A pipe can fill an empty tank in 5 h and there is an emptying pipe which can empty the full tank in 7 h. If both the pipes are kept open simultaneously, then in how many hours will the tank be filled?

Answer

To solve the problem of how long it will take to fill the tank when both the filling and emptying pipes are open simultaneously, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the rates of the pipes**:
- The filling pipe can fill the tank in 5 hours. Therefore, its rate of filling is:
\[ \text{Rate of filling pipe} = \frac{1 \text{ tank}}{5 \text{ hours}} = \frac{1}{5} \text{ tanks/hour} \]
- The emptying pipe can empty the tank in 7 hours. Therefore, its rate of emptying is:
\[ \text{Rate of emptying pipe} = \frac{1 \text{ tank}}{7 \text{ hours}} = \frac{1}{7} \text{ tanks/hour} \]

2. **Combine the rates**:
- When both pipes are open, the effective rate of filling the tank is the rate of the filling pipe minus the rate of the emptying pipe:
\[ \text{Effective rate} = \frac{1}{5} - \frac{1}{7} \]

3. **Find a common denominator**:
- The least common multiple (LCM) of 5 and 7 is 35. We can rewrite the rates with a common denominator:
\[ \frac{1}{5} = \frac{7}{35}, \quad \frac{1}{7} = \frac{5}{35} \]
- Now, substituting these into the effective rate:
\[ \text{Effective rate} = \frac{7}{35} - \frac{5}{35} = \frac{2}{35} \text{ tanks/hour} \]

4. **Calculate the time to fill the tank**:
- To find the time taken to fill 1 tank at the effective rate, we take the reciprocal of the effective rate:
\[ \text{Time} = \frac{1 \text{ tank}}{\frac{2}{35} \text{ tanks/hour}} = \frac{35}{2} \text{ hours} = 17.5 \text{ hours} \]

### Final Answer:
The tank will be filled in **17.5 hours** when both pipes are open simultaneously.