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Math Help - application problem

  1. #1
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    application problem

    One maid can clean the house three times faster than another. working together they can clean the entire house in 3 hours. How long would it take the faster maid cleaning alone?

    Help!
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  2. #2
    Super Member Anonymous1's Avatar
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    f = faster maid.
    s = slower maid.

    Quote Originally Posted by kelsikels View Post
    One maid can clean the house three times faster than another.
    3s = f \implies s = \frac{f}{3}

    Quote Originally Posted by kelsikels View Post
    Working together they can clean the entire house in 3 hours.
    s + f = \color{red}{\frac{1}{3}}

    ...
    Last edited by Anonymous1; June 2nd 2010 at 03:40 PM.
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  3. #3
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by kelsikels View Post
    One maid can clean the house three times faster than another. working together they can clean the entire house in 3 hours. How long would it take the faster maid cleaning alone?

    Help!
    We can measure house-cleaning speed with the unit: houses/hour. (We consider the house in question to be the "standard" house.)

    Let s be the slower speed, and f be the faster speed.

    Then s+f = 1/3.

    We also know that f = 3s.

    So s + 3s = 1/3.

    s = 1/12 houses/hour.

    f = 3s = 3/12 = 1/4 houses/hour.

    So it would take the faster maid 4 hours.
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