One tap can fill a water tank four times as fast as another tap. If together the two taps can fill the water tank in 30 minutes then the slower tap alone will be able to fill the water tank in _____

Options

A. 81 min
B. 108 min
C. 150 min
D. 192 min

Answer

To solve the problem, we need to determine how long the slower tap (let's call it Tap B) will take to fill the water tank on its own.

### Step-by-Step Solution:

1. **Define the Rates of the Taps**:
- Let the rate of the slower tap (Tap B) be \( x \) tanks per minute.
- Since Tap A is 4 times faster than Tap B, the rate of Tap A will be \( 4x \) tanks per minute.

2. **Combine the Rates**:
- When both taps are working together, their combined rate is:
\[ \text{Combined Rate} = x + 4x = 5x \text{ tanks per minute} \]

3. **Use the Given Information**:
- We know that together, both taps can fill the tank in 30 minutes. Therefore, their combined rate can also be expressed as:
\[ \text{Combined Rate} = \frac{1 \text{ tank}}{30 \text{ minutes}} = \frac{1}{30} \text{ tanks per minute} \]

4. **Set the Rates Equal**:
- Now we can set the two expressions for the combined rate equal to each other:
\[ 5x = \frac{1}{30} \]

5. **Solve for \( x \)**:
- To find \( x \), we divide both sides by 5:
\[ x = \frac{1}{30} \div 5 = \frac{1}{150} \text{ tanks per minute} \]

6. **Find the Rate of Tap B**:
- The rate of Tap B (the slower tap) is \( x = \frac{1}{150} \) tanks per minute.

7. **Calculate the Time for Tap B to Fill the Tank Alone**:
- To find out how long it takes for Tap B to fill the tank alone, we take the reciprocal of its rate:
\[ \text{Time for Tap B} = \frac{1 \text{ tank}}{\frac{1}{150} \text{ tanks per minute}} = 150 \text{ minutes} \]

### Final Answer:
The slower tap (Tap B) alone will be able to fill the water tank in **150 minutes**.