document.write( "Question 993323: Adding 25 gallons of acid to a mixture, makes it 80% Acid solution, while adding 25 gallons of water makes it 60% Acid solution. What is the original acid percentage of the mixture? \n" ); document.write( "
Algebra.Com's Answer #612590 by josgarithmetic(39617)![]() ![]() ![]() You can put this solution on YOUR website! You are not given the amount of original acid mixture, so give this a variable for its volume. \n" ); document.write( "v, the initial volume in gallons of the starting mixture of unknown percent concentration.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let p be the percent concentration of your original acid mixture. \n" ); document.write( "Additionally, you are forced to assume that your acid to add in pure form, the 100%, is a liquid.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "ADDING 25 GALLONS \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "ADDING 25 GALLONS OF WATER \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The system of equations slightly reworked is \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "STEPS SOLVING THE SYSTEM \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( "Next use the simpler, dilution equation. \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( " \n" ); document.write( " |