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Math

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A cyclist and a jogger start from a town at the same time and head for a destination 11 mi away. The rate of the cyclist is twice the rate of the jogger. The cyclist arrives 1.1 h before the jogger. Find the rate of the cyclist.

  • Math -

    Jogger's Rate = r mi/h.
    Cyclist Rate = 2r mi/h.

    Cyclist time = t hrs.
    Jogger's time = (t + 1.1) hrs.

    Eq1: 2r*t = 11 mi.

    Eq2: r(t + 1.1) = 11 mi.

    Solve Eq1 for r:

    2r*t = 11,
    r = 11 / 2t.

    In Eq2, substitute 11/2t for r:
    11/2t(t + 1.1) = 11,
    (5.5/t)(t + 1.1) = 11,
    5.5 + 6.05 / t = 11.
    Common denominator = t.
    (5.5t + 6.05) / t = 11,
    Cross multiply:
    5.5t + 6.05 = 11t,
    11t - 5.5t = 6.05,
    5.5t = 6.05,

    t = 6.05 / 5.5 = 1.1 h.

    Solve Eq1 for 2r;
    2r * 1.1 = 11,

    2r = 11 / 1.1 = 10 mi / h = cyclist rate.

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