Asked by Anonymous
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.
Answers
Answered by
Reiny
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
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
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