algebra

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alberto drove to the lake and back. the trip there took two hours and the trip back took three hours. he averaged 20 km/h faster on the trip there than on the return trip. What was alberto's average speed on the outbound trip?

  • algebra -

    speed on first trip --- x km/h
    speed on return trip --- x-20
    let the distance be d
    time for first part = d/x
    time for 2nd part = d/(x-20)

    d/x = 2 ---> d = 2x
    d/(x-20) = 3 ---> d = 3x - 60

    3x-60 = 2x
    x = 60
    So the distance each way was 120 km

    average speed = 240/5 = 48 km/h

    Notice that the two speeds were 60 km/h and 40 km/h, and to get the average speed we CANNOT just average the two speeds.

  • algebra -

    since distance = speed * time, if the outbound speed was x, then

    2x = 3(x-20)

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