Question
Two trains started from stations βAβ and βBβ at
same time and started travelling towards each other at speeds of 55 km/hr and 45 km/hr, respectively. At the time of their meeting, the faster train has travelled 116 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 + 116β km ATQ, (x/45) = {(x + 116)/55} Or, 55x = 45x + 5220 Or, 10x = 5220 So, x = 522 Total distance between station βAβ and station βBβ = (522Β + 522Β + 116) = 1160 km
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