To solve this, we first find the daily work rate of Sam, Mohit, and Ayana and then track their work over the described days.
1. Sam finishes the job in 20 days. His daily work rate is \( \frac{1}{20} \).
2. Mohit is twice as fast as Sam, meaning he finishes the job in \( \frac{20}{2} = 10 \) days. His daily work rate is \( \frac{1}{10} \).
3. Mohit is thrice as fast as Ayana. This means Ayana takes \( 10 \times 3 = 30 \) days to finish the job, and her daily work rate is \( \frac{1}{30} \).
Next, we calculate the work done on each day of a 3-day cycle:
- Day 1: Sam and Mohit work together. Their combined rate is \( \frac{1}{20} + \frac{1}{10} = \frac{1}{20} + \frac{2}{20} = \frac{3}{20} \).
- Day 2: Sam and Ayana work together. Their combined rate is \( \frac{1}{20} + \frac{1}{30} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12} \).
- Day 3: Mohit and Ayana work together. Their combined rate is \( \frac{1}{10} + \frac{1}{30} = \frac{3}{30} + \frac{1}{30} = \frac{4}{30} = \frac{2}{15} \).
The total work completed in one 3-day cycle is:
\( \frac{3}{20} + \frac{1}{12} + \frac{2}{15} \)
To add these fractions, we find the least common multiple of 20, 12, and 15, which is 60:
\(\frac{3}{20} = \frac{9}{60}, \frac{1}{12} = \frac{5}{60}, \frac{2}{15} = \frac{8}{60}\)
Total work in 3 days: \( \frac{9}{60} + \frac{5}{60} + \frac{8}{60} = \frac{22}{60} = \frac{11}{30} \)
Since the entire task is 1 unit of work:
The number of cycles needed is \( \frac{1}{\frac{11}{30}} = \frac{30}{11} \), which is approximately \( 2.727 \) cycles.
This means 2 full cycles are completed, followed by a portion of the third cycle. Let's calculate Sam's total work.
In each full cycle, Sam works on two days (one with Mohit, one with Ayana).
Sam's work per cycle: \( \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \).
Sam's total work in 2 full cycles: \( 2 \times \frac{1}{10} = \frac{1}{5} \).
For the partial third cycle, Sam works on Day 1 with Mohit, contributing \( \frac{1}{20} \) of the day's work.
In the partial cycle, the work done on Day 1 is \( \frac{9}{60} \). Sam's portion of this is \( \frac{1}{20} \).
Sam's total contribution is his work in the 2 full cycles plus his work in the partial third cycle. The fraction of work completed by Sam is approximately \( \frac{1}{5} + \frac{1}{20} \times \frac{30}{11} \), which simplifies to:
\(= \frac{1}{5} + \frac{9}{110} = \frac{22}{110} + \frac{9}{110} = \frac{31}{110} \approx \frac{3}{10} \).
Therefore, Sam completes \( \frac{3}{10} \) of the total work.