Two pipes are running continuously to fill the tank. The 1st pipe has filled it in 5 hrs by itself and 2nd in 20 hrs. But a 3rd pipe was outlet pipe and the operator did not notice it due to which it caused a delay of 1 hour in filling the tank. Find the time in which the 3rd pipe would empty the filled tank?

Options

A. 15
B. 20
C. 25
D. 30

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the filling rates of the pipes.
- The first pipe can fill the tank in 5 hours. Therefore, its rate of filling is:
\[ \text{Rate of Pipe 1} = \frac{1}{5} \text{ tank per hour} \]
- The second pipe can fill the tank in 20 hours. Therefore, its rate of filling is:
\[ \text{Rate of Pipe 2} = \frac{1}{20} \text{ tank per hour} \]

### Step 2: Calculate the combined filling rate of the two pipes.
- The combined rate of both pipes working together is:
\[ \text{Combined Rate} = \text{Rate of Pipe 1} + \text{Rate of Pipe 2} = \frac{1}{5} + \frac{1}{20} \]
- To add these fractions, we need a common denominator. The least common multiple of 5 and 20 is 20:
\[ \frac{1}{5} = \frac{4}{20} \]
Therefore,
\[ \text{Combined Rate} = \frac{4}{20} + \frac{1}{20} = \frac{5}{20} = \frac{1}{4} \text{ tank per hour} \]

### Step 3: Calculate the time taken to fill the tank with the outlet pipe.
- Let the time taken to fill the tank without the outlet pipe be \( t \) hours. The combined rate of the two pipes filling the tank is \( \frac{1}{4} \text{ tank per hour} \).
- The equation for filling the tank is:
\[ \text{Total Work} = \text{Rate} \times \text{Time} \]
Thus,
\[ 1 = \frac{1}{4} \times t \]
Solving for \( t \):
\[ t = 4 \text{ hours} \]

### Step 4: Account for the delay caused by the outlet pipe.
- The problem states that there was a delay of 1 hour due to the outlet pipe. Therefore, the effective time taken to fill the tank with the outlet pipe is:
\[ t + 1 = 4 + 1 = 5 \text{ hours} \]

### Step 5: Calculate the effective rate of the outlet pipe.
- Let the rate of the outlet pipe be \( R \) (in tanks per hour). The effective rate when all three pipes are working together is:
\[ \text{Effective Rate} = \text{Combined Rate} - R \]
We know that the effective rate is:
\[ \frac{1}{5} \text{ tank per hour} = \frac{1}{4} - R \]
Rearranging gives:
\[ R = \frac{1}{4} - \frac{1}{5} \]
Finding a common denominator (20):
\[ R = \frac{5}{20} - \frac{4}{20} = \frac{1}{20} \text{ tank per hour} \]

### Step 6: Calculate the time taken by the outlet pipe to empty the tank.
- If the outlet pipe empties at a rate of \( \frac{1}{20} \text{ tank per hour} \), the time taken to empty the entire tank is:
\[ \text{Time} = \frac{1 \text{ tank}}{\frac{1}{20} \text{ tank per hour}} = 20 \text{ hours} \]

### Final Answer:
The time in which the 3rd pipe would empty the filled tank is **20 hours**.

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