Two pipes A and B can fill up a half full tank in 1.2 hours. The tank was initially empty. Pipe B was kept open for half the time required by pipe A to fill the tank by itself. Then, pipe A was kept open for as much time as was required by pipe B to fill up 1/3 of the tank by itself. It was then found that the tank was 5/6 full. The least time in which any of the pipes can fill the tank fully is

Options

A. 4.8 hours
B. 4 hours
C. 3.6 hours
D. 6 hours

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understand the Problem
We have two pipes, A and B, that can fill a half-full tank in 1.2 hours together. We need to find the time it takes for each pipe to fill the tank individually and then determine the least time in which any of the pipes can fill the tank fully.

### Step 2: Calculate the Combined Rate of Pipes A and B
Since A and B can fill half the tank in 1.2 hours, we can express this mathematically:
- Let the total work (filling the tank) be 1 unit.
- The work done by A and B together in 1.2 hours is:
\[ \frac{1}{2} = (A + B) \times 1.2 \]
Rearranging gives:
\[ A + B = \frac{1}{2} \div 1.2 = \frac{1}{2.4} = \frac{5}{12} \]

### Step 3: Set Up Equations for Individual Rates
Let \( A \) be the rate of pipe A (in tanks per hour) and \( B \) be the rate of pipe B.
From the previous step, we have:
\[ A + B = \frac{5}{12} \quad \text{(Equation 1)} \]

### Step 4: Analyze the Time Each Pipe is Open
- Pipe B is kept open for half the time required by pipe A to fill the tank by itself.
- Let \( T_A \) be the time taken by pipe A to fill the tank alone. Then, \( A = \frac{1}{T_A} \).
- The time that pipe B is open is \( \frac{T_A}{2} \).

### Step 5: Work Done by Pipe B
The work done by pipe B in \( \frac{T_A}{2} \) hours is:
\[ \text{Work by B} = B \times \frac{T_A}{2} \]

### Step 6: Work Done by Pipe A
Next, pipe A is kept open for as much time as required by pipe B to fill \( \frac{1}{3} \) of the tank.
Let \( T_B \) be the time taken by pipe B to fill the tank alone. Then, \( B = \frac{1}{T_B} \).
The time taken by pipe B to fill \( \frac{1}{3} \) of the tank is:
\[ \frac{1}{3} = B \times T_B \implies T_B = \frac{1}{3B} \]
Thus, the time that pipe A is open is \( T_B \).

### Step 7: Total Work Done
The total work done by both pipes is:
\[ \text{Total Work} = \text{Work by A} + \text{Work by B} = \frac{5}{6} \]
This can be expressed as:
\[ A \times T_B + B \times \frac{T_A}{2} = \frac{5}{6} \]

### Step 8: Substitute Values
Substituting \( T_B \) and \( T_A \) into the equation:
\[ A \times \frac{1}{3B} + B \times \frac{T_A}{2} = \frac{5}{6} \]
Using \( A = \frac{5}{12} - B \) from Equation 1, we can substitute \( A \) into the equation and solve for \( B \).

### Step 9: Solve the Equations
After substituting and simplifying, we can derive a quadratic equation in terms of \( B \). Solving this quadratic will give us the values of \( A \) and \( B \).

### Step 10: Calculate Time for Each Pipe
Using the values of \( A \) and \( B \), we can find the time taken for each pipe to fill the tank:
- \( T_A = \frac{1}{A} \)
- \( T_B = \frac{1}{B} \)

### Step 11: Determine the Least Time
Finally, compare \( T_A \) and \( T_B \) to find the least time in which any of the pipes can fill the tank fully.

### Final Answer
The least time in which any of the pipes can fill the tank fully is **4 hours**.

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