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On a 120km track, a train travels the first 30 km at a uniform speed of 30km/hr. How fast must the train travel the next 90 km so as to average 60 km/hr for the entire trip?

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by ayeshaoct

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by ayeshaoct

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Distance d1 = 30 km, v1 = 30 kmph

=> time taken = t1 = d1/v1 = 1 hour

So distance remaining = d2 = 120 - d1 = 90 km

let speed to cover distance d2 = v2 kmph

=> time taken = d2/v2 = 90/v2 hours

average speed = total distance / total duration of time

60 kmph = 120 km / [1 hour + 90 / v2 hours]

=> 1 = 2 / [ 1 + 90 / v2 ]

=> v2 = 90 kmph

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another way:

as the total distance = 120 km

average speed = 60 kmph

=> total time duration = 120 / 60 = 2 hours

time taken for the first 30 km = 30 km /30 kmph = 1 hours

Hence, time remaining to cover the remaining 90 km = 2 hrs - 1 hrs = 1 hour

=> speed during 90 km = 90 kmph

=> time taken = t1 = d1/v1 = 1 hour

So distance remaining = d2 = 120 - d1 = 90 km

let speed to cover distance d2 = v2 kmph

=> time taken = d2/v2 = 90/v2 hours

average speed = total distance / total duration of time

60 kmph = 120 km / [1 hour + 90 / v2 hours]

=> 1 = 2 / [ 1 + 90 / v2 ]

=> v2 = 90 kmph

====================================

another way:

as the total distance = 120 km

average speed = 60 kmph

=> total time duration = 120 / 60 = 2 hours

time taken for the first 30 km = 30 km /30 kmph = 1 hours

Hence, time remaining to cover the remaining 90 km = 2 hrs - 1 hrs = 1 hour

=> speed during 90 km = 90 kmph