13.083. A freight train was delayed on the way for 12 minutes, and then at a distance of 60 km, it made up for the lost time by increasing its speed by 15 km/h. Find the original speed of the train.
## Solution.

Let the speed of the train before the delay be $x$ km/h, and after the delay $(x+15)$ km/h. Then (Fig. 13.5) $A B=\frac{x}{5}, C E=60, C D=60-\frac{x}{5}$, $B D=\frac{60-\frac{x}{5}}{x}, A E=\frac{60}{x+15}$. Since $B D=A E$, then $\frac{60-\frac{x}{5}}{x}=\frac{60}{x+15}$, from which $x=60($ km $/ \mathbf{4})$.

Answer: $60 \mathrm{km} / \mathrm{h}$.