A can finish `(2)/(5)` of work done by B in `(1)/(4)` of time taken by B. B can finish `(4)/(3)` of work done by C in `(3)/(5)` of time taken by C. Working together they can finish the work in 60 days. In how many days they will finish the work, working alone?

Answer

To solve the problem step by step, we will analyze the work done by A, B, and C based on the information provided.

### Step 1: Establish the relationships between A, B, and C

1. **A's work in relation to B**:
- A can finish \( \frac{2}{5} \) of the work done by B in \( \frac{1}{4} \) of the time taken by B.
- Let's assume B takes \( t \) days to complete the work. Then A takes \( \frac{t}{4} \) days to do \( \frac{2}{5} \) of B's work.
- Therefore, in one day, A does:
\[ \text{Work done by A in one day} = \frac{2/5}{t/4} = \frac{2 \times 4}{5 \times t} = \frac{8}{5t} \]
- In one day, B does:
\[ \text{Work done by B in one day} = \frac{1}{t} \]

2. **Efficiency Ratio of A to B**:
- The efficiency ratio of A to B is:
\[ \text{Efficiency of A} : \text{Efficiency of B} = \frac{8}{5t} : \frac{1}{t} = 8 : 5 \]

### Step 2: Establish the relationship between B and C

1. **B's work in relation to C**:
- B can finish \( \frac{4}{3} \) of the work done by C in \( \frac{3}{5} \) of the time taken by C.
- Let C take \( s \) days to complete the work. Then B takes \( \frac{3s}{5} \) days to do \( \frac{4}{3} \) of C's work.
- Therefore, in one day, C does:
\[ \text{Work done by C in one day} = \frac{1}{s} \]
- In one day, B does:
\[ \text{Work done by B in one day} = \frac{4/3}{3s/5} = \frac{4 \times 5}{3 \times 3s} = \frac{20}{9s} \]

2. **Efficiency Ratio of B to C**:
- The efficiency ratio of B to C is:
\[ \text{Efficiency of B} : \text{Efficiency of C} = \frac{20}{9s} : \frac{1}{s} = 20 : 9 \]

### Step 3: Combine the ratios

1. **Combine A, B, and C's efficiencies**:
- We have:
- \( A : B = 8 : 5 \)
- \( B : C = 20 : 9 \)
- To combine these ratios, we need to make the B parts equal. We can multiply the first ratio by 4:
\[ A : B = 32 : 20 \]
- Now we can write:
\[ A : B : C = 32 : 20 : 9 \]

### Step 4: Calculate the total efficiency

1. **Total efficiency**:
- Total efficiency = \( 32 + 20 + 9 = 61 \)

### Step 5: Calculate total work done

1. **Work done together in 60 days**:
- If they can finish the work together in 60 days, then the total work is:
\[ \text{Total Work} = \text{Total Efficiency} \times \text{Time} = 61 \times 60 = 3660 \text{ units} \]

### Step 6: Calculate individual work done by A, B, and C

1. **Work done by A in one day**:
- A's efficiency = 32 units/day
- Days taken by A to finish the work:
\[ \text{Days taken by A} = \frac{3660}{32} = 114.375 \text{ days} \]

2. **Work done by B in one day**:
- B's efficiency = 20 units/day
- Days taken by B to finish the work:
\[ \text{Days taken by B} = \frac{3660}{20} = 183 \text{ days} \]

3. **Work done by C in one day**:
- C's efficiency = 9 units/day
- Days taken by C to finish the work:
\[ \text{Days taken by C} = \frac{3660}{9} = 406.67 \text{ days} \]

### Final Answer
- A can finish the work alone in **114.375 days**.
- B can finish the work alone in **183 days**.
- C can finish the work alone in **406.67 days**.