$28 \cdot 32$ From the sum
$$
\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}
$$

remove some terms so that the sum of the remaining terms equals 1. The terms removed are
(A) $\frac{1}{4}$ and $\frac{1}{8}$.
(B) $\frac{1}{4}$ and $\frac{1}{12}$.
(C) $\frac{1}{8}$ and $\frac{1}{12}$.
(D) $\frac{1}{6}$ and $\frac{1}{10}$.
(E) $\frac{1}{8}$ and $\frac{1}{10}$.
(36th American High School Mathematics Examination, 1985)
[Solution] Since $\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$, and $\frac{1}{6}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}$, then $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=1$.
Therefore, the answer is $(E)$.