document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=830613>Question 830613</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A 18 gallon salt-water solution contains 15% pure salt. how much pure salt should be added to produce the solution to 20% salt? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #500977 by </B><A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><B>josgarithmetic(39618)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><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=500977\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Best judgement about this is that an estimate is possible, assuming 18 gallons of water as \"pounds\" of water and that 15% of this value is the amount of salt dissolved in it.\r<BR>\n" );
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document.write( "You must use the density of water, 8.345 pounds per gallon; or whatever value you can best accept.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=18%2Agallons%288.345%2Apound%2Fgallon%29=2703.78 ALIGN=MIDDLE  ALT=\"18%2Agallons%288.345%2Apound%2Fgallon%29=2703.78\"   DISABLED_style= \"max-width:90%; height:auto;\" > pounds\r<BR>\n" );
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document.write( "Next, assuming that concentration is MASS per MASS, we can continue...\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%280.15%29%282704%29%2B1.00%2Am%29%2F%282704%2Bm%29=0.20 ALIGN=MIDDLE  ALT=\"%28%280.15%29%282704%29%2B1.00%2Am%29%2F%282704%2Bm%29=0.20\"   DISABLED_style= \"max-width:90%; height:auto;\" >, using m for the mass of salt to add, in \"pounds\".<BR>\n" );
document.write( "Solve for m. </SPAN>\n" );
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