document.write( "Question 1133876: How many liters each of a 45% acid solution and a 70% acid solution must be used to produce 50 liters of a
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document.write( "65% acid solution? (Round to two decimal places if necessary.) \n" );
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Algebra.Com's Answer #751182 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "The standard algebraic method for solving a mixture problem like this involving two ingredients is to write and solve an equation that says the total amount of acid in the two ingredients is equal to the amount of acid in the mixture. \n" ); document.write( "Let x be the number of liters of the 45% acid solution; then, since the mixture is to be 50 liters, the number of liters of the 70% acid solution is (50-x). Then \n" ); document.write( "the amount of acid in x liters of 45% acid solution is 0.45(x) liters; \n" ); document.write( "the amount of acid in (50-x) liters of 70% acid solution is 0.70(50-x) liters; and \n" ); document.write( "the amount of acid in the mixture is 0.65(50) liters \n" ); document.write( "Then \n" ); document.write( " \n" ); document.write( "You can do the work to solve that equation to find the answer to the problem. \n" ); document.write( "If an algebraic solution is not required, here is a method that will get you to the answer with far less work. \n" ); document.write( "The desired 65% is 4/5 of the way from 45% to 70%; that means 4/5 of the mixture must be the 70% acid solution. \n" ); document.write( "4/5 of 50 liters is 40 liters. \n" ); document.write( "So 40 liters of the 70% acid solution and 10 liters of the 45% acid solution. \n" ); document.write( " |