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

#1**0 **

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

CPhill
Mar 17, 2017