document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1062560>Question 1062560</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If 20&#8453; of A= B & 40&#8453; of B=C then 60&#8453; of (A+B)=? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #677484 by </B><A HREF=/tutors/aboutme.mpl?userid=math_helper><B>math_helper(2461)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_helper><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=math_helper><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=677484\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 0.2A = B  —> A = 5B      <BR>\n" );
document.write( "0.4B = C  —> B = 2.5C \r<BR>\n" );
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document.write( "I assume the problem wants the answer in terms of C:<BR>\n" );
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document.write( "(0.6)(A+B) = <BR>\n" );
document.write( "(0.6)(5B + 2.5C) =           (substituted for A and B)<BR>\n" );
document.write( "(0.6)(5(2.5C) + 2.5C) =    (substituted for B)<BR>\n" );
document.write( "(0.6)(15C) =<BR>\n" );
document.write( "9C\r<BR>\n" );
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document.write( " </SPAN>\n" );
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