Capacity of tap B is 80% more than that of A.If both the taps are opened simultaneously, they take 45 h to fill the tank. How long will B take to fill the tank alone?

Options

A. 72h
B. 48h
C. 66h
D. 70h

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Define the capacities of taps A and B
Let the capacity of tap A be \( x \) units per hour. Since tap B is 80% more than tap A, the capacity of tap B will be:
\[ \text{Capacity of B} = x + 0.8x = 1.8x \]

### Step 2: Combine the capacities of taps A and B
When both taps are opened together, their combined capacity is:
\[ \text{Combined capacity} = x + 1.8x = 2.8x \]

### Step 3: Calculate the total capacity of the tank
According to the problem, both taps together fill the tank in 45 hours. Therefore, the total capacity of the tank can be calculated as:
\[ \text{Total capacity of the tank} = \text{Combined capacity} \times \text{Time taken} = 2.8x \times 45 \]
\[ = 126x \text{ units} \]

### Step 4: Determine the time taken by tap B to fill the tank alone
Now, we need to find out how long tap B will take to fill the tank alone. The capacity of tap B is \( 1.8x \) units per hour. The time taken by tap B to fill the tank can be calculated using the formula:
\[ \text{Time} = \frac{\text{Total capacity of the tank}}{\text{Capacity of B}} = \frac{126x}{1.8x} \]
Here, \( x \) cancels out:
\[ = \frac{126}{1.8} \]

### Step 5: Simplify the calculation
To simplify \( \frac{126}{1.8} \):
\[ = \frac{1260}{18} = 70 \text{ hours} \]

### Conclusion
Thus, tap B alone will take **70 hours** to fill the tank.

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