document.write( "Question 1119209: A plumber works twice as fast as his apprentice. After the plumber has worked alone for 3 hours, his apprentice joins him and working together they complete the job 4 hours later. How many hours would it have taken the plumber to do the entire job by himself? \n" ); document.write( "

Algebra.Com's Answer #734689 by ikleyn(52798) $\"\"$ $\"About$

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\n" ); document.write( "The correct answer is 9 hours for the plumber to make the job working alone.\r
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document.write( "Let  r be the rate of of work of the apprentice.\r\n" );
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document.write( "Then the plumber's rate of work is 2r.\r\n" );
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document.write( "After working 3 hours alone, the plumber completed  3*(2r)  pars of the entire job.\r\n" );
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document.write( "When the apprentice joined him, they worked 4 more hours and made 4*(2r+r) = 12r  parts of the job.\r\n" );
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document.write( "For the entire job you have this equation\r\n" );
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document.write( "    3*(2r) + 12r = 1,   or\r\n" );
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document.write( "    18r  = 1.\r\n" );
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document.write( "It means  r = ,  i.e.  apprentice can make the job in 18 hours working alone.\r\n" );
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document.write( "Since the plumber works two times faster, he needs only 9 hours.\r\n" );
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