document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=931706>Question 931706</A>: <I><SPAN STYLE=\"word-break: break-all;\"> It takes an older pump <BR>\n" );
document.write( "5<BR>\n" );
document.write( " times as long to drain a certain pool as it does a newer pump. Working together, it takes the two pumps <BR>\n" );
document.write( "3<BR>\n" );
document.write( " hours to drain the pool. How long will it take the newer pump to drain the pool working alone? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #565753 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=565753\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let t=time it takes newer pump to drain the pool<BR>\n" );
document.write( "Newer pump drains at the rate of 1/t of the pool per hour<BR>\n" );
document.write( "Older pump drains the pool in 5t hours<BR>\n" );
document.write( "Older pump drains at the rate of 1/5t of the pool per hour<BR>\n" );
document.write( "Our equation to solve is:<BR>\n" );
document.write( "1/t + 1/5t =1/3  multiply each term by 15t<BR>\n" );
document.write( "15+3=5t<BR>\n" );
document.write( "5t=18<BR>\n" );
document.write( "t=18/5=3.6  hours---time it takes newer pump to drain the pool\r<BR>\n" );
document.write( "\n" );
document.write( "CK<BR>\n" );
document.write( "1/3.6 + 1/18=1/3<BR>\n" );
document.write( "5/18+1/18=1/3<BR>\n" );
document.write( "1/3=1/3<BR>\n" );
document.write( "Hope this helps---ptaylor </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );