3. As shown in Figure 1, points $D$ and $E$ are on the sides $AC$ and $AB$ of $\triangle ABC$, respectively. $BD$ and $CE$ intersect at point $O$. The areas of $\triangle OBE$, $\triangle OBC$, and $\triangle OCD$ are $15$, $30$, and $24$, respectively. Then $AE: BE=(\quad)$.
(A) $5: 2$
(B) $2: 1$
(C) $5: 4$
(D) $9: 5$
3. B.

As shown in Figure 3, draw $D F / / E C$ intersecting $A B$ at point $F$.

Since $\triangle O B E$ and $\triangle O B C$ have the same height, we have
$$
\frac{O E}{O C}=\frac{S_{\triangle O B E}}{S_{\triangle O H C}}=\frac{1}{2}.
$$

Similarly, $\frac{O B}{O D}=\frac{S_{\triangle O B C}}{S_{\triangle D D}}=\frac{5}{4}$.
Let $O E=m, O D=4 n$, then $O C=2 m, O B=5 n$.
Since $D F / / E C$, we have
$\triangle B O E \backsim \triangle B D F, \triangle A D F \backsim \triangle A C E$.
Thus, $\frac{B E}{B F}=\frac{O E}{D F}=\frac{O B}{B D}=\frac{5 n}{9 n}=\frac{5}{9}$,
$\frac{A F}{A E}=\frac{D F}{C E}$.
From equation (1), we get $B E=\frac{5}{4} E F, D F=\frac{9}{5} O E=\frac{9}{5} m$.
From equation (2), we get $\frac{A F}{A E}=\frac{\frac{9}{5} m}{m+2 m}=\frac{3}{5}$.
Therefore, $A E=\frac{5}{2} E F$.
Hence, $A E: B E=\frac{5}{2} E F: \frac{5}{4} E F=2: 1$.