document.write( "Question 98712: A tank holds 80 liters of a chemical solution. Currently, the solution has a strength of 30%. How much of this should be drained and replaced with a 70% solution to get a final strength of 40%? \n" ); document.write( "

Algebra.Com's Answer #71876 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let x= amount drained and replaced by a 70% solution\r
\n" ); document.write( "\n" ); document.write( "Amount of pure chemical remaining after x amount was drained out is 0.30(80-x)\r
\n" ); document.write( "\n" ); document.write( "Amount of pure chemical added back in is 0.70x\r
\n" ); document.write( "\n" ); document.write( "Amount of pure chemical in the final mixture is 0.40*80\r
\n" ); document.write( "\n" ); document.write( "Amount of pure chemical in our solution before they are mixed together=amount of pure chemical after they are mixed so our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "0.30(80-x)+0.70x=0.40*80 get rid of parens
\n" ); document.write( "24-0.30x+0.70x=32 subtract 24 from both sides\r
\n" ); document.write( "\n" ); document.write( "24-24-0.30x+0.70x=32-24 collect like terms\r
\n" ); document.write( "\n" ); document.write( "0.40x=8 divide both sides by 0.40\r
\n" ); document.write( "\n" ); document.write( "x=20 liters------------amount drained and replaced by a 70% solution\r
\n" ); document.write( "\n" ); document.write( "CK\r
\n" ); document.write( "\n" ); document.write( "60*0.30+0.70*20=0.40*80
\n" ); document.write( "18+14=32
\n" ); document.write( "32=32\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor\r
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