Question 1156867
Same idea, different values;  see #1156869:
<a href="https://www.algebra.com/algebra/homework/word/mixtures/Mixture_Word_Problems.faq.question.1156869.html">https://www.algebra.com/algebra/homework/word/mixtures/Mixture_Word_Problems.faq.question.1156869.html</a>
https://www.algebra.com/algebra/homework/word/mixtures/Mixture_Word_Problems.faq.question.1156869.html


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This example, he is supplied with 48% and 26% solutions; and he wants to make 44 liters of 32% solution.


Final concentration needs to be  32%, or  0.32 in decimal fraction.

If variable v is for the amount of 48% solution to use, then  44-v  is the amount of the 26% solution to use.


To account for the PURE amount of substance in the solutions:

{{{highlight_green(48v+26(44-v)=32*44)}}}
Now solve for v.


{{{48v+26*44-26v=32*44}}}
{{{(48-26)v+26*44=32*44}}}
{{{(48-26)v=32*44-26*44}}}
{{{highlight_green(v=44(32-26)/(48-26))}}}

{{{v=44(6/22)=highlight_green(44(3/11)=v)}}}


{{{highlight(v=12)}}}-----------liters of the 48%
-
{{{highlight(32)}}}-------------liters of the 26%