यदि (n(A-B)=36), (n(B-A)=28), (n\(A\cap B\)=22) और (n(U)=120) है, तो (n\(A^c\cup B^c\)) कितना होगा?
If (n(A-B)=36), (n(B-A)=28), (n\(A\cap B\)=22), and (n(U)=120), what is (n\(A^c\cup B^c\))?
Explanation opens after your attempt
D. (98)
Concept
(A^c\cup B^c=\(A\cap B\)^c), so (120-22=98). The complement of the common part is found by subtracting it from (U).
Why this answer is correct
The correct answer is D. (98). (A^c\cup B^c=\(A\cap B\)^c), so (120-22=98). The complement of the common part is found by subtracting it from (U).
Exam Tip
(A^c\cup B^c=\(A\cap B\)^c), इसलिए (120-22=98)। साझा भाग का पूरक पूरे (U) से साझा भाग घटाकर मिलता है।
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