document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1199176>Question 1199176</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A biologist has two brine solutions, one containing 2% salt and another containing 8% salt. How many milliliters of each solution should she mix to obtain 1 L of a solution that contains 5.6% salt?<BR>\n" );
document.write( "2% solution    	  mL<BR>\n" );
document.write( "8% solution    	  mL<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 #833090 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=833090\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "You have received three responses, all showing variations of the standard formal algebraic method for solving 2-part mixture problems like this.<br><BR>\n" );
document.write( "Here is an informal method that can be used to solve this kind of problem quickly -- if a formal algebraic solution is not required.<br><BR>\n" );
document.write( "(1) Look at the three percentages -- 2, 5.6, and 8 -- (on a number line, if it helps)<BR>\n" );
document.write( "(2) observe/calculate that 5.6 is 3.6/6 = 0.6 = 3/5 of the way from 2 to 8<BR>\n" );
document.write( "(3) that means 3/5 of the mixture must be the solution with the higher percentage<br><BR>\n" );
document.write( "3/5 of the total 1 liter (1000 ml) is 600ml<br><BR>\n" );
document.write( "ANSWER: 600 ml of the 8% solution and 400 ml of the 2% solution<br><BR>\n" );
document.write( "CHECK:<BR>\n" );
document.write( ".08(600)+.02(400) = 48+8 = 56<BR>\n" );
document.write( ".056(1000) = 56<br><BR>\n" );
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