If the speed of a train is increased by 20%, its travel time between two stations reduces by 2 hrs. If its speed is decreased by 20%, the travel time increases by 3 hrs. What is the normal duration of travel (in hrs)?

Concept:

distance travel by a train with speed s for time t is d = s × t

Explanation:

Let d be the distance, s be the speed and t be the time

If the speed of a train is increased by 20%, its travel time between two stations reduces by 2 hrs.

So now speed = s + \(\frac{20s}{100}\) = \(\frac{120s}{100}\) and time = t - 2

Therefore now distance traveled = \(\frac{120s}{100}\) × (t - 2)

 If its speed is decreased by 20%, the travel time increases by 3 hrs.

So now speed = s - \(\frac{20s}{100}\) = \(\frac{80s}{100}\) and time = t +3

Therefore now distance traveled = \(\frac{80s}{100}\) × (t + 3)

Hence by given condition

\(\frac{120s}{100}\) × (t - 2) = \(\frac{80s}{100}\)× (t + 3)

12(t - 2) = 8(t + 3)

3(t - 2) = 2(t + 3)

3t - 6 = 2t + 6

t = 12

the normal duration of travel = 12 hr

Option (2) is correct