Bob can finish a job in 40 days, if he works alone. Alex is twice as fast as Bob and thrice as fast as Cole in the same job. Suppose Alex and Bob work together on the first day, Bob and Cole work together on the second day, Cole and Alex work together on the third day, and then, they continue the work by repeating this three-day roster, with Alex and Bob working together on the fourth day, and so on. Then, the total number of days Alex would have worked when the job gets finished, is
Started 1 month ago by Shashank in
Explanatory Answer
Since Alex is twice as fast as Bob and thrice as fast as Cole, let us assume that the work done by Alex in a day as 6x.
This means that the work done by Bob and Cole in a day is 3x and 2x respectively.
| Alex | Bob | Cole |
| 6x | 3x | 2x |
Since Bob can finish the job in 40 days working alone, the quantum of work to finish the job = 40 ×× 3x = 120x
The work done on the first day, the second day, and the third day is as follows.
| Day 1 (Alex & Bob) |
Day 2 (Bob & Cole) |
Day 3 (Cole & Alex) |
| 9x | 5x | 8x |
The total work done in one cycle of 3 days = 9x + 5x + 8x = 22x
120x = 22x(5) + 9x + 1x
This means, it takes 5 complete cycles of 3 days, a full Day 1 and Day 2 to finish the job.
In one cycle Alex works twice, on Day 1 and Day 3.
So the total number of days that Alex works = 2(5) = 1 = 11 days
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