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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.

Guest Mar 17, 2017
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By " the middle of the journey," I'm assuming this is 60km]

 

Theoretically......the time for the first half of the trip should = time for second half of trip......but....10 minutes  [1/6 hr]  must be added to the second half

 

Time of the 1st half of trip   =  Time of 2nd half of trip + 1/6 hr

 

D/R     =  D/(R + 12)  + 1/6         d = 60km   and R is the original rate

 

60/R  =  60/[R + 12] + 1/6

 

60/R - 60/[R + 12]  = 1/6

 

[60R + 720 - 60R] / [ R (R + 12)]  = 1/6

 

720   =  [ R (R + 12)] / 6

 

4320  = R^2 + 12R

 

R^2 + 12r - 4320  =  0       factor

 

(R - 60) (R + 72)  = 0

 

Set the factors to 0  and R = -72 km/hr   (reject) ....   or   R = 60km/hr

 

So....the whole trip should take 2 hours....check

 

60/60  + 60/72  + 1/6   = 2 hrs

 

So....the original rate was 60km/hr

 

 

 

cool cool cool

CPhill  Mar 17, 2017

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