document.write( "Question 765298: A 5 gallon radiator containing a mixture of water and antifreeze solution. When tested , it was found to have only 40% antifreeze. How much must be drained out so that the radiator will then contain the desire 50% antifreeze solution. \n" ); document.write( "
Algebra.Com's Answer #466065 by josgarithmetic(39620)\"\" \"About 
You can put this solution on YOUR website!
The question really should be changed. How about this way:\r
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\n" ); document.write( "\n" ); document.write( "A 5 gallon radiator containing a mixture of water and antifreeze solution. When tested , it was found to have only 40% antifreeze. How much must be drained out AND REPLACED WITH PURE ANTIFREEZE so that the radiator will then contain 5 GALLONS OF THE desired 50% antifreeze solution?\r
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\n" ); document.write( "\n" ); document.write( "Let v = the amount of solution to remove and then replaced with pure antifreeze\r
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\n" ); document.write( "\n" ); document.write( "\"%285%2A40-v%2A40%2Bv%2A100%29%2F5=50\"
\n" ); document.write( "Note that the desired final volume must be the same as the starting volume in the five-gallon tank.\r
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\n" ); document.write( "\n" ); document.write( "Arithmetic steps to solving:
\n" ); document.write( "(5*40+(100-40)v)/5=50
\n" ); document.write( "\"5%2A40%2B60v=5%2A50\"
\n" ); document.write( "\"60v=5%2A50-5%2A40\"
\n" ); document.write( "\"60v=50\"
\n" ); document.write( "\"highlight%28v=5%2F6%29\", meaning, take out five-sixths of a gallon and replace with 100% antifreeze.
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