Two cyclists set out at the same time from the same place on a trip. They traveled the same route and returned together. On the way, both took breaks. The first cycled twice as long as the second rested. The second cycled four times as long as the first rested. Who of them cycles faster and by how many times?

(Černek)

#
Let

$T_{1} \cdots$ the riding time of the 1st cyclist,

$T_{2} \cdots$ the riding time of the 2nd cyclist,

$o_{1} \cdots$ the resting time of the 1st cyclist,

$\mathrm{O}_{2} \cdots$ the resting time of the 2nd cyclist.

Then we have:

$$
\begin{aligned}
o_{1}+T_{1} & =o_{2}+T_{2} \\
T_{1} & =2 o_{2} \\
T_{2} & =4 o_{1}
\end{aligned}
$$

(3 points)

Substituting $T_{1}, T_{2}$ into the first equation, we get

$$
o_{2}=3 o_{1} \text {. }
$$

Let $o_{1}=t$. Then we immediately get that $o_{2}=3 t, T_{1}=6 t$ and $T_{2}=4 t$. We see that the 2nd cyclist rode 1.5 times faster than the first and rested three times as long.