यदि (n\(A\cup B\)=96), (n\(A\cap B\)=18) और (n(A-B)=41) है, तो (n(B)) कितना है?
If (n\(A\cup B\)=96), (n\(A\cap B\)=18), and (n(A-B)=41), what is (n(B))?
Explanation opens after your attempt
A. (,55,)
Concept
The sum of (A-B), \(A\cap B\), and (B-A) is (96), so (B-A=96-41-18=37). Thus (n(B)=37+18=55).
Why this answer is correct
The correct answer is A. (,55,). The sum of (A-B), \(A\cap B\), and (B-A) is (96), so (B-A=96-41-18=37). Thus (n(B)=37+18=55).
Exam Tip
(A-B), \(A\cap B\) और (B-A) का योग (96) है, इसलिए (B-A=96-41-18=37)। अतः (n(B)=37+18=55)।
Login to save your score, XP, coins and progress.
