n sets each with three elements.
Therefore, the total number of elements in\[{B_j}\]’s is given by:-
\[ = n \times 3\]
\[ = 3n\]elements
Therefore, total elements in\[{B_j}\]’s are 3n elements.
Each element of S belongs to exactly 9 of the A.
Therefore, the elements in S is given by:-
\[ = \dfrac{{3n}}{9}\]
\[ = \dfrac{n}{3}\]
Therefore, number of elements in S are \[\dfrac{n}{3}\]elements…………………………….(2)
Equating the values in equation 1 and 2 we get:-
\[15 = \dfrac{n}{3}\]
Solving for n we get:-
\[n = 15 \times 3\]
\[n = 45\]
Therefore, the value of n is 45

Therefore, option B is correct.

Note: Students here might mistake in determining the elements in S.
So here we need to find the number of elements from both A and B which belong to S separately and then equate them as the elements in S will be the same.