Q167 / 270 · NDA 2012 (Phase 2) · Mathematics

Let $\mathrm{A}=\{x \in \mathrm{~W}$, the set of whole numbers and $x \lt 3\}$, $\mathrm{B}=\{x \in \mathrm{~N}$, the set of natural numbers and $2 \leq x \lt 4\}$ and $C=\{3,4\}$, then how many elements will $(A \cup B) \times C$ contain?

  • A
    $6$
  • Correct answer
    B
    $8$
  • C
    $10$
  • D
    $12$

Solution

$\begin{aligned} & A=\{0,1,2\}, B=\{2,3\}, C=\{3,4\} \\ & \therefore A \cup B=\{0,1,2,3\}\end{aligned}$ $\therefore$ no. of elements in $(A \cup B) \times C$ will be 8 .

Hence, the correct answer is option: 'B'.

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