4. A motorist set off from point A to point B, the distance between which is 100 km. At the moment when the navigator showed that 30 minutes of travel remained, the motorist first reduced the speed by 10 km/h, and at the moment when the navigator showed that 20 km remained to travel, the motorist reduced the speed by the same 10 km/h again. (The navigator determines the remaining time and distance based on the current speed of movement.) Determine the initial speed of the car, given that the car traveled 5 minutes longer at the reduced speed than at the twice-reduced speed.

NOTE. In the 9th grade, the same problem was given in a slightly different version.
Solution. According to the condition, the distance from C to D is 20 km. Let the distance from A to B be denoted as

![](https://cdn.mathpix.com/cropped/2024_05_06_63cad4cd842f2f58c3e5g-2.jpg?height=120&width=737&top_left_y=874&top_left_x=471)

$x$ (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, the speed is $(v-10) \frac{\text{km}}{4}$ 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,

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

Furthermore, the time spent on the journey from B to C is 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: $100 \frac{\text{km}}{4}$.