C is `20%` less efficient than A. A and B together can finish a piece of work in 16 days. B and C together can do it in `18 6/13` days. In how many days can A alone finish the same piece of work?

Options

A. 24
B. 25
C. 20
D. 30

Answer

To solve the problem, we need to determine how many days A alone can finish the work, given the efficiencies of A, B, and C.

### Step-by-Step Solution:

1. **Understanding Efficiency**:
- Let the efficiency of A be \( E_A = 100 \) (arbitrary units).
- Since C is 20% less efficient than A, the efficiency of C will be:
\[ E_C = E_A - 0.2 \times E_A = 100 - 20 = 80 \]

2. **Finding Efficiency of B**:
- A and B together can finish the work in 16 days. Therefore, their combined efficiency is:
\[ E_{A+B} = \frac{1}{16} \text{ (work done per day)} \]
- Thus, the equation for their efficiencies is:
\[ E_A + E_B = \frac{1}{16} \]
- Substituting \( E_A = 100 \):
\[ 100 + E_B = \frac{1}{16} \]

3. **Finding B's Efficiency**:
- Rearranging gives:
\[ E_B = \frac{1}{16} - 100 \]

4. **Finding Efficiency of B and C**:
- B and C together can finish the work in \( 18 \frac{6}{13} \) days, which is equal to \( \frac{240}{13} \) days. Therefore, their combined efficiency is:
\[ E_{B+C} = \frac{1}{\frac{240}{13}} = \frac{13}{240} \]
- Thus, the equation for their efficiencies is:
\[ E_B + E_C = \frac{13}{240} \]

5. **Substituting C's Efficiency**:
- We know \( E_C = 80 \), so:
\[ E_B + 80 = \frac{13}{240} \]

6. **Solving for B's Efficiency**:
- Rearranging gives:
\[ E_B = \frac{13}{240} - 80 \]

7. **Finding A's Work Alone**:
- Now, we have the efficiencies of A, B, and C. To find how many days A can finish the work alone, we use:
\[ \text{Days taken by A} = \frac{1}{E_A} \]
- Since \( E_A = 100 \):
\[ \text{Days taken by A} = \frac{1}{100} \]

### Final Calculation:
To find the total days A can finish the work alone:
\[ \text{Days taken by A} = 100 \text{ days} \]

### Conclusion:
A alone can finish the work in **100 days**.