document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=218863>Question 218863</A>: <I><SPAN STYLE=\"word-break: break-all;\"> This is a real question I need answered.  I have 1 gallon of grape juice with 20% sugar content. I have another gallon of grape juice that has 68% sugar content. How much of the 68% sugar juice do I add to the 20% sugar juice to get a Juice that has 24% sugar content. Please help if you can.\r<BR>\n" );
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document.write( "Thanks,\r<BR>\n" );
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document.write( "Jim </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #164734 by </B><A HREF=/tutors/aboutme.mpl?userid=ptaylor><B>ptaylor(2198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ptaylor><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=ptaylor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=164734\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> OK.\r<BR>\n" );
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document.write( "Let x=amount of 68% juice that is needed\r<BR>\n" );
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document.write( "Now we know that the amount of pure sugar in the 20% juice (0.20*1) plus the amount of pure sugar in the 68% juice that is added(0.68x) has to equal the amount of pure sugar in the final mixture (0.24(1+x)), so our equation to solve is:\r<BR>\n" );
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document.write( "0.20*1+0.68x=0.24(1+x)  get rid of parens<BR>\n" );
document.write( "0.20*1+0.68x=0.24+0.24x or<BR>\n" );
document.write( "0.20+0.68x=0.24+0.24x  subtract 0.24x and also 0.20 from each side of the equation<BR>\n" );
document.write( "0.20-0.20+0.68x-0.24x=0.24-0.20 combine like terms\r<BR>\n" );
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document.write( "0.44x=0.04  divide each side by 0.44<BR>\n" );
document.write( "x=0.091 gal-------------amount of 68% juice that is needed\r<BR>\n" );
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document.write( "CK<BR>\n" );
document.write( "0.20*1 +0.091*0.68=(1.091)*0.24<BR>\n" );
document.write( "0.20+0.062=0.262<BR>\n" );
document.write( "0.262=0.262\r<BR>\n" );
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document.write( "Hope this helps---ptaylor </SPAN>\n" );
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