Question

A train is running at a constant speed from station A to station B. If the train runs 10kms faster than its normal speed, its covers the same distance in 2 hours less time. But if it runs 10kms slower, it takes 3 hours more. What is the distance between the stations?

Answer 1

Let 'd' be the distance between the stations and 's' be the normal speed.
Time taken to travel distance 'd' at speed 's' is: d / s

If the speed is increased by 10, time taken to travel distance 'd' is: d / (s+10)
The difference in travel time = d / s - d / (s+10) = 2 hours
d{ 1/s - 1/(s+10) } = 2
d{ s+10 - s } / { s(s+10) } = 2
10d/(s^2+10s) = 2
10d = 2(s^2+10s)
5d = s^2 + 10s ---- (1)

If the speed is decreased by 10, time taken to travel distance 'd' is: d / (s-10)
The difference in travel time = d / (s-10) - d / s = 3 hours
d{ 1/(s-10) - 1/s } = 3
d{ s - s + 10 } / { s(s-10) } = 3
10d/(s^2-10s) = 3
10d = 3s^2 - 30s ------ (2)

Multiply (1) by 2:
10d = 2s^2 + 20s ---- (3)

From (2) and (3): 3s^2 - 30s = 2s^2 + 20s
s^2 - 50s = 0
s(s-50) = 0
s = 0 or s = 50
s = 0 is an extraneous solution and can be ignored.

s = 50. Put it in (1)
5d = s^2 + 10s = 50^2 + 10*50 = 2500 + 500 = 3000
d = 3000/5 = 600

Distance between the stations = 600 Km.

---- (2)