document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1184883>Question 1184883</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A chemist has three different acid solutions. The first acid solution contains 25 % acid, the second contains 40 % and the third contains 85 % . He wants to use all three solutions to obtain a mixture of 135 liters containing 35 % acid, using 2 times as much of the 85 % solution as the 40 % solution. How many liters of each solution should be used?\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 #815554 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=815554\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "You have received responses showing three different algebraic solutions -- one using three variables, one using two, and one using one.<br><BR>\n" );
document.write( "Use the solution from the tutor who uses one variable.  Taking the little extra time to figure out how to set up the problem using a single variable will save you a lot of time in solving the problem.<br><BR>\n" );
document.write( "Here then is a completely different method for solving this kind of problem, without formal algebra.  The words of explanation make it sound long and difficult; but the calculations are relatively simple, so if your mental math is good you can solve the problem quickly by this method.<br><BR>\n" );
document.write( "First consider mixing the 40% and 85% acid, using twice as much of the 85% acid.  That means 2/3 of this mixture will be the 85% acid; and that means the percentage of this mixture will be 2/3 of the way from 40% to 85% -- which is 70%.<br><BR>\n" );
document.write( "Then consider mixing this 70% acid solution with the 25% acid solution to get a 35% acid solution.  35% is (35-25)/(70-25)=10/45=2/9 of the way from 25% to 70%, so 2/9 of the 135 liters total, or 30 liters, is the 70% acid solution.<br><BR>\n" );
document.write( "Then, since the 70% acid solution contains twice as much 85% acid as 40% acid, the mixture contains 10 liters of 40% acid and 20 liters of 85% acid.<br><BR>\n" );
document.write( "Then the amount of 25% acid solution used in the mixture is 135-30=105 liters.<br><BR>\n" );
document.write( "ANSWERS:<BR>\n" );
document.write( "25%: 105 liters<BR>\n" );
document.write( "40%: 10 liters<BR>\n" );
document.write( "85%: 20 liters<br><BR>\n" );
document.write( "CHECK:<BR>\n" );
document.write( ".25(105)+.40(10)+.85(20)=26.25+4+17=47.25<BR>\n" );
document.write( ".35(135)=47.25<br>\r<BR>\n" );
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