A moving staircase has 125 steps. On the evenly ascending moving staircase, we start moving upwards,

and after climbing 45 steps, we reach the top. Next time, we can manage to climb 55 steps this way. (The speed of the moving staircase has not changed.) By what factor did we manage to increase our speed?
I. solution. In the first case, the moving staircase covers 80 steps in $t_{1}$ time units, and in the second case, 70 steps in $t_{2}$ time units. For its speed (which we know is constant), or the number of steps covered per time unit:

$$
\frac{80}{t_{1}}=\frac{70}{t_{2}}
$$

from which $t_{2}=\frac{7}{8} t_{1}$

Our speed in the first case is $v_{1}=\frac{45}{t_{1}}$, and in the second case $v_{2}=\frac{55}{t_{2}}$.

Therefore,

$$
\frac{v_{2}}{v_{1}}=\left(\frac{55}{t_{2}} / \frac{45}{t_{1}}\right)=\frac{11 t_{1}}{9 t_{2}}=\frac{11}{9 \cdot \frac{7}{8}}=\frac{88}{64} \approx 1.397
$$

Krisztián Balog (Miskolc, Földes F. Gymn., 12th grade)

II. solution. The first time, the "missing" $125-45=80$ steps were covered by the moving staircase in the time it took us to cover 45 steps. Thus, our speed relative to the moving staircase was $\frac{45}{80}$. The second time, this speed ratio is $\frac{55}{70}$. Therefore, the ratio of the speed increase is:

$$
\left(\frac{55}{70}\right) /\left(\frac{45}{80}\right)=\frac{55 \cdot 80}{70 \cdot 45}=\frac{88}{63} \approx 1.397
$$

(This is much larger than the incorrectly calculated $\frac{55}{45} \approx 1.222$ value.)