13.129. At 9 AM, a self-propelled barge left $A$ upstream and arrived at point $B$; 2 hours after arriving at $B$, the barge set off on the return journey and arrived back at $A$ at 7:20 PM on the same day. Assuming the average speed of the river current is 3 km/h and the barge's own speed is constant throughout, determine when the barge arrived at point $B$. The distance between $A$ and $B$ is 60 km.
Solution.

Let $v$ km/h be the own speed of the barge. From $A$ to $B$, the barge sailed $\frac{60}{v-3}$ hours, and on the way back $-\frac{60}{v+3}$ hours. According to the condition, $\frac{60}{v-3}+2+\frac{60}{v+3}=10 \frac{1}{3}$, from which $v=15$ km/h. The barge arrived in $B$ at $9+\frac{60}{15-3}=14$ hours.

Answer: at 14 hours.