Question 167380
A&B work at the rate of 1/42 job per day
B&C work at the rate of 1/31 job per day
A&C work at the rate of 1/20 job per day
Let t=time it takes all workers, working together, to do the job
Let 1/A=rate at which A works
1/B=rate that B works
1/C=rate that C works

Now we know that:
1/A + 1/B=1/42-------------------eq1
1/B + 1/C=1/31---------------------eq2
1/A + 1/C=1/20---------------------eq3

subtract eq3 from eq1:
1/B - 1/C=1/42 - 1/20------------------eq3a
add eq3a and eq2:
2/B=1/42 - 1/20 + 1/31 (13020 is LCM)
2/B=(310-651+420)/13020=79/13020
1/B=79/26040 job per day-------------------------rate at which B works

substitute the value for 1/B into eq1:
1/A + 79/26040=1/42
1/A=1/42 - 79/26040=(620-79)/26040=541/26040 job per day----rate at which A works
substitute the value for 1/A into eq3:
541/26040 + 1/C=1/20
1/C = 1/20 - 541/26040=(1302-541)/26040=761/26040 job per day----rate at which C works

Together, A, B, & C work at the the rate of:
541/26040 +79/26040 + 761/26040= 1381/26040 job per day
Now, our final equation to solve is:
(1381/26040)t=1 (1 job that is) multiply each side by 26040
(rate at which they work * time it take to complete the job=1 job)

1381t=26040  divide by 1381
t= 18.86 days----------------time it takes all the workers working together

CK
A:  (541/26040)*18.86=0.3917 of the job
B:  (79/26040)*18.86=0.0572 of the job
C:  (761/26040)*18.86=0.5512 of the job
0.3917 + 0.0572 + 0.5512=1.000069
1~~~1

Hope this helps---sorry about the first effort----ptaylor