A train is traveling from City A to City B. If the train travels at a speed of $80 \mathrm{~km} / \mathrm{h}$, it will arrive 24 minutes late. If it travels at a speed of $90 \mathrm{~km} / \mathrm{h}$, it will arrive 32 minutes early. At what speed in $\mathrm{km} / \mathrm{h}$ should the train travel in order to arrive on time?
Let $t$ be the amount of time in hours that the train will take to make the trip if it is on time and let $d$ be the distance in kilometres between City A and City B.

If the train is 24 minutes late, then it is $\frac{24}{60}$ hours late, which means it takes $t+\frac{24}{60}=t+\frac{2}{5}$ hours to reach City B.

Similarly, if the train is 32 minutes early, then it takes $t-\frac{32}{60}=t-\frac{8}{15}$ hours to reach City B.

From the given assumptions, this means $80=\frac{d}{t+\frac{2}{5}}$ and $90=\frac{d}{t-\frac{8}{15}}$.

Cross multiplying these equations, we get $80\left(t+\frac{2}{5}\right)=d$ and $90\left(t-\frac{8}{15}\right)=d$.

Equating the values of $d$ and solving for $t$, we get

$$
\begin{aligned}
80\left(t+\frac{2}{5}\right) & =90\left(t-\frac{8}{15}\right) \\
80 t+32 & =90 t-48 \\
80 & =10 t \\
8 & =t
\end{aligned}
$$

Substituting $t=8$ into $d=80\left(t+\frac{2}{5}\right)$ gives $d=80\left(8+\frac{2}{5}\right)=672$.

The speed in $\mathrm{km} / \mathrm{h}$ that the train should travel in order to arrive on time is $\frac{672}{8}=84$.

ANsWER: 84