document.write( "Question 507422: Pure acid is to be added to a 10% acid solution to obtain 54L of a 20% acid solution. What amount of each should be used? \n" ); document.write( "
Algebra.Com's Answer #340661 by ankor@dixie-net.com(22740)\"\" \"About 
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Pure acid is to be added to a 10% acid solution to obtain 54L of a 20% acid solution.
\n" ); document.write( " What amount of each should be used?
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\n" ); document.write( "Let x = amt of pure acid required
\n" ); document.write( "Since the resulting amt is to be 54L:
\n" ); document.write( "(54-x) = amt of 10% acid required
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\n" ); document.write( "A typical mixture equation
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\n" ); document.write( "x + .10(54-x) = .20(54)
\n" ); document.write( "x + 5.4 - .10x = 10.8
\n" ); document.write( "x - .1x = 10.8 - 5.4
\n" ); document.write( ".9x = 5.4
\n" ); document.write( "x = \"5.4%2F.9\"
\n" ); document.write( "x = 6 L of pure acid to be combined with: 54-6 = 48 L of 10% solution
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