. B can finish a job in 40 days if he works alone. A is twice as fast as B and thrice as fast as C in the same job. Suppose A and B work together on the first day, B and C work together on the second day, C and A work together on the third day, and then, they continue the work by repeating this same three-day roster. What is the total number of days A would have worked when the job gets finished?

Using the efficiency method, let B's rate be 1/40 of the job per day. A is twice as fast as B, so A's rate is 1/20, and A is thrice as fast as C, making C's rate 1/60. In a 3-day cycle, A and B work 1/20 + 1/40 = 3/40, B and C work 1/40 + 1/60 = 5/120, and C and A work 1/60 + 1/20 = 4/60, totaling exactly 11/60 of the job per 3 days. After 5 full cycles (15 days), they complete 55/60 of the job, leaving 5/60. On day 16, A and B work together and complete 3/40 (which is 4.5/60), finishing the job with a surplus. Since A works on the first and third day of every cycle, A works 2 days per cycle, meaning A worked 10 days during the 5 cycles plus 1 day on day 16, totaling 11 days.

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