यदि (A), (B) और (C) परस्पर असंबद्ध हैं, (n(A)=27), (n(B)=34), (n(C)=41) और (n(U)=125) है, तो (n(\(A\cup B\cup C\)')) कितना है?

If (A), (B) and (C) are mutually disjoint, (n(A)=27), (n(B)=34), (n(C)=41) and (n(U)=125), then what is (n(\(A\cup B\cup C\)'))?

Explanation opens after your attempt
Correct Answer

A. (23)

Step 1

Concept

For mutually disjoint sets, the union is (27+34+41=102), so the complement is (125-102=23). No common part needs to be subtracted.

Step 2

Why this answer is correct

The correct answer is A. (23). For mutually disjoint sets, the union is (27+34+41=102), so the complement is (125-102=23). No common part needs to be subtracted.

Step 3

Exam Tip

परस्पर असंबद्ध होने पर संघ (27+34+41=102) है, इसलिए पूरक (125-102=23) है। साझा भाग घटाने की जरूरत नहीं होती।

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 (A), (B) और (C) परस्पर असंबद्ध हैं, (n(A)=27), (n(B)=34), (n(C)=41) और (n(U)=125) है, तो (n(\(A\cup B\cup C\)')) कितना है? / If (A), (B) and (C) are mutually disjoint, (n(A)=27), (n(B)=34), (n(C)=41) and (n(U)=125), then what is (n(\(A\cup B\cup C\)'))?

Correct Answer: A. (23). Explanation: परस्पर असंबद्ध होने पर संघ (27+34+41=102) है, इसलिए पूरक (125-102=23) है। साझा भाग घटाने की जरूरत नहीं होती। / For mutually disjoint sets, the union is (27+34+41=102), so the complement is (125-102=23). No common part needs to be subtracted.

Which concept should I revise for this Mathematics MCQ?

For mutually disjoint sets, the union is (27+34+41=102), so the complement is (125-102=23). No common part needs to be subtracted.

What exam hint can help solve this Mathematics question?

परस्पर असंबद्ध होने पर संघ (27+34+41=102) है, इसलिए पूरक (125-102=23) है। साझा भाग घटाने की जरूरत नहीं होती।