After travelling 30 min, a train met with an accident and was stopped there for 45 min. Due to the accident, its speed reduced to 2/3 of its former speed and the train reached its destination 1 hr 30 min late. Had the accident occurred 60 km after the point it occurred earlier, the train would have reached 30 min earlier. The length of the journey is?

Speed 2/3 ho re mtlb..1/2 time bdhana hoga, 1 1.5hr k brabr h to 2 3hr(usual time) k brabr h...
Wese he 30 min phle phauchta agr accident 60km baad hota, 1 30min k brabr h to 2 1hr k brabr... Eska mtlb 1 hr m 60km chla.. Mtlb speed 60km h... 60*3(ususl time) =180km

Let D be the distance from point of 1st accident to destination & S be speed of train.

Case 1:

D/S + 3/4 =D/(2S/3)
On Solving we get,
D/S=3/2
Here, 3/4 is taken for 45min train stopped at the spot and train got late by 90 min so actual delay due to speed reduction is 90-45=45 mins.

Case 2:

30 min have been saved cause of train travelling 60 km more with speed S. So time difference if it would have travelled this 60 km with 2S/3 speed is nothing but the save of 30 min.
So, 60/S - 60/(2S/3)= 1/2

Solving S=60km/h.
Now initially train travelled for 30 min before accident so distance travelled before accident = 60/2= 30km.
And after accident it travelled D km.
So from case 1, D/S=3/2===> D=90.
So total distance = 30 +90=120 kms.