यदि (n(A)=86), (n(B)=79), (n(C)=71), (n\(A\cap B\)=34), (n\(B\cap C\)=29), (n\(C\cap A\)=27) और (n\(A\cap B\cap C\)=12) है, तो केवल (C) में कितने तत्व हैं?

If (n(A)=86), (n(B)=79), (n(C)=71), (n\(A\cap B\)=34), (n\(B\cap C\)=29), (n\(C\cap A\)=27) and (n\(A\cap B\cap C\)=12), then how many elements are only in (C)?

Explanation opens after your attempt
Correct Answer

A. (27)

Step 1

Concept

Only (C=71-29-27+12=27). When two common parts are subtracted, the centre is removed twice, so add it back.

Step 2

Why this answer is correct

The correct answer is A. (27). Only (C=71-29-27+12=27). When two common parts are subtracted, the centre is removed twice, so add it back.

Step 3

Exam Tip

केवल (C=71-29-27+12=27) है। दो साझा भाग घटाने पर केंद्र दो बार घटता है, इसलिए उसे जोड़ें।

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Mathematics Answer, Explanation and Revision Hints

यदि (n(A)=86), (n(B)=79), (n(C)=71), (n\(A\cap B\)=34), (n\(B\cap C\)=29), (n\(C\cap A\)=27) और (n\(A\cap B\cap C\)=12) है, तो केवल (C) में कितने तत्व हैं? / If (n(A)=86), (n(B)=79), (n(C)=71), (n\(A\cap B\)=34), (n\(B\cap C\)=29), (n\(C\cap A\)=27) and (n\(A\cap B\cap C\)=12), then how many elements are only in (C)?

Correct Answer: A. (27). Explanation: केवल (C=71-29-27+12=27) है। दो साझा भाग घटाने पर केंद्र दो बार घटता है, इसलिए उसे जोड़ें। / Only (C=71-29-27+12=27). When two common parts are subtracted, the centre is removed twice, so add it back.

Which concept should I revise for this Mathematics MCQ?

Only (C=71-29-27+12=27). When two common parts are subtracted, the centre is removed twice, so add it back.

What exam hint can help solve this Mathematics question?

केवल (C=71-29-27+12=27) है। दो साझा भाग घटाने पर केंद्र दो बार घटता है, इसलिए उसे जोड़ें।