A tap can empty a tank in 30 minutes. A second tap can
empty it in 45 minutes. If both the taps operate
simultaneously, how muchh time is needed to empty the tank?

Options

A. 18 minutes
B. 14 minutes
C. 15 minutes
D. 30 minutes

Explanation

To solve the problem of how much time is needed to empty the tank when both taps operate simultaneously, we can follow these steps:

### Step 1: Determine the rate of each tap
- The first tap can empty the tank in 30 minutes. Therefore, its rate of work is:
\[ \text{Rate of Tap 1} = \frac{1 \text{ tank}}{30 \text{ minutes}} = \frac{1}{30} \text{ tanks per minute} \]

- The second tap can empty the tank in 45 minutes. Therefore, its rate of work is:
\[ \text{Rate of Tap 2} = \frac{1 \text{ tank}}{45 \text{ minutes}} = \frac{1}{45} \text{ tanks per minute} \]

### Step 2: Combine the rates of both taps
When both taps are opened simultaneously, their combined rate of work is the sum of their individual rates:
\[ \text{Combined Rate} = \text{Rate of Tap 1} + \text{Rate of Tap 2} = \frac{1}{30} + \frac{1}{45} \]

### Step 3: Find a common denominator
To add the fractions, we need a common denominator. The least common multiple (LCM) of 30 and 45 is 90. We can express both rates with this common denominator:
\[ \frac{1}{30} = \frac{3}{90} \quad \text{and} \quad \frac{1}{45} = \frac{2}{90} \]

### Step 4: Add the rates
Now, we can add the two rates:
\[ \text{Combined Rate} = \frac{3}{90} + \frac{2}{90} = \frac{5}{90} = \frac{1}{18} \text{ tanks per minute} \]

### Step 5: Calculate the time to empty the tank
To find the time taken to empty one tank, we take the reciprocal of the combined rate:
\[ \text{Time} = \frac{1 \text{ tank}}{\frac{1}{18} \text{ tanks per minute}} = 18 \text{ minutes} \]

### Final Answer
Thus, the time needed to empty the tank when both taps operate simultaneously is **18 minutes**.

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