Subject
Quantitative Aptitude
Topic
Speed & Distance
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?
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?
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320Explanation
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 travRelated questions
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