The volume of the tetrahedron $ABCD$ is $V$. Points $K, L$, and $M$ are taken on the edges $CD, DB$, and $AB$ respectively, such that $2CK = CD$, $3DL = DB$, and $5BM = 2AB$. Find the volume of the tetrahedron $KLM D$.

#
$$
\begin{gathered}
V_{M B C D}=\frac{M B}{A B} \cdot V_{A B C D}=\frac{2}{5} V \\
V_{K L M D}=\frac{D K}{D C} \cdot \frac{D L}{D B} \cdot \frac{D M}{D M} \cdot V_{M B C D}=\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{1} \cdot \frac{2}{5} V=\frac{1}{15} V
\end{gathered}
$$

## Answer

$\frac{V}{15}$