document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=677743>Question 677743</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A metallurgist wants to make 150 pounds of an alloy that is 55% aluminum. If he mixes an alloy that is 20% aluminum with an alloy that is 70% aluminum, how much of each should he use? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #421056 by </B><A HREF=/tutors/aboutme.mpl?userid=ptaylor><B>ptaylor(2198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ptaylor><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=ptaylor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=421056\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x=amount of 70% aluminum<BR>\n" );
document.write( "Then 150-x=amount of 20% aluminum<BR>\n" );
document.write( "Now we know that the amount of pure aluminum that exists before the mixture takes place(0.70x+0.20(150-x)) has to equal the amount of pure aluminum that exists after the mixture takes place(0.55*150).<BR>\n" );
document.write( "Sooooooo<BR>\n" );
document.write( "0.70x+0.20(150-x)=0.55*150<BR>\n" );
document.write( "0.70x+30-0.20x=82.5<BR>\n" );
document.write( "0.50x=52.5<BR>\n" );
document.write( "x=105 lb-------------------amount of 70% aluminum needed<BR>\n" );
document.write( "150-x=150-105=45 lb---------amount of 20% aluminum needed<BR>\n" );
document.write( "CK<BR>\n" );
document.write( "0.7*105+0.2*45=0.55*150<BR>\n" );
document.write( "73.5+9=82.5<BR>\n" );
document.write( "82.5=82.5\r<BR>\n" );
document.write( "\n" );
document.write( "Hope this helps---ptaylor\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );