document.write( "Question 709907: A radiator contains 10 gal of a 20% antifreeze solution. How many gallons must be drained from the radiator and replaced by pure antifreeze so that the radiator will contain 10 gal of a 50% antifreeze solution? \n" ); document.write( "
Algebra.Com's Answer #436816 by josgarithmetic(39618)![]() ![]() ![]() You can put this solution on YOUR website! Try to start with the expression for what is in the radiator at start.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "The righthand member is 20% antifreeze, and the left hand member is the quantity of pure antifreeze over 10 gallons of the liquid in the radiator.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "In the numerator, you want to SUBTRACT g gallons of the 20% liquid and ADD g gallons of the 100% material. \n" ); document.write( " \n" ); document.write( "The denominator will stay unchanged, 10, for 10 gallons. Also, you want this used as a ratio equalling 50, for 50% antifreeze.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "Solve for g.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "g=30/8=15/4 \n" ); document.write( " |