document.write( "Question 1097530: David needs to have a 20% solution of antifreeze in his truck's radiator. If the radiator is now full and contains 8 quarts of 10% antifreeze solution, how much of that solution must be drained and replaced with pure antifreeze to create the desired solution? \n" ); document.write( "
Algebra.Com's Answer #711961 by math_helper(2461)\"\" \"About 
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Let x be the amount of 10% mix to be drained and replaced with 100% antifreeze.\r
\n" ); document.write( "\n" ); document.write( "For every x quarts drained, that drains 0.10*x quarts of (pure) antifreeze. \r
\n" ); document.write( "\n" ); document.write( "Note that we want 8 quarts *0.2 = 1.6 quarts of antifreeze in the final mixture.
\n" ); document.write( "And, we start with 8*0.1 quarts of antifreeze.
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\n" ); document.write( "\n" ); document.write( "Expressing the amount of antifreeze drained and added back gives us:
\n" ); document.write( "(0.10)*(8) - (0.1)(x) + x = 1.6
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\n" ); document.write( "\n" ); document.write( "0.9x = 0.8
\n" ); document.write( " x = 0.8/0.9 = 8/9\r
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\n" ); document.write( "Ans: 8/9 of a quart (approx 0.88889 quart) must be drained and replaced with pure antifreeze to get to a final mix that is 20% antifreeze.
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\n" ); document.write( "\n" ); document.write( "Check (how much antifreeze… it should be 1.6quarts):\r
\n" ); document.write( "\n" ); document.write( " (8)*(0.1) - (8/9)*(0.1) + 8/9 =
\n" ); document.write( " 0.8 - 0.8/9 + 8/9 =
\n" ); document.write( " 72/90 - 8/90 + 80/90 = 144/90 = 1.6 (ok)\r
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