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.
+128407 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
