Question 1053307
.
how many ounces of a 100% juice concentrate must be added to a 20% juice concentrate to yield 128 ounces of a 50% juice 
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<U>Solution 1</U>.  Two equations


<pre>
x +    y = 128,       (1)   (x is the volume of the 100% juice)
x + 0.2y = 0.5*128.   (2)

Simplify

x +    y = 128,       (1)
x + 0.2y =  64.       (2)

To solve it, distract (2) from (1). You will get

0.8y = 128-64  --->  0.8y = 64.

From this point, please complete the solution on your own.
</pre>


<U>Solution 2</U>. One equation


<pre>
x + 0.5(128-x) = 0.5*128.    (x is the volume of the 100% juice)

Simplify and solve for x.
</pre>

There is entire bunch of the 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 the mixture word problems.


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It contains many other solved word problems, as well as many other interesting and useful things.