In the second case, the speed is \(\frac{5}{4}\) times the speed in the first case.
Consequently, the time would be \(\frac{4}{5}\) times the time, which is equivalent to being \(\frac{1}{5}\) less. This one-fifth is 20 minutes.
Therefore, the time taken in the first case is 100 minutes.
The distance \(=12\times\frac{5}{3}\ km=20\ km\)