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 1: Determine the rate of each tap
- **First Tap**: It can empty the tank in 1 hour. Therefore, its rate of emptying is:
\[ \text{Rate of First Tap} = \frac{1 \text{ tank}}{1 \text{ hour}} = 1 \text{ tank/hour} \]

- **Second Tap**: It can empty the tank in 30 minutes. Since 30 minutes is half an hour, its rate of emptying is:
\[ \text{Rate of Second Tap} = \frac{1 \text{ tank}}{0.5 \text{ hour}} = 2 \text{ tanks/hour} \]

### Step 2: Combine the rates of both taps
When both taps are operating simultaneously, their combined rate of emptying the tank is:
\[ \text{Combined Rate} = \text{Rate of First Tap} + \text{Rate of Second Tap} = 1 \text{ tank/hour} + 2 \text{ tanks/hour} = 3 \text{ tanks/hour} \]

### Step 3: Calculate the time taken to empty the tank
To find the time taken to empty one tank when both taps are working together, we use the formula:
\[ \text{Time} = \frac{\text{Total Work}}{\text{Combined Rate}} \]
Here, the total work is 1 tank:
\[ \text{Time} = \frac{1 \text{ tank}}{3 \text{ tanks/hour}} = \frac{1}{3} \text{ hours} \]

### Step 4: Convert the time into minutes
Since \(\frac{1}{3}\) of an hour is equal to 20 minutes (because \(60 \text{ minutes} \div 3 = 20 \text{ minutes}\)):
\[ \text{Time} = 20 \text{ minutes} \]

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

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