Pipes A,B and C are attached to an empty cistern. The first two can fill the cistern in 7 and 10.5 hours respectively whereas the third can drain the cistern when completely filled, in 6 hours IF all the three pipes are turned on simultaneously when the cistern is half full how many hours will be needed to fill the cistern

Options

A. 8
B. 14
C. 7.5
D. 7

Answer

To solve the problem step by step, we need to analyze the rates at which the pipes fill or drain the cistern.

### Step 1: Determine the filling rates of pipes A and B
- Pipe A can fill the cistern in 7 hours. Therefore, its rate of filling is:
\[ \text{Rate of A} = \frac{1 \text{ cistern}}{7 \text{ hours}} = \frac{1}{7} \text{ cistern per hour} \]

- Pipe B can fill the cistern in 10.5 hours. Therefore, its rate of filling is:
\[ \text{Rate of B} = \frac{1 \text{ cistern}}{10.5 \text{ hours}} = \frac{1}{10.5} \text{ cistern per hour} = \frac{2}{21} \text{ cistern per hour} \]

### Step 2: Determine the draining rate of pipe C
- Pipe C can drain the cistern in 6 hours. Therefore, its rate of draining is:
\[ \text{Rate of C} = -\frac{1 \text{ cistern}}{6 \text{ hours}} = -\frac{1}{6} \text{ cistern per hour} \]

### Step 3: Calculate the net rate of filling when all pipes are opened
- The net rate when all three pipes are opened is given by the sum of the rates of A and B minus the rate of C:
\[ \text{Net Rate} = \text{Rate of A} + \text{Rate of B} + \text{Rate of C} \]
\[ \text{Net Rate} = \frac{1}{7} + \frac{2}{21} - \frac{1}{6} \]

### Step 4: Find a common denominator to add the rates
- The least common multiple (LCM) of 7, 21, and 6 is 42. Now we convert each rate to have a denominator of 42:
\[ \frac{1}{7} = \frac{6}{42}, \quad \frac{2}{21} = \frac{4}{42}, \quad -\frac{1}{6} = -\frac{7}{42} \]
\[ \text{Net Rate} = \frac{6}{42} + \frac{4}{42} - \frac{7}{42} = \frac{3}{42} = \frac{1}{14} \text{ cistern per hour} \]

### Step 5: Calculate the time to fill the remaining half of the cistern
- Since the cistern is half full, we need to fill only half of it. The time required to fill half of the cistern at the net rate of \(\frac{1}{14}\) cistern per hour is:
\[ \text{Time} = \frac{\text{Volume}}{\text{Rate}} = \frac{0.5 \text{ cistern}}{\frac{1}{14} \text{ cistern per hour}} = 0.5 \times 14 = 7 \text{ hours} \]

### Final Answer
The time needed to fill the cistern when it is half full is **7 hours**.