A tank has three pipes. The first one fills up the tank in 30 min and the second one can fill it up in 45 min. The third one drains out the tank. With all the pipes open it takes 27 min to fill up the tank. Find out how long it takes the third pipe alone to empty a full tank?

Options

A. 56min
B. 50min
C. 52min
D. 54min

Answer

To solve the problem, we need to determine how long it takes for the third pipe, which drains the tank, to empty a full tank. We will follow these steps:

### Step 1: Determine the filling rates of the first two pipes.
- The first pipe fills the tank in 30 minutes. Therefore, its rate is:
\[ \text{Rate of first pipe} = \frac{1 \text{ tank}}{30 \text{ min}} = \frac{1}{30} \text{ tanks/min} \]

- The second pipe fills the tank in 45 minutes. Therefore, its rate is:
\[ \text{Rate of second pipe} = \frac{1 \text{ tank}}{45 \text{ min}} = \frac{1}{45} \text{ tanks/min} \]

### Step 2: Determine the combined filling rate of all pipes when they are open.
- When all pipes are open, it takes 27 minutes to fill the tank. Therefore, the combined rate of all pipes is:
\[ \text{Combined rate} = \frac{1 \text{ tank}}{27 \text{ min}} = \frac{1}{27} \text{ tanks/min} \]

### Step 3: Set up the equation for the combined rates.
Let the rate of the third pipe (which drains the tank) be \( -x \) (negative because it is draining). The equation for the combined rates of the pipes is:
\[ \frac{1}{30} + \frac{1}{45} - x = \frac{1}{27} \]

### Step 4: Find a common denominator and solve for \( x \).
The least common multiple (LCM) of 30, 45, and 27 is 270. We can rewrite the rates with a common denominator:
- For the first pipe:
\[ \frac{1}{30} = \frac{9}{270} \]
- For the second pipe:
\[ \frac{1}{45} = \frac{6}{270} \]
- For the combined rate:
\[ \frac{1}{27} = \frac{10}{270} \]

Substituting these into the equation gives:
\[ \frac{9}{270} + \frac{6}{270} - x = \frac{10}{270} \]

### Step 5: Simplify the equation.
Combine the fractions:
\[ \frac{15}{270} - x = \frac{10}{270} \]

### Step 6: Isolate \( x \).
Rearranging the equation gives:
\[ -x = \frac{10}{270} - \frac{15}{270} \]
\[ -x = -\frac{5}{270} \]
\[ x = \frac{5}{270} \]

### Step 7: Determine the time taken by the third pipe to empty the tank.
The rate of the third pipe is \( \frac{5}{270} \) tanks per minute. To find the time taken to empty one full tank, we take the reciprocal of the rate:
\[ \text{Time} = \frac{1 \text{ tank}}{\frac{5}{270} \text{ tanks/min}} = \frac{270}{5} = 54 \text{ minutes} \]

### Final Answer:
The third pipe alone takes **54 minutes** to empty a full tank.

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