Question 1100733
.
You need 600 mL of a 45% alcohol solution. On hand, you have a 30% alcohol mixture. You also have a 70% alcohol mixture. 
How much of each mixture will you need to add to obtain the desired solution?
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<pre>
Let X = "How much of the 30% mixture to mix", in mL.

Then the volume of the 70% mixture to mix is (600-X) mL.


X mL of the 30% mixture contribute 0.3*X mL of the pure alcohol to the new mixture.

(600-X) mL of the 70% mixture contribute 0.7*(600-X) mL of the pure alcohol to the new mixture.

In all, two input mixtures provide  0.3*X + 0.7*(600-X)  mL of the pure alcohol in the output mixture.


The output volume  is 600 mL,  and it contains 0.3*X + 0.7*(600-X)  mL of the pure alcohol, as we find out above.


Hence, the concentration of the new mixture is  {{{(0.3*X + 0.7*(600-X))/600}}} as the fraction.


According to the condition, it must be  45% or 0.45. It gives you an equation

{{{(0.3*X + 0.7*(600-X))/600}}} = 0.45.


It is you major/basic equation, so called "concentration" equation.

As soon you got it (and understand it), the setup part is completed.


To solve the equation, first multiply both sides by 600. You will get

0.3*X + 0.7*(600-X) = 0.45*600.


Simplify and solve it:

0.3X + 0.7*600 - 0.7X = 0.45*600  ====>  -0.4X = 0.45*600-0.7*600 = -0.25*600  ====>  X = {{{((-0.25)*600)/(-0.4)}}} = 375.


Thus you need 375 mL of the 30% mixture and  600-375 = 225 mL of the 70% mixture.


<U>Check</U>.  {{{(0.3*375 + 0.7*225)/600}}} = 0.45  ! correct concentration !


<U>Answer</U>.  You need 375 mL of the 30% mixture and  225 mL of the 70% mixture.
</pre>

Solved.


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There is entire bunch of introductory lessons covering various types of mixture problems

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/Mixture-problems.lesson>Mixture problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/More-Mixture-problems.lesson>More Mixture problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/Solving-typical-mixture-problems.lesson>Solving typical word problems on mixtures for solutions</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Word-problems-on-mixtures-for-antifreeze-solutions.lesson>Word problems on mixtures for antifreeze solutions</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Word-problems-on-mixtures-for-alloys.lesson>Word problems on mixtures for alloys</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Typical-word-problems-on-mixtures-from-the-archive.lesson>Typical word problems on mixtures from the archive</A>

in this site.


Read them and become an expert in solution mixture word problems.


Also, you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this textbook in the section "<U>Word problems</U>" under the topic "<U>Mixture problems</U>".



Save the link to this online textbook together with its description


Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson


to your archive and use it when it is needed.