Pipes P and Q can fill a completely empty tank in 24 minutes and in 36 minutes, respectively. when opened independently. Pipe R can empty the completely filled tank in 30 minutes. Initially all the three pipes P, Q and R, are opened simultaneously. After how many minutes should pipe Q be closed so that the tank is filled in 1 hour.?

Options

A. 18
B. 15
C. 24
D. 30

Answer

To solve the problem step by step, we will first determine the rates at which each pipe fills or empties the tank, then calculate how long pipe Q should remain open to fill the tank in 1 hour.

### Step 1: Determine the filling and emptying rates of the pipes.

- Pipe P fills the tank in 24 minutes. Therefore, its rate is:
\[ \text{Rate of P} = \frac{1}{24} \text{ tanks per minute} \]

- Pipe Q fills the tank in 36 minutes. Therefore, its rate is:
\[ \text{Rate of Q} = \frac{1}{36} \text{ tanks per minute} \]

- Pipe R empties the tank in 30 minutes. Therefore, its rate is:
\[ \text{Rate of R} = -\frac{1}{30} \text{ tanks per minute} \]

### Step 2: Calculate the combined rate when all pipes are open.

The combined rate of pipes P, Q, and R when all are open is:
\[ \text{Combined Rate} = \text{Rate of P} + \text{Rate of Q} + \text{Rate of R} \]
\[ = \frac{1}{24} + \frac{1}{36} - \frac{1}{30} \]

### Step 3: Find a common denominator to simplify the rates.

The least common multiple (LCM) of 24, 36, and 30 is 360. We will convert each rate to have a denominator of 360:
- Rate of P: \(\frac{1}{24} = \frac{15}{360}\)
- Rate of Q: \(\frac{1}{36} = \frac{10}{360}\)
- Rate of R: \(-\frac{1}{30} = -\frac{12}{360}\)

Now, combine these rates:
\[ \text{Combined Rate} = \frac{15}{360} + \frac{10}{360} - \frac{12}{360} = \frac{13}{360} \text{ tanks per minute} \]

### Step 4: Determine the time for which pipe Q should remain open.

Let \( t \) be the time in minutes that all pipes are open, and pipe Q is closed after \( t \) minutes. The tank needs to be filled in 60 minutes. Therefore, the tank will be filled by:
1. Pipes P and Q for \( t \) minutes.
2. Only pipe P and R for the remaining \( 60 - t \) minutes.

The amount filled in \( t \) minutes is:
\[ \text{Amount filled in } t \text{ minutes} = \left(\frac{15}{360} + \frac{10}{360}\right) t = \frac{25}{360} t \]

The amount filled in the remaining \( 60 - t \) minutes is:
\[ \text{Amount filled in } (60 - t) \text{ minutes} = \left(\frac{15}{360} - \frac{12}{360}\right)(60 - t) = \frac{3}{360}(60 - t) = \frac{1}{120}(60 - t) \]

### Step 5: Set up the equation for the total amount filled.

The total amount filled in 60 minutes must equal 1 tank:
\[ \frac{25}{360} t + \frac{1}{120}(60 - t) = 1 \]

### Step 6: Solve for \( t \).

First, convert \(\frac{1}{120}\) to a fraction with a denominator of 360:
\[ \frac{1}{120} = \frac{3}{360} \]

Now substitute back into the equation:
\[ \frac{25}{360} t + \frac{3}{360}(60 - t) = 1 \]
Multiply through by 360 to eliminate the denominator:
\[ 25t + 3(60 - t) = 360 \]
\[ 25t + 180 - 3t = 360 \]
\[ 22t + 180 = 360 \]
\[ 22t = 360 - 180 \]
\[ 22t = 180 \]
\[ t = \frac{180}{22} = \frac{90}{11} \approx 8.18 \text{ minutes} \]

### Step 7: Determine how long pipe Q should be closed.

Since pipe Q should be closed after approximately \( 8.18 \) minutes, it will remain open for this duration.

### Final Answer:
Pipe Q should be closed after approximately **8.18 minutes**.