After travelling 50km, a train is meeting with an accident and travels at (3/4)th of the usual Speed and reaches 45 min late. Had the accident happened 10km further on it would have reached 35 min late. Find the usual Speed?

Here there are 2 cases
Case 1: accident happens at 50 km
Case 2: accident happens at 60 km
Difference between two cases is only for the 10 kms between 50 and 60.
Time difference of 10 minutes is only due to these 10 kms.
In case 1, 10 kms between 50 and 60 is covered at (3/4)^th Speed.
In case 2, 10 kms between 50 and 60 is covered at usual Speed.
So the usual Time 't' taken to cover 10 kms, can be found out as below.
(4/3)t – t = 10 mins [s = vt]
⇒ t = 30 mins, d = 10 kms
so usual
Speed = Distance/Time
= 10/30min
= 10/(1/2) km/hr.
= 20 km/hr.