Clara leaves home by bike at 1:00 p.m. for a meeting scheduled with Quinn later that afternoon. If Clara travels at an average of $20 \mathrm{~km} / \mathrm{h}$, she would arrive half an hour before their scheduled meeting time. If Clara travels at an average of $12 \mathrm{~km} / \mathrm{h}$, she would arrive half an hour after their scheduled meeting time. At what average speed, in $\mathrm{km} / \mathrm{h}$, should Clara travel to meet Quinn at the scheduled time?
Suppose that the distance that Clara has to travel is $d \mathrm{~km}$ and that the time from when she starts to the scheduled meeting time is $T$ hours.

When she travels at an average speed of $20 \mathrm{~km} / \mathrm{h}$, she arrives half an hour early.

Thus, she travels for $T-\frac{1}{2}$ hours and so $\frac{d}{20}=T-\frac{1}{2}$.

When she travels at an average speed of $12 \mathrm{~km} / \mathrm{h}$, she arrives half an hour late.

Thus, she travels for $T+\frac{1}{2}$ hours and so $\frac{d}{12}=T+\frac{1}{2}$.

Adding these two equations, we obtain $\frac{d}{20}+\frac{d}{12}=2 T$ or $\frac{3 d}{60}+\frac{5 d}{60}=2 T$.

Thus, $\frac{8 d}{60}=2 T$ or $\frac{d}{15}=T$.

This means that if Clara travels at an average speed of $15 \mathrm{~km} / \mathrm{h}$, then it will take her exactly $T$ hours to travel the $d \mathrm{~km}$, and so she will arrive at the scheduled time.

ANSWER: 15