Question

After traveling exactly half 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.

Answer 1

Let's define variables:
s = original speed
s + 12 = faster speed
The time for the half of the route is:
60 / s
The time for the second half of the route is:
60 / (s + 12)
The equation for the time of the trip is:
60 / s + 60 / (s + 12) + 1/6 = 120 / s
Where,
1/6: held up for 10 minutes (in hours).
Rewriting the equation we have:
6s (60) + s (s + 12) = 60 * 6 (s + 12)
360s + s ^ 2 + 12s = 360s + 4320
s ^ 2 + 12s = 4320
s ^ 2 + 12s - 4320 = 0
We factor the equation:
(s + 72) (s-60) = 0
We take the positive root so that the problem makes physical sense.
s = 60 Km / h
Answer:
The original speed of the train before it was held up is:
s = 60 Km / h