Problem 10.1. A bus leaves city $A$ for city $B$, which is 240 km away from $A$, at a speed of 40 km/h. At the same time, a car leaves $B$ for $A$ at a speed of $v$ km/h. Half an hour after meeting the bus, the car, without reaching city $A$, turns around and heads back towards $B$ at the same speed. Determine all values of $v$ for which the car arrives in $B$ before the bus.
Answer: $v \in(56 ; 120)$.

Solution. Let $t$ be the time of the meeting, measured in hours. Then, by the condition, $t=\frac{240}{v+40}$. Further, the condition states that the car, having driven for half an hour, did not reach point $A$, which leads to the inequality $v(t+0.5) < 240 - 0.5v$.

Considering the positivity of the denominators, we obtain the equivalent quadratic inequality $v^{2}+40 v-19200 < 0$.

## Criteria

One of the largest suitable criteria is used:

76 . Any correct solution to the problem.

## 2 6. Only the lower estimate is obtained.

## 2 6. Only the upper estimate is obtained.

Points are deducted for the following error in the solution, which is otherwise correct:

-1 6. One or both ends of the interval are incorrectly included in the answer.