A cistern can be filled by two pipes in 30 min and 36 min respectively. Both pipes are opened simultaneously to fill a cistern but due to ob struction, `5//6` of first pipe and `9//10` of second pipe is filling the cistern. The obstruction is removed, the cistern get filled in `15(1)/(2)` min from that moment. After how much time the obstruction was removed?

Options

A. 1 min.
B. 2 min.
C. `(1)/(2)` min.
D. 3 min.

Answer

To solve the problem, we need to follow these steps:

### Step 1: Determine the rates of the pipes
The first pipe can fill the cistern in 30 minutes, so its rate of filling is:
\[ \text{Rate of Pipe 1} = \frac{1}{30} \text{ cisterns per minute} \]

The second pipe can fill the cistern in 36 minutes, so its rate of filling is:
\[ \text{Rate of Pipe 2} = \frac{1}{36} \text{ cisterns per minute} \]

### Step 2: Calculate the effective rates with obstructions
Due to obstructions, only \(\frac{5}{6}\) of the first pipe and \(\frac{9}{10}\) of the second pipe are effective. Therefore, the effective rates become:
\[ \text{Effective Rate of Pipe 1} = \frac{5}{6} \times \frac{1}{30} = \frac{5}{180} = \frac{1}{36} \text{ cisterns per minute} \]
\[ \text{Effective Rate of Pipe 2} = \frac{9}{10} \times \frac{1}{36} = \frac{9}{360} = \frac{1}{40} \text{ cisterns per minute} \]

### Step 3: Calculate the combined effective rate
Now, we can find the combined effective rate of both pipes:
\[ \text{Combined Rate} = \frac{1}{36} + \frac{1}{40} \]
To add these fractions, we need a common denominator, which is 720:
\[ \frac{1}{36} = \frac{20}{720}, \quad \frac{1}{40} = \frac{18}{720} \]
Thus,
\[ \text{Combined Rate} = \frac{20 + 18}{720} = \frac{38}{720} = \frac{19}{360} \text{ cisterns per minute} \]

### Step 4: Determine the total time to fill the cistern after obstruction is removed
After the obstruction is removed, the cistern is filled in \(15 \frac{1}{2}\) minutes, which is \(15.5\) minutes or \(\frac{31}{2}\) minutes. The rate of both pipes without obstruction is:
\[ \text{Rate of Pipe 1} + \text{Rate of Pipe 2} = \frac{1}{30} + \frac{1}{36} \]
Finding a common denominator (which is 180):
\[ \frac{1}{30} = \frac{6}{180}, \quad \frac{1}{36} = \frac{5}{180} \]
Thus,
\[ \text{Total Rate without obstruction} = \frac{6 + 5}{180} = \frac{11}{180} \text{ cisterns per minute} \]

### Step 5: Calculate the amount filled after obstruction is removed
The amount filled after the obstruction is removed in \(15.5\) minutes is:
\[ \text{Amount filled} = \text{Rate} \times \text{Time} = \frac{11}{180} \times \frac{31}{2} = \frac{11 \times 31}{360} = \frac{341}{360} \text{ cisterns} \]

### Step 6: Calculate the amount filled before the obstruction was removed
The total amount filled by both pipes before the obstruction was removed is:
\[ \text{Total Amount} = 1 \text{ cistern} \]
Thus, the amount filled before the obstruction was removed is:
\[ \text{Amount filled before} = 1 - \frac{341}{360} = \frac{19}{360} \text{ cisterns} \]

### Step 7: Calculate the time taken before the obstruction was removed
Let \(t\) be the time taken before the obstruction was removed. The amount filled in that time is given by:
\[ \text{Amount filled} = \text{Combined Rate} \times t \]
So,
\[ \frac{19}{360} = \frac{19}{360} \times t \]
This implies:
\[ t = 1 \text{ minute} \]

### Final Answer
The obstruction was removed after **1 minute**.