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Math Help - Mixing 2 liquids of different concentration

  1. #1
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    Mixing 2 liquids of different concentration

    Hi,
    If I had 500 litres of a solution that is 10% concentration and I want to add another solution to it which is 20% concentration to have a total concentration of 14%, how much of the 20% solution do I need to add? I think the answer is 334 litres but I am not sure how I came to this answer can anyone teach me a formula?

    Thanks
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  2. #2
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    Re: Mixing 2 liquids of different concentration

    Quote Originally Posted by bronson View Post
    Hi,
    If I had 500 litres of a solution that is 10% concentration
    So you have 500(.10)= 50 litres of concentrate.

    and I want to add another solution to it which is 20% concentration to have a total concentration of 14%, how much of the 20% solution do I need to add? I think the answer is 334 litres but I am not sure how I came to this answer can anyone teach me a formula?
    334 litres of a 20% solution would give .2(334)= 66.8 litres of concentrate. Adding that to the original 50 litres gives 116.8 litres of concentrate in a total of 500+ 334= 834 litres. That is a concentration of 116.8/834= .14 or 14%. Looks good!

    Thanks
    Let "x" be the amount added. That will be .2x of concentrate giving a total of 50+ .2x concentrate in 500+ x litres. That is a concentration of (50+ .2x)/(500+ x)= .14. Solve that for x.
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    Re: Mixing 2 liquids of different concentration

    Hi,
    Thanks for the reply, your answer looks great but without trying to sound dumb or ask for all the answers how would I go about solving that equation?
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  4. #4
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    Re: Mixing 2 liquids of different concentration

    Hello, bronson!

    I have 500 litres of a solution that is 10% acid.
    I want to add another solution which is 20% acid to have a total concentration of 14% acid.
    How much of the 20% solution do I need to add?
    I think the answer is 334 litres . Quite close!
    but I am not sure how I came to this answer. . If you don't know, we sure can't guess!

    We start with 500 liters of a solution which is 10% acid.
    . . It contains: . (0.10)(500) \:=\:50 liters of acid.

    We add x liters of a solution which is 20% acid.
    . . It contains: . 0.20x liters of acid.

    Hence, the mixture will contain: . 50 + 0.20x liters of acid.


    But we know that the mixure will be: 50+x liters which will be 14% acid.
    . . The mixture will contain: . 0.14(50+x) liters of acid.


    We just described the amount of acid in the final mixture in two ways.

    There is our equation! . . . . 50 + 0.20x \;=\;0.14(50 + x)

    Got it?
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  5. #5
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    Re: Mixing 2 liquids of different concentration

    Hi,
    thanks for your reply, I understand how both equations where figured out but now I am having problems solving them.

    I cannot find any info on how to solve (50+ .2x)/(500+ x)= .14

    and the info i found on solving 50 + 0.20x = 0.14(50 + x) gives me the wrong answer here is what I did:

    first I simplified the equation
    50 + 0.20x = 0.14(50 + x)
    50 + 0.20x = (50 * 0.14 + x * 0.14)
    50 + 0.20x = 7 + 0.14x

    Add -0.14x
    50 + 0.20x + -0.14x = 7 + 0.14x + -0.14x

    Combine like terms:
    50 + 0.06x = 7 + 0.14x + -0.14x
    50 + 0.06x = 7 + 0.00
    50 + 0.06x = 7

    Add -50
    50 + -50 + 0.06x = 7 + -50

    Combine like terms:
    0 + 0.06x = 7 + -50
    0.06x = 7 + -50
    0.06x = -43

    Divide by 0.06
    x = -716.6666667
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  6. #6
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    Re: Mixing 2 liquids of different concentration

    If I had 500 litres of a solution that is 10% concentration and I want to add another solution to it which is 20% concentration to have a total concentration of 14%, how much of the 20% solution do I need to add?
    500(10) + x(20) = (500+x)(14)

    5000 + 20x = 7000 + 14x

    6x = 2000

    x = 333 L
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  7. #7
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    Re: Mixing 2 liquids of different concentration

    Thanks to everyone that has helped, I can now easily work this out using the method that skeeter posted.

    I would still be interested in finding out how to solve it using the equation from HallsofIvy's post which is:

    (50+ .2x)/(500+ x)= .14
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