10,11 |

On the edges $A B, B C$, and $A D$ of the tetrahedron $A B C D$ with volume $V$, points $K, L$, and $N$ are taken respectively, such that $2 A K=A B, 3 B L=B C, 5 D N=A D$. Find the volume of the tetrahedron $N K L B$.
Let $D D_1$ be the height of the tetrahedron $A B C D$. Then the height $N N_1$ of the tetrahedron $N K L B$ is $\frac{\mathbf{4}}{5} D D_1$. Therefore,

$$
\begin{aligned}
& V_{N K L B}=\frac{1}{3} S_{\triangle K L B} \cdot N N_1=\frac{1}{3} \cdot \frac{B K}{B A} \cdot \cdot \frac{B L}{B C} \cdot S_{\triangle A B C} \cdot \frac{4}{5} D D_1= \\
& =\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{2} S_{\triangle A B C} \cdot \frac{4}{5 D D_1}=\frac{1}{6} \cdot \frac{4}{5}\left(\frac{1}{3} S_{\triangle A B C} \cdot D D_1\right)=\frac{2}{15 \mathrm{~V}}
\end{aligned}
$$