SOLUTION: A train, after travelling for 3 hours, was detained 40 minutes. Then it proceeded at 6/5 its former rate and arrived 10 minutes late. If the delay had occurred at a point 15 kilome

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Question 903383: A train, after travelling for 3 hours, was detained 40 minutes. Then it proceeded at 6/5 its former rate and arrived 10 minutes late. If the delay had occurred at a point 15 kilometers farther from its destination, the train would have been only 5 minutes late. Find the rate of the train and the total distance traveled.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
let s = the normal speed of the train
let d = total distance of the trip
:
A train, after travelling for 3 hours, was detained 40 minutes.
3s = train distance up to this point
then
d%2Fs = normal time for the trip
:
Then it proceeded at 6/5 its former rate and arrived 10 minutes late.
faster speed = 1.2s
at this speed it cut half an hour off the normal travel time (40-10)
d%2Fs - (3+%28%28d-3s%29%29%2F%28%281.2s%29%29) = .5 hrs
d%2Fs - 3 - %28%28d-3s%29%29%2F%28%281.2s%29%29 = .5 hrs
add 3 to both sides
d%2Fs - %28%28d-3s%29%29%2F%28%281.2s%29%29 = 3.5
multiply by 1.2s, resulting in
1.2d - d + 3s = 3.5(1.2s)
.2d = 4.2s - 3s
.2d = 1.2s
d = 1.2s/.2
d = 6s
then
3s = halfway point, where the 1st delay occurred
:
If the delay had occurred at a point 15 kilometers farther from its destination, the train would have been only 5 minutes late.
This time the train gained 40-5 = 35 minutes over the original undelayed time
35 min = 7/12 hrs
:
the distance traveled at the faster speed this time; (3s+15)
The distance traveled at the normal speed: (3s-15)
:
time at normal speed + time at faster speed = normal time - 35 min
%283s-15%29%2Fs + %283s%2B15%29%2F1.2s = 6s%2Fs - 7%2F12
multiply equation by 12s
12(3s-15) + 10(3s+15) = 12(6s) - 7s
36s - 180 + 30s + 150 = 72s - 7s
66s - 30 = 65s
66s - 65s = 30
s = 30 mph is the normal speed of the train
Find the distance
6(30) = 180 mi is the distance