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Thread: May I have some guidance on this homework problem?

  1. #1 May I have some guidance on this homework problem? 
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    These are my achilles heel...and I plan to get into engineering, so...I really need to get better at these word problems. I am very frustrated.

    What quantity of 60% acid solution must be mixed with a 30% solution to produce 300mL of a 50% solution?

    What I have tried so far is:

    I have 300mL total, of a 30% solution, which means that I currently have 90mL of acid in the 300mL solution.

    I need to add x amount of 60% acid solution, which is .6x.

    300 + x is what I get when I add the current 300mL to x amount of acid.

    It needs to be a 50% solution, so the expression is .5(300 + x) = .3(300) + .6x, right?

    I come up with 150mL, but the correct answer is 200mL. What am I missing? I have tried and tried, and even started switching out the numbers, but I cannot come up with the right solution. This is frustrating me to no end...help please !

    EDIT: I got it!

    Suddenly it just clicked for some reason...I had been trying to solve this for hours.

    I remembered that I am adding mixing it with a 30% solution...so that isn't .5(300 + x), it's 150 + .3x, right?

    150 + .3x = 90 + .6x

    x = 200mL


    Last edited by bowlslaw; October 23rd, 2011 at 03:52 PM.
    I've always been crazy, and it's kept me from going insane.
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  3. #2  
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    Not really... Look at the question again: "What quantity of 60% acid solution must be mixed with a 30% solution to produce 300mL of a 50% solution?"

    More precisely: You are mixing an unspecified quantity of a 60% solution, with an unspecified quantity of a 30% solution to get 300ml of a 50% solution. How many unknowns do you have here? How many equations do you think you'll need? (The homework assignment may only explicitly ask for one of the unknowns, but that doesn't mean it's the only unknown).

    200ml is the right answer, but your solution approach is not right.


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  4. #3  
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    Yes, I think you did figure it out correctly, though I would have done it the way M did it.

    Your method was as follows. The quantity of 60 % solution is x and this contains .6x milliliters of acid. The quantity of 30 % solution is 300-x and contains .3*(300-x) ml of acid. Taken together you have 0.5*300=150 ml of acid so

    .6x+90-.3x=150

    which is your equation.

    The way M did it was to write two equations.
    x= the quantity of 60 percent solution
    y= the quantity of 30 percent solution
    You have a total of 300 ml, so
    x+y=300
    The amount of acid in x added to the amount of acid in y = the total amount of acid:
    .6x + .3y = .5*300

    If you solve those equations simultaneously, you get the same answer.
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