document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=146093>Question 146093</A>: <I><SPAN STYLE=\"word-break: break-all;\"> We have similar problems to this in our textbook, but this is off a handwritten midterm review from our \"teacher.\" It's an online course. \r<BR>\n" );
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document.write( "A 2004 Pontiac Montana has an engine coolant capacity of 10.1 liters. If the system is currently filled with a mixture that is 30% antifreeze, how much should be drained and replaced with pure antifreeze so the system is filled with  a mix that is 50% pure antifreeze? <BR>\n" );
document.write( "The answer we have is 2.89. I do not understand how to get to this answer! HELP! :) Thanks.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #106643 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=106643\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A 2004 Pontiac Montana has an engine coolant capacity of 10.1 liters. <BR>\n" );
document.write( "If the system is currently filled with a mixture that is 30% antifreeze, how much should be drained and replaced with pure antifreeze so the system is filled with a mix that is 50% pure antifreeze? <BR>\n" );
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document.write( "30% DATA:<BR>\n" );
document.write( "Amt. = 10.1 liters ; Amt of active ingredient = 0.3(10.1) = 3.03 liters<BR>\n" );
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document.write( "After removing x liters:<BR>\n" );
document.write( "Amt = 10.1-x liters ; Amt of active ingredient = 0.3(10.1-x) = 3.03-0.3x liters<BR>\n" );
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document.write( "After add x liters of 100% active ingredient:<BR>\n" );
document.write( "Amt = 10.1 liters ; Amt of active ingredient = 0.5*10.1 = 5.05 liters<BR>\n" );
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document.write( "EQUATION:<BR>\n" );
document.write( "active + active = active<BR>\n" );
document.write( "3.03-0.03x + x + 5.05<BR>\n" );
document.write( "0.97x = 2.02<BR>\n" );
document.write( "x = 2.0825 liters (amount that is subtracted and then added)<BR>\n" );
document.write( "===============<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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