Pleeeeeease answer me immmediately.
Asked by vidyatmika | 26th Jan, 2010, 10:16: PM
Let the original speed and time required be V and T.
The the total distance is D = VT
Time to cover distance of 30 km = 30/V
Time to cover remaining distance of D - 30 or VT - 30 = (VT-30)/(4V/5) = 5(VT-30)/4V
And it reaches 45 min late i.e. T + 45.
30/V + 5(VT-30)/4V = T + 45
This simplifies to,
VT - 180V = 30 .....(1)
Similarly for the second case, it reaches 9 min earlier, had the defect occured covering extra 18 km, hence T+45-9 = T+36
48/V +5(VT-48)/4V = T+36
VT - 144V = 48 .........(2)
Solving (1) and (2),
V = 1/3 = 0.33 km/min = 0.3333x60 = 20 km/hr
and D = VT = 96 km
Hence the original speed of the train is 20 km/hr and the distance is 96 km.
Answered by | 26th Jan, 2010, 10:40: PM
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