Question 1205616
.
Manuel has 400 mL of a 100% antifreeze solution. 
how many mL of a 30% antifreeze solution should be added 
to get a 50% solution of the antifreeze?
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<pre>
Let x be the volume of the 30% antifreeze solution to add.


Then you have this equation

    {{{(400 + 0.3x)/(400+x)}}} = 0.5.


The denominator id the volume of the mixture; the numerator is the volume of the pure antifreeze.

Simplify and solve to find x

   400 + 0.3x = 0.5*(400+x)

   400 + 0.3x = 200 + 0.5x

   400 - 200 = 0.5x - 0.3x

      200    =    0.2x

        x    =    200/0.2 = 2000/2 = 1000.


<U>ANSWER</U>.  1000 mL of the 30% antifreeze solution should be added.
</pre>

Solved.


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It is a standard and typical mixture word problem. 


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> 

in this site.


You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization, &nbsp;from very detailed to very short.