3. A motorist set off from point A to point D, the distance between which is 100 km. The road from A to D passes through points B and C. At point B, the navigator showed that 30 minutes of driving remained, and the motorist immediately reduced the speed by 10 km/h. At point C, the navigator showed that 20 km remained to drive, and the motorist immediately reduced the speed by the same 10 km/h again. (The navigator determines the remaining time based on the current speed of movement.) Determine the initial speed of the car, given that the motorist spent 5 minutes more on the journey from B to C than on the journey from C to D.
Solution. According to the condition, the distance from C to D is 20 km. Let the distance from A to B be $x$ (km), then the distance from B to C will be ( $80-x$ ) km.

Let $v \frac{\text { km }}{4}$ be the initial speed of the car. Then on the segments BC and CD, it is ( $v-$

![](https://cdn.mathpix.com/cropped/2024_05_06_d272e1fd1357fc2601b4g-1.jpg?height=128&width=739&top_left_y=2329&top_left_x=724)

10) $\frac{\text { km }}{4}$ and $(v-20) \frac{\text { km }}{4}$, respectively. According to the condition, the journey from B to D would have taken half an hour if the car had continued to move at a speed of $v \frac{\text { km }}{4}$, that is,

$$
100-x=\frac{v}{2}
$$

Interregional Olympiad for Schoolchildren based on Departmental Educational Organizations in Mathematics

Further, the time spent on the journey from B to C was 5 minutes more than the time spent on the journey from C to D:

$$
\frac{80-x}{v-10}-\frac{20}{v-20}=\frac{1}{12}
$$

Expressing $x$ from the first equation and substituting it into the second, we get an equation to determine $v$:

$$
\frac{\frac{v}{2}-20}{v-10}-\frac{20}{v-20}=\frac{1}{12}
$$

which has roots 14 and 100. The root 14 is obviously extraneous, as $v>20$.

Answer: $100 \frac{\mathrm{KM}}{\mathrm{q}}$.