document.write( "Question 56504: a tank holds 80 liters of a chemical solution. currently, the solution has a strenght of 30%. how much of this solution must be drained and replaced with a 70% solution to get a strenght of 40%? \n" ); document.write( "
Algebra.Com's Answer #38371 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
a tank holds 80 liters of a chemical solution. currently, the solution has a strenght of 30%. how much of this solution must be drained and replaced with a 70% solution to get a strenght of 40%?
\n" ); document.write( ":
\n" ); document.write( "Let x = amt to be drained and it's also the amt to be added
\n" ); document.write( ":
\n" ); document.write( "The final solution will be still be 80 liters
\n" ); document.write( ":
\n" ); document.write( "30% Solution after x is drained: (80-x)
\n" ); document.write( ":
\n" ); document.write( "70% Solution added = x
\n" ); document.write( ":
\n" ); document.write( "A simple equation:
\n" ); document.write( ".30(80-x) + .70x = .40(80)
\n" ); document.write( "24 - .3x + .7x = 32
\n" ); document.write( ".4x = 32 - 24
\n" ); document.write( ".4x = 8
\n" ); document.write( " x = 8/.4
\n" ); document.write( "x = 20 gallons drained and replaced
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check:
\n" ); document.write( " .30(60) + .7(20) = .4(80)
\n" ); document.write( " 18 + 14 = 32
\n" ); document.write( " \n" ); document.write( "
\n" );