An aquarium 75cm long, 50cm wide and 40cm high is half filled with water. Ten identical metal cubes of edge 15cm are completely immersed in the water. Calculate the new height of water in the aquarium 

To find the new height of water in the aquarium after the cubes are immersed, we'll proceed step by step:

### 1. Volume of the aquarium:
The dimensions of the aquarium are given as:
- Length \( L = 75 \, \text{cm} \)
- Width \( W = 50 \, \text{cm} \)
- Height \( H = 40 \, \text{cm} \)

The total volume of the aquarium is:
\[
V_{\text{aquarium}} = L \times W \times H = 75 \times 50 \times 40 = 150,000 \, \text{cm}^3
\]

Since the aquarium is **half-filled with water**, the volume of water initially present is:
\[
V_{\text{water initial}} = \frac{1}{2} \times V_{\text{aquarium}} = \frac{1}{2} \times 150,000 = 75,000 \, \text{cm}^3
\]

### 2. Volume of the cubes:
The volume of one cube is:
\[
V_{\text{cube}} = \text{edge}^3 = 15^3 = 3375 \, \text{cm}^3
\]

Since there are **10 identical cubes**, the total volume of all the cubes is:
\[
V_{\text{cubes total}} = 10 \times V_{\text{cube}} = 10 \times 3375 = 33,750 \, \text{cm}^3
\]

### 3. Total volume of water after adding cubes:
After the cubes are completely immersed in the water, the new total volume of water (water + displaced water by cubes) is:
\[
V_{\text{water final}} = V_{\text{water initial}} + V_{\text{cubes total}} = 75,000 + 33,750 = 108,750 \, \text{cm}^3
\]

### 4. New height of the water:
The base area of the aquarium remains the same, which is:
\[
A_{\text{base}} = L \times W = 75 \times 50 = 3750 \, \text{cm}^2
\]

To find the new height of the water \( h_{\text{new}} \), we use the formula for volume:
\[
V_{\text{water final}} = A_{\text{base}} \times h_{\text{new}}
\]
Substitute the known values:
\[
108,750 = 3750 \times h_{\text{new}}
\]

Solving for \( h_{\text{new}} \):
\[
h_{\text{new}} = \frac{108,750}{3750} = 29 \, \text{cm}
\]

### Conclusion:
The new height of the water in the aquarium after immersing the cubes is **29 cm**.

Would you like more details or have any questions?

Here are 5 related questions to expand on this problem:
1. What would happen if only 5 cubes were immersed instead of 10?
2. How would the final water height change if the aquarium were fully filled at the start?
3. What is the maximum number of cubes that could be added before the water overflows?
4. How much water is displaced by each cube in percentage of the aquarium's volume?
5. What if the dimensions of the aquarium were increased while keeping the cubes' size the same?

**Tip:** Remember that the volume of displaced water is always equal to the volume of the object fully submerged in it.

An aquarium 75cm long, 50cm wide and 40cm high is half filled with water. Ten identical metal cubes of edge 15cm are completely immersed in the water. Calculate the new height of water in the aquarium.

This problem involves calculating the new water height in an aquarium after immersing 10 metal cubes with 15cm edges into a half-filled aquarium with dimensions 75cm x 50cm x 40cm. Explore the concept of water displacement and volume calculation in this detailed step-by-step solution, suitable for students in Grades 9-11.

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