Question

Sam can complete a job in 20 days when working alone. Mohit is twice as fast as Sam and thrice as fast as Ayana in the same job. They undertake a job with an arrangement where Sam and Mohit work together on the first day, Sam and Ayana on the second day, Mohit and Ayana on the third day, and this three-day pattern is repeated till the work gets completed. Then, the fraction of total work done by Sam is

• A. \(\frac{3}{20}\)
• B. \(\frac{3}{10}\)
• C. \(\frac{1}{5}\)
• D. \(\frac{1}{20}\)

Answer 1

The Correct Option is B

To solve the problem, we start by determining the work rates of Sam, Mohit, and Ayana and follow the sequence of work distribution over the days mentioned:

1. Sam completes the entire job in 20 days, so his work rate is \( \frac{1}{20} \) per day.

2. Mohit is twice as fast as Sam, meaning Mohit completes the job in \( \frac{20}{2} = 10 \) days. Therefore, Mohit's work rate is \( \frac{1}{10} \) per day.

3. Since Mohit is thrice as fast as Ayana, Ayana's time to complete the job is \( 10 \times 3 = 30 \) days, giving Ayana a work rate of \( \frac{1}{30} \) per day.

Next, we calculate the amount of work done each day for the three-day cycle:

• Day 1: Sam and Mohit work together: \( \frac{1}{20} + \frac{1}{10} = \frac{1}{20} + \frac{2}{20} = \frac{3}{20} \)
• Day 2: Sam and Ayana work together: \( \frac{1}{20} + \frac{1}{30} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12} \)
• Day 3: Mohit and Ayana work together: \( \frac{1}{10} + \frac{1}{30} = \frac{3}{30} + \frac{1}{30} = \frac{4}{30} = \frac{2}{15} \)

The total work completed in a 3-day cycle is:

\( \frac{3}{20} + \frac{1}{12} + \frac{2}{15} \)

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: \( \frac{9}{60} + \frac{5}{60} + \frac{8}{60} = \frac{22}{60} = \frac{11}{30} \)

Given that the total task equals one whole work unit:

Number of cycles required: \( \frac{1}{\frac{11}{30}} = \frac{30}{11} \approx 2.727 \)

The cycles cover \( 2 \) complete cycles plus a fraction of the third cycle. Let's calculate Sam's contribution:

In each cycle, Sam works \( 1 \) day with Mohit and \( 1 \) day with Ayana:

Sam's work per cycle: \( \frac{1}{20} \times 2 = \frac{2}{20} = \frac{1}{10} \)

Total work by Sam in 2 cycles: \( 2 \times \frac{1}{10} = \frac{1}{5} \)

In the partial cycle contribution, consider Sam's daily work with Mohit, completing \( \frac{9}{60} \) for day 1 of cycle 3:

Sam's individual contribution for Day 1 of cycle 3: \( \frac{1}{20} = \frac{3}{60} \)

Hence, Sam's total contribution: \( \frac{1}{5} + \left(\frac{3}{60} \approx \frac{1}{20}\times\frac{30}{11}\right)\)

\(= \frac{1}{5} + \frac{9}{110} = \frac{22}{110} + \frac{9}{110} = \frac{31}{110} \approx \frac{3}{10} \)

Thus, the fraction of the total work done by Sam is \(\frac{3}{10}\).

Answer 2

Let's first calculate the rates at which Sam, Mohit, and Ayna work.
 Sam's rate = $\frac{1}{20}$ (jobs per day)

 Mohit is twice as fast as Sam, so Mohit's rate of work is:
   Mohit's rate = $2 \times \frac{1}{20} = \frac{1}{10}$
   Ayna is thrice as slow as Mohit, so Ayna's rate of work is:
   Ayna's rate = $\frac{1}{3} \times \frac{1}{10} = \frac{1}{30}$

Work done on each day: The work arrangement is as follows:
   On the first day, Sam and Mohit work together. The total work done on the first day is:
   Work on day 1 = $\frac{1}{20} + \frac{1}{10} = \frac{1+2}{20} = \frac{3}{20}$

On the second day, Sam and Ayna work together. The total work done on the second day is:
   Work on day 2 = $\frac{1}{20} + \frac{1}{30} = \frac{3+2}{60} = \frac{5}{60} = \frac{1}{12}$

On the third day, Mohit and Ayna work together. The total work done on the third day is:
   Work on day 3 = $\frac{1}{10} + \frac{1}{30} = \frac{3+1}{30} = \frac{4}{30} = \frac{2}{15}$

Total work done in 3 days: The total work done in one complete cycle (3 days) is:
\[ \text{Total work in 3 days} = \frac{3}{20} + \frac{1}{12} + \frac{2}{15} \]

To add these fractions, we need to find the least common denominator (LCD). The LCD of 20, 12, and 15 is 60.
\[ \frac{3}{20} = \frac{9}{60}, \quad \frac{1}{12} = \frac{5}{60}, \quad \frac{2}{15} = \frac{8}{60} \]

Thus, the total work done in one cycle is:
\[ \frac{9}{60} + \frac{5}{60} + \frac{8}{60} = \frac{22}{60} = \frac{11}{30} \]

So, in every 3-day period, $\frac{11}{30}$ of the total work is completed.

Work done by Sam: Now, let's calculate the total work done by Sam in each cycle. Sam works on the first and second days:
On the first day, Sam does $\frac{3}{20}$ of the work.
On the second day, Sam does $\frac{5}{60} = \frac{1}{12}$ of the work.

Thus, the total work done by Sam in one cycle is:
\[ \frac{3}{20} + \frac{1}{12} = \frac{14}{60} = \frac{7}{30} \]

Sam's total work in 3 days = $\frac{3}{20} + \frac{1}{12} = \frac{4}{15} + \frac{1}{12} = \frac{11}{30}$

Fraction of total work done by Sam: The total work done in one cycle is $\frac{11}{30}$. Therefore, the fraction of the total work done by Sam in one cycle is:
\[ \frac{\frac{7}{30}}{\frac{11}{30}} = \frac{7}{11} \]
Thus, the fraction of total work done by Sam is $\frac{7}{11}$. Therefore, the correct answer is Option (2).