If A & B two pipes can fill a tank in 10 hr. when 'A' pipe can fill a tank in 6 hr. alone then in how much time will be taken to fill/empty the tank when pipes 'B' open alone ?

Answer

To solve the problem step by step, we need to determine how long it will take for pipe B to empty the tank alone, given the information about pipes A and B.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Pipe A can fill the tank alone in 6 hours.
- Pipes A and B together can fill the tank in 10 hours.
- We need to find out how long it will take for pipe B to empty the tank alone.

2. **Calculate the Work Done**:
- The total work to fill the tank can be represented in terms of "units". If we assume the total capacity of the tank is 30 units (the least common multiple of 10 and 6), we can use this for our calculations.

3. **Efficiency of Pipe A**:
- Since pipe A fills the tank in 6 hours, its efficiency is:
\[ \text{Efficiency of A} = \frac{30 \text{ units}}{6 \text{ hours}} = 5 \text{ units/hour} \]

4. **Efficiency of Pipes A and B Together**:
- Since A and B together fill the tank in 10 hours, their combined efficiency is:
\[ \text{Efficiency of A and B} = \frac{30 \text{ units}}{10 \text{ hours}} = 3 \text{ units/hour} \]

5. **Calculate the Efficiency of Pipe B**:
- To find the efficiency of pipe B, we can use the equation:
\[ \text{Efficiency of A} + \text{Efficiency of B} = \text{Efficiency of A and B} \]
Substituting the known values:
\[ 5 + \text{Efficiency of B} = 3 \]
Rearranging gives us:
\[ \text{Efficiency of B} = 3 - 5 = -2 \text{ units/hour} \]
- The negative sign indicates that pipe B is emptying the tank rather than filling it.

6. **Time Taken by Pipe B to Empty the Tank**:
- Since pipe B has an efficiency of 2 units/hour (as it is emptying), we can find the time taken to empty the entire tank:
\[ \text{Time} = \frac{\text{Total Work}}{\text{Efficiency of B}} = \frac{30 \text{ units}}{2 \text{ units/hour}} = 15 \text{ hours} \]

### Final Answer:
Pipe B will take **15 hours** to empty the tank alone.