document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1135946>Question 1135946</A>: <I><SPAN STYLE=\"word-break: break-all;\"> It takes an older pump twice as long to drain a certain pool as it does a newer pump. Working together, it takes the two pumps 2 hours to drain the pool. How long will it take the older 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 #753658 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=753658\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> let a = the rate of the older pump.<BR>\n" );
document.write( "let b = the rate of the newer pump.\r<BR>\n" );
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document.write( "general formula is rate * time = quantity of work\r<BR>\n" );
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document.write( "quantity of work = 1 drained pool.\r<BR>\n" );
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document.write( "older pump takes twice as long to drain the pool than the newer pump.\r<BR>\n" );
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document.write( "that means that the rate of the newer pump is 2 times the rate of the older pump.\r<BR>\n" );
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document.write( "that gets you b = 2a.\r<BR>\n" );
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document.write( "when they work together, their rates are aqditive.\r<BR>\n" );
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document.write( "combined rate = a + b<BR>\n" );
document.write( "time = 2<BR>\n" );
document.write( "quantity of work = 1\r<BR>\n" );
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document.write( "formula of rate * time = quantity becomes (a + b) * 2 = 1\r<BR>\n" );
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document.write( "since b = 2a, the formula becomes (a + 2a) * 2 = 1\r<BR>\n" );
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document.write( "combine like terms to get 3a * 2 = 1\r<BR>\n" );
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document.write( "simplify to get 6a = 1\r<BR>\n" );
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document.write( "divide both sides by 6 to get a = 1/6\r<BR>\n" );
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document.write( "that's the rate of the older pump.\r<BR>\n" );
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document.write( "since b = 2a, then b = 2/6.\r<BR>\n" );
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document.write( "that's the rate of the newer pump.\r<BR>\n" );
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document.write( "formula of (a + b) * 2 = 1 becomes (1/6 + 2/6) * 2 = 1 which becomes 3/6 * 2 = 1 which becomes 1 = 1 which confirms the solution is correct.\r<BR>\n" );
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document.write( "rate * time = quantity for the older pump becomes 1/6 * time = 1.\r<BR>\n" );
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document.write( "this results in time = 6.\r<BR>\n" );
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document.write( "rate * time = quantity for the newer pump becomes 2/6 * time = 1.\r<BR>\n" );
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document.write( "this results in time = 3.\r<BR>\n" );
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document.write( "your solution is that it will take the older pump 6 hours to drain the pool by itself.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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