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)![]() ![]() ![]() You can put this solution on YOUR website! You need more information. The description is incomplete.\r \n" ); document.write( " \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 \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \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 \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "Solve for x. \n" ); document.write( " |