After travelling 30 minutes a train met with an accident and was stopped there for 45 minutes due to the accident its speed reduced to 2 by 3 of its former speed and then the train reached its destination 1 hour 30 minutes late had the accident occurred 60 km after.
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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.
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.
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