document.write( "Question 788263: A 90% acid solution is diluted with water to make a solution of 60% acid. When liters of water is added to dilute it again, the solution becomes 40% acid. How much water is added to the solution on the first instance? How much 90% acid was available at the start?\r
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Algebra.Com's Answer #478467 by josgarithmetic(39618)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "Let f=how much volume of 90% acid
\n" ); document.write( "Let y=how much water to add to get 60% acid.\r
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\n" ); document.write( "\n" ); document.write( "\"90%2Af%2F%28y%2Bf%29=60\"
\n" ); document.write( "\"90f=60%28y%2Bf%29\"
\n" ); document.write( "\"90f=60y%2B60f\"
\n" ); document.write( "\"9f-6f=6y\"
\n" ); document.write( "\"3f=6y\"
\n" ); document.write( "\"highlight%28y=f%2F2%29\" liters of water\r
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\n" ); document.write( "\n" ); document.write( "The second dilution now starts with (f+f/2) liters of 60% acid.
\n" ); document.write( "You want now to dilute to 40% acid.
\n" ); document.write( "Let x=volume of water to add for diluting to 40%.\r
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\n" ); document.write( "\n" ); document.write( "\"60%2A%28f%2Bf%2F2%29%2F%28x%2Bf%2Bf%2F2%29=40\"
\n" ); document.write( "Solve for x.
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