BECE Mathematics 2020 Paper

Concept:The problem uses sets of factors and multiples, then asks for their common elements.Explanation:$N=\{x: x \text{ is a factor of } 18\}$The factors of $18$ are $1, 2, 3, 6, 9$ and $18$.Hence $N=\{1, 2, 3, 6, 9, 18\}$.$M=\{x: x \text{ is a multiple of } 12\}$The multiples of $12$ are $12, 24, 36, 48, 60, 72, \ldots$Hence $M=\{12, 24, 36, 48, 60, 72, \ldots\}$.The symbol $\cap$ means intersection, or the elements common to both sets.No number from $N$ appears in $M$, because every element of $N$ is smaller than $12$ except $18$, and $18$ is not a multiple of $12$.Therefore $N \cap M$ has no elements.Answer:$N \cap M=\{\}$The correct option is D.