document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=197840>Question 197840</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A solution containing 28% acid is to be mixed with a solution containing 40% acid to make a 300 L solution which contains 36% acid. How of each solution should be used to make the desired mixture? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #148357 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=148357\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> .40x+.28(300-x)=.36*300<BR>\n" );
document.write( ".40x+84-.28x=108<BR>\n" );
document.write( ".12x=108-84<BR>\n" );
document.write( ".12x=24<BR>\n" );
document.write( "x=24/.12<BR>\n" );
document.write( "x=200 L of 40% solution is used.<BR>\n" );
document.write( "300-200=100 L of 28% solution is used.<BR>\n" );
document.write( "Proof:<BR>\n" );
document.write( ".40*200+.28*100=.36*300<BR>\n" );
document.write( "80+28=108<BR>\n" );
document.write( "108=108 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );