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????
Thanks in advance.
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????
Thanks in advance.
-
Let x = distance between stations
v = normal speed
t = normal time
t = x/v
t+2 = 5(x-v)/3v + 1
t+1.333 = 5(x-v-50)/3v + (v+50)/v
x/v + 1 = 5(x-v)/3v
x/v +1.333 = 5(x-v-50)/3v + (v+50)/v
3x+3v = 5x - 5v
2x = 8v
x = 4v
3x+4v=5x-5v-250+3v+150
6v = 2x -100
6v = 8v - 100
2v = 100
v = 50 mph
x = 200 miles
v = normal speed
t = normal time
t = x/v
t+2 = 5(x-v)/3v + 1
t+1.333 = 5(x-v-50)/3v + (v+50)/v
x/v + 1 = 5(x-v)/3v
x/v +1.333 = 5(x-v-50)/3v + (v+50)/v
3x+3v = 5x - 5v
2x = 8v
x = 4v
3x+4v=5x-5v-250+3v+150
6v = 2x -100
6v = 8v - 100
2v = 100
v = 50 mph
x = 200 miles