तीन समुच्चयों में (n(A)=72), (n(B)=66), (n(C)=60), ठीक एक समुच्चय में (84) और तीनों में (10) हैं। ठीक दो समुच्चयों में कितने तत्व हैं?

In three sets (n(A)=72), (n(B)=66), (n(C)=60), (84) elements are in exactly one set, and (10) are in all three. How many elements are in exactly two sets?

Explanation opens after your attempt
Correct Answer

B. (42)

Step 1

Concept

Total memberships are (72+66+60=198). \(198=84+2x+3\times10\), so (x=42).

Step 2

Why this answer is correct

The correct answer is B. (42). Total memberships are (72+66+60=198). \(198=84+2x+3\times10\), so (x=42).

Step 3

Exam Tip

कुल सदस्यता (72+66+60=198) है। \(198=84+2x+3\times10\), इसलिए (x=42)।

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

तीन समुच्चयों में (n(A)=72), (n(B)=66), (n(C)=60), ठीक एक समुच्चय में (84) और तीनों में (10) हैं। ठीक दो समुच्चयों में कितने तत्व हैं? / In three sets (n(A)=72), (n(B)=66), (n(C)=60), (84) elements are in exactly one set, and (10) are in all three. How many elements are in exactly two sets?

Correct Answer: B. (42). Explanation: कुल सदस्यता (72+66+60=198) है। \(198=84+2x+3\times10\), इसलिए (x=42)। / Total memberships are (72+66+60=198). \(198=84+2x+3\times10\), so (x=42).

Which concept should I revise for this Mathematics MCQ?

Total memberships are (72+66+60=198). \(198=84+2x+3\times10\), so (x=42).

What exam hint can help solve this Mathematics question?

कुल सदस्यता (72+66+60=198) है। \(198=84+2x+3\times10\), इसलिए (x=42)।