1. From the set of fractions $\left\{\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \frac{1}{12}\right\}$, remove two fractions so that the sum of the remaining numbers is 1. The two numbers to be removed are ( ).
(A) $\frac{1}{4}$ and $\frac{1}{8}$
(B) $\frac{1}{4}$ and $\frac{1}{10}$
(C) $\frac{1}{8}$ and $\frac{1}{10}$
(D) $\frac{1}{8}$ and $\frac{1}{12}$
$-1 . C$
From $\frac{1}{4}+\frac{1}{12}=\frac{1}{3}$, and $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$, thus by removing $\frac{1}{8}$ and $\frac{1}{10}$, the sum of the remaining numbers can be 1.