document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=70275>Question 70275</A>: <I><SPAN STYLE=\"word-break: break-all;\"> (3)A 100% concetrate is to be mixed with a mixture having a concentrate of 40% to obtain 55 gallons of a mixture with a concentrate of 75%. how much of the 100% concentrate will be needed?\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #50131 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley75><B>checkley75(3666)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley75><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=50131\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1X+.4(55-X)=.75*55<BR>\n" );
document.write( "X+22-.4X=41.25<BR>\n" );
document.write( ".6X=41.25-22<BR>\n" );
document.write( ".6X=19.25<BR>\n" );
document.write( "X=19.25/.6<BR>\n" );
document.write( "X=32.0833 GALLONS  OF 100% SOLUTION IS NEEDED.<BR>\n" );
document.write( "PROOF<BR>\n" );
document.write( "32.0833+.4(55-32.08333)=41.25<BR>\n" );
document.write( "32.08333+22-12.8333=41.25<BR>\n" );
document.write( "54.08333-12.8333=41.25<BR>\n" );
document.write( "41.25=41.25\r<BR>\n" );
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