यदि (n(A)=70), (n(B)=65), (n(C)=60), (n\(A\cup B\cup C\)=128), (n\(A\cap B\)=27), (n\(B\cap C\)=24), (n\(C\cap A\)=22) है, तो (n\(A\cap B\cap C\)) कितना है?
If (n(A)=70), (n(B)=65), (n(C)=60), (n\(A\cup B\cup C\)=128), (n\(A\cap B\)=27), (n\(B\cap C\)=24), (n\(C\cap A\)=22), then what is (n\(A\cap B\cap C\))?
Explanation opens after your attempt
A. (6)
Concept
Using (128=70+65+60-27-24-22+x), we get (x=6). Let the unknown centre be (x) and form an equation.
Why this answer is correct
The correct answer is A. (6). Using (128=70+65+60-27-24-22+x), we get (x=6). Let the unknown centre be (x) and form an equation.
Exam Tip
सूत्र में (128=70+65+60-27-24-22+x), इसलिए (x=6) है। अज्ञात केंद्र को (x) मानकर समीकरण बनाएं।
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