document.write( "Question 373433: pure acid is to be added to a 5% acid solution to obtain 95L of 68% solution. what amounts of each should be used?\r
\n" ); document.write( "\n" ); document.write( "How many Liter of 100% pure acid should be used to make the solution? ____L\r
\n" ); document.write( "\n" ); document.write( "How many liters of the 5% solution should be in the mixture? ___L \n" ); document.write( "
Algebra.Com's Answer #265889 by ankor@dixie-net.com(22740)\"\" \"About 
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pure acid is to be added to a 5% acid solution to obtain 95L of 68% solution.
\n" ); document.write( " what amounts of each should be used?
\n" ); document.write( "How many Liter of 100% pure acid should be used to make the solution? ____L
\n" ); document.write( ":
\n" ); document.write( "Let x = amt of pure acid to be used
\n" ); document.write( "then
\n" ); document.write( "(95-x) = amt of 5% acid needed
\n" ); document.write( ":
\n" ); document.write( ".05(95-x) + x = .68(95)
\n" ); document.write( "4.75 - .05x + x = 64.6
\n" ); document.write( "-.05x = x = 64.6 - 4.75
\n" ); document.write( ".95x = 59.85
\n" ); document.write( "x = \"59.85%2F.95\"
\n" ); document.write( "x = 63 L of pure acid required to make 95 liters of a 68% solution
\n" ); document.write( ":
\n" ); document.write( "Check solution in original equation (32 L of 5% solution required)
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