document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=992764>Question 992764</A>: <I><SPAN STYLE=\"word-break: break-all;\"> X can do a certain job in 10 hours and Y can do it in 12 hours. After X and Y have been working together for 1 hour, Z joins them, and the three completed the job in 3 more hours, How long will it take Z to do the job alone? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #612263 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=612263\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let t = time it takes z to do the job working alone.  Then z works at the rate of 1/t of the job per hour<BR>\n" );
document.write( "x works at the rate of 1/10 of the job per hr<BR>\n" );
document.write( "y works at the rate of 1/12 of the job per hour<BR>\n" );
document.write( "Together x and y works at the rate of 1/10 + 1/12 =12/120 + 10/120=22/120=11/60 of the job per hour<BR>\n" );
document.write( "After 1 hour, together they complete (11/60)*1=11/60 of the job leaving 49/60 of the job yet to be done<BR>\n" );
document.write( "Soooo, (11/60)*3 + (1/t)*3= 49/60, or<BR>\n" );
document.write( "33/60 + 3/t=49/60. multiply each term by 60t<BR>\n" );
document.write( "33t + 180=49t<BR>\n" );
document.write( "16t=180<BR>\n" );
document.write( "t=11.25 hrs<BR>\n" );
document.write( "CK 33/60 + 3/11.25 =49/60<BR>\n" );
document.write( "99/180 + 48/180 = 147/180<BR>\n" );
document.write( "147/180=174/180<BR>\n" );
document.write( "Hope this helps ----ptaylor </SPAN>\n" );
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