document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1034412>Question 1034412</A>: <I><SPAN STYLE=\"word-break: break-all;\"> solve. Round to the nearest tenth of necessary\r<BR>\n" );
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document.write( "Mr. Endicott, Mr. Bookout, and Mr. Piatt are going to paint a house together. Mr. Endicott can paint one side of the house in 4 hrs. to paint an equal area, Mr. Bookout takes 3 hrs and Mr. Piatt takes 2 hrs. If the men work together, how long will it take them to paint one side of the house? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #649050 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=649050\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> e = the rate that mr. endicott can paint the side of a house.<BR>\n" );
document.write( "b = the rate that mr. bookout can paint the side of a house.<BR>\n" );
document.write( "p = the rate that mr. piatt can paint the side of a house.\r<BR>\n" );
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document.write( "e = 1/4<BR>\n" );
document.write( "b = 1/3<BR>\n" );
document.write( "p = 1/2\r<BR>\n" );
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document.write( "this is derived from the general formula of r * t = q<BR>\n" );
document.write( "r is the rate of work.<BR>\n" );
document.write( "t is the time.<BR>\n" );
document.write( "q is the quantity of work.\r<BR>\n" );
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document.write( "q = 1 side of a house.<BR>\n" );
document.write( "t = number of hours to paint the side of the house.<BR>\n" );
document.write( "r is the rate at which the side of the house is painted.\r<BR>\n" );
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document.write( "for mr. endicott, the formula becomes r * 4 = 1<BR>\n" );
document.write( "solve for r to get 1/4.\r<BR>\n" );
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document.write( "do the same for mr. bookman and mr. piatt and you get the rates shown above.\r<BR>\n" );
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document.write( "when they work together, their rates are additive.\r<BR>\n" );
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document.write( "you get (r1 + r2 + r2) * t = q<BR>\n" );
document.write( "q = 1<BR>\n" );
document.write( "r1 = e = 1/4<BR>\n" );
document.write( "r2 = b = 1/3<BR>\n" );
document.write( "r3 = p = 1/2\r<BR>\n" );
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document.write( "you get (1/4 + 1/3 + 1/2) * t = 1\r<BR>\n" );
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document.write( "least common denominator looks to be 12.\r<BR>\n" );
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document.write( "you get (3/12 + 4/12 + 6/12) * t = 1\r<BR>\n" );
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document.write( "combine like terms to get 13/12 * t = 1\r<BR>\n" );
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document.write( "divide both sides of the equation by 13/12 to get t = 1 / (13/12)\r<BR>\n" );
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document.write( "this is the same as t = 1 * (12/13) = 12/13\r<BR>\n" );
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document.write( "it would take them 12/13 of an hour to paint the side of the house if they all work together.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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