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 step by step, we will determine the net effect of the two pipes when they are opened simultaneously.

### Step 1: Determine the filling rate of the first pipe (Pipe A)
Pipe A can fill the tank in 4 hours. Therefore, in one hour, it fills:
\[ \text{Filling rate of Pipe A} = \frac{1}{4} \text{ of the tank} \]

### Step 2: Determine the emptying rate of the second pipe (Pipe B)
Pipe B can empty the tank in 6 hours. Therefore, in one hour, it empties:
\[ \text{Emptying rate of Pipe B} = \frac{1}{6} \text{ of the tank} \]

### Step 3: Calculate the net effect when both pipes are opened
When both pipes are opened simultaneously, the net effect on the tank will be the filling from Pipe A minus the emptying from Pipe B:
\[ \text{Net effect} = \text{Filling rate of Pipe A} - \text{Emptying rate of Pipe B} \]
Substituting the values we found:
\[ \text{Net effect} = \frac{1}{4} - \frac{1}{6} \]

### Step 4: Find a common denominator
To subtract the fractions, we need a common denominator. The least common multiple (LCM) of 4 and 6 is 12. We convert both fractions:
\[ \frac{1}{4} = \frac{3}{12} \quad \text{and} \quad \frac{1}{6} = \frac{2}{12} \]
Now, substituting these values back into the equation:
\[ \text{Net effect} = \frac{3}{12} - \frac{2}{12} = \frac{1}{12} \]

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

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

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