Question
Two trains started from stations βAβ and βBβ at
same time and started travelling towards each other at speeds of 30 km/hr and 20 km/hr, respectively. At the time of their meeting, the faster train has travelled 200 km more than the slower train. Find the distance between the stations βA and βBβ.Solution
Let the distance travelled by slower train be βxβ km So, distance travelled by faster train = βx + 200β km ATQ, (x/20) = {(x + 200)/30} Or, 30x = 20x + 4000 Or, 10x = 4000 So, x = 400 Total distance between station βAβ and station βBβ = (400 + 400 + 200) = 1000 km
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