## Task B-2.5.

A bus left from place $A$ to place $B$. 50 minutes later, a car left from place $A$ and arrived at place $B$ 10 minutes before the bus. If they had left simultaneously, one from place $A$ and the other from place $B$ (one heading towards the other), they would have met after one hour and 12 minutes. If traveling the same route and at the same speed, how long will it take for the bus to return from place $B$ to place $A$?
## Solution.

The bus spent $x$ hours on the journey between locations $A$ and $B$, while the car spent $x-1$ hours, so it must be that $x>1$.

The speed of the bus is $\frac{s}{x}$ (where $s$ is the distance between locations $A$ and $B$).

The speed of the car is $\frac{s}{x-1}$.

Starting from location $A$, the bus traveled $\frac{6}{5} \cdot \frac{s}{x} \mathrm{~km}$ in 1 hour and 12 minutes (which is $\frac{6}{5}$ hours).

Starting from location $B$, the car traveled $\frac{6}{5} \cdot \frac{s}{x-1} \mathrm{~km}$ in 1 hour and 12 minutes. Therefore,

$$
\frac{6}{5} \cdot \frac{s}{x} + \frac{6}{5} \cdot \frac{s}{x-1} = s
$$

or

$$
\frac{1}{x} + \frac{1}{x-1} = \frac{5}{6}
$$

By simplifying, we get the equation $5 x^{2} - 17 x + 6 = 0$, whose solutions are 3 and $\frac{2}{5}$.

Since $x$ must be greater than 1, we conclude that the time required for the bus to travel the distance from $B$ to $A$ (or vice versa) is 3 hours.

## NATIONAL MATHEMATICS COMPETITION

## 3rd grade - high school - B variant  Šibenik, April 29, 2010.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER.