A container has two holes. The `1^(st)` hole alone empties the container in 15 minutes and `2^(nd)` hole alone empties the container in 10 minutes. If water leaks out at a constant rate, how long (in minutes) does it take if both the holes together empty the container ?

Options

A. 6
B. `(1)/(6)`
C. `(1)/(7)`
D. 7

Answer

To solve the problem of how long it takes for both holes to empty the container together, we can follow these steps:

### Step 1: Determine the rate of each hole
- The first hole empties the container in 15 minutes. Therefore, its rate of emptying is:
\[ \text{Rate of Hole 1} = \frac{1 \text{ container}}{15 \text{ minutes}} = \frac{1}{15} \text{ containers per minute} \]

- The second hole empties the container in 10 minutes. Therefore, its rate of emptying is:
\[ \text{Rate of Hole 2} = \frac{1 \text{ container}}{10 \text{ minutes}} = \frac{1}{10} \text{ containers per minute} \]

### Step 2: Combine the rates of both holes
- When both holes are open, their rates combine. Thus, the combined rate of emptying the container is:
\[ \text{Combined Rate} = \text{Rate of Hole 1} + \text{Rate of Hole 2} = \frac{1}{15} + \frac{1}{10} \]

### Step 3: Find a common denominator
- The least common multiple (LCM) of 15 and 10 is 30. We can rewrite the rates with a common denominator:
\[ \frac{1}{15} = \frac{2}{30} \quad \text{and} \quad \frac{1}{10} = \frac{3}{30} \]

### Step 4: Add the rates
- Now, we can add the two rates:
\[ \text{Combined Rate} = \frac{2}{30} + \frac{3}{30} = \frac{5}{30} = \frac{1}{6} \text{ containers per minute} \]

### Step 5: Calculate the time to empty the container
- Since the combined rate is \(\frac{1}{6}\) containers per minute, it will take:
\[ \text{Time} = \frac{1 \text{ container}}{\frac{1}{6} \text{ containers per minute}} = 6 \text{ minutes} \]

### Conclusion
Thus, it takes **6 minutes** for both holes together to empty the container.

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