To solve this problem:

Step 1: Volume of the rectangular tank

The dimensions of the rectangular tank are:

• Length = 35 cm
• Width = 28 cm
• Height = 16 cm

The total volume of the tank: $V_{\text{rectangular}} = 35 \times 28 \times 16 = 15,680 \, \text{cm}^3$

Step 2: Volume of water in the rectangular tank

The tank is $\frac{2}{5}$ filled with water. So, the volume of water: $V_{\text{water}} = \frac{2}{5} \times 15,680 = 6,272 \, \text{cm}^3$

Step 3: Volume of the cubical tank

The cubical tank has a side length of 18 cm. The volume of the cube: $V_{\text{cube}} = 18 \times 18 \times 18 = 5,832 \, \text{cm}^3$

Step 4: Remaining water in the rectangular tank

When the cubical tank is filled with water, it will consume 5,832 cm³ of the 6,272 cm³ initially in the rectangular tank. The remaining volume of water in the rectangular tank is: $V_{\text{remaining}} = V_{\text{water}} - V_{\text{cube}} = 6,272 - 5,832 = 440 \, \text{cm}^3$

Step 5: Convert to milliliters

Since 1 cm³ = 1 mL, the remaining water in milliliters is: $440 \, \text{mL}$

Final Answer:

The amount of water left in the rectangular tank is 440 milliliters.

Do you want a detailed explanation of any step or related assistance?

Here are 5 related questions:

1. How do we calculate the volume of a cylinder if its dimensions are given?
2. What would happen if the cubical tank was larger than the initial water volume?
3. How can this problem be adjusted to include a rate of water transfer?
4. Can you calculate the volume of an irregular shape with different dimensions?
5. How do we handle conversions between liters, milliliters, and cubic centimeters?

Tip: Always ensure the units are consistent when solving volume and capacity problems!