document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=851616>Question 851616</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A man undertakes to do a piece of work in 150 days. He employs 200 men. He finds that only a quarter of work done in 50 days. How many additional men must be employed so that the work may be finished on time?<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 #512899 by </B><A HREF=/tutors/aboutme.mpl?userid=richwmiller><B>richwmiller(17219)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richwmiller><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=richwmiller><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=512899\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 200*50=10000 man days for 1/4 of the job<BR>\n" );
document.write( "He needs 30000 man days to finish the job.<BR>\n" );
document.write( "He has 100 more days to finish the job<BR>\n" );
document.write( "30000/100=300 men<BR>\n" );
document.write( "He needs 100 more plus his original 200\r<BR>\n" );
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