यदि (n\(A\cup B\)=98) और (n\(A\cap B\)=35) है, तो (n(A-B)+n(B-A)) कितना होगा?

If (n\(A\cup B\)=98) and (n\(A\cap B\)=35), then what is (n(A-B)+n(B-A))?

Explanation opens after your attempt
Correct Answer

A. (63)

Step 1

Concept

The sum of the two only regions is (98-35=63). This is also the size of \(A\triangle B\).

Step 2

Why this answer is correct

The correct answer is A. (63). The sum of the two only regions is (98-35=63). This is also the size of \(A\triangle B\).

Step 3

Exam Tip

दोनों केवल क्षेत्रों का योग (98-35=63) है। यह \(A\triangle B\) की संख्या भी होती है।

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Mathematics Answer, Explanation and Revision Hints

यदि (n\(A\cup B\)=98) और (n\(A\cap B\)=35) है, तो (n(A-B)+n(B-A)) कितना होगा? / If (n\(A\cup B\)=98) and (n\(A\cap B\)=35), then what is (n(A-B)+n(B-A))?

Correct Answer: A. (63). Explanation: दोनों केवल क्षेत्रों का योग (98-35=63) है। यह \(A\triangle B\) की संख्या भी होती है। / The sum of the two only regions is (98-35=63). This is also the size of \(A\triangle B\).

Which concept should I revise for this Mathematics MCQ?

The sum of the two only regions is (98-35=63). This is also the size of \(A\triangle B\).

What exam hint can help solve this Mathematics question?

दोनों केवल क्षेत्रों का योग (98-35=63) है। यह \(A\triangle B\) की संख्या भी होती है।