A cistern which has a leak in the bottom is filled in 15 h. Had there been no leak it can be filled in 12 h. If the cistern is full, the leak can empty it in

Options

A. 20 h
B. 60 h
C. 45 h
D. 46 h

Answer

To solve the problem, we need to determine how long it takes for the leak to empty the full cistern. We will proceed step by step.

### Step-by-Step Solution:

1. **Understanding the Filling Rates:**
- The cistern can be filled completely in 12 hours without any leak.
- Therefore, the filling rate without leak = \( \frac{1 \text{ cistern}}{12 \text{ hours}} = \frac{1}{12} \text{ cistern per hour} \).

2. **Understanding the Effect of the Leak:**
- With the leak, the cistern takes 15 hours to fill completely.
- Therefore, the effective filling rate with the leak = \( \frac{1 \text{ cistern}}{15 \text{ hours}} = \frac{1}{15} \text{ cistern per hour} \).

3. **Calculating the Rate of the Leak:**
- Let the rate at which the leak empties the cistern be \( L \) (in cisterns per hour).
- The effective filling rate with the leak can be expressed as:
\[ \text{Filling rate without leak} - \text{Leak rate} = \text{Effective filling rate with leak} \]
- Thus, we have:
\[ \frac{1}{12} - L = \frac{1}{15} \]

4. **Finding the Leak Rate (L):**
- Rearranging the equation gives:
\[ L = \frac{1}{12} - \frac{1}{15} \]
- To subtract these fractions, we need a common denominator. The least common multiple of 12 and 15 is 60.
- Converting the fractions:
\[ L = \frac{5}{60} - \frac{4}{60} = \frac{1}{60} \text{ cistern per hour} \]

5. **Calculating the Time Taken by the Leak to Empty the Full Cistern:**
- If the leak empties \( \frac{1}{60} \) of the cistern in one hour, then to empty the entire cistern (1 cistern), the time taken will be:
\[ \text{Time} = \frac{1 \text{ cistern}}{L} = \frac{1}{\frac{1}{60}} = 60 \text{ hours} \]

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

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