document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=304823>Question 304823</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A 200-gram solution is 25% acid. How much pure acid must be added to produce a solution that is 40% acid? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #218318 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=218318\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> .25*200+X=.40(200+X)<BR>\n" );
document.write( "50+X=80+.40X<BR>\n" );
document.write( "X-.4X=80-50<BR>\n" );
document.write( ".6X=30<BR>\n" );
document.write( "X=30/.6<BR>\n" );
document.write( "X=50 GRAMS OF PURE ACID IS USED.<BR>\n" );
document.write( "PROOF:<BR>\n" );
document.write( ".25*200+50=.40(200+50)<BR>\n" );
document.write( "50+50=.40*250<BR>\n" );
document.write( "100=100<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );