If the time taken to go from one station to another is required to be decreased by $20\%$, then by what percent must the speed of the train be increased?

Distance ($D$) is constant, so speed ($S$) and time ($T$) are inversely proportional to each other ($S\propto \frac{1}{T}$).
Let the original time taken be $T_1=100$.
The time is decreased by $20\%$, so the new time taken is:

$$T_2=100-20=80$$

Since $D=S_1\times T_1=S_2\times T_2$:

$$\frac{S_2}{S_1}=\frac{T_1}{T_2}=\frac{100}{80}=\frac{5}{4}$$

The fractional increase in speed is:

$$\frac{S_2-S_1}{S_1}=\frac{5-4}{4}=\frac{1}{4}$$

Percentage increase in speed:

$$\text{Percentage Increase}=\frac{1}{4}\times 100\%=25\%$$