Cátia leaves school every day at the same time and returns home by bicycle. When she cycles at $20 \mathrm{~km} / \mathrm{h}$, she arrives home at $4: 30$ PM. If she cycles at $10 \mathrm{~km} / \mathrm{h}$, she arrives home at $5: 15$ PM. At what speed should she cycle to arrive home at 5: 00 PM?
Let $t$ be the time she spends cycling at $20 \mathrm{~km} / \mathrm{h}$. Cycling at $10 \mathrm{~km} / \mathrm{h}$, she takes double the time compared to cycling at $20 \mathrm{~km} / \mathrm{h}$, which is $2 t$. However, since she takes 45 minutes longer, we have:

$$
2 t - t = 45 \Longrightarrow t = 45 \mathrm{~min}
$$

Therefore, she leaves school daily at $4: 30 \mathrm{~h} - 45 \mathrm{~min} = 3: 15 \mathrm{~h}$, and the distance to her home is $45 \mathrm{~min} \times 20 \mathrm{~km} / \mathrm{h} = \frac{3}{4} \times 20 = 15 \mathrm{~km}$. To cover $15 \mathrm{~km}$ in $5: 00 \mathrm{~h} - 3: 15 \mathrm{~h} = 1: 45 \mathrm{~h} = \frac{5}{4} \mathrm{~h}$, she needs to maintain a speed of

$$
\frac{15 \mathrm{~km}}{\frac{5}{4} \mathrm{~h}} = 12 \mathrm{~km} / \mathrm{h}
$$