document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=40186>Question 40186</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Joe's Dairy Farm has 100 gallons of milk that is 4.6% butterfat. How much skimmilk (no butterfat) should be mixed with it to make milk that is 3.2% fat?<BR>\n" );
document.write( "________gallons of milk should be added.<BR>\n" );
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document.write( "A chemist has one solution that is 80% acid and another soulution that is 30% acid. How much of the first (80%) solution is needed to make a 400L solution that is 62% acid?<BR>\n" );
document.write( "__________liters of the 80% solution is required.<BR>\n" );
document.write( "Thanx </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #25619 by </B><A HREF=/tutors/aboutme.mpl?userid=longjonsilver><B>longjonsilver(2297)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=longjonsilver><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=longjonsilver><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=25619\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> read my Lesson on Mixtures. That should help you understand this type of question.\r<BR>\n" );
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document.write( "Anyway:<BR>\n" );
document.write( "1. We cannot think about \"amounts of butterfat, since the skimmed has none: its percentage is zero. So i have turned it around and considered the percentage of non-fat milk in the milk: the skimmed will be 100%.\r<BR>\n" );
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document.write( "100*0.954 + x*1.00 = (100+x)*0.968<BR>\n" );
document.write( "95.4 + x = 96.8 + 0.968x<BR>\n" );
document.write( "x - 0.968x = 96.8 - 95.4<BR>\n" );
document.write( "0.032x = 1.4<BR>\n" );
document.write( "x = (1.4)/(0.032)<BR>\n" );
document.write( "x = 43.75 gallons\r<BR>\n" );
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document.write( "2. Start with 80% acid and a 30% acid. <BR>\n" );
document.write( "--> Let there be x litres of the 80% solution acid. So, how much of the 30%? Answer is \"the rest of the 400 litres ie (400-x). This is the tricky bit you need to \"see\".\r<BR>\n" );
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document.write( "So, 0.80x + 0.30(400-x) = 0.62*400<BR>\n" );
document.write( "0.80x + 120 - 0.30x = 256<BR>\n" );
document.write( "0.80x - 0.30x = 256 - 120<BR>\n" );
document.write( "0.50x = 136<BR>\n" );
document.write( "x = 272\r<BR>\n" );
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document.write( "So we need 272 litres of the 80% acid<BR>\n" );
document.write( "and we need 128 litres of the 30% acid\r<BR>\n" );
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document.write( "jon.<BR>\n" );
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