GRE® Practice | Solid Geometry

This GRE quant practice question is a solid geometry problem solving question. A classic example of equating volume of water in a vessel getting displaced to the volume of the object immersed in the vessel. Volumes of cylinder and sphere.

Question 2 : A cylindrical vessel is filled with water up to some height. If a sphere of diameter 8 cm is dropped into the cylinder, the water level rises by half of the initial level. Instead, if a sphere of diameter 16 cm is dropped, the water level rises to a height h2. What percentage of this new height h2 is the initial level of water?

1. 33\\frac{1}{3})%
2. 40%
3. 20%
4. 25%
5. 16 \\frac{1}{6})%

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Video Explanation

Play Video: GRE Cylinder Sphere Volume Video Solution

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Explanatory Answer | GRE Geometry Practice Question 2

Given Data & Variables Assigned

Let the radius of the cylinder be 'r'.
Let the initial height up to which water is filled be 'h'.
Let the height up to which water is filled after the sphere of diameter 8 cm is dropped be h₁.
Height of water displaced after sphere 1 is dropped = h₁ - h
The height of water in the vessel after sphere of diameter 16 cm is dropped = h₂
Height of water displaced after sphere 2 is dropped = h₂ - h.

Step 1 of solving this GRE question : Volume of water displaced when sphere 1 is dropped

According to the question, the water level rises by 50% after sphere 1 is dropped into the cylinder.
Therefore, height of water displaced after sphere 1 is dropped, h₁ - h = 0.5h
The water is stored in a cylindrical vessel. So, it will take the shape of the vessel that holds it.
The volume of water displaced = π r² (height of water displaced) = π r² (0.5h)
The volume of water displaced is the same as the volume of sphere 1 = \\frac{4}{3}) × π × (4)³
Equating the two volumes we get π r² (0.5h) = \\frac{4}{3}) × π × (4)³...... (1)

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Step 2 of solving this GRE question : Volume of water displaced when sphere 2 is dropped

The height of water in the vessel after sphere of diameter 16 cm is dropped = h₂
Height of water displaced after sphere 2 is dropped = h₂ - h.
Volume of water displaced after sphere 2 is dropped = π r² (height of water displaced)
Volume of water displaced = π r² (h2 - h)
The volume of water displaced is the same as the volume of sphere 2 = \\frac{4}{3}) × π × 8³
Equating the two volumes we get π r² (h2 - h) = \\frac{4}{3}) × π × 8³...... (2)

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Step 3 of solving this question : Solve the 2 equations obtained in step 1 and step 2

Divide equation (2) by equation (1): \\frac{πr^2 (h_2−h)}{πr^2(0.5h)}) = \\frac{\frac{4}{3} π8^3}{\frac{4}{3} π4^3})
\\frac{(h_2−h)}{(0.5h)}) = 8 → (h2 - h) = 4h
h₂ = 5h
Or h = \\frac{1}{5}) h₂ → i.e., h = 20% of h₂

Choice C is the correct answer