A pipe can fill a tank in 4 hours and another pipe can empty the same tank in 6 hours. When both the pipes are opened simultaneously, the part of tank filled in 1 hour is …………..

Answer

To solve the problem, we need to determine how much of the tank is filled when both pipes are opened simultaneously. Let's break it down step by step.

### Step 1: Determine the filling rate of the first pipe.
The first pipe can fill the tank in 4 hours. Therefore, the part of the tank filled by the first pipe in 1 hour is:
\[ \text{Filling rate of Pipe A} = \frac{1}{4} \text{ of the tank per hour} \]

### Step 2: Determine the emptying rate of the second pipe.
The second pipe can empty the tank in 6 hours. Therefore, the part of the tank emptied by the second pipe in 1 hour is:
\[ \text{Emptying rate of Pipe B} = \frac{1}{6} \text{ of the tank per hour} \]

### Step 3: Calculate the net effect when both pipes are opened.
When both pipes are opened simultaneously, the filling and emptying rates will affect each other. The net part of the tank filled in 1 hour can be calculated as:
\[ \text{Net filling rate} = \text{Filling rate of Pipe A} - \text{Emptying rate of Pipe B} \]
Substituting the values we found:
\[ \text{Net filling rate} = \frac{1}{4} - \frac{1}{6} \]

### Step 4: Find a common denominator and perform the subtraction.
To subtract these fractions, we need a common denominator. The least common multiple of 4 and 6 is 12. We convert the fractions:
\[ \frac{1}{4} = \frac{3}{12} \quad \text{and} \quad \frac{1}{6} = \frac{2}{12} \]
Now, we can perform the subtraction:
\[ \text{Net filling rate} = \frac{3}{12} - \frac{2}{12} = \frac{1}{12} \]

### Step 5: Conclusion
Thus, when both pipes are opened simultaneously, the part of the tank filled in 1 hour is:
\[ \frac{1}{12} \]

### Final Answer
The part of the tank filled in 1 hour is \(\frac{1}{12}\).

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