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Between a cube and a right circular cylinder, does the cube have a

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Between a cube and a right circular cylinder, does the cube have a  [#permalink]

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New post 23 Jul 2017, 22:26
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Between a cube and a right circular cylinder, does the cube have a higher lateral surface area? Lateral surface area refers to the area of the sides (the two base areas, one on top and one on bottom, are not taken into account). Assume π = (3/2)^3

(1) The height of the cylinder is twice its radius.
(2) Both the cube and the cylinder have the same volume.

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Re: Between a cube and a right circular cylinder, does the cube have a  [#permalink]

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New post 24 Jul 2017, 01:11
Bunuel wrote:
Between a cube and a right circular cylinder, does the cube have a higher lateral surface area? Lateral surface area refers to the area of the sides (the two base areas, one on top and one on bottom, are not taken into account). Assume π = (3/2)^3

(1) The height of the cylinder is twice its radius.
(2) Both the cube and the cylinder have the same volume.


Call the side of the cube is \(a\), the volume of the cube is \(a^3\) and the lateral surface area of the cube is \(4a^2\)

Call the height and the radius of the cylinder are \(h\) and \(b\). The volume of the cylinder is \(\pi hb^2\) and the lateral surface area of the cylinder is \(2\pi bh\)

(1) Since we could know \(h=2b\), however we can't know the value of \(a\). Insufficient.

(2) We have \(a^3 = \pi hb^2\). We have to compare \(4a^2\) and \(2\pi bh\)

Since \(a^3 = \pi hb^2 \implies 2 \pi bh = \frac{2a^3}{b}\) . We now have to compare \(\frac{2a^3}{b}\) and \(4a^2\) or we have to compare \(\frac{a}{b}\) and 2. However, we have no information about \(\frac{a}{b}\), insufficient.

Combine (1) and (2):

From (1) have \(h=2b\).

From (2) have \(a^3 = \pi hb^2 = 2\pi b^3 \implies \frac{a}{b} = \sqrt[3]{2\pi}\)

Now, from given condition we have \(\pi = (\frac{3}{2})^3 \\
\implies 2\pi = 2 * \frac{27}{8} = \frac{27}{4} < 8 \\
\implies \sqrt[3]{2\pi} < 2 \)

From this, we could compare \(\frac{a}{b}\) and 2 or we could compare \(\frac{2a^3}{b}\) and \(4a^2\). Sufficient.

The answer is C
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Re: Between a cube and a right circular cylinder, does the cube have a  [#permalink]

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New post 30 Oct 2018, 02:55
Bunuel wrote:
Between a cube and a right circular cylinder, does the cube have a higher lateral surface area? Lateral surface area refers to the area of the sides (the two base areas, one on top and one on bottom, are not taken into account). Assume π = (3/2)^3

(1) The height of the cylinder is twice its radius.
(2) Both the cube and the cylinder have the same volume.


Let's formalize the surface area for both the cube and the cylinder:

- A cube has 4 side faces with each face having an area of \(a^2\) (assuming that a cube's side's length is "a"). Thus: \(CubeLatArea = 4*a^2\)
- For the cylinder, its lateral area is that of a rectangle with a length equal to the height of the cylinder and a width equal to the circumference of its base. Thus: \(CylinderLatArea = 2*π*r*h\)

Therefore, the question asks us if \(4*a^2 - 2*π*r*h > 0\) (Eq)

(1) The height of the cylinder is twice its radius.

This statement does not provide any data regarding the cube itself. Therefore, by itself, it's insufficient (cross off A and D).

(2) Both the cube and the cylinder have the same volume.

Translated, this statement means that: \(a^3 = π*r^2*h\). This does not help us in answering the question stated by the problem. Therefore, by itself, this statement is insufficient (cross off B).

(1)+(2) The height of the cylinder is twice its radius and both the cube and the cylinder have the same volume

In this case, we have:
- \(h = 2*r\)
- \(a^3 = π*r^2*h\)

Thus: \(a^3 = 2*π*r^3\) meaning that \(a = (2*π)^(\frac{1}{3})*r\)

Injecting the above in (Eq) and keeping in mind that \(h = 2*r\) yields:

\(4*(2*π)^(\frac{2}{3})*r^2 - 4*π*r^2\)

After simplifying it, we get: \(4*r^2*π*(\frac{2}{π^(1/3)}-1)\) which is negative. Thus, both statements together are sufficient.

Answer: C.
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Re: Between a cube and a right circular cylinder, does the cube have a &nbs [#permalink] 30 Oct 2018, 02:55
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