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B

no height in statement A, therefore it is useless.

looking at 2, imagine the cylinder having a right angled triangle with in it. we have the hypotenuse, 13, and only one pair of 5, 12 satisfies this. (we know height is greater than base)
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The height of a certain right circular cylinder is greater than its diameter, and both numbers are positive integers. What is the volume of the cylinder?

(1) The radius of the base is 2.5.
Still height can not be determined as h>d so it can be anything above 5 (Insufficient)

(2) The longest distance between any two points on the cylinder is 13.

From the longest distance we can find the height and radius also

Hence Sufficient B!

Ans

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Bunuel
The height of a certain right circular cylinder is greater than its diameter, and both numbers are positive integers. What is the volume of the cylinder?

(1) The radius of the base is 2.5.
(2) The longest distance between any two points on the cylinder is 13.


Kudos for a correct solution.

The OA will be revealed on Sunday

Reworded, the question is asking for V= π(r^2)*h , or what is r and h equal to?

Statement 1:
r=2.5
We still need h, and we only know that h > 5
Insufficient

Statement 2:
The diagonal is 13. 13 is the hypotenuse of the right triangle (5-12-13) with base 5, and height 12. 5/2=2.5=radius
V= π(r^2)*h
V= π(2.5^2)*12
Sufficient

Answer: B
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Why can't the diameter be 6? And then you can solve for the height 169-36 ^ .5?
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bankerboy30
Why can't the diameter be 6? And then you can solve for the height 169-36 ^ .5?

The problem states that both the diameter and the height are integers. For the height to be an integer, only the 5-12-13 triplet would work here.
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Bunuel
The height of a certain right circular cylinder is greater than its diameter, and both numbers are positive integers. What is the volume of the cylinder?

(1) The radius of the base is 2.5.
(2) The longest distance between any two points on the cylinder is 13.


Kudos for a correct solution.

The OA will be revealed on Sunday

VERITAS PREP OFFICIAL SOLUTION:

The correct response is (B). Start by drawing the figure:
Attachment:
Picture_2.png
Picture_2.png [ 15.65 KiB | Viewed 7026 times ]
The “longest distance between any two points” will be the hypotenuse of a right triangle formed by the height and the diameter as the “legs.” If that hypotenuse is 13 and both the height and diameter are positive integers where h > d, the value of the legs can only be 5 and 12 (the ratio 5:12:13).

To find the volume of a right cylinder we use the formula πr^2h. If the diameter is 5, the radius is 2.5. The height is 12. We’ll be able to find the volume. Sufficient.

If you chose (A), this tells us the base is 5, but we still don’t know the height. We cannot assume it is 12 because the question doesn’t stipulate that the “longest distance between any two points” be an integer as well. For example, the height could be 10.

If you chose (C), we can find Statement (1) based solely on the information provided in Statement (2). The first statement is redundant.

If you chose (D), only Statement (2) is independently sufficient. We cannot determine the height from Statement (1) and we need BOTH the height and the radius (the base of the triangle) to find the volume of the cylinder.

If you chose (E), you may want to review the common right triangle ratios such as 3:4:5 and 5:12:13, and/or the Pythagorean theorem. The GMAT will sometimes hide right triangles (2-D shapes) inside boxes (rectangular prisms) and cylinders (3-D shapes).
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The explanation seems ok but I have a question, and I would really appreciate it if someone could help me understand.

Can I not create a really thin cylinder, with the diagonal as 13, and height as !=12 and base != 5? compared to the normal 5-12-13 cylinder? Is it just 1 possible cylinder with diagonal as 13?

For example:
The diagonal in both is 13, but the height and diameter are different?
Cylinder 1 Cylinder 2. Pardon the horrible diagrams. :roll:
| | | /|
| | | / |
| | | / |
| | | / |
---- |/ |

Sorry if it may sound stupid, it just seems like a possibility in real life to me to have 2 such cylinders with diagonal 13 and different areas, both with height > diameter.

Thanks in advance!
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