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C is twice efficient as A, B takes thrice as many days as C. A takes 12 days to finish the work alone. If they work in pairs (i.e., AB, BC, CA) starting with AB on the first day then BC on the second day and AC on the third and so on, then how many days are required to finish the work?
6days
5 1/9days
7 2/3days
8days
A takes 12 days. C=A/2=6days. B=3*6 =18days. (A+B)’s 1 day work =1/12+1/18 =5/36 (B+C)’s 1 day work =1/18 +1/6 =8/36 (C+A)’s 1 day work =1/12+1/6 =9/36 In 5 days they worked =(5+8+9+5+8)/36 =35/36 Remaining 1/36 work done by CA 1/36*36/9=1/9days. Total days=5 1/9.
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