A tap can empty a tank in one hour. A second tap can empty It in 30 minutes. If both the taps operate simultaneously, how much time is needed to empty the tank?

Options

A. 20 minutes
B. 30 minutes
C. 40 minutes
D. 45 minutes

Answer

To solve the problem of how long it takes for two taps to empty a tank when operating simultaneously, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Rates of Each Tap:**
- The first tap can empty the tank in 1 hour (60 minutes). Therefore, its rate of emptying is:
\[ \text{Rate of Tap A} = \frac{1 \text{ tank}}{60 \text{ minutes}} = \frac{1}{60} \text{ tanks per minute} \]
- The second tap can empty the tank in 30 minutes. Therefore, its rate of emptying is:
\[ \text{Rate of Tap B} = \frac{1 \text{ tank}}{30 \text{ minutes}} = \frac{1}{30} \text{ tanks per minute} \]

2. **Combine the Rates of Both Taps:**
- When both taps are operating simultaneously, their rates add together:
\[ \text{Combined Rate} = \text{Rate of Tap A} + \text{Rate of Tap B} = \frac{1}{60} + \frac{1}{30} \]
- To add these fractions, find a common denominator (which is 60):
\[ \frac{1}{60} + \frac{2}{60} = \frac{3}{60} = \frac{1}{20} \text{ tanks per minute} \]

3. **Calculate the Time to Empty the Tank:**
- If both taps together can empty \(\frac{1}{20}\) of the tank in one minute, then the time taken to empty 1 full tank is the reciprocal of the combined rate:
\[ \text{Time} = \frac{1 \text{ tank}}{\frac{1}{20} \text{ tanks per minute}} = 20 \text{ minutes} \]

### Final Answer:
The time needed to empty the tank when both taps operate simultaneously is **20 minutes**.

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