SOLUTION: In the middle of its journey from station A to station B, a train was held up for 10 minutes. In order to arrive at City B on schedule, the engine driver had to increase the speed

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Question 1019612: In the middle of its journey from station A to station B, a train was held up for 10 minutes. In order to arrive at City B on schedule, the engine driver had to increase the speed of the train by 12 km/hour. Find the original speed of the train before it was held up, if it is known that the distance between the two stations is 120 km.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
In the middle of its journey from station A to station B, a train was held up for 10 minutes.
In order to arrive at City B on schedule, the engine driver had to increase the speed of the train by 12 km/hour.
Find the original speed of the train before it was held up, if it is known that the distance between the two stations is 120 km.
:
Change 10 min to 1%2F6 hr
:
let s = original speed
then
(s+12) = faster speed
And
120%2Fs = the normal time for the journey
:
Assume that is was exactly in the middle of the journey, where
60%2Fs = time for the 1st 60 mi
and
60%2F%28%28s%2B12%29%29 = time for 2nd 60 mi
:
A time equation;
normal speed time + faster speed time + 10 min = normal time
60%2Fs + 60%2F%28%28s%2B12%29%29 + 1%2F6 = 120%2Fs
60%2F%28%28s%2B12%29%29 + 1%2F6 = 120%2Fs - 60%2Fs
60%2F%28%28s%2B12%29%29 + 1%2F6 = 60%2Fs
multiply equation by 6s(s+12), cancel the denominators and we have
6s(60) + s(s+12) = 60*6(s+12)
360s + s^2 + 12s = 360s + 4320
Subtract 360s from both side
s^2 + 12s = 4320
a quadratic equation
s^2 + 12s - 4320 = 0
you can use the quadratic formula, a=1; b=12; c=-4320; but this will factor to
(s+72)(s-60) = 0
the positive solution
s = 60 mph is the original speed
:
:
Check this by finding the time, normal time 120/60 = 2hr
60/60 = 1.00 hr
60/72 = 0.833 hr
1/6 hr = .167 (10 min wait)
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total time: 2hrs