document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=993597>Question 993597</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How much of an alloy that is 20% copper should be mixed with 600 ounces of an alloy that is 70% copper in order to get an alloy that is 30% copper? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #612886 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=612886\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x = the weight of the 20% alloy<BR>\n" );
document.write( "Therefore the total weight of the mixture is x + 600<BR>\n" );
document.write( "We can write the following equation which equates the amount of copper from each of the two alloys to the total amount contained in the mixture:<BR>\n" );
document.write( "0.2x + 0.7*600 = 0.3(x+600)<BR>\n" );
document.write( "Solve for x:<BR>\n" );
document.write( "0.1x = = 240<BR>\n" );
document.write( "x = 2400 oz<BR>\n" );
document.write( "Check:<BR>\n" );
document.write( "0.2*2400 + 0.7*600 = 480 + 420 = 0.3(3000) = 900 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );