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)\"\" \"About 
You can put this solution on YOUR website!
Try to start with the expression for what is in the radiator at start.\r
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\n" ); document.write( "\n" ); document.write( "\"%2810%2A20%29%2F10=20\"
\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
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\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( "\"10%2A20-g%2A20%2Bg%2A100\".
\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
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\n" ); document.write( "\n" ); document.write( "\"%2810%2A20-g%2A20%2Bg%2A100%29%2F10=50\"
\n" ); document.write( "Solve for g.\r
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\n" ); document.write( "\n" ); document.write( "\"%28200%2B%28100-20%29g%29=50%2A10\"
\n" ); document.write( "\"200%2B80g=500\"
\n" ); document.write( "\"80g=300\"
\n" ); document.write( "g=30/8=15/4
\n" ); document.write( "\"highlight%28g=3%263%2F4%29\" gallons to remove and replace with 100% antifreeze.
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