To solve the problem step-by-step, we need to determine the value of \( x \) for the emptying pipe D, given the filling rates of pipes A, B, and C.
### Step 1: Determine the filling rates of pipes A, B, and C
- Pipe A can fill the tank in 10 hours, so its rate is \( \frac{1}{10} \) of the tank per hour.
- Pipe B can fill the tank in 15 hours, so its rate is \( \frac{1}{15} \) of the tank per hour.
- Pipe C can fill the tank in 30 hours, so its rate is \( \frac{1}{30} \) of the tank per hour.
### Step 2: Calculate the combined filling rate of A, B, and C
To find the combined filling rate of A, B, and C, we add their individual rates:
\[
\text{Combined rate} = \frac{1}{10} + \frac{1}{15} + \frac{1}{30}
\]
Finding a common denominator (which is 30):
\[
\frac{1}{10} = \frac{3}{30}, \quad \frac{1}{15} = \frac{2}{30}, \quad \frac{1}{30} = \frac{1}{30}
\]
So,
\[
\text{Combined rate} = \frac{3}{30} + \frac{2}{30} + \frac{1}{30} = \frac{6}{30} = \frac{1}{5}
\]
This means A, B, and C together can fill \( \frac{1}{5} \) of the tank in one hour.
### Step 3: Calculate the amount filled in 3 hours
In 3 hours, the amount filled by A, B, and C is:
\[
\text{Amount filled in 3 hours} = 3 \times \frac{1}{5} = \frac{3}{5}
\]
### Step 4: Determine the amount left to fill the tank
After 3 hours, the amount left to fill the tank is:
\[
\text{Amount left} = 1 - \frac{3}{5} = \frac{2}{5}
\]
### Step 5: Determine the emptying rate of pipe D
Pipe D can empty the tank in \( x \) hours, so its rate is \( -\frac{1}{x} \) of the tank per hour. We also know that pipe D can empty the tank in 30 hours, so its rate is:
\[
\text{Rate of D} = -\frac{1}{30}
\]
### Step 6: Set up the equation for the emptying process
When pipe D is opened, it will empty the remaining \( \frac{2}{5} \) of the tank. The equation for the time taken to empty the remaining tank is:
\[
\frac{2}{5} = \left(-\frac{1}{30}\right) \times t
\]
Where \( t \) is the time taken to empty the remaining \( \frac{2}{5} \) of the tank.
### Step 7: Solve for \( t \)
Rearranging the equation gives:
\[
t = \frac{2}{5} \times (-30) = -12
\]
This means it takes 12 hours for pipe D to empty the remaining \( \frac{2}{5} \) of the tank.
### Step 8: Relate the time to the emptying rate
Since we have established that the emptying rate of D is \( -\frac{1}{x} \), we can set up the equation:
\[
\frac{2}{5} = \frac{t}{x}
\]
Substituting \( t = 12 \) gives:
\[
\frac{2}{5} = \frac{12}{x}
\]
Cross-multiplying gives:
\[
2x = 60 \implies x = 30
\]
### Conclusion
Thus, the value of \( x \) is \( 30 \) hours.
