I added a screenshot with the complete question.
Part (1):First, we need to get the volume of the empty aquarium.
Volume of empty aquarium = length * width * height
Volume of empty aquarium = 30 * 10 * 20
Volume of empty aquarium = 6000 in³
Now, we know that the water fills 3/4 of the total volume of the aquarium.
This means that:
volume of water = 0.75 * 6000
volume of water = 4500 in³
The length and width of the volume of water are the same as the length and width of the aquarium The only parameter that is tunable is the height.
Therefore, we can get the height of water in the aquarium as follows:
height of water = volume of water / (length*width)
height of water =
= 15 in
Now, we are given that the height of water increased 0.25 in after dropping the rock.
This means that:
height of water after dropping rock = 15 + 0.25 = 15.25 in
Now, we will get the volume of both the water and the rock combined.
volume of water and rock combined = 30 * 10 * 15.25 = 4575 in³
Finally, the volume of the rock can be obtained by subtracting the volume of water from the volume of water and rock combined as follows:
volume of rock = 4575 - 4500 = 75 in³
Based on the above, the volume of the rock is 75 in³
Part (2):We know that the total volume of the aquarium is 6000 in³ and that water fills 4500 in³ from this volume.
This means that:
empty volume of aquarium = 6000 - 4500 = 1500 in³
Rocks can only fill this empty volume before the water starts spilling out of the aquarium.
We know that the volume of one rock is 75 in³ and that the volume of the total number of rocks is 1500 in³.
This means that:
number of rocks = 1500 / 75 = 20 rocks
Based on the above, 20 rocks can be added to the aquarium before the water start spilling out.
Hope this helps :)