Let \(U=\{1,2,\ldots,60\}\), \(A=\{x\in U:x\text{ is a multiple of }6\}\), and \(B=\{x\in U:x\text{ is a multiple of }18\}\). Then \(A'\cap B'\) is equal to which of the following?
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Answer and explanation
Correct answer: \(A'\)
Every multiple of 18 is also a multiple of 6, so \(B\subseteq A\). Hence \(A\cup B=A\). Applying De Morgan’s law gives \(A'\cap B'=(A\cup B)'=A'\). Notice that although \(A'\subseteq B'\), the intersection of the two complements is the smaller complement, namely \(A'\).
Frequently asked questions
What is the correct answer to this question?
\(A'\)
Why is this the correct answer?
Every multiple of 18 is also a multiple of 6, so \(B\subseteq A\). Hence \(A\cup B=A\). Applying De Morgan’s law gives \(A'\cap B'=(A\cup B)'=A'\). Notice that although \(A'\subseteq B'\), the intersection of the two complements is the smaller complement, namely \(A'\).
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.
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