document.write( "Question 1144712: A chemical company mixes pure water with their premium antifreeze solution to create an inexpensive antifreeze mixture. The premium antifreeze solution contains 85% pure antifreeze. The company wants to obtain 170 gallons of a mixture that contains 15% pure antifreeze. How many gallons of water and how many gallons of the premium antifreeze solution must be mixed?
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Algebra.Com's Answer #765873 by ikleyn(52786) $\"\"$ $\"About$

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document.write( "Let x be the volume of the premium 85% antifreeze (in gallons) and\r\n" );
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document.write( "let y be the volume of water.\r\n" );
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document.write( "        x + y  = 170       gallons     (1)   (total volume equation)\r\n" );
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document.write( "    0.85x      = 0.15*170   gallons    (2)   (pure antifreeze volume equation)\r\n" );
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document.write( "From equation (2),  x =  = 30.\r\n" );
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document.write( "Then from (1),  y = 170-30 = 140 gallons.\r\n" );
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document.write( "ANSWER.  140 gallons of water.\r\n" );
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document.write( "Since they simply DILUTE the existing mixture, the content of the pure antifreeze remains the same: it is   = 25.5 gallons.\r\n" );
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document.write( "Hence, the 85% premium antifreeze should have the volume of   = 30 gallons to be 85% mixture.\r\n" );
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document.write( "The complement to 170 gallons, i.e.  170 - 30 = 140 gallons, is the water volume, which should be added.\r\n" );
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