document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=263828>Question 263828</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Initially, a tank contains 20 gallons of 30% antifreeze solution. How many gallons of an 80% antifreeze solution should be added to the tank in order to have the concentration of the antifreeze in the tank increase to 50%? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #194387 by </B><A HREF=/tutors/aboutme.mpl?userid=ptaylor><B>ptaylor(2198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ptaylor><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/cgi-bin/show-face.mpl?user=ptaylor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=194387\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <BR>\n" );
document.write( "Let x=amount of 80% antifreeze needed\r<BR>\n" );
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document.write( "Initially the amount of pure antifreeze in the tank =0.30*20=6 gal. If we increase the pure antifreeze in the tank by 50% we will need to have have 6+3=9 gal of pure antifreeze((9-6)/6=50% increase).  In other words, we need to use enough of the 80% antifreeze to get 3 gal of pure antifreeze.  So our equation to solve is:<BR>\n" );
document.write( " 0.80x=3<BR>\n" );
document.write( "x=3.75 gal---needs to be added to the tank\r<BR>\n" );
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document.write( "CK<BR>\n" );
document.write( "20*0.30+3.75*0.80=9<BR>\n" );
document.write( "6+3=9<BR>\n" );
document.write( "9=9\r<BR>\n" );
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document.write( "Hope this helps---ptaylor </SPAN>\n" );
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