document.write( "Question 559380: A 8-quart radiator contains a 60% antifreeze solution. How much of the solution needs to be drained out and replaced with pure antifreeze in order to raise the solution to 80% antifreeze? \n" ); document.write( "
Algebra.Com's Answer #363402 by mananth(16946)\"\" \"About 
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you have 60% solution
\n" ); document.write( "you have to add 100% antifreeze to get 80% solution
\n" ); document.write( "so we find out how much antifreeze is required to raise ti 80%
\n" ); document.write( "This quantity is to be replaced\r
\n" ); document.write( "\n" ); document.write( "percent ---------------- quantity
\n" ); document.write( "Antifreeze 100 ----------------x quarts
\n" ); document.write( "solution 60 ----------------5-x quarts
\n" ); document.write( "Mixture 80.00% ---------------- 5
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\n" ); document.write( "100x+60(5-x)=80*5
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\n" ); document.write( "100x+300-60x=400
\n" ); document.write( "100x-60 x=400-300
\n" ); document.write( "40x= 100
\n" ); document.write( "/ 40
\n" ); document.write( "x=2.5 quarts 100.00% Antifreeze
\n" ); document.write( "2.5 quarts 60.00% solution\r
\n" ); document.write( "\n" ); document.write( "2.5 quarts has to be removed
\n" ); document.write( "m.ananth@hotmail.ca
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