Question 272631
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How many quarts of pure antifreeze must be added to 6 quarts of a 10% solution to obtain a 20% antifreeze soltuion? 
(Round to the nearest tenth.)
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        The solution in the post by @mananth is INCORRECT: its governing equation

        is written incorrectly.


        I came to provide a correct solution.



<pre>
Let x be the volume of pure antifreeze to add, in quarts.


Then the governing equation is 

    x + 0.1*6 = 0.2(x+6)    <<<---===  the amount of the pure antifreeze in ingredients (left side)
                                       and in the mixture (right side)


Simplify and find x

    x + 0.6 = 0.2x + 1.2,

    x - 0.2x = 1.2 - 0.6,

      0.8x   =    0.6,

         x   =    0.6/0.8 = 6/8 = 3/4 = 0.75.


<U>ANSWER</U>.  0.75 quarts of pure antifreeze should be added.
</pre>

Solved correctly.


Notice that the instruction for rounding is not appropriate to the problem.


It tells me that the problem's composer was not focused on the problem when created it.
Or simply does not understand what he/she composes.