document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=167380>Question 167380</A>: <I><SPAN STYLE=\"word-break: break-all;\">  A and B can do a piece of work in 42 days, B and C in 31 days, and A and C in 20 days. Working together, how many days can all of them finish the work? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #123283 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=123283\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A&B work at the rate of 1/42 job per day<BR>\n" );
document.write( "B&C work at the rate of 1/31 job per day<BR>\n" );
document.write( "A&C work at the rate of 1/20 job per day<BR>\n" );
document.write( "Let t=time it takes all workers, working together, to do the job<BR>\n" );
document.write( "Let 1/A=rate at which A works<BR>\n" );
document.write( "1/B=rate that B works<BR>\n" );
document.write( "1/C=rate that C works\r<BR>\n" );
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document.write( "Now we know that:<BR>\n" );
document.write( "1/A + 1/B=1/42-------------------eq1<BR>\n" );
document.write( "1/B + 1/C=1/31---------------------eq2<BR>\n" );
document.write( "1/A + 1/C=1/20---------------------eq3\r<BR>\n" );
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document.write( "subtract eq3 from eq1:<BR>\n" );
document.write( "1/B - 1/C=1/42 - 1/20------------------eq3a<BR>\n" );
document.write( "add eq3a and eq2:<BR>\n" );
document.write( "2/B=1/42 - 1/20 + 1/31 (13020 is LCM)<BR>\n" );
document.write( "2/B=(310-651+420)/13020=79/13020<BR>\n" );
document.write( "1/B=79/26040 job per day-------------------------rate at which B works\r<BR>\n" );
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document.write( "substitute the value for 1/B into eq1:<BR>\n" );
document.write( "1/A + 79/26040=1/42<BR>\n" );
document.write( "1/A=1/42 - 79/26040=(620-79)/26040=541/26040 job per day----rate at which A works<BR>\n" );
document.write( "substitute the value for 1/A into eq3:<BR>\n" );
document.write( "541/26040 + 1/C=1/20<BR>\n" );
document.write( "1/C = 1/20 - 541/26040=(1302-541)/26040=761/26040 job per day----rate at which C works\r<BR>\n" );
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document.write( "Together, A, B, & C work at the the rate of:<BR>\n" );
document.write( "541/26040 +79/26040 + 761/26040= 1381/26040 job per day<BR>\n" );
document.write( "Now, our final equation to solve is:<BR>\n" );
document.write( "(1381/26040)t=1 (1 job that is) multiply each side by 26040<BR>\n" );
document.write( "(rate at which they work * time it take to complete the job=1 job)\r<BR>\n" );
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document.write( "1381t=26040  divide by 1381<BR>\n" );
document.write( "t= 18.86 days----------------time it takes all the workers working together\r<BR>\n" );
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document.write( "CK<BR>\n" );
document.write( "A:  (541/26040)*18.86=0.3917 of the job<BR>\n" );
document.write( "B:  (79/26040)*18.86=0.0572 of the job<BR>\n" );
document.write( "C:  (761/26040)*18.86=0.5512 of the job<BR>\n" );
document.write( "0.3917 + 0.0572 + 0.5512=1.000069<BR>\n" );
document.write( "1~~~1\r<BR>\n" );
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document.write( "Hope this helps---sorry about the first effort----ptaylor </SPAN>\n" );
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