Pipes A and B together can fill an empty tank in `6 2/3` minutes. If A takes `3` minutes more than B to fill the tank, then the time (in minutes) in which A alone would fill one-third part of the tank is:

Options

A. `6`
B. `5`
C. `5.5`
D. `4.5`

Answer

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

### Step 1: Understand the problem
We know that pipes A and B together can fill a tank in \(6 \frac{2}{3}\) minutes, which can be converted to an improper fraction:
\[ 6 \frac{2}{3} = \frac{20}{3} \text{ minutes} \]

### Step 2: Calculate the combined rate of A and B
If A and B together fill the tank in \(\frac{20}{3}\) minutes, their combined rate of filling the tank is:
\[ \text{Rate of A + Rate of B} = \frac{1 \text{ tank}}{\frac{20}{3} \text{ minutes}} = \frac{3}{20} \text{ tanks per minute} \]

### Step 3: Define the time taken by each pipe
Let the time taken by pipe B to fill the tank be \(x\) minutes. Then, pipe A takes \(x + 3\) minutes to fill the tank.

### Step 4: Write the rates for A and B
The rates of A and B can be expressed as:
- Rate of B = \(\frac{1}{x}\) tanks per minute
- Rate of A = \(\frac{1}{x + 3}\) tanks per minute

### Step 5: Set up the equation
Since the combined rate of A and B is equal to their individual rates, we can write:
\[ \frac{1}{x} + \frac{1}{x + 3} = \frac{3}{20} \]

### Step 6: Solve the equation
To solve the equation, find a common denominator:
\[ \frac{x + 3 + x}{x(x + 3)} = \frac{3}{20} \]
This simplifies to:
\[ \frac{2x + 3}{x^2 + 3x} = \frac{3}{20} \]

Cross-multiplying gives:
\[ 20(2x + 3) = 3(x^2 + 3x) \]
Expanding both sides:
\[ 40x + 60 = 3x^2 + 9x \]
Rearranging the equation:
\[ 3x^2 - 31x - 60 = 0 \]

### Step 7: Factor the quadratic equation
To factor \(3x^2 - 31x - 60 = 0\), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \(a = 3\), \(b = -31\), and \(c = -60\):
\[ x = \frac{31 \pm \sqrt{(-31)^2 - 4 \cdot 3 \cdot (-60)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ x = \frac{31 \pm \sqrt{961 + 720}}{6} = \frac{31 \pm \sqrt{1681}}{6} = \frac{31 \pm 41}{6} \]
This gives:
\[ x = \frac{72}{6} = 12 \quad \text{or} \quad x = \frac{-10}{6} \text{ (not valid)} \]

### Step 8: Find the time for A
Now, since \(x = 12\), pipe B takes 12 minutes, and pipe A takes:
\[ x + 3 = 12 + 3 = 15 \text{ minutes} \]

### Step 9: Calculate the time for A to fill one-third of the tank
To find the time taken by A to fill one-third of the tank:
\[ \text{Time for A to fill } \frac{1}{3} \text{ of the tank} = 15 \times \frac{1}{3} = 5 \text{ minutes} \]

### Final Answer
The time in which A alone would fill one-third part of the tank is **5 minutes**.

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