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  1. #1
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    Word problem help

    could you please tell me how I set this up to solve? thanks



    a veterinarian has two solutions that contain different concentrations of a certain medicine. One is 15% concentration and the other is 5 % concentration. How many cubic centimeters of each should the veterinarian mix to get 20 cc of a 6 % solution?
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  2. #2
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    Hello Stephen

    Welcome to Math Help Forum!
    Quote Originally Posted by Stephen View Post
    could you please tell me how I set this up to solve? thanks



    a veterinarian has two solutions that contain different concentrations of a certain medicine. One is 15% concentration and the other is 5 % concentration. How many cubic centimeters of each should the veterinarian mix to get 20 cc of a 6 % solution?
    Suppose he uses \displaystyle x cc of the first solution, and \displaystyle y cc of the second. Then we can set up two simultaneous equations as follows:

    The total volume must be 20 cc. Therefore
    \displaystyle x+y = ... ?
    This is equation (1).

    The quantity of the medicine in \displaystyle x cc of the first solution is \displaystyle \frac{15x}{100}.

    The quantity of the medicine in \displaystyle y cc of the second solution is \displaystyle \frac{5y}{100}.

    The total quantity of medicine in the mixture is therefore ... ?

    This must be 6% of the total volume, \displaystyle 20 cc, which is \displaystyle \frac{6\times 20}{100} = ... ?

    The second equation is therefore
    \displaystyle \frac{15x}{100}+\frac{5y}{100}=... ?
    This is equation (2).

    Can you complete what I have started, and then solve the simultaneous equations to find \displaystyle x and \displaystyle y?

    Grandad
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  3. #3
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    Hello, Stephen!

    Here's a one-variable approach . . .


    A vet has two solutions that contain different concentrations of a certain medicine.
    One is 15% concentration and the other is 5% concentration.
    How many cc's of each should the vet mix to get 20 cc of a 6% solution?

    We will use x cc's of solution A

    . . It will contain: . 15\%\times x \:=\:0.15x cc's of the med.


    We will use (20-x) cc's of solution B.

    . . It will contain: . 5\%\times(20-x) \:=\:0.05(20-x) cc's of the med.


    The mixture will contain: . 0.15x + 0.05(20-x) cc's of the med.



    But we know that the mixture will be 20 cc's with 6% med.

    . . It contains: . 6\% \times 20 \:=\:0.06(20) \:=\:1.2 cc's of the med.



    We just described the final amount of med in two ways.

    There is our equation! . . . 0.15x + 0.05(20-x) \;=\;1.2

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  4. #4
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    Well...

    To each his own; I find this way removes confusion:
    Code:
       x  :  15
    20-x  :   5
    ===========
      20  :   6
    15x + 5(20-x) = 6(20) ; solve: x = 2

    In general:
    Code:
     x  :  a
     y  :  b
    ========
    x+y :  c
    ax + by = c(x+y)
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  5. #5
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    Thanks so much for the help! Ive been taking a algebra course over the summer and Ive been doing well. But im studying for a cumulative end of course exam and i have a hard time retaining some of the stuff i learned earlier on. thanks again for the help.
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