Login

A train after travelling 150 km meets with an accident and then proceed with 3 / 5 of its former speed it arrives at its destination 8 hours late. had the accident occur 360 km further, it would have reached the destination 4 hours late. what is the total distance travelled by the train?

Dear Student,


let s be the normal speed of the train then0.6s = the speed after the accident (35the normal speed)let d = total distance Then d-150km = distance traveled at the slower speed, first scenario and d - 150+360 =distance traveled at the slower speed (360 km further) d - 510 km Write a time equation for the 1st scenario slower time - normal time = 8 hrs 150s+d-1500.6s-ds=8 Multiply by 0.6s to clear the denominators .6×150 + (d-150) - .6d = 0.6s×8 90+d-150-0.6d=4.8s .4d - 60 = 4.8s0.4d - 4.8s = 60....i  Same with the 2nd scenario, (accident 360 km further down the road) 150s+d-5100.6s-ds=4 Multiply by .6s to clear the denominators (.6×510 + (d-510)) - .6d =.6s×4 306 + d - 510 - .6d = 2.4s .4d - 204 = 2.4s .4d - 2.4s = 204......ii on subtracting i from iiwe get.4d-2.4s-.4d+4.8s=204-60 s= 1442.4= 60 km/hr is the normal speed then0.6(60) = 36 km/hr is the "after accident speed" Find the distance using the 1st scenario equation, replace s with 60 .4d - 4.8(60) = 60 .4d - 288 = 60 .4d = 60 + 288 .4d = 348 d=348.4 = 870 km is the total distancehence total distance is 870 km

Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards
 

  • 3
View Full Answer
What are you looking for?