document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=6203>Question 6203</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solution A is 12% acid and Solution B is 4% acid. If a technician wants to mix them to make 60 liters of Solutin C which is 10% acid, how many liters of each should be mixed together? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #3302 by </B><A HREF=/tutors/aboutme.mpl?userid=roy_algebra2006><B>roy_algebra2006(2)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=roy_algebra2006><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=3302\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let us assume that solution A required is 'x' litres & solution B required is 'y' litres.<BR>\n" );
document.write( "we can see that x+y=60---------(1)<BR>\n" );
document.write( "now 10% acid =10*60/100 litres=6 litres.<BR>\n" );
document.write( "now the tricky part....<BR>\n" );
document.write( " 12% 0f 'x' =12*x/100 & 4% of 'y'=4*y/100 ......(all in litres)<BR>\n" );
document.write( "  thus we get 12*x/100+4*y/100=6<BR>\n" );
document.write( "             =>3*x/25+y/25=6<BR>\n" );
document.write( "             =>3*x+y=150--------(2)<BR>\n" );
document.write( " solving the above simultaneous equations numbered (1)& (2)......<BR>\n" );
document.write( "  we get<BR>\n" );
document.write( "  x=45  &  y=15<BR>\n" );
document.write( " hence solution A required is 45 litres & solution B required is 15 litres. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );