Question 427227
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a radiator contains 25 quartz of water and antifreeze solution, which of 60 percent (by volume) is antifreeze. 
how much of this solution should be drained and replaced with water for the new solution to be 40 percent antifreeze?
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        The solution in the post by @mananth is incorrect.

        I came to bring a correct solution.



<pre>
Let x be the volume of the original 60% antifreeze solution to partly drain
and to replace by pure water to get the 40% antifreeze solution.


After draining, we then have (25-x) quartz of the 60% antifreeze solution.
It contains 0.6*(25-x) quartz of the pure antifreeze.


Adding water does not change the amount of the antifreeze in the solution.


At the end, the volume of the pure antifreeze in the radiator after adding x quartz of water is 0.4*25 quartz.


So, we equate these two expressions for the pure antifreeze amount

    0.6*(25-x) = 0.4*25  quartz.    (1)


Simplify and find x

    15 - 0.6x = 10,

    15 - 10 = 0.6x,

       5    = 0.6x

       x    = {{{5/0.6}}} = {{{50/6}}} = 8{{{1/3}}}.


<U>ANSWER</U>.  8{{{1/3}}}  quartz of the original 60% solution should be drained and replaced by pure water to get the 40% antifreeze solution.
</pre>

Solved correctly.