यदि \(U={x:x\in\mathbb{N},x\le 96}\), \(A={x:x\in U,8\mid x}\) और \(B={x:x\in U,12\mid x}\), तो (n\(A'\cap B'\)) क्या है?

If \(U={x:x\in\mathbb{N},x\le 96}\), \(A={x:x\in U,8\mid x}\), and \(B={x:x\in U,12\mid x}\), what is (n\(A'\cap B'\))?

Explanation opens after your attempt
Correct Answer

A. (80)

Step 1

Concept

By De Morgan's law, (A'\cap B'=\(A\cup B\)'). Since (n\(A\cup B\)=12+8-4=16), the complement has (96-16=80) elements.

Step 2

Why this answer is correct

The correct answer is A. (80). By De Morgan's law, (A'\cap B'=\(A\cup B\)'). Since (n\(A\cup B\)=12+8-4=16), the complement has (96-16=80) elements.

Step 3

Exam Tip

डी मॉर्गन से (A'\cap B'=\(A\cup B\)') है। (n\(A\cup B\)=12+8-4=16), इसलिए पूरक में (96-16=80) सदस्य हैं।

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Mathematics Answer, Explanation and Revision Hints

यदि \(U={x:x\in\mathbb{N},x\le 96}\), \(A={x:x\in U,8\mid x}\) और \(B={x:x\in U,12\mid x}\), तो (n\(A'\cap B'\)) क्या है? / If \(U={x:x\in\mathbb{N},x\le 96}\), \(A={x:x\in U,8\mid x}\), and \(B={x:x\in U,12\mid x}\), what is (n\(A'\cap B'\))?

Correct Answer: A. (80). Explanation: डी मॉर्गन से (A'\cap B'=\(A\cup B\)') है। (n\(A\cup B\)=12+8-4=16), इसलिए पूरक में (96-16=80) सदस्य हैं। / By De Morgan's law, (A'\cap B'=\(A\cup B\)'). Since (n\(A\cup B\)=12+8-4=16), the complement has (96-16=80) elements.

Which concept should I revise for this Mathematics MCQ?

By De Morgan's law, (A'\cap B'=\(A\cup B\)'). Since (n\(A\cup B\)=12+8-4=16), the complement has (96-16=80) elements.

What exam hint can help solve this Mathematics question?

डी मॉर्गन से (A'\cap B'=\(A\cup B\)') है। (n\(A\cup B\)=12+8-4=16), इसलिए पूरक में (96-16=80) सदस्य हैं।