A train after traveling at 50kmph meets with an accident and then

here is less mathematical approach.

let the two end points between which this train is running be A and B. Also suppose that accident occurs at C which is at a distance of 50km from A. thus we have

A---------50km----------C----------------------------B 1)

Time taken to travel the distance after the accident i.e. CB is 35 minutes more than the usual time.

now, if the accident were to occur at 24 km farther, then train would have reached only 25 minutes late. i.e.

A--------50 km----------C---------24--------D---------------B 2)

if we look at the situation depicted in 1) and 2) we will see that 10 minute difference in time is occurring because of this 24 Km. which is traveled by its usual speed (=x) in the later case. thus we have

\(\frac{24*4}{3x}\) - \(\frac{24}{x}\) = \(\frac{10}{60}\)
\(\frac{32}{x}\) - \(\frac{24}{x}\) = \(\frac{1}{6}\)
solving this we get x= 48km/h