document.write( "Question 782484: How much of an alloy that is 10% copper should be mixed with 600 oz of an alloy that is 60% copper in order to get an alloy that is 20% cooper \n" ); document.write( "
Algebra.Com's Answer #476415 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
--- percent ---------------- Amount
\n" ); document.write( "Alloy I 60 ---------------- 600 oz
\n" ); document.write( "Alloy II 10 ---------------- x oz
\n" ); document.write( "Mixture 20 ---------------- 600 + x oz
\n" ); document.write( "60% * 600 + 10% x = 20% ( 600 + x )
\n" ); document.write( "multiply by 100
\n" ); document.write( "
\n" ); document.write( "60 * 600 + 10 x = 20 ( 600 + x )
\n" ); document.write( "36000 + 10 x = 12000 + 20 x
\n" ); document.write( "10 x -20 x = -36000 + 12000
\n" ); document.write( "-10 x = -24000
\n" ); document.write( "/ -10
\n" ); document.write( "x = 2400 oz Alloy II
\n" ); document.write( "
\n" ); document.write( "m.ananth@hotmail.ca
\n" ); document.write( "
\n" ); document.write( "
\n" );