Problem Solving

A train after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late. Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late. What is the speed of the train?

A. 45
B. 33
C. 48
D. 55
E. 61

The OA is C.

Please, can someone assist me with this PS question? I appreciate your help. Thanks!

Let the two endpoints 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 the 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,
$$\frac{24\cdot4}{3x}-\frac{24}{x}=\frac{10}{60}$$
$$\frac{32}{x}-\frac{24}{x}=\frac{1}{6}$$
Solving this we get x = 48 km/h. Regards!