यदि \(U={1,2,\ldots,90}\), \(A={x:x \in U,6\mid x}\), \(B={x:x \in U,10\mid x}\), तो (n(\(A\cap B\)')) क्या है?

If \(U={1,2,\ldots,90}\), \(A={x:x \in U,6\mid x}\), and \(B={x:x \in U,10\mid x}\), what is (n(\(A\cap B\)'))?

Explanation opens after your attempt
Correct Answer

A. (87)

Step 1

Concept

\(A\cap B\) contains multiples of (30), and there are (3) up to (90). So the complement has (90-3=87) elements.

Step 2

Why this answer is correct

The correct answer is A. (87). \(A\cap B\) contains multiples of (30), and there are (3) up to (90). So the complement has (90-3=87) elements.

Step 3

Exam Tip

\(A\cap B\) में (30) के गुणज होंगे, जो (90) तक (3) हैं। इसलिए पूरक में (90-3=87) सदस्य हैं।

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Mathematics Answer, Explanation and Revision Hints

यदि \(U={1,2,\ldots,90}\), \(A={x:x \in U,6\mid x}\), \(B={x:x \in U,10\mid x}\), तो (n(\(A\cap B\)')) क्या है? / If \(U={1,2,\ldots,90}\), \(A={x:x \in U,6\mid x}\), and \(B={x:x \in U,10\mid x}\), what is (n(\(A\cap B\)'))?

Correct Answer: A. (87). Explanation: \(A\cap B\) में (30) के गुणज होंगे, जो (90) तक (3) हैं। इसलिए पूरक में (90-3=87) सदस्य हैं। / \(A\cap B\) contains multiples of (30), and there are (3) up to (90). So the complement has (90-3=87) elements.

Which concept should I revise for this Mathematics MCQ?

\(A\cap B\) contains multiples of (30), and there are (3) up to (90). So the complement has (90-3=87) elements.

What exam hint can help solve this Mathematics question?

\(A\cap B\) में (30) के गुणज होंगे, जो (90) तक (3) हैं। इसलिए पूरक में (90-3=87) सदस्य हैं।