The average speed of a train in the onward journey is 25% more than that in the return journey. The train halts for one hour on reaching the destination. The total time taken for the complete to and from journey is 17 hours, covering a distance of 800 km
The average speed of a train in the onward journey is 25% more than that in the return journey. The train halts for one hour on reaching the destination. The total time taken for the complete to and from journey is 17 hours, covering a distance of 800 km. The speed of the train in the onward journey is:
[A] 45 km/hr
[B] 47.5 km/hr
[C] 52 km/hr
[D] 56.25 km/hr
ANS: D
Explanation:
Let the speed in return journey be x km/hr.
Then, speed in onward journey =(125/100)x = (5/4)x km/hr
So. Speed in onward journey = [(5/4)*45] km/hr = 56.25 km/hr
[A] 45 km/hr
[B] 47.5 km/hr
[C] 52 km/hr
[D] 56.25 km/hr
ANS: D
Explanation:
Let the speed in return journey be x km/hr.
Then, speed in onward journey =(125/100)x = (5/4)x km/hr
So. Speed in onward journey = [(5/4)*45] km/hr = 56.25 km/hr
Let the speed in the return journey be x km/hr
ReplyDeleteThrn speed in onward journey=(125/100)x=(5/4)x km/hr
Here time taken =17-1=16 hrs
Then 400/x + 400×4/5x =16
x=45
SO, speed in onward journey={5/4*45}km/hr
=56.25 Km/hr
Thanks
DeleteThanks
DeleteThis comment has been removed by the author.
ReplyDeleteThanku more appropriate
ReplyDelete