Step 1: Understanding the Concept:
This problem deals with the rates of work and time. A single tap has a constant filling rate. When multiple identical taps are opened simultaneously, their combined filling rate increases proportionally to the number of taps running, which significantly shortens the remaining time required to fill the rest of the tank.
Step 2: Key Formula or Approach:
- Total Work = Rate $\times$ Time
- Remaining Work = Total Work $-$ Completed Work
- Combined Rate = Number of identical active taps $\times$ Rate of a single tap
Step 3: Detailed Explanation:
Let the total capacity of the tank be 6 units.
Since one tap can fill the tank in 6 hours, its filling rate ($R_1$) is:
\[ R_1 = \frac{6 \text{ units}}{6 \text{ hours}} = 1 \text{ unit/hour} \]
The first half of the tank (which is $\frac{6}{2} = 3$ units) is filled by this single tap. The time taken ($T_1$) for this part is:
\[ T_1 = \frac{3 \text{ units}}{1 \text{ unit/hour}} = 3 \text{ hours} \]
Now, the remaining volume to be filled is:
\[ \text{Remaining Work} = 6 - 3 = 3 \text{ units} \]
At this point, 3 more similar taps are opened, making a total of $1 + 3 = 4$ identical taps working together. The new total combined rate ($R_{\text{total}}$) becomes:
\[ R_{\text{total}} = 4 \times 1 \text{ unit/hour} = 4 \text{ units/hour} \]
The time taken ($T_2$) to fill the remaining 3 units with all 4 taps open is:
\[ T_2 = \frac{3 \text{ units}}{4 \text{ units/hour}} = \frac{3}{4} \text{ hours} \]
Convert $\frac{3}{4}$ hours into minutes:
\[ \frac{3}{4} \times 60 \text{ minutes} = 45 \text{ minutes} \]
Calculate the total time required to fill the tank completely:
\[ \text{Total Time} = T_1 + T_2 = 3 \text{ hours} + 45 \text{ minutes} = 3 \text{ hours 45 minutes} \]
Step 4: Final Answer:
The total time taken to fill the tank completely is 3 hours 45 minutes.