document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=269791>Question 269791</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How many grams of 60% silver should be added to 2 grams of 20% silver to produce a 35% silver alloy?<BR>\n" );
document.write( " <BR>\n" );
document.write( " so i tried to make a table <BR>\n" );
document.write( "sliver    amount   %      total quantity<BR>\n" );
document.write( "silver 1   x       60%      .6x<BR>\n" );
document.write( "silver 2    </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #197758 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley77><B>checkley77(12844)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley77><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=197758\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> .60X+.20*2=.35(X+2)<BR>\n" );
document.write( ".60X+.40=.35X+.70<BR>\n" );
document.write( ".60X-35X=.7-.4<BR>\n" );
document.write( ".25X=.3<BR>\n" );
document.write( "X=.3/.25<BR>\n" );
document.write( "X=1.2 G. OF 60% SILVER IS USED.<BR>\n" );
document.write( "PROOF:<BR>\n" );
document.write( ".60*1.2+.2*2=.35(1.2+2)<BR>\n" );
document.write( ".72+.4=.35*3.2<BR>\n" );
document.write( "1.12=1.12<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );