document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1153659>Question 1153659</A>: <I><SPAN STYLE=\"word-break: break-all;\">  metallurgist has one alloy containing 23% titanium and another containing 51% titanium. How many pounds of each alloy must he use to make 56 pounds of a third alloy containing 47% titanium? (Round to two decimal places if necessary.)\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 #775958 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=775958\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "A quick and easy non-algebraic path to the answer to this problem and a wide variety of similar \"mixture\" problems....<br><BR>\n" );
document.write( "Look at the percentages 23, 47 and 51 on a number line.<BR>\n" );
document.write( "47 is 24/28 = 6/7 of the way from 23 to 51.<BR>\n" );
document.write( "That means 6/7 of the mixture is the 51% alloy.<br><BR>\n" );
document.write( "With a total of 56 pounds, that means 48 pounds of the 51% alloy an 8 pounds of the 23% alloy.<br><BR>\n" );
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