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.
original speed --- x mph
new speed ---- x+12 km/h
time at original speed = 60/x
time at new speed = 60/(x+12)
but the difference in those times is 10 minutes or 1/6 hr
60/x - 60/(x+12) = 1/6
times 6x(x+12) , the LCD
360(x+12) - 360x = x(x+12)
360x + 4320 - 360x = x^2 + 12x
x^2 + 12x - 4320 = 0
(x-60)(x+72) = 0
x = 60 or x is a negative
original speed was 60 km/h