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.

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