SOLUTION: A train after traveling 150 km meets with an accident and then proceeds with 3/5 of its former speed and arrives its destination 8 hours late. Had the accident occurred 360 km furt

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Question 634535: A train after traveling 150 km meets with an accident and then proceeds with 3/5 of its former speed and arrives its destination 8 hours late. Had the accident occurred 360 km further, it would have reached the destination 4 hours late. What is the total distance traveled by the train?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A train after traveling 150 km meets with an accident and then proceeds
with 3/5 of its former speed and arrives its destination 8 hours late.
Had the accident occurred 360 km further, it would have reached the
destination 4 hours late.
What is the total distance traveled by the train?
:
let s = the normal speed of the train
then
.6s = the speed after the accident (3/5 the normal speed)
:
let d = total distance
Then
(d-150)km = 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
(150%2Fs + %28d-150%29%2F.6s) - d%2Fs = 8
Multiply by .6s to clear the denominators
(.6(150) + (d-150)) - .6d = .6s(8)
90 + d - 150 - .6d = 4.8s
.4d - 60 = 4.8s
.4d - 4.8s = 60
:
Same with the 2nd scenario, (accident 360 km further down the road)
(510%2Fs + %28d-510%29%2F.6s) - d%2Fs = 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
:
Find the speed using elimination with these two equations
.4d - 2.4s = 204
.4d - 4.8s = 60
------------------subtraction eliminates d find s
0 + 2.4s = 144
s = 144/2.4
s = 60 km/hr is the normal speed
then
.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
d = 870 km is the total distance
:
:
Check solution using the 2nd scenario
510%2F60 + %28870-510%29%2F36 - 870%2F60 =
8.5 + 10 - 14.5 = 4 hrs, confirms our solution of d=870km