document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=129615>Question 129615</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How many gallons each of 25% alcohol and 35% alcohol should be mixed to get 20 gallons of 32% alcohol? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #94620 by </B><A HREF=/tutors/aboutme.mpl?userid=kev82><B>kev82(151)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=kev82><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=94620\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If we take X gallons of the 25% liquid, and Y gallons of the 35% liquid, then it is quite obvious that we will have X+Y gallons of liquid in total. We need to make 20 gallons  of liquid so X+Y=20.\r<BR>\n" );
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document.write( "Let us now consider the amount of alcohol in our solution. Taking X gallons of the first liquid will give us 0.25X gallons of alcohol because 25% of the liquid is alcohol. Similarly taking Y gallons of the second liquid gives us 0.35Y gallons of alcohol, so in total there are 0.25X+0.35Y gallons of alcohol in the final mixture.\r<BR>\n" );
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document.write( "The final mixture is 20 gallons as we know, so the percentage of alcohol in it is (0.25X+0.35Y)/20. This must be 0.32 (32%) This gives us the second equation (0.25X+0.35Y)/20=0.32. So we now have:\r<BR>\n" );
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document.write( "X+Y=20<BR>\n" );
document.write( "0.25X+0.35Y=6.4\r<BR>\n" );
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document.write( "These equations can be solved for X and Y which should give your answer. </SPAN>\n" );
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