यदि केवल (A) में (12), केवल (B) में (15), केवल (C) में (18), केवल \(A\cap B\) में (9), केवल \(B\cap C\) में (7), केवल \(C\cap A\) में (6), और \(A\cap B\cap C\) में (4) हैं, तो (n(A\cap\(B\cup C\))) कितना है?

If only (A) has (12), only (B) has (15), only (C) has (18), only \(A\cap B\) has (9), only \(B\cap C\) has (7), only \(C\cap A\) has (6), and \(A\cap B\cap C\) has (4), then what is (n(A\cap\(B\cup C\)))?

Explanation opens after your attempt
Correct Answer

A. (19)

Step 1

Concept

(A\cap\(B\cup C\)) includes \(A\cap B\), \(A\cap C\), and the centre, so (9+6+4=19). The only (A) region is not included.

Step 2

Why this answer is correct

The correct answer is A. (19). (A\cap\(B\cup C\)) includes \(A\cap B\), \(A\cap C\), and the centre, so (9+6+4=19). The only (A) region is not included.

Step 3

Exam Tip

(A\cap\(B\cup C\)) में \(A\cap B\), \(A\cap C\) और केंद्र आते हैं, इसलिए (9+6+4=19) है। केवल (A) इसमें शामिल नहीं होगा।

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FAQs

Mathematics Answer, Explanation and Revision Hints

यदि केवल (A) में (12), केवल (B) में (15), केवल (C) में (18), केवल \(A\cap B\) में (9), केवल \(B\cap C\) में (7), केवल \(C\cap A\) में (6), और \(A\cap B\cap C\) में (4) हैं, तो (n(A\cap\(B\cup C\))) कितना है? / If only (A) has (12), only (B) has (15), only (C) has (18), only \(A\cap B\) has (9), only \(B\cap C\) has (7), only \(C\cap A\) has (6), and \(A\cap B\cap C\) has (4), then what is (n(A\cap\(B\cup C\)))?

Correct Answer: A. (19). Explanation: (A\cap\(B\cup C\)) में \(A\cap B\), \(A\cap C\) और केंद्र आते हैं, इसलिए (9+6+4=19) है। केवल (A) इसमें शामिल नहीं होगा। / (A\cap\(B\cup C\)) includes \(A\cap B\), \(A\cap C\), and the centre, so (9+6+4=19). The only (A) region is not included.

Which concept should I revise for this Mathematics MCQ?

(A\cap\(B\cup C\)) includes \(A\cap B\), \(A\cap C\), and the centre, so (9+6+4=19). The only (A) region is not included.

What exam hint can help solve this Mathematics question?

(A\cap\(B\cup C\)) में \(A\cap B\), \(A\cap C\) और केंद्र आते हैं, इसलिए (9+6+4=19) है। केवल (A) इसमें शामिल नहीं होगा।