Question
A square-based rectangular pool is filled with water. The side length of the base is x+20 ft. The height of the pool is 2x+10 ft. A solid cube with side length x+5 ft is then thrown into the pool and splashes exactly its volume of water out of the pool. How much water is left in the pool?
Answers
GPT-4o mini
To find the amount of water left in the pool after the cube is thrown in, we need to calculate the volume of the pool and the volume of the cube.
1. **Volume of the pool:**
The pool has a square base with side length \( x + 20 \) ft and a height of \( 2x + 10 \) ft.
The volume \( V_{\text{pool}} \) of the pool can be calculated using the formula for the volume of a rectangular prism:
\[
V_{\text{pool}} = \text{base area} \times \text{height} = (x + 20)^2 \cdot (2x + 10)
\]
First, we need to compute the base area:
\[
\text{base area} = (x + 20)^2 = x^2 + 40x + 400
\]
Now substitute this into the volume formula:
\[
V_{\text{pool}} = (x^2 + 40x + 400)(2x + 10)
\]
Now, we will expand this:
\[
V_{\text{pool}} = (x^2 + 40x + 400)(2x) + (x^2 + 40x + 400)(10)
\]
Expanding each term:
\[
= 2x^3 + 80x^2 + 800x + 10x^2 + 400x + 4000
\]
Now combine like terms:
\[
= 2x^3 + (80x^2 + 10x^2) + (800x + 400x) + 4000
\]
\[
= 2x^3 + 90x^2 + 1200x + 4000
\]
2. **Volume of the cube:**
The cube has a side length of \( x + 5 \) ft. The volume \( V_{\text{cube}} \) of the cube can be calculated using the formula for the volume of a cube:
\[
V_{\text{cube}} = (x + 5)^3
\]
Expanding this using the binomial cube formula:
\[
= (x + 5)(x + 5)(x + 5)
\]
\[
= (x^2 + 10x + 25)(x + 5)
\]
Expanding this:
\[
= x^3 + 5x^2 + 10x^2 + 50x + 25x + 125
\]
\[
= x^3 + 15x^2 + 75x + 125
\]
3. **Amount of water left in the pool:**
When the cube is placed in the water, it splashes out a volume of water equal to its own volume. Thus, the volume of water left in the pool \( V_{\text{remaining}} \) can be calculated as:
\[
V_{\text{remaining}} = V_{\text{pool}} - V_{\text{cube}}
\]
\[
= (2x^3 + 90x^2 + 1200x + 4000) - (x^3 + 15x^2 + 75x + 125)
\]
Now perform the subtraction:
\[
= (2x^3 - x^3) + (90x^2 - 15x^2) + (1200x - 75x) + (4000 - 125)
\]
\[
= x^3 + 75x^2 + 1125x + 3875
\]
Thus, the amount of water left in the pool after the cube is thrown in is:
\[
\boxed{x^3 + 75x^2 + 1125x + 3875}
\]
1. **Volume of the pool:**
The pool has a square base with side length \( x + 20 \) ft and a height of \( 2x + 10 \) ft.
The volume \( V_{\text{pool}} \) of the pool can be calculated using the formula for the volume of a rectangular prism:
\[
V_{\text{pool}} = \text{base area} \times \text{height} = (x + 20)^2 \cdot (2x + 10)
\]
First, we need to compute the base area:
\[
\text{base area} = (x + 20)^2 = x^2 + 40x + 400
\]
Now substitute this into the volume formula:
\[
V_{\text{pool}} = (x^2 + 40x + 400)(2x + 10)
\]
Now, we will expand this:
\[
V_{\text{pool}} = (x^2 + 40x + 400)(2x) + (x^2 + 40x + 400)(10)
\]
Expanding each term:
\[
= 2x^3 + 80x^2 + 800x + 10x^2 + 400x + 4000
\]
Now combine like terms:
\[
= 2x^3 + (80x^2 + 10x^2) + (800x + 400x) + 4000
\]
\[
= 2x^3 + 90x^2 + 1200x + 4000
\]
2. **Volume of the cube:**
The cube has a side length of \( x + 5 \) ft. The volume \( V_{\text{cube}} \) of the cube can be calculated using the formula for the volume of a cube:
\[
V_{\text{cube}} = (x + 5)^3
\]
Expanding this using the binomial cube formula:
\[
= (x + 5)(x + 5)(x + 5)
\]
\[
= (x^2 + 10x + 25)(x + 5)
\]
Expanding this:
\[
= x^3 + 5x^2 + 10x^2 + 50x + 25x + 125
\]
\[
= x^3 + 15x^2 + 75x + 125
\]
3. **Amount of water left in the pool:**
When the cube is placed in the water, it splashes out a volume of water equal to its own volume. Thus, the volume of water left in the pool \( V_{\text{remaining}} \) can be calculated as:
\[
V_{\text{remaining}} = V_{\text{pool}} - V_{\text{cube}}
\]
\[
= (2x^3 + 90x^2 + 1200x + 4000) - (x^3 + 15x^2 + 75x + 125)
\]
Now perform the subtraction:
\[
= (2x^3 - x^3) + (90x^2 - 15x^2) + (1200x - 75x) + (4000 - 125)
\]
\[
= x^3 + 75x^2 + 1125x + 3875
\]
Thus, the amount of water left in the pool after the cube is thrown in is:
\[
\boxed{x^3 + 75x^2 + 1125x + 3875}
\]