Tap A can fill a tank in 10 hours and tap B can fill same tank in 12 hours .Tap C can empty the same full tank in 6 hours .If all the taps are opened together,then what what portion of the tank will be filled after 6 hours ?

Options

A. A)`1//5`
B. B)`1//15`
C. C)`1//10`
D. D)`1//18`

Answer

To solve the problem step by step, we will first determine the filling and emptying rates of the taps and then calculate the portion of the tank filled after 6 hours when all taps are opened together.

### Step 1: Determine the rates of each tap
- **Tap A** can fill the tank in 10 hours. Therefore, its rate of filling is:
\[ \text{Rate of A} = \frac{1}{10} \text{ tank per hour} \]

- **Tap B** can fill the tank in 12 hours. Therefore, its rate of filling is:
\[ \text{Rate of B} = \frac{1}{12} \text{ tank per hour} \]

- **Tap C** can empty the tank in 6 hours. Therefore, its rate of emptying is:
\[ \text{Rate of C} = -\frac{1}{6} \text{ tank per hour} \]
(The negative sign indicates that it is emptying the tank.)

### Step 2: Calculate the combined rate of all taps
Now, we will find the combined rate of all three taps when they are opened together:
\[ \text{Combined Rate} = \text{Rate of A} + \text{Rate of B} + \text{Rate of C} \]
Substituting the rates we calculated:
\[ \text{Combined Rate} = \frac{1}{10} + \frac{1}{12} - \frac{1}{6} \]

### Step 3: Find a common denominator
The least common multiple (LCM) of 10, 12, and 6 is 60. We will convert each rate to have a denominator of 60:
\[ \frac{1}{10} = \frac{6}{60}, \quad \frac{1}{12} = \frac{5}{60}, \quad \frac{1}{6} = \frac{10}{60} \]
Now substituting these values:
\[ \text{Combined Rate} = \frac{6}{60} + \frac{5}{60} - \frac{10}{60} = \frac{6 + 5 - 10}{60} = \frac{1}{60} \text{ tank per hour} \]

### Step 4: Calculate the portion filled in 6 hours
If the combined rate is \(\frac{1}{60}\) tank per hour, then in 6 hours the amount filled will be:
\[ \text{Amount filled in 6 hours} = \text{Combined Rate} \times \text{Time} = \frac{1}{60} \times 6 = \frac{6}{60} = \frac{1}{10} \text{ of the tank} \]

### Conclusion
Thus, the portion of the tank that will be filled after 6 hours is:
\[ \frac{1}{10} \]