Question

Let U be the universal set for all sets $A$ and $B$ such that $n(A) = 200,n(B) = 300$ and \[\left( {A \cap B} \right) = 100\] , then \[n\left( {A' \cap B'} \right)\] is equal to $300$ provided that $n(U)$ is equal to
A.$600$
B.$700$
C.$800$
D.$900$

Answer 1

Hint: In this question two sets are given set $A$ and set $B$ . It is mentioned in the question that $n(A) = 200,n(B) = 300$ and \[\left( {A \cap B} \right) = 100\] . So, find out the pure number of $A$ and pure number of $B$ , then try to solve it. Since the value (number of components) \[n\left( {A' \cap B'} \right)\] is given so it will also help to solve it very easily, putting all these things properly.

Complete step-by-step answer:
Given: $n(A) = 200,n(B) = 300,\left( {A \cap B} \right) = 100$ and \[n\left( {A' \cap B'} \right) = 300\] . Then find out $n(U)$ =?
According to the number of sets $A$ and $B$ are clearly mentioned so first try to find out only in $A$ and only in $B$ removing common parts. So,
$\therefore $ Number that only present only in $A$ not in common $ = n(A) - n(A \cap B)$
$\
   = 200 - 100 \\
   = 100 \\
\ $
Similarly, number present only in \[B = n(B) - n(A \cap B)\]
$\
   = 300 - 100 \\
   = 200 \\
\ $
Now, $n(U) = n(onlyA) + n(onlyB) + n(A \cap B) + n(A' \cap B')$
$\
   = 100 + 200 + 100 + 300 \\
   = 700 \\
\ $
Hence, $n(U) = 700$ . Therefore, the correct answer will be option B. $700$ .

Note: In this type of question students should be careful about common parts of the set that overlap each other. The common value that is given of the set $A$ and $B$ is to separate it wisely otherwise if we include the common values of both the sets the number of values of the sets will overflow and a student must try to understand all the given information very carefully.