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

Question

Accepted Answer

1. **Efficiency of Workers:**  
   - Assume Alex's daily work efficiency is $$6x$$.  
   - Since Alex is twice as fast as Bob, Bob's efficiency is $$3x$$.  
   - Since Alex is thrice as fast as Cole, Cole's efficiency is $$2x$$.

2. **Total Work to be Done:**  
   - Bob can finish the job alone in 40 days. Hence, total work = $$40 	imes 3x = 120x$$.

3. **Work Done in Each Day of the Cycle:**  
   - **Day 1 (Alex & Bob):** Work done = $$6x + 3x = 9x$$.  
   - **Day 2 (Bob & Cole):** Work done = $$3x + 2x = 5x$$.  
   - **Day 3 (Cole & Alex):** Work done = $$2x + 6x = 8x$$.  
   - Total work done in one cycle of three days = $$9x + 5x + 8x = 22x$$.

4. **Number of Complete Cycles Required:**  
   - Total work = $$120x$$.  
   - Number of complete cycles required = $$120x \div 22x = 5 	ext{ complete cycles}$$, with a remainder of $$10x$$ to be completed after these cycles.

5. **Distribution of Remaining Work (10x):**  
   - After completing five cycles, Day 1 (Alex & Bob) contributes $$9x$$, leaving $$1x$$ to be completed on Day 2 (Bob & Cole).

6. **Days Worked by Alex:**  
   - In one cycle, Alex works two days: Day 1 and Day 3. Therefore, in five cycles, Alex works $$5 	imes 2 = 10 	ext{ days}$$.  
   - Additionally, Alex works on Day 1 of the sixth partial cycle. Thus, total days worked by Alex = $$10 + 1 = 11 	ext{ days}$$.

Hence, the total number of days Alex worked is **11 days**.