4. In $\triangle A B C$, $P, Q$ are points on $A B$ and $A C$ respectively, and the median $A M$ intersects $P Q$ at $N$. If $A B$ : $A P=5: 2, A C: A Q=4: 3$, then $A M: A N$ $=$ $\qquad$
4. As shown in the figure, draw $C D / /$ $P Q$ intersecting $A B$ at $D$, and $M K / /$ $P Q$ intersecting $A B$ at $K$, then $M K / / C D$.
$$
\begin{array}{c}
\because B M=M C, \\
\therefore B K=K D . \\
\therefore A K=\frac{1}{2}(A D+A B) . \\
\text { Therefore, } \frac{A K}{A P}=\frac{1}{2}\left(\frac{A D}{A P}+\frac{A B}{A P}\right) . \\
\text { And } \frac{A K}{A P}=\frac{A M}{A N}, \frac{A D}{A P}=\frac{A C}{A Q}, \\
\text { hence } \frac{A M}{A N}=\frac{1}{2}\left(\frac{A B}{A P}+\frac{A C}{A Q}\right) \\
\quad=\frac{1}{2}\left(\frac{5}{2}+\frac{4}{3}\right)=\frac{23}{12} .
\end{array}
$$