document.write( "Question 420867: If I have a 20 qt radiator containing a 80% antifreeze solution. How much of the solution should I drain and replace with pure water to get a solution that is 50% antifreeze? \n" ); document.write( "
Algebra.Com's Answer #853222 by ikleyn(53431)\"\" \"About 
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\n" ); document.write( "If I have a 20 qt radiator containing a 80% antifreeze solution, how much of the solution
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\n" ); document.write( "\n" ); document.write( "        The solution in the post by @mananth is incorrect.\r
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document.write( "Let x be the volume of the original 80% antifreeze solution to partly drain\r\n" );
document.write( "and to replace with pure water to get the 40% antifreeze solution.\r\n" );
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document.write( "After draining, we then have (20-x) quartz of the 80% antifreeze solution.\r\n" );
document.write( "It contains 0.8*(20-x) quartz of the pure antifreeze.\r\n" );
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document.write( "Adding water does not change the amount of the antifreeze in the solution.\r\n" );
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document.write( "At the end, the volume of the pure antifreeze in the radiator after adding x quartz of water is 0.5*20 quartz.\r\n" );
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document.write( "So, we equate these two expressions for the pure antifreeze amount\r\n" );
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document.write( "    0.8*(20-x) = 0.5*20  quartz.    (1)\r\n" );
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document.write( "Simplify and find x\r\n" );
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document.write( "    16 - 0.8x = 10,\r\n" );
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document.write( "    16 - 10 = 0.8x,\r\n" );
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document.write( "       6    = 0.8x\r\n" );
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document.write( "       x    = \"6%2F0.8\" = \"60%2F8\" = 7.5.\r\n" );
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document.write( "ANSWER.  7.5  quartz of the original 80% solution should be drained and replaced by pure water to get the 50% antifreeze solution.\r\n" );
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