5. From the series
$$
\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}
$$

which terms should be removed so that the sum of the remaining terms equals 1?
(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 $\frac{1}{10}$.
5. Since $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}=\frac{60}{120}$ $+\frac{30}{120}+\frac{20}{120}+\frac{15}{120}+\frac{12}{120}+\frac{10}{120}$, is - 5 . One possibility is to remove $\frac{15}{120}$ and $\frac{12}{120}$, i.e., $\frac{1}{8}$ and $\frac{1}{10}$. Moreover, from the above expression, it can be seen that the sum of any other terms does not equal
(E) True.