यदि (n(A)=82), (n(B)=77), (n(C)=69), (n\(A\cup B\cup C\)=151), (n\(A\cap B\)=32), (n\(B\cap C\)=28), (n\(C\cap A\)=25) है, तो (n\(A\cap B\cap C\)) कितना है?
If (n(A)=82), (n(B)=77), (n(C)=69), (n\(A\cup B\cup C\)=151), (n\(A\cap B\)=32), (n\(B\cap C\)=28), (n\(C\cap A\)=25), then what is (n\(A\cap B\cap C\))?
Explanation opens after your attempt
A. (8)
Concept
Using (151=82+77+69-32-28-25+x), we get (x=8). Let the unknown centre be (x) and form an equation.
Why this answer is correct
The correct answer is A. (8). Using (151=82+77+69-32-28-25+x), we get (x=8). Let the unknown centre be (x) and form an equation.
Exam Tip
सूत्र में (151=82+77+69-32-28-25+x), इसलिए (x=8) है। अज्ञात केंद्र को (x) मानकर समीकरण बनाएं।
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