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

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.