Question

Let \[{A_1},{\text{ }}{A_2},{\text{ }}{A_3}, \ldots ..{A_{40}}\] are \[40\] sets each with \[7\] elements and \[{B_1},{\text{ }}{B_2}, \ldots .{B_n}\] are n sets each with \[7\] elements. If $ U_{i = 1}^{40}{A_i} = U_{j = 1}^n{B_j} = S $ and each element of S belongs to exactly ten of Ai’s and exactly \[9\] of Bj’s, then n equals
A) \[42\]
B) $ 35 $
C) $ 28 $
D) $ 36 $

Answer 1

Hint: To solve this problem, we will first take the \[{A_1},{\text{ }}{A_2},{\text{ }}{A_3}, \ldots ..{A_{40}}\] , i.e.,\[40\] sets each with \[7\] elements then we will find the sum of total elements of it. Then we will take the value of n(S). Then we will take \[{B_1},{\text{ }}{B_2}, \ldots .{B_n}\]i.e., n sets each with \[7\] elements, afterwards we will use the given condition, and get the value of n, and hence with that we will get our required answer.

Complete step-by-step answer:
We have been given that \[{A_1},{\text{ }}{A_2},{\text{ }}{A_3}, \ldots ..{A_{40}}\] are \[40\] sets each with \[7\] elements and also \[{B_1},{\text{ }}{B_2}, \ldots .{B_n}\] are n sets each with \[7\] elements. If $ U_{i = 1}^{40}{A_i} = U_{j = 1}^n{B_j} = S $ and each element of S belongs to exactly ten of Ai’s and exactly \[9\] of Bj’s, then we need to find the value of n.
So, we have \[{A_1},{\text{ }}{A_2},{\text{ }}{A_3}, \ldots ..{A_{40}}\] i.e., \[40\] sets each with \[7\] elements, then, \[\sum n\left( {{A_i}} \right) = 7 \times 40 = 280\]
But, it is given in the question that each element of S occurs in \[10\] of Ai's,
\[ \Rightarrow n\left( S \right) = \dfrac{{280}}{{10}} = 28\]
Also, we have \[{B_1},{\text{ }}{B_2}, \ldots .{B_n}\] are n sets each with \[7\] elements, then, \[\sum n\left( {{B_j}} \right) = 7n\]
But, it is given in the question that each element of S occurs in \[9\] of Bj's,
\[28 = \dfrac{{7n}}{9}\]
\[
\Rightarrow n = \dfrac{{28 \times 9}}{7} = 9 \times 4 \\
\Rightarrow n = 36 \\
 \]
So, n equals to \[36.\].
So, the correct answer is “Option D”.

Note: Students should carefully solve these types of questions, as these types of questions are simple but we need to carefully check the conditions given to us. As in this problem, we were given that, $ U_{i = 1}^{40}{A_i} = U_{j = 1}^n{B_j} = S $ and each element of S belongs to exactly ten of Ai’s and exactly \[9\] of Bj’s, without using this we won’t be able to get the value of n.