Question

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

Answer 1

The correct answer is 11
Bob's productivity should be 3 units each day. Alex will therefore be 6 units/day efficient, while Cole will be 2 units/day.
Bob can complete the task in 40 days, hence the total amount of work will be $40\times3=120 units$.
Because Alex and Bob worked on the first day, the total amount of work completed was 3+6=9.
Likewise, it will be 5 and 8 units on days 2 and 3, respectively.
So, the total amount of work completed throughout the first three days was 9+5+8=22 units.
110 pieces were completed in the first 15 days of work, or $22 \times 5$.
As a result, the project will be completed on day 17 (since the remaining effort is larger than 9+5=14 units).
Since Alex works two out of every three days, he will be employed for ten of the first fifteen days.
He will then work on the sixteenth day as well. There have been 11 days in all.