The problem involves calculating the volume of water filled into a cylindrical tank based on the flow rates of two taps. Here's how we'll solve it step by step:

Understanding the Problem

1. Given Data:

   • Tap A increases the water level by 8 cm per minute.
   • Tap B increases the water level by 6 cm per minute.
   • Tap A can fill 20% of the tank in 6 minutes and 15 seconds (or 375 seconds).
   • Initially, the tank is empty.
   • Tap A is opened for 5 minutes, and then Tap B is opened simultaneously for an additional 15 minutes.

2. Required:

   • Find the total volume of water in the tank as a fraction of its full capacity.

Step 1: Determine the Capacity of the Tank

Tap A fills 20% of the tank in 375 seconds.
This means: $\text{Flow rate of Tap A} = \frac{20}{375} \% \text{of the tank per second}.$ Simplify: $\text{Flow rate of Tap A} = \frac{20}{375} = \frac{4}{75} \% \text{of the tank per second.}$

In terms of full tank capacity per minute: $\text{Flow rate of Tap A (per minute)} = \frac{4}{75} \times 60 = \frac{16}{25} \% \text{of the tank per minute.}$

Step 2: Determine the Total Time Contribution of Tap A

Tap A is open for 5 minutes. During this time: $\text{Volume filled by Tap A in 5 minutes} = 5 \times \frac{16}{25} = \frac{80}{25} = 3.2 \% \text{of the tank.}$

Step 3: Determine the Flow Rate of Tap B

From the problem, Tap B raises the water level by 6 cm per minute, while Tap A raises it by 8 cm per minute. Thus, the flow rate of Tap B (in terms of tank capacity) is proportional: $\text{Flow rate of Tap B} = \frac{6}{8} \times \text{Flow rate of Tap A}.$

Substituting Tap A's flow rate: $\text{Flow rate of Tap B} = \frac{6}{8} \times \frac{16}{25} = \frac{12}{25} \% \text{of the tank per minute.}$

Step 4: Combined Contribution of Taps A and B

After the first 5 minutes, both taps are opened for an additional 15 minutes. During this time, the combined flow rate of A and B is: $\text{Combined flow rate} = \frac{16}{25} + \frac{12}{25} = \frac{28}{25} \% \text{of the tank per minute.}$

In 15 minutes, the total volume filled by both taps is: $\text{Volume filled by A and B in 15 minutes} = 15 \times \frac{28}{25} = \frac{420}{25} = 16.8 \% \text{of the tank.}$

Step 5: Total Volume

The total volume of water filled in the tank is the sum of the contributions: $\text{Total volume} = 3.2 + 16.8 = 20 \% \text{of the tank.}$

Final Answer

The total volume of water filled is 20% of the tank's capacity.

Would you like further clarification or a breakdown of any step?
Here are 5 related questions for practice:

1. How do you calculate the flow rate of a tap in terms of tank capacity?
2. If Tap A worked alone for 10 minutes, how much of the tank would it fill?
3. How does the ratio of flow rates between two taps affect the filling process?
4. What happens if both taps are opened simultaneously from the start?
5. How can we model the water level rise mathematically using a formula?

Tip: Always check units (minutes, seconds, percent, etc.) when dealing with flow rate problems!