यदि (n\(A\cap B^c\)=41), (n\(A^c\cap B\)=34), (n\(A\cap B\)=26) और (n(U)=130) है, तो (n\(A^c\cap B^c\)) कितना है?

If (n\(A\cap B^c\)=41), (n\(A^c\cap B\)=34), (n\(A\cap B\)=26), and (n(U)=130), what is (n\(A^c\cap B^c\))?

Explanation opens after your attempt
Correct Answer

A. (29)

Step 1

Concept

The sum of the three inside regions is (41+34+26=101), so outside is (130-101=29). The four regions together form the whole (U).

Step 2

Why this answer is correct

The correct answer is A. (29). The sum of the three inside regions is (41+34+26=101), so outside is (130-101=29). The four regions together form the whole (U).

Step 3

Exam Tip

तीन अंदरूनी भागों का योग (41+34+26=101) है, इसलिए बाहर (130-101=29)। चार क्षेत्र मिलकर पूरा (U) बनाते हैं।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

यदि (n\(A\cap B^c\)=41), (n\(A^c\cap B\)=34), (n\(A\cap B\)=26) और (n(U)=130) है, तो (n\(A^c\cap B^c\)) कितना है? / If (n\(A\cap B^c\)=41), (n\(A^c\cap B\)=34), (n\(A\cap B\)=26), and (n(U)=130), what is (n\(A^c\cap B^c\))?

Correct Answer: A. (29). Explanation: तीन अंदरूनी भागों का योग (41+34+26=101) है, इसलिए बाहर (130-101=29)। चार क्षेत्र मिलकर पूरा (U) बनाते हैं। / The sum of the three inside regions is (41+34+26=101), so outside is (130-101=29). 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 (41+34+26=101), so outside is (130-101=29). The four regions together form the whole (U).

What exam hint can help solve this Mathematics question?

तीन अंदरूनी भागों का योग (41+34+26=101) है, इसलिए बाहर (130-101=29)। चार क्षेत्र मिलकर पूरा (U) बनाते हैं।