document.write( "Question 64839This question is from textbook An Incremental Development
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document.write( ": A druggist needs 25 liters of a solution that is 16% iodine. She has one solution that is 12% iodine and another solution that is 37% iodine. How many liters of each solution should the druggiest use? \n" );
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Algebra.Com's Answer #45345 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! Let x=number of liters of 37% iodine \n" ); document.write( "Then 25-x=number of liters of 12% iodine\r \n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure iodine in the 37% solution (.37)(x) plus the amount of pure iodine in the 12% solution .12(25-x) must equal the amount of pure iodine in the final solution (25)(.16) so our equation to solve:\r \n" ); document.write( "\n" ); document.write( ".37x+.12(25-x)=25(.16) simplifying we get:\r \n" ); document.write( "\n" ); document.write( ".37x-.12x+3=4 collecting like terms we have: \n" ); document.write( ".25x=4-3 or \n" ); document.write( ".25x=1 \n" ); document.write( "x=4 liters of the 37% solution \n" ); document.write( "25-x=25-4=21 liters of 12% solution\r \n" ); document.write( "\n" ); document.write( "ck\r \n" ); document.write( "\n" ); document.write( ".37(4)+.12(25-4)=25(.16) \n" ); document.write( "1.48+2.52=4 \n" ); document.write( "4=4\r \n" ); document.write( "\n" ); document.write( "Hope this helps. Happy holidays.-----ptaylor \n" ); document.write( " |