यदि (n\(A\cap B^c\)=48), (n\(A^c\cap B\)=44), (n\(A\cap B\)=28) और (n(U)=160) है, तो (n\(A^c\cap B^c\)) कितना है?
If (n\(A\cap B^c\)=48), (n\(A^c\cap B\)=44), (n\(A\cap B\)=28), and (n(U)=160), what is (n\(A^c\cap B^c\))?
Explanation opens after your attempt
B. (40)
Concept
The sum of the three inside regions is (48+44+28=120), so outside is (160-120=40). The four regions together form the whole (U).
Why this answer is correct
The correct answer is B. (40). The sum of the three inside regions is (48+44+28=120), so outside is (160-120=40). The four regions together form the whole (U).
Exam Tip
तीन अंदरूनी भागों का योग (48+44+28=120) है, इसलिए बाहर (160-120=40)। चारों क्षेत्र मिलकर पूरा (U) बनाते हैं।
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