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Math Help - Work problems

  1. #1
    m58
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    Work problems

    More SAT practice problems that are beyond me...

    John can complete a job in 1/2 the time it takes Harry to complete the same job. If they both work together they can complete the job in 4 hours. How long does it take John to do the job alone?

    Sam can do a job in x hours. Denise can do the same job in x + 2 hours. How long will it take Sam and Denise to do to the job together? (Answer must be expressed in terms of x)
    This is what I have so far: (1/x) + (1/2+x) = 1/x
    Where do I go from here, assuming that this is correct?
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  2. #2
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    Quote Originally Posted by m58 View Post
    More SAT practice problems that are beyond me...

    John can complete a job in 1/2 the time it takes Harry to complete the same job. If they both work together they can complete the job in 4 hours. How long does it take John to do the job alone?

    Sam can do a job in x hours. Denise can do the same job in x + 2 hours. How long will it take Sam and Denise to do to the job together? (Answer must be expressed in terms of x)
    This is what I have so far: (1/x) + (1/2+x) = 1/x
    Where do I go from here, assuming that this is correct?
    These aren't too bad, how far in math have you gone?

    Really all you need to do is to define the the variables. For the first one, we'll let the time for John to complete a job be T_J and we'll call the time for harry to complete the same job T_H.

    Now just write down what your told in terms of an equation. John can complete the job in 1/2 the time it takes harry. Therefore:

    T_J=\frac{1}{2}T_H (the time it takes john is equal to 1/2 the time it takes harry.)

    If they both work together, they can complete the job in four hours:

    T_J+T_H=4 (the time it takes harry + the time it takes john, is equal to 4)

    Now just solve the system of equations for the time it takes john:

    T_J+T_H=4
    T_J=\frac{1}{2}T_H
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  3. #3
    m58
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    I'm only in Algebra II, so not too far.

    Here's what I did:

    1/x + 1/.5x = 1/4 -> 2x(1/x + 1/.5x = 1/4) -> 4x+4x=8x -> 2x/8x = .25 = x -> 1/.25 + 1/(.25)(.5) = 12/2 = 6

    The answer I got matched the one in the book, though I'm not positive if my work is correct. Your method seems far simpler, though. Thanks for the input.
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  4. #4
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    Quote Originally Posted by m58 View Post
    I'm only in Algebra II, so not too far.

    Here's what I did:

    1/x + 1/.5x = 1/4 -> 2x(1/x + 1/.5x = 1/4) -> 4x+4x=8x -> 2x/8x = .25 = x -> 1/.25 + 1/(.25)(.5) = 12/2 = 6

    The answer I got matched the one in the book, though I'm not positive if my work is correct. Your method seems far simpler, though. Thanks for the input.
    You're right,

    My method was actually wrong.

    It's one job every x units of time + one job every .5x units of time is equal to 1 job every 4 units of time. That was a careless mistake on my part, and your method is correct. So obviously these problems aren't "beyond you."

    For your other problem, your method is also correct so far. Just solve for x now, and the total time should be (x+2)+x hours, or 2x+2 hours. Plug in the x, and that should be your answer.
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  5. #5
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    When two or more people work together (or pipes fill a tank, etc.) their speeds add.

    You are told "John can complete a job in 1/2 the time it takes Harry to complete the same job" so if we let "T" be the time it takes John to complete a job in hours, then 2T is the time it take Harry. There speeds are \frac{1}{T} "jobs per hour" and \frac{1}{2T} "jobs per hour". Together, then, they work at the rate of \frac{1}{T}+ \frac{1}{2T}= \frac{3}{2T} jobs per hour and so can complete one job in \frac{2T}{3} hours. We are told that together they complete a certain job in 4 hours: \frac{2T}{3}= 4. Solve that for T.
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  6. #6
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    Work problems

    Hello m58,
    In problem 3 your equation is incorrect . To illustrate the method assume x=4 and x+2=6
    Both working together
    1/4 + 1/6 = 5/12 the fraction of work in one hr so 12/5 = the total time required.
    Working only with x and x+2 perform the same operations

    bjh
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