document.write( "Question 869338: How much of a 50% antifreeze solution must a mechanic mix with a 70% antifreeze solution if 20 gallons of a 65% antifreeze solution are needed?
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Algebra.Com's Answer #524134 by mananth(16946)\"\" \"About 
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Percent ---------------- quantity
\n" ); document.write( "Antifreeze I 50.00% ---------------- x gallons
\n" ); document.write( "Antifreeze II 70.00% ------ 20 - x gallons
\n" ); document.write( "Mixture 65.00% ---------------- 20
\n" ); document.write( "Total 20 gallons
\n" ); document.write( "50.00% x + 70.00% ( 20 - x ) = 65.00% * 20
\n" ); document.write( "50 x + 70 ( 20 - x ) = 1300
\n" ); document.write( "50 x + 1400 - 70 x = 1300
\n" ); document.write( "50 x - 70 x = 1300 - -1400
\n" ); document.write( "-20 x = -100
\n" ); document.write( "/ -20
\n" ); document.write( " x = 5 gallons 50.00% Antifreeze I
\n" ); document.write( " 15 gallons 70.00% Antifreeze II
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\n" ); document.write( "m.ananth@hotmail.ca
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