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Question 678628: A,B and C working together can do a certain task in 6 hours. A & B can do the task in 9 hours, and B & C can do the task in 12 hours. How long would it take each working alone to do the task?
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Let x=amount of time needed for A do do the task working alone
Then A works at the rate of 1/x of the task per hour
Let y=amount of time needed for B to do the task working alone
Then B works at the rate of 1/y of the task per hour
Let z=amount of time needed for C to do the task working alone
Then C works at the rate of 1/z of the task per hour
A,B,C working together works at the rate of (1/x)+(1/y)+(1/z) of the task per hour and this equals (1/6)
(1/x)+(1/y)+(1/z)=(1/6)------------------eq 1
(1/x)+(1/y)=(1/9)------------------------eq2
(1/y)+(1/z)=(1/12)-------------------------eq3
Substitute (1/9) from eq2 for ((1/x)+(1/y)) in eq1
(1/9)+(1/z)=(1/6)
(1/z)=(3/18)-(2/18)=(1/18); z=18 hr----time needed for C working alone
Substitute (1/18) for (1/z) in eq3
(1/y)+(1/18)=(1/12)
(1/y)=(3/36)-(2/36)=(1/36); y=36 hr-----time needed for B working alone
Substitute (1/36) for (1/y) in eq2
(1/x)+(1/36)=(1/9)
(1/x)=(4/36)-(1/36)=(3/36)=(1/12); x=12 hrs---time needed for A working alone
CK
(1/12)+(1/36)+(1/18)=(1/6)=(6/36)
(3/36)+(1/36)+(2/36)=(6/36)
(6/36)=(6/36)
Hope this helps---ptaylor
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