C is twice efficient as A. B takes thrice as many days as C. A takes 12 days to finish the work alone. If they work in pairs (ie, AB, BC, CA) starting with AB on the first day then BC on the second day and AC on the third day and so on, then how many days are required to finish the work?

Options

A. `6 1/5` days
B. 4.5 days
C. `5 1/9` days
D. 8 days

Answer

To solve the problem step by step, let's break down the information given and calculate the required work completion time.

### Step 1: Determine the efficiencies of A, B, and C

1. **Efficiency of A**: A takes 12 days to complete the work alone. Therefore, A's efficiency is:
\[ \text{Efficiency of A} = \frac{1 \text{ work}}{12 \text{ days}} = \frac{1}{12} \text{ work/day} \]

2. **Efficiency of C**: C is twice as efficient as A. Therefore, C's efficiency is:
\[ \text{Efficiency of C} = 2 \times \text{Efficiency of A} = 2 \times \frac{1}{12} = \frac{1}{6} \text{ work/day} \]

3. **Efficiency of B**: B takes thrice as many days as C. Since C takes 6 days to complete the work alone, B takes:
\[ \text{Days taken by B} = 3 \times 6 = 18 \text{ days} \]
Thus, B's efficiency is:
\[ \text{Efficiency of B} = \frac{1 \text{ work}}{18 \text{ days}} = \frac{1}{18} \text{ work/day} \]

### Step 2: Calculate the total work

The total work can be calculated using A's efficiency:
\[ \text{Total Work} = \text{Efficiency of A} \times \text{Days taken by A} = \frac{1}{12} \times 12 = 1 \text{ work unit} \]

### Step 3: Calculate the work done in pairs

1. **Day 1 (A and B)**:
\[ \text{Work done} = \text{Efficiency of A} + \text{Efficiency of B} = \frac{1}{12} + \frac{1}{18} \]
To add these fractions, find a common denominator (36):
\[ = \frac{3}{36} + \frac{2}{36} = \frac{5}{36} \text{ work units} \]

2. **Day 2 (B and C)**:
\[ \text{Work done} = \text{Efficiency of B} + \text{Efficiency of C} = \frac{1}{18} + \frac{1}{6} \]
Common denominator (18):
\[ = \frac{1}{18} + \frac{3}{18} = \frac{4}{18} = \frac{2}{9} \text{ work units} \]

3. **Day 3 (C and A)**:
\[ \text{Work done} = \text{Efficiency of C} + \text{Efficiency of A} = \frac{1}{6} + \frac{1}{12} \]
Common denominator (12):
\[ = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \text{ work units} \]

### Step 4: Total work done in three days

Now, let's calculate the total work done in three days:
\[ \text{Total work in 3 days} = \frac{5}{36} + \frac{2}{9} + \frac{1}{4} \]
Finding a common denominator (36):
\[ = \frac{5}{36} + \frac{8}{36} + \frac{9}{36} = \frac{22}{36} = \frac{11}{18} \text{ work units} \]

### Step 5: Calculate remaining work and days needed

After 3 days, the remaining work is:
\[ \text{Remaining work} = 1 - \frac{11}{18} = \frac{7}{18} \text{ work units} \]

### Step 6: Work done in the next cycles

1. **Day 4 (A and B)**:
\[ \text{Work done} = \frac{5}{36} \]
Remaining work after Day 4:
\[ \frac{7}{18} - \frac{5}{36} = \frac{14}{36} - \frac{5}{36} = \frac{9}{36} = \frac{1}{4} \]

2. **Day 5 (B and C)**:
\[ \text{Work done} = \frac{2}{9} \]
Remaining work after Day 5:
\[ \frac{1}{4} - \frac{2}{9} = \frac{9}{36} - \frac{8}{36} = \frac{1}{36} \]

3. **Day 6 (C and A)**:
\[ \text{Work done} = \frac{1}{4} \]
On Day 6, they will complete the remaining work of \(\frac{1}{36}\).

### Final Calculation of Total Days

Total days taken:
- 5 full days + a fraction of the 6th day.

Thus, the total time taken to complete the work is:
\[ \text{Total Days} = 6 \text{ days} \]

### Final Answer
The total number of days required to finish the work is **6 days**.