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 $16 \mathrm{~h} 30 \mathrm{~m}$. If she cycles at $10 \mathrm{~km} / \mathrm{h}$, she arrives home at $17 \mathrm{~h} 15 \mathrm{~m}$. At what speed should she cycle to arrive home at $17 \mathrm{~h}$?
Let $t$ be the time Cátia spends cycling at $20 \mathrm{~km} / \mathrm{h}$. Cycling at $10 \mathrm{~km} / \mathrm{h}$, she takes twice the time she spends cycling at $20 \mathrm{~km} / \mathrm{h}$, which is $2 t$. However, since she takes 45 minutes longer, we have $2 t - t = 45$, so $t = 45 \mathrm{~min}$. Therefore, she leaves school 45 minutes before $16 \mathrm{~h} 30 \mathrm{~m}$, which is at $15 \mathrm{~h} 45 \mathrm{~m}$. The distance from school to her home, which is covered in $45 \mathrm{~min}$ at $20 \mathrm{~km} / \mathrm{h}$, is $\frac{3}{4} \times 20 = 15 \mathrm{~km}$. To leave school at $15 \mathrm{~h} 45 \mathrm{~m}$ and arrive home at $17 \mathrm{~h}$, she must cover these $15 \mathrm{~km}$ between school and her home in $1 \mathrm{~h} 15 \mathrm{~m}$, which corresponds to $5 / 4$ of an hour. Therefore, Cátia must maintain a speed of

$$
\frac{15 \mathrm{~km}}{5 / 4 \mathrm{~h}} = \frac{60}{5} \mathrm{~km} / \mathrm{h} = 12 \mathrm{~km} / \mathrm{h}
$$