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3 September, 05:27

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.

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  1. 3 September, 06:26
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    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
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