Two pipes X and Y can fill an empty tank in 36 and 45 hours respectively. Ravi opens these two pipes simultaneously. After 6 hours45min he comes back and sees that there was a leak in the tank. He stops the leakage and thus tank took 15.5 hours to be filled completely after closing the leak. In what time leakage will empty the filled tank?

Options

A. 54 hrs
B. 60 hrs
C. 72 hrs
D. 81 hrs

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Determine the filling rates of pipes X and Y
- Pipe X can fill the tank in 36 hours, so its rate is \( \frac{1}{36} \) of the tank per hour.
- Pipe Y can fill the tank in 45 hours, so its rate is \( \frac{1}{45} \) of the tank per hour.

### Step 2: Find the combined filling rate of both pipes
- The combined rate of both pipes is:
\[ \text{Combined rate} = \frac{1}{36} + \frac{1}{45} \]
- To add these fractions, we need a common denominator. The least common multiple (LCM) of 36 and 45 is 180.
- Converting both rates:
\[ \frac{1}{36} = \frac{5}{180}, \quad \frac{1}{45} = \frac{4}{180} \]
- Thus, the combined rate is:
\[ \frac{5}{180} + \frac{4}{180} = \frac{9}{180} = \frac{1}{20} \]
- Therefore, both pipes together can fill \( \frac{1}{20} \) of the tank in one hour.

### Step 3: Calculate the amount filled in 6 hours and 45 minutes
- 6 hours and 45 minutes is \( 6 + \frac{45}{60} = 6.75 \) hours.
- The amount filled in this time is:
\[ \text{Amount filled} = \text{Combined rate} \times \text{Time} = \frac{1}{20} \times 6.75 = \frac{6.75}{20} = \frac{27}{80} \text{ of the tank} \]

### Step 4: Determine the remaining part of the tank to be filled
- The remaining part of the tank after 6 hours and 45 minutes is:
\[ 1 - \frac{27}{80} = \frac{80 - 27}{80} = \frac{53}{80} \]

### Step 5: Calculate the time taken to fill the remaining part after stopping the leak
- After stopping the leak, the pipes continue to fill the tank at a combined rate of \( \frac{1}{20} \) of the tank per hour.
- The time taken to fill the remaining \( \frac{53}{80} \) of the tank is:
\[ \text{Time} = \frac{\text{Remaining part}}{\text{Combined rate}} = \frac{\frac{53}{80}}{\frac{1}{20}} = \frac{53}{80} \times 20 = \frac{53 \times 20}{80} = \frac{1060}{80} = 13.25 \text{ hours} \]

### Step 6: Set up the equation to find the leak rate
- Let the leak rate be \( L \) (the fraction of the tank emptied by the leak in one hour).
- The effective filling rate while the leak was present is:
\[ \frac{1}{20} - L \]
- The total time taken to fill the tank with the leak was:
\[ 6.75 + 15.5 = 22.25 \text{ hours} \]
- The equation for the total amount filled while the leak was present can be set up as:
\[ \left( \frac{1}{20} - L \right) \times 22.25 = 1 \]

### Step 7: Solve the equation for L
- Rearranging gives:
\[ \frac{22.25}{20} - 22.25L = 1 \]
\[ 1.1125 - 22.25L = 1 \]
\[ 22.25L = 1.1125 - 1 = 0.1125 \]
\[ L = \frac{0.1125}{22.25} = 0.00505 \text{ (approx)} \]

### Step 8: Find the time taken by the leak to empty the tank
- The leak empties the tank at a rate of \( L \) per hour. Therefore, the time taken to empty the full tank is:
\[ \text{Time} = \frac{1}{L} = \frac{1}{0.00505} \approx 198.02 \text{ hours} \]

### Final Answer
The time taken by the leak to empty the filled tank is approximately **198 hours**.

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