document.write( "Question 24270: An automobile radiatior contains 16 liters of antifreeze and water. This mix is 30% antifreeze. how much of this mix should be drained and replaced with pure antifreeze so that there will be 50% antifreeze? \n" ); document.write( "
Algebra.Com's Answer #12906 by venugopalramana(3286)\"\" \"About 
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An automobile radiatior contains 16 liters of antifreeze and water.
\n" ); document.write( " This mix is 30% antifreeze.
\n" ); document.write( " how much of this mix should be drained and replaced with pure antifreeze so that there will be 50% antifreeze?
\n" ); document.write( "LET US REMOVE X LITRES OF ORIGINAL MIX AND REPLACE IT WITH X LITRES OF PURE ANTIFREEZE .
\n" ); document.write( "SO WE HAVE NOW 16-X LITRES OF ORIGINAL MIX PLUS X LITRES OF PURE ANTIFREEZE.
\n" ); document.write( "ORIGINAL MIX HAS 30 % ANTIFREEZE
\n" ); document.write( "16-X LITRES WILL HAVE (16-X)*30/100=4.8-0.3X LITRES ANTIFREEZE. THIS PLUS X LITRES OF PURE ANTIFREEXE MAKE IT 4.8-0.3X+X =4.8 + 0.7X LITRES OF ANTIFREEZE
\n" ); document.write( "TOTAL MIX =16 LITRES WITH 50 % ANTIFREEZE =16*50/100=8 LITRES OF ANTIFREEZE
\n" ); document.write( "SO 4.8+0.7X=8
\n" ); document.write( "0.7X=8-4.8=3.2
\n" ); document.write( "X=3.2/0.7=32/7=4.57 LITRES.\r
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