यदि (n(A)=72), (n(B)=68) और ठीक एक समुच्चय में (82) तत्व हैं, तो (n\(A\cap B\)) कितना है?

If (n(A)=72), (n(B)=68), and exactly one set has (82) elements, what is (n\(A\cap B\))?

Explanation opens after your attempt
Correct Answer

A. (29)

Step 1

Concept

Exactly one equals (n(A)+n(B)-2n\(A\cap B\)), so (82=140-2x) and (x=29). The common part is removed twice for exactly one.

Step 2

Why this answer is correct

The correct answer is A. (29). Exactly one equals (n(A)+n(B)-2n\(A\cap B\)), so (82=140-2x) and (x=29). The common part is removed twice for exactly one.

Step 3

Exam Tip

ठीक एक (=n(A)+n(B)-2n\(A\cap B\)), इसलिए (82=140-2x) और (x=29)। ठीक एक में साझा भाग दो बार घटता है।

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यदि (n(A)=72), (n(B)=68) और ठीक एक समुच्चय में (82) तत्व हैं, तो (n\(A\cap B\)) कितना है? / If (n(A)=72), (n(B)=68), and exactly one set has (82) elements, what is (n\(A\cap B\))?

Correct Answer: A. (29). Explanation: ठीक एक (=n(A)+n(B)-2n\(A\cap B\)), इसलिए (82=140-2x) और (x=29)। ठीक एक में साझा भाग दो बार घटता है। / Exactly one equals (n(A)+n(B)-2n\(A\cap B\)), so (82=140-2x) and (x=29). The common part is removed twice for exactly one.

Which concept should I revise for this Mathematics MCQ?

Exactly one equals (n(A)+n(B)-2n\(A\cap B\)), so (82=140-2x) and (x=29). The common part is removed twice for exactly one.

What exam hint can help solve this Mathematics question?

ठीक एक (=n(A)+n(B)-2n\(A\cap B\)), इसलिए (82=140-2x) और (x=29)। ठीक एक में साझा भाग दो बार घटता है।