If the universal set is \(U=\{1,2,\ldots,27\}\), \(A=\{x\in U:x\text{ is a multiple of }3\}\), and \(B=\{x\in U:x\text{ is a multiple of }9\}\), how many elements are in \(A'\cup B\)?
Answer and explanation
Correct answer: 21
The multiples of 3 in \(U\) are \(3,6,9,12,15,18,21,24,27\), so \(|A|=9\) and \(|A'|=27-9=18\). The multiples of 9 are \(9,18,27\), so \(|B|=3\). Because every multiple of 9 is also a multiple of 3, \(B\subseteq A\), which means \(A'\cap B=\varnothing\). Therefore, \(|A'\cup B|=|A'|+|B|=18+3=21\). Thus, option D is correct.
Frequently asked questions
What is the correct answer to this question?
21
Why is this the correct answer?
The multiples of 3 in \(U\) are \(3,6,9,12,15,18,21,24,27\), so \(|A|=9\) and \(|A'|=27-9=18\). The multiples of 9 are \(9,18,27\), so \(|B|=3\). Because every multiple of 9 is also a multiple of 3, \(B\subseteq A\), which means \(A'\cap B=\varnothing\). Therefore, \(|A'\cup B|=|A'|+|B|=18+3=21\). Thus, option D is correct.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Complement of a Set and Its Properties.
