To solve this problem, we use the concept of proportionality since the flow rate of the hose remains constant.

Given:

• Time to fill a 1638 in³ aquarium = 4 minutes
• Volume of larger aquarium = 20790 in³

Find:

Time to fill the 20790 in³ aquarium.

Calculation:

1. Find the flow rate of the hose: $\text{Flow rate} = \frac{\text{Volume of smaller aquarium}}{\text{Time for smaller aquarium}} = \frac{1638 \, \text{in}^3}{4 \, \text{min}} = 409.5 \, \text{in}^3/\text{min}.$

2. Calculate the time to fill the larger aquarium: $\text{Time for larger aquarium} = \frac{\text{Volume of larger aquarium}}{\text{Flow rate}} = \frac{20790 \, \text{in}^3}{409.5 \, \text{in}^3/\text{min}} = 50.75 \, \text{minutes}.$

3. Round to the nearest minute: $50.75 \, \text{minutes} \approx 51 \, \text{minutes}.$

Answer:

• Numerical answer: 51
• Unit abbreviation: min

Would you like further details or explanations? Here are five related questions for practice:

1. If the flow rate of the hose is halved, how long will it take to fill the 20790 in³ aquarium?
2. What is the time required to fill a 10000 in³ aquarium with the same hose?
3. How many gallons of water (1 gallon = 231 in³) does the 1638 in³ aquarium hold?
4. If a different hose fills at 500 in³/min, how long will it take to fill a 20790 in³ aquarium?
5. How would the time change if the larger aquarium were twice its current size?

Tip: Always ensure your units match when solving proportional problems!