Question 275795
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Beth wants to make 11.6 fl. oz. of a 45% acid solution by mixing together a 22% acid solution and a 80% acid solution. 
How much of each solution must she use?
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        Calculations in the post by @mananth are incorrect, leading to wrong answer.

        I came to bring a correct solution.



<pre>
Let x be the volume of the 80% solution, in oz.

Then the volume of the 22% solution is  (11.6-x) oz.



The balance equation is

    0.8x + 0.22(11.6-x) = 0.45*11.6.    (1)


It says that the combined mass of pure acid in ingredients (left side) 
is the same as the mass of pure acid in the mixture (right side).


Simplify and find 'x'

    0.8x + 0.22*11.6 - 0.22x = 0.45*11.6,

    0.8x - 0.22x = 0.45*11.6 - 0.22*11.6

        0.58x    =        2.668

            x    =        2.668/0.58 = 4.6.


<U>ANSWER</U>.  4.6 oz of the 80% solution is needed and  (11.6-4.6) = 7 oz of the 22% solution.


<U>CHECK</U>.   Let's check the final solution for its concentration  {{{(0.8*4.6 + 0.22*7)/11.6}}} = 0.45,
         which is precisely correct.
</pre>

Solved.