document.write( "Question 507710: How many liters of a 10% alcohol solution must be mixed with 40 liters of a 50% solution to get a 40% solution? \r
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document.write( "I got the answer 13.3 L, but I'm not sure that it's right. \n" );
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Algebra.Com's Answer #340704 by oberobic(2304) ![]() You can put this solution on YOUR website! With mixture problems, you need to pay attention to how much 'pure' stuff you have and how much you need. \n" ); document.write( ". \n" ); document.write( "40 liters of 50% solution means you have 20 liters of 'pure' alcohol \n" ); document.write( ". \n" ); document.write( "x = liters of 10% solution is to be added until you have a final solution that is 40% alcohol. \n" ); document.write( ". \n" ); document.write( ".5(40) + .1(x) = .4(40+x) \n" ); document.write( ". \n" ); document.write( "multiply by 10 to remove decimals \n" ); document.write( ". \n" ); document.write( "5(40) + x = 4(40+x) \n" ); document.write( ". \n" ); document.write( "200 +x = 160 + 4x \n" ); document.write( "40 = 3x \n" ); document.write( ". \n" ); document.write( "x = 40/3 = 13 1/3 liters \n" ); document.write( ". \n" ); document.write( "Check this to be sure it is the answer. \n" ); document.write( ". \n" ); document.write( "40/3 * .1 = 4/3 liters of alcohol \n" ); document.write( "120/3 * .5 = 20 liters of alcohol \n" ); document.write( "= 21 1/3 liters of alcohol. \n" ); document.write( ". \n" ); document.write( "40 + 40/3 = 120/3 + 40/3 = 160/3 \n" ); document.write( ". \n" ); document.write( "160/3*.4 = 64/3 = 21 1/3 liters of alcohol \n" ); document.write( "which is correct. \n" ); document.write( ". \n" ); document.write( "Answer: Add 13 1/3 liters of 10% alcohol solution. \n" ); document.write( ". \n" ); document.write( "Done. \n" ); document.write( " |