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algebra

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lean can paint her new apartment 3 hours longer than his friend,carlos.if both can do painting in 2 hours,how long will it take each one to finish the job alone?

  • algebra -

    labour*time=job time=job/labour
    time=3 and 3x=1 x=1/3
    If both can do painting in 2 hours,So
    (x+y)2=1 x=1/3 hence (1/3+y)*2=1
    And y is 1/6.We can say that it take 6hours to finish the job alone.

  • algebra -

    let Carlos' time be x hrs
    so rate for Carlos = 1/x
    Lean's time is x+3
    Lean's rate = 1/(x+3)

    combined rate = 1/x + 1/(x+3)
    = (x+3 + x)/(x(x+3)) = (2x+3)/(x(x+3))

    1/[ (2x+3)/(x(x+3))] = 2
    x(x+3)/(2x+3) = 2
    x^2 + 3x = 4x + 6
    x^2 - x - 6 = 0
    (x-3)(x+2) = 0
    x = 3 or x=-2, the latter is not possible

    Carlos would take 3 hrs, and Lean would take 6 hrs alone

    check:
    combined rate = 1/3 + 1/6 = 1/2
    combined time = 1/(1/2) = 2

  • algebra -

    Keiko bicycles 4km/h faster than Carlos. In the same time it takes Carlos to Bicycle 45km, Keiko can bicycle 57km. How fast does each bicyclist travel?

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