Two cyclists: $A$ and $B$ start from two locations, $M$ and $N$, heading towards $P$ at the same time. The distance $NP$ is $30 \mathrm{~km}$ longer than the distance $MP$. Each cyclist takes 10 hours to complete their journey.

What is the distance $MP$, if for every $42 \mathrm{~km}$ that $B$ travels, it takes $B$ 20 minutes less than it takes $A$?
Let the sought $M P=x \mathrm{~km}$. For a $1 \mathrm{~km}$ distance, it takes $A \frac{10}{x}$ hours and $B \frac{10}{x+30}$ hours. For 42 $\mathrm{km}$, $A$ takes $\frac{420}{x}$ hours, and $B$ takes $\frac{420}{x+30}$ hours, which, according to the problem, is $\frac{1}{3}$ hour less than the former. Therefore,

$$
\frac{420}{x}=\frac{420}{x+30}+\frac{1}{3}
$$

which simplifies to

$$
420 \cdot 3(x+30)=420 \cdot 3 x+x(x+30)
$$

Rearranging gives

$$
x^{2}+30 x-37800=0
$$

from which

$$
x_{1}=180 \quad\left[x_{2}=-210\right] .
$$

Thus, $M P=180 \mathrm{~km}$.

Imre Jámbor (Pannonhalma Abbey II. Class)