SOLUTION: I am doing test corrections on our chapter test of volume and density, and this is the only problem that I still do not understand. I know that the ratio we have to figure out will

Question 196428: I am doing test corrections on our chapter test of volume and density, and this is the only problem that I still do not understand. I know that the ratio we have to figure out will be a comparison between the different bases, and it would be figured out with something along the lines of B(1)=B(.5), but I don't understand how you would figure out the exact numbers of the ratio.
A filter apparatus is submerged in fish tank A, raising the water level 1 cm. When the same filter is submerged in fish tank B, the water level is raised only 0.5 cm. If the fish tanks have the same water level without a filter, what is the ratio of the area of the water surface of tank A to the area of the water surface of tank B?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A filter apparatus is submerged in fish tank A, raising the water level 1 cm. When the same filter is submerged in fish tank B, the water level is raised only 0.5 cm. If the fish tanks have the same water level without a filter, what is the ratio of the area of the water surface of tank A to the area of the water surface of tank B?
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The volume of a tank is the area of its base times its height,
V = A*h --> h = V/A
The change in level, Ch = CV/A (change in volume/Area)
The filter displaces a volume of liquid equal to its volume in both tanks.
The increase in level is the same as if that volume of liquid is added to the tanks.
Ch1 = 2*Ch2
Ch1 = CV/A1 = 2*CV/A2
Only the Areas are different, so the cross-sectional Areas differ by a factor of 2.
The smaller Area will change in level (h) twice as much.

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