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Math Help - Word Problem....ahhh

  1. #1
    Junior Member CONFUSED_ONE's Avatar
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    Word Problem....ahhh

    A chemist at a university needs to mix a 25% ethanol solution with a 40%ethanol solution to obtain 120mL of a 30% ethanol solution. How much of each kind is needed?

    okay yeah, the two equations i got out of this was
    .25a+.40b=120--->25a+40b=12000
    a+b=120(.30)---->a+b=36
    ______________________________
    yeah? no objections??? okay.......
    ______________________________
    yeah so i times the bottom equation by (-25) and that cancelled out the A's so i was left with
    15b=11100
    so i divided that out and got B=740...
    i plugged that into the original equation and got A=(-704)

    (-704)mL of 25% ethanol solution and 740 mL of 40% ethanol solution...

    THAT CANT BE RIGHT! A NEGATIVE AMOUNT OF SOLUTION?
    can someone please please please help me figure out what i did wrong?
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  2. #2
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    Quote Originally Posted by CONFUSED_ONE
    A chemist at a university needs to mix a 25% ethanol solution with a 40%ethanol solution to obtain 120mL of a 30% ethanol solution. How much of each kind is needed?

    okay yeah, the two equations i got out of this was
    .25a+.40b=120--->25a+40b=12000
    a+b=120(.30)---->a+b=36
    ______________________________
    yeah? no objections??? okay.......
    ______________________________>3
    yeah so i times the bottom equation by (-25) and that cancelled out the A's so i was left with
    15b=11100
    so i divided that out and got B=740...
    i plugged that into the original equation and got A=(-704)

    (-704)mL of 25% ethanol solution and 740 mL of 40% ethanol solution...

    THAT CANT BE RIGHT! A NEGATIVE AMOUNT OF SOLUTION?
    can someone please please please help me figure out what i did wrong?
    I would say the equations should be
    x+y=120
    .25x+.4y=.3(120)=36
    Thus,
    D=\left|\begin{array}{cc}1&1\\.25&.4\end{array}\ri  ght|=.4-.25=.15
    D_x=\left|\begin{array}{cc}120&1\\36&.4\end{array}  \right|=12
    D_y=\left|\begin{array}{cc}1&120\\.25&36\end{array  }\right|=6
    Thus,
    x=\frac{D_x}{D}=12/.15=80
    y=\frac{D_y}{D}=6/.15=40
    Q.E.D.

    Note, I used determinants because I was practing my LaTeX coding for arrays of numbers. But apparently you made a mistake.
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  3. #3
    Junior Member CONFUSED_ONE's Avatar
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    thank you for the help. could you point out to me what mistake i made?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by CONFUSED_ONE
    thank you for the help. could you point out to me what mistake i made?
    .25a+.40b=120--->25a+40b=12000
    The RHS of the first of these equations should be 0.3\times 120 as you end with 120 ml of 30% solution.

    RonL
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  5. #5
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    [QUOTE=CONFUSED_ONE][center]A chemist at a university needs to mix a 25% ethanol solution with a 40%ethanol solution to obtain 120mL of a 30% ethanol solution. How much of each kind is needed?

    The amount of 25% is x, so the amount of 40% is 120 -x
    .25x + .40(120-x) = .30(120)
    .25x + 48 -.40x = 36
    -.15x = -12
    x = 80
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  6. #6
    Junior Member CONFUSED_ONE's Avatar
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    Thank You So Very Much!
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