A train covered a distance at a uniform speed .if the train had been 6 km/hr faster it would have been 4 hour less than schedule time and if the train were slower by 6 km/hr it would have been 6 hrs more.find the distance.
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Let t be the usual time taken by the train to cover the distance
Let d be the distance, s be the usual speed
Usual time taken => d/s = t => d=ts
d/(s+6) = t-4
ts/(s+6) = t-4
ts = ts+6t-4s-24
6t - 4s - 24 = 0 --> (1)
d/(s-6) = t+6
ts = ts-6t+6s-36
-6t + 6s - 36=0 --->(2)
Solving (1) nd (2), v get
s = 30 km/h
t = 24 hrs
d = t * s
d = 30 * 24 = 720 km
Ans : 720 km
Let d be the distance, s be the usual speed
Usual time taken => d/s = t => d=ts
d/(s+6) = t-4
ts/(s+6) = t-4
ts = ts+6t-4s-24
6t - 4s - 24 = 0 --> (1)
d/(s-6) = t+6
ts = ts-6t+6s-36
-6t + 6s - 36=0 --->(2)
Solving (1) nd (2), v get
s = 30 km/h
t = 24 hrs
d = t * s
d = 30 * 24 = 720 km
Ans : 720 km
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