यदि (n\(A\cup B\)=115) और (n\(A\cap B\)=42) है, तो (n(A-B)+n(B-A)) कितना होगा?

If (n\(A\cup B\)=115) and (n\(A\cap B\)=42), then what is (n(A-B)+n(B-A))?

Explanation opens after your attempt
Correct Answer

A. (73)

Step 1

Concept

The sum of the two only-side regions is (115-42=73). This is also the size of \(A\triangle B\).

Step 2

Why this answer is correct

The correct answer is A. (73). The sum of the two only-side regions is (115-42=73). This is also the size of \(A\triangle B\).

Step 3

Exam Tip

केवल दोनों ओर के भागों का योग (115-42=73) है। यही \(A\triangle B\) की संख्या भी है।

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Mathematics Answer, Explanation and Revision Hints

यदि (n\(A\cup B\)=115) और (n\(A\cap B\)=42) है, तो (n(A-B)+n(B-A)) कितना होगा? / If (n\(A\cup B\)=115) and (n\(A\cap B\)=42), then what is (n(A-B)+n(B-A))?

Correct Answer: A. (73). Explanation: केवल दोनों ओर के भागों का योग (115-42=73) है। यही \(A\triangle B\) की संख्या भी है। / The sum of the two only-side regions is (115-42=73). This is also the size of \(A\triangle B\).

Which concept should I revise for this Mathematics MCQ?

The sum of the two only-side regions is (115-42=73). This is also the size of \(A\triangle B\).

What exam hint can help solve this Mathematics question?

केवल दोनों ओर के भागों का योग (115-42=73) है। यही \(A\triangle B\) की संख्या भी है।