If $\frac{1}{2 n}+\frac{1}{4 n}=\frac{3}{12}$, then $n$ equals
(A) 6
(B) $\frac{1}{2}$
(C) $\frac{1}{3}$
(D) 2
(E) 3
Since $\frac{1}{2 n}+\frac{1}{4 n}=\frac{2}{4 n}+\frac{1}{4 n}=\frac{3}{4 n}$, then the given equation becomes $\frac{3}{4 n}=\frac{3}{12}$ or $4 n=12$.

Thus, $n=3$.

Answer: (E)