Pipe A can fill a tank in 4 hours and pipe B can fill it in 6 hours. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full ?

Options

A. `4(1)/(2)`
B. `3(1)/(2)`
C. `3(1)/(4)`
D. `4(2)/(3)`

Answer

To solve the problem step by step, we will analyze the filling rates of both pipes and how they work together when opened alternately.

### Step 1: Determine the filling rates of the pipes
- **Pipe A** can fill the tank in 4 hours. Therefore, in 1 hour, Pipe A fills:
\[ \text{Rate of Pipe A} = \frac{1}{4} \text{ tank/hour} \]

- **Pipe B** can fill the tank in 6 hours. Therefore, in 1 hour, Pipe B fills:
\[ \text{Rate of Pipe B} = \frac{1}{6} \text{ tank/hour} \]

### Step 2: Calculate the amount filled in two hours (one cycle)
Since the pipes are opened alternately, in a 2-hour cycle:
- In the first hour, Pipe A fills:
\[ \text{Amount filled by A in 1 hour} = \frac{1}{4} \text{ tank} \]

- In the second hour, Pipe B fills:
\[ \text{Amount filled by B in 1 hour} = \frac{1}{6} \text{ tank} \]

- Therefore, in 2 hours, the total amount filled is:
\[ \text{Total amount filled in 2 hours} = \frac{1}{4} + \frac{1}{6} \]

### Step 3: Find a common denominator and add the fractions
To add the fractions, we find the least common multiple (LCM) of 4 and 6, which is 12:
- Convert \(\frac{1}{4}\) and \(\frac{1}{6}\) to have a common denominator:
\[ \frac{1}{4} = \frac{3}{12}, \quad \frac{1}{6} = \frac{2}{12} \]

- Now add them:
\[ \text{Total amount filled in 2 hours} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \text{ tank} \]

### Step 4: Determine how many cycles are needed to fill the tank
To fill the entire tank (1 tank), we need to find out how many cycles of 2 hours are required:
- In 2 hours, \(\frac{5}{12}\) of the tank is filled.
- To find out how many cycles are needed to fill 1 tank, we set up the equation:
\[ n \cdot \frac{5}{12} = 1 \implies n = \frac{12}{5} = 2.4 \text{ cycles} \]

### Step 5: Calculate the total time taken
Since each cycle takes 2 hours:
- The total time for 2 full cycles (which is 4 hours) fills:
\[ 2 \cdot \frac{5}{12} = \frac{10}{12} = \frac{5}{6} \text{ tank} \]

- After 4 hours, \(\frac{5}{6}\) of the tank is filled. We need to fill the remaining:
\[ 1 - \frac{5}{6} = \frac{1}{6} \text{ tank} \]

### Step 6: Determine the time taken to fill the remaining \(\frac{1}{6}\) tank
In the next hour (the 5th hour), Pipe A will fill \(\frac{1}{4}\) of the tank. We need to find out how much time it takes for Pipe A to fill \(\frac{1}{6}\) of the tank:
- Since Pipe A fills \(\frac{1}{4}\) tank in 1 hour, the time to fill \(\frac{1}{6}\) tank is:
\[ \text{Time} = \frac{\frac{1}{6}}{\frac{1}{4}} = \frac{1}{6} \cdot \frac{4}{1} = \frac{4}{6} = \frac{2}{3} \text{ hours} \]

### Step 7: Calculate the total time
The total time taken to fill the tank is:
- Time for 4 hours (2 cycles) + Time for \(\frac{2}{3}\) hours:
\[ \text{Total time} = 4 + \frac{2}{3} = \frac{12}{3} + \frac{2}{3} = \frac{14}{3} \text{ hours} = 4 \text{ hours and } 40 \text{ minutes} \]

### Final Answer
Thus, the tank will be full in **4 hours and 40 minutes**.