A train which travels at a uniform speed due to some mechanical fault after traveling for an hour goes at 3/5th of the original speed and reaches the destination 2 hrs late.If the fault had occurred after traveling another 50 miles the train would have reached 40 min earlier. What is distance between the two stations.

Let original speed be $v$ and distance be $D$. The delay of 2 hours is caused by traveling the remaining distance at $3/5v$. If the fault occurs 50 miles later, the delay is reduced by 40 minutes (2/3 hours). This means traveling 50 miles at $3/5v$ instead of $v$ takes 2/3 hours longer. Thus, $50/(3v/5) - 50/v = 2/3$, which gives $v = 50$ mph. Substituting $v$ back into the first scenario yields a total distance of 300 miles.