document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1134842>Question 1134842</A>: <I><SPAN STYLE=\"word-break: break-all;\"> pure acid is to be added to a 10% acid solution to obtain 90L of 32% soulution. what amounts of each should be used? how many liters of 100% pure acid should be used to make the solution? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #752348 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=752348\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "There is an alternative to the standard algebraic method for solving mixture problems like this, as shown in the solution from tutor @ikleyn.<br><BR>\n" );
document.write( "The alternative method is based on where the mixture percentage lies between the percentages of the two ingredients.<br><BR>\n" );
document.write( "The given numbers in this problem make the solution almost trivial.  Here is what the method looks like for this problem.<br><BR>\n" );
document.write( "(1) The 32% target percentage is 22/90 of the way from the 10% of the first ingredient to the 100% of the second ingredient. (100-10 = 90; 32-10 = 22; so 22/90)<BR>\n" );
document.write( "(2) That means 22/90 of the mixture has to be the second ingredient.<BR>\n" );
document.write( "(3) 22/90 of 90L is 22L<br><BR>\n" );
document.write( "So 22L of the 100% pure acid, 90-22 = 68L of the original 10% acid. </SPAN>\n" );
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