22. Geraint always cycles to work, leaving at 8am every morning. When he averages $15 \mathrm{~km} / \mathrm{h}$, he arrives 10 minutes late. However, when he averages $30 \mathrm{~km} / \mathrm{h}$, he arrives 10 minutes early. What speed should he average to arrive on time?
A $20 \mathrm{~km} / \mathrm{h}$
B $21 \mathrm{~km} / \mathrm{h}$
C $22.5 \mathrm{~km} / \mathrm{h}$
D $24 \mathrm{~km} / \mathrm{h}$
E $25 \mathrm{~km} / \mathrm{h}$
22. A Let $x \mathrm{~km}$ be the distance Geraint cycles and let $t$ hours be the time his journey should take if he is to be on time. Since $\frac{\text { distance }}{\text { speed }}=$ time, the information in the question tells us that $\frac{x}{15}=t+\frac{1}{6}$ and that $\frac{x}{30}=t-\frac{1}{6}$. When we subtract the second equation from the first, we obtain $\frac{x}{30}=\frac{2}{6}$ and so $x=10$. Hence, from the second equation, $\frac{10}{30}=t-\frac{1}{6}$ and so $t=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$. Therefore, to arrive on time, Geraint needs to travel $10 \mathrm{~km}$ in $\frac{1}{2}$ hour, which is an average speed of $20 \mathrm{~km} / \mathrm{h}$.