A cistern normally takes 10 hours to be filled by a tap. But because of a leak, it takes 2 hours more. In how many hours will the leak empty a full cistern?
एक सिस्टर्न को नल से भरने में आम तौर पर 10 घंटे लगते हैं। लेकिन रिसाव होने के कारण इसे भरने में 2 घंटे अधिक लगते हैं। भरा हुआ सिस्टर्न रिसाव होने के कारण कितने घंटे में खाली हो जाएगा?

Options

A. A)45
B. B)48
C. C)30
D. D)60

Explanation

To solve the problem, we need to find out how long it will take for the leak to empty a full cistern. Let's break it down step by step.

### Step 1: Understand the filling and leaking rates
The tap fills the cistern in 10 hours. Therefore, the rate of the tap filling the cistern is:
\[ \text{Rate of tap} = \frac{1 \text{ cistern}}{10 \text{ hours}} = 0.1 \text{ cisterns per hour} \]

### Step 2: Determine the time taken with the leak
With the leak, it takes 2 hours longer to fill the cistern, so it takes 12 hours to fill it with the leak. Therefore, the combined rate of the tap and the leak is:
\[ \text{Rate of tap + leak} = \frac{1 \text{ cistern}}{12 \text{ hours}} = \frac{1}{12} \text{ cisterns per hour} \]

### Step 3: Set up the equation for the leak's rate
Let the rate of the leak be \( L \) (in cisterns per hour). The equation representing the situation is:
\[ \text{Rate of tap} - \text{Rate of leak} = \text{Rate of tap + leak} \]
Substituting the values we have:
\[ 0.1 - L = \frac{1}{12} \]

### Step 4: Solve for the leak's rate
Rearranging the equation gives:
\[ L = 0.1 - \frac{1}{12} \]
To perform the subtraction, we need a common denominator. The common denominator of 10 and 12 is 60. Thus:
\[ 0.1 = \frac{6}{60} \quad \text{and} \quad \frac{1}{12} = \frac{5}{60} \]
Now substituting these into the equation:
\[ L = \frac{6}{60} - \frac{5}{60} = \frac{1}{60} \text{ cisterns per hour} \]

### Step 5: Calculate the time taken by the leak to empty the cistern
If the leak's rate is \( \frac{1}{60} \) cisterns per hour, then the time taken to empty one full cistern is the reciprocal of the leak's rate:
\[ \text{Time to empty} = \frac{1 \text{ cistern}}{L} = \frac{1}{\frac{1}{60}} = 60 \text{ hours} \]

### Final Answer
The leak will empty the full cistern in **60 hours**.

---