A courier travels from location $A$ to location $B$ in 14 hours: another courier starts at the same time as the first, from a place 10 km behind $A$, and arrives at $B$ at the same time as the first. The latter covers the last $20 \mathrm{~km}$ in half an hour less time than the first. What is the distance between $A$ and $B$?
Let the distance $AB$ be denoted by $x$ (in kilometers). The first runner covers $x \, \text{km}$, and the second runner covers $x+10 \, \text{km}$ in 14 hours; the first runner covers $\frac{x}{14}$, and the second runner covers $\frac{x+10}{14} \, \text{km}$ per hour. The first runner takes $20 \div \frac{x}{14} = \frac{280}{x}$ hours, and the second runner takes $20 \div \frac{x+10}{14} = \frac{280}{x+10}$ hours to cover 20 km. According to the problem,

$$
\frac{280}{x} = \frac{280}{x+10} + \frac{1}{2} \quad \text{or} \quad 560(x+10) = 560x + x(x+10)
$$

From this:

$$
\begin{aligned}
& x^2 + 10x - 5600 = 0 \\
& x = \frac{-10 \pm \sqrt{100 + 22400}}{2} = \frac{-10 \pm 150}{2}; \quad x_1 = -80, \quad x_2 = 70
\end{aligned}
$$

Only the positive root can satisfy our problem, so $AB = 70 \, \text{km}$.

László Csatlós (State Electrical Engineering Vocational School, 1st year, Budapest)

Note. The question may arise as to what meaning we should attribute to the negative root of the equation

$$
\frac{280}{x} = \frac{280}{x+10} + \frac{1}{2} \quad \text{or} \quad x^2 + 10x - 5600 = 0
$$

If we replace $x$ with $-x$ in these equations, then the

$$
\begin{aligned}
& \frac{280}{-x} = \frac{280}{-x+10} + \frac{1}{2} \quad \text{or} \\
& x^2 - 10x - 5600 = 0 \ldots
\end{aligned}
$$

equation has a positive root of 80. However, if we start solving the problem by considering $CB$ as the unknown, where $C$ is the point behind $A$, then we arrive at the equation

$$
\frac{280}{x-10} = \frac{280}{x} + \frac{1}{2}
$$

which is nothing but the equation (a).