3.20 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.
3.20 Let the speed of the train before the delay be $x$ km/h, and after the delay $(x+15)$ km/h. Then (Fig. 3.5)

$A B=\frac{x}{5}, C E=60$,

$C D=60-\frac{x}{5}$,

$B D=\frac{60-\frac{x}{5}}{x}$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-100.jpg?height=509&width=460&top_left_y=153&top_left_x=689)

Fig. 3.5

$A E=\frac{60}{x+15}$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-100.jpg?height=141&width=312&top_left_y=741&top_left_x=563)

from which $x=60$ (km/h).

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