13.085. Two brothers took their bicycles and set off simultaneously with the intention of riding $42 \mathrm{km}$. The older brother maintained the same speed throughout the journey; while the younger brother fell behind by 4 km every hour. However, since the older brother rested for a whole hour during the trip, and the younger brother only rested for 20 minutes, they arrived at the finish line simultaneously. How long did the trip last?
## Solution.

Let $x$ km/h be the speed of the older brother, then $(x-4)$ km/h is the speed of the younger brother. The older brother was on the road for $t_{\text {old }}=\frac{42}{x}+1$, and the younger brother for $t_{\text {young }}=\frac{42}{x-4}+\frac{1}{3}$. According to the problem, $t_{\text {old }}=t_{\text {young }}$, that is, $\frac{42}{x}+1=\frac{42}{x-4}+\frac{1}{3}$, from which $x=18$ and $t=1+\frac{42}{18}=3 \frac{1}{3}$.

Answer: 3 hours 20 minutes.