$\left[\begin{array}{ll}{[\text { Ratio of volumes }}\end{array}\right]$

On the edges $B C, C D$ and $A D$ of the tetrahedron $A B C D$ with volume $V$, points $L, M$ and $N$ are taken respectively, such that $3 B L=B C$, $4 C M=C D$ and $5 D N=A D$. Find the volume of the tetrahedron $N M L B$.
Let $A A 1$ be the height of the tetrahedron $A B C D$, and $D P$ and $M Q$ be the heights of triangles $B C D$ and $B C M$. Then the height $N N 1$ of the tetrahedron $N M L B$ is $\frac{1}{5} A A 1$, and $M Q=\frac{1}{4} D P$. Therefore,

$$
\begin{aligned}
V_{N M L B} & =\frac{1}{3} S_{\triangle M L B} \cdot N N 1=\frac{1}{3} \cdot\left(\frac{1}{2} B L \cdot M Q\right) \cdot \frac{1}{5} A A 1=\frac{1}{3} \cdot\left(\frac{1}{2} \cdot \frac{1}{3} B C \cdot \frac{1}{4} D P\right) \cdot \frac{1}{5} A A 1= \\
& =\left(\frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{5}\right) \cdot \frac{1}{3} \cdot\left(\frac{1}{2} B C \cdot D P\right) \cdot A A 1=\frac{1}{60} \cdot\left(\frac{1}{3} S_{\triangle B C D} \cdot A A 1\right)=\frac{1}{60} V .
\end{aligned}
$$

## Answer

$\frac{v}{60}$.