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

Let the normal speed be s, the normal time be t and the distance be d. s is in miles/hour. t in hours. d in miles. Then d=st --------------------------------------... Now suppose the train travels at s for an hour, and then completes the rest of the journey at (3s/5). Distance travelled in the first hour = s*1=s Distance remaining = d-s Time to cover remaining distance = (d-s)/ (3s/5) Total time = (d-s)/ (3s/5) +1 We are told the train is 2h late, so total time = t + 2 Therefore: 1+ (d-s)/(3s/5)=t+2 (d-s)/(3s/5)=t+1 (5d-5s)/3s=t+1 5d-5s=3st+3s 5d=3st+8s --------------------------------------... Now suppose the train travels for an hour at speed s, covering a distance s. The train then travels a further 50 miles at speed s. Total distance before fault = (s+50) Time taken to trav