A car traveling at $66 \mathrm{~km} /$ hour arrives at its destination at 6:53, while a truck traveling at $42 \mathrm{~km} /$ hour on the same route arrives at the common destination at 7:11. By how many kilometers did the car overtake the truck before the destination?
I. solution: The distance to be found is covered by a car traveling at a speed of $\frac{66}{60}=1.1 \mathrm{~km} /$ minute in $x$ minutes, and by a truck traveling at a speed of $\frac{42}{60}=0.7 \mathrm{~km} /$ minute in $x+18$ minutes. Therefore,

$$
1.1 x=0.7(x+18)
$$

which simplifies to

$$
0.4 x=12.6
$$

from which

$$
x=\frac{126}{4}=31.5 \text { minutes }
$$

and thus the lead occurred $31.5 \cdot 1.1=34.65 \mathrm{~km}$ before the finish line.

Heckenast Tamás (Szombathely, Nagy Lajos g. II.o.t.)

II. solution: Let the distance to be found be $x \mathrm{~km}$. According to the problem, the car covers this $x$ distance in $\frac{x}{66}$ hours, while the truck covers it in $\frac{x}{42}$ hours. The latter time is 18 minutes $=0.3$ hours longer than the former, so

$$
\frac{x}{66}=\frac{x}{42}-0.3
$$

which simplifies to

$$
\frac{x}{11}=\frac{x}{7}-1.8
$$

or

$$
7 x=11 x-138.6
$$

from which

$$
x=\frac{138.6}{4}=34.65 \mathrm{~km}
$$