document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1069274>Question 1069274</A>: <I><SPAN STYLE=\"word-break: break-all;\"> in a mixture we know that we want 1% of a product that is made up of 50% of a particular  ingredient. If another product were substituted that had 75% of that ingredient, what would we reduce the 1% of the new product to. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #684544 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=684544\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Suppose the entire product weighs 1000 pounds, so the 1% additive of a product weighs 10 pounds, with 5 pounds of ingredient X.  <BR>\n" );
document.write( "Now, suppose 5 pounds of ingredient X is 75% of the additive.  How much does the additive weigh?  <BR>\n" );
document.write( "5=(3/4)x<BR>\n" );
document.write( "multiply both sides by 4/3<BR>\n" );
document.write( "x=20/3 pounds, or 6 2/3 pounds.<BR>\n" );
document.write( "Instead of 10 pounds we now need 6 2/3 pounds. And that is 20/3/1000=20/3000 of the entire product or 2/300 or 0.6667%, reduced from 1.000%.<BR>\n" );
document.write( "------------------------<BR>\n" );
document.write( ".01*0.50=x*0.75<BR>\n" );
document.write( ".005=0.75x<BR>\n" );
document.write( "x=0.006667 or 0.6667%<BR>\n" );
document.write( " </SPAN>\n" );
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