A train completes a journey with a few stoppages in between at an average speed of 40 km per hour. If the train had not stopped anywhere, it would have completed the journey at an average speed of 60 km per hour. On an average, how many minutes per hour does the train stop during the journey?

Let the train move (excluding stoppages) for time Tm and stop for Ts while covering distance S. Without stoppages S = 60·Tm, and with stoppages average speed 40 = S/(Tm+Ts), so 40(Tm+Ts)=60·Tm. Rearranging gives 40Ts = 20Tm, hence Ts = (1/2)Tm. Thus stoppage time is half the running time, so fraction of total time spent stopped = Ts/(Tm+Ts) = (0.5Tm)/(1.5Tm) = 1/3. Therefore the train stops for one-third of each hour, i.e., (1/3)·60 = 20 minutes per hour. The setup (using S/60 + T) and algebraic approach follow standard formulations for speed-with/without-stoppage problems.