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4 November, 05:49

Dos motos, A y B, toman la salida en una carrera de 90 km. La moto A realiza el recorrido con una velocidad media inferior en 20 km/h a la de la moto B, con lo que llega a la meta 3 minutos después que B. Calcula la velocidad de cada moto.

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  1. 4 November, 05:55
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    motorcycle A: 180 km/h

    motorcycle B: 200 km/h

    Step-by-step explanation:

    To solve this question we need to write a system of equations for A and B, using the equation:

    distance = speed * time

    The speed of A is 20 less than the speed of B, so:

    speedA = speedB - 20

    And the time A traveled is 3 minutes (0.05 hours) more than B's time, so:

    timeA = timeB + 0.05

    Then, using the distance equation, we have that:

    distanceB = speedB * timeB

    90 = speedB * timeB

    distanceA = speedA * timeA

    90 = (speedB - 20) * (timeB + 0.05)

    90 = speedB * timeB + 0.05 * speedB - 20*timeB - 1

    90 = 90 + 0.05 * speedB - 20*timeB - 1

    20*timeB = 0.05 * speedB - 1

    timeB = (speedB - 20) / 400

    using this timeB in the distance equation, we have:

    90 = speedB * (speedB - 20) / 400

    90 * 400 = speedB^2 - 20*speedB

    speedB^2 - 20*speedB - 36000 = 0

    Solving this quadratic equation, we have speedB = 200 km/h

    And the speedA is:

    speedA = speedB - 20 = 200 - 20 = 180 km/h
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