document.write( "Question 1112892: A chemical company makes two brands of antifreeze. The first brand is
\n" ); document.write( "70
\n" ); document.write( "%\r
\n" ); document.write( "\n" ); document.write( "pure antifreeze, and the second brand is
\n" ); document.write( "95
\n" ); document.write( "%\r
\n" ); document.write( "\n" ); document.write( "pure antifreeze. In order to obtain
\n" ); document.write( "140\r
\n" ); document.write( "\n" ); document.write( "gallons of a mixture that contains
\n" ); document.write( "80
\n" ); document.write( "%\r
\n" ); document.write( "\n" ); document.write( "pure antifreeze, how many gallons of each brand of antifreeze must be used?
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Algebra.Com's Answer #727944 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "The desired percentage of the mixture, 80, is 2/5 of the way from 70 to 95. (95-70 = 25; 80-70 = 10; 10/25 = 2/5)

\n" ); document.write( "That means 2/5 of the mixture must be the 95% antifreeze.

\n" ); document.write( "95% antifreeze: 2/5 of 140 gallons = 56 gallons
\n" ); document.write( "70% antifreeze: 3/5 of 140 gallons = 84 gallons

\n" ); document.write( "Answer: 56 gallons of 95%, 84 gallons of 70%

\n" ); document.write( "or you can use the much more difficult traditional algebraic solution method:

\n" ); document.write( "Let x be the number of gallons of 95% antifreeze
\n" ); document.write( "Then 140-x = number of gallons of 70% antifreeze

\n" ); document.write( "The 140 gallons of the mixture is 80% antifreeze:
\n" ); document.write( "\".95%28x%29%2B.70%28140-x%29+=+.80%28140%29\"

\n" ); document.write( "A relatively easy equation to solve; but far more work than is required by the first method....
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