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
Correct Answer

B. (40)

Step 1

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).

Step 2

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).

Step 3

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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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\(A^c\cap B^c\))?

Correct Answer: B. (40). Explanation: तीन अंदरूनी भागों का योग (48+44+28=120) है, इसलिए बाहर (160-120=40)। चारों क्षेत्र मिलकर पूरा (U) बनाते हैं। / 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).

Which concept should I revise for this Mathematics MCQ?

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).

What exam hint can help solve this Mathematics question?

तीन अंदरूनी भागों का योग (48+44+28=120) है, इसलिए बाहर (160-120=40)। चारों क्षेत्र मिलकर पूरा (U) बनाते हैं।