document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1013775>Question 1013775</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 40% and the third contains 80%. He wants to use all three solutions to obtain a mixture of 90 liters containing 60% acid, using 3 times as much of the 80% solution as the 40% solution. How many liters of each solution should be used?<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #630232 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=630232\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Call the amount used of the 40% solution, x.<BR>\n" );
document.write( "Then the amount of 80% used would be 3x.<BR>\n" );
document.write( "Thus the amount of 20% acid would be 90 - 3x - x = 90 - 4x.<BR>\n" );
document.write( "The set up is like this<BR>\n" );
document.write( ".40x + .80(3x) + .20(90-4x) = .60(90)<BR>\n" );
document.write( ".4x + 2.4x + 18 - .8x = 54<BR>\n" );
document.write( "2x = 36<BR>\n" );
document.write( "x = 18 L at 40%<BR>\n" );
document.write( "3x = 54 L at 80%<BR>\n" );
document.write( "90-4x = 18 L at 20% </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );