यदि (n(A)=78), (n(B)=69), (n(C)=63), (n\(A\cap B\)=30), (n\(B\cap C\)=25), (n\(C\cap A\)=21) और (n\(A\cap B\cap C\)=9) है, तो केवल (A) में कितने तत्व हैं?

If (n(A)=78), (n(B)=69), (n(C)=63), (n\(A\cap B\)=30), (n\(B\cap C\)=25), (n\(C\cap A\)=21) and (n\(A\cap B\cap C\)=9), then how many elements are only in (A)?

Explanation opens after your attempt
Correct Answer

A. (36)

Step 1

Concept

Only (A=78-30-21+9=36). In three sets, the centre is subtracted twice, so add it back.

Step 2

Why this answer is correct

The correct answer is A. (36). Only (A=78-30-21+9=36). In three sets, the centre is subtracted twice, so add it back.

Step 3

Exam Tip

केवल (A=78-30-21+9=36) है। तीन समुच्चयों में केंद्र दो बार घटता है इसलिए उसे वापस जोड़ें।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

यदि (n(A)=78), (n(B)=69), (n(C)=63), (n\(A\cap B\)=30), (n\(B\cap C\)=25), (n\(C\cap A\)=21) और (n\(A\cap B\cap C\)=9) है, तो केवल (A) में कितने तत्व हैं? / If (n(A)=78), (n(B)=69), (n(C)=63), (n\(A\cap B\)=30), (n\(B\cap C\)=25), (n\(C\cap A\)=21) and (n\(A\cap B\cap C\)=9), then how many elements are only in (A)?

Correct Answer: A. (36). Explanation: केवल (A=78-30-21+9=36) है। तीन समुच्चयों में केंद्र दो बार घटता है इसलिए उसे वापस जोड़ें। / Only (A=78-30-21+9=36). In three sets, the centre is subtracted twice, so add it back.

Which concept should I revise for this Mathematics MCQ?

Only (A=78-30-21+9=36). In three sets, the centre is subtracted twice, so add it back.

What exam hint can help solve this Mathematics question?

केवल (A=78-30-21+9=36) है। तीन समुच्चयों में केंद्र दो बार घटता है इसलिए उसे वापस जोड़ें।