\(यदि (U={1,2,\ldots,49}), (A={x:x\) पूर्ण वर्ग है\(}) और (B={x:x\) विषम है\(}), तो (|A'\cap B|) कितना है\)?
\(If (U={1,2,\ldots,49}), (A={x:x\) is a perfect square\(}) and (B={x:x\) is odd\(}), what is (|A'\cap B|)\)?
Explanation opens after your attempt
Correct Answer
B. (21)
Concept
There are (25) odd numbers from (1) to (49), and the odd perfect squares are (1,9,25,49), so there are (4). Hence \(A'\cap B\) has (25-4=21) elements.
Why this answer is correct
The correct answer is B. (21). There are (25) odd numbers from (1) to (49), and the odd perfect squares are (1,9,25,49), so there are (4). Hence \(A'\cap B\) has (25-4=21) elements.
Exam Tip
(1) से (49) तक (25) विषम संख्याएँ हैं और विषम पूर्ण वर्ग (1,9,25,49) यानी (4) हैं। इसलिए \(A'\cap B\) में (25-4=21) अवयव हैं।
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