document.write( "Question 264461: A chemist has one solution that is 60% chlorinated and another solution that is 40% chlorinated. How much of the first (60%) solution is needed to make a 100 L solution that is 50% chlorinated? \n" ); document.write( "

Algebra.Com's Answer #194701 by Earlsdon(6294) $\"\"$ $\"About$

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Let x = the number of liters of the 60% solution. After changing the percentages to their decimal equivalents, we can write:
\n" ); document.write( " $\"x%280.6%29%2B%28100-x%29%280.4%29+=+100%280.5%29\"$ Simplify and solve for x.
\n" ); document.write( "[{{0.6x+40-0.4x = 50}}} Subtract 40 from both sides and combine the x's.
\n" ); document.write( " $\"0.2x+=+10\"$ Divide both sides by 0.2
\n" ); document.write( " $\"highlight%28x+=+50%29\"$
\n" ); document.write( "The chemist will need to mix 50 liters of the 60% solution with 40 (100-x) liters of the 40% solution to obtain 100 liters of 50% solution. \n" ); document.write( "

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