Question

Two trains start simultaneously with uniform speeds from two station 270 km apart, each to the opposition they reach their destination in 6.25 hrs and 4 hrs after they meet.The rate at which the slower train travels is?

Answer 1

Answer:

The time taken by each of the train from the point to reach the the destination will be 25/4 and 4 hrs respectively.

Now, we know the formulae of ratio of speeds = √(time1/time2).

So, we will get the ratio of speeds = √{4/(25/4)} which on solving we will get the ratio to be 4/5.

So, we let the speed of train A = 4x and speed of train B is 5x.

Then, the relative speed will be (4x+5x) = 9x.

Hence, the time taken by both the trains to meet each other = 270/9x =30/x.

So, time taken by slower train = 270/4x and the other time will be 30/x and 25/4 which on solving we will get the value of the x to be 6hr.

So, the speed of the slower train will be 4x = 4*6 = 24Km/hr.