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Question: Let U be the universal set for all sets \(A\) and \(B\) such that \(n(A) = 200,n(B) = 300\) and \[\l...

Let U be the universal set for all sets AA and BB such that n(A)=200,n(B)=300n(A) = 200,n(B) = 300 and (AB)=100\left( {A \cap B} \right) = 100 , then n(AB)n\left( {A' \cap B'} \right) is equal to 300300 provided that n(U)n(U) is equal to
A.600600
B.700700
C.800800
D.900900

Explanation

Solution

In this question two sets are given set AA and set BB . It is mentioned in the question that n(A)=200,n(B)=300n(A) = 200,n(B) = 300 and (AB)=100\left( {A \cap B} \right) = 100 . So, find out the pure number of AA and pure number of BB , then try to solve it. Since the value (number of components) n(AB)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,(AB)=100n(A) = 200,n(B) = 300,\left( {A \cap B} \right) = 100 and n(AB)=300n\left( {A' \cap B'} \right) = 300 . Then find out n(U)n(U) =?
According to the number of sets AA and BB are clearly mentioned so first try to find out only in AA and only in BB removing common parts. So,
\therefore Number that only present only in AA not in common =n(A)n(AB) = n(A) - n(A \cap B)
 =200100 =100  \ = 200 - 100 \\\ = 100 \\\ \
Similarly, number present only in B=n(B)n(AB)B = n(B) - n(A \cap B)
 =300100 =200  \ = 300 - 100 \\\ = 200 \\\ \
Now, n(U)=n(onlyA)+n(onlyB)+n(AB)+n(AB)n(U) = n(onlyA) + n(onlyB) + n(A \cap B) + n(A' \cap B')
 =100+200+100+300 =700  \ = 100 + 200 + 100 + 300 \\\ = 700 \\\ \
Hence, n(U)=700n(U) = 700 . Therefore, the correct answer will be option B. 700700 .

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 AA and BB 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.