Cylinder

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N:Dure: Master | Next Rank: 500 Posts; Posts: 181; Joined: Sat Oct 16, 2010 1:57 pm; Thanked: 4 times

by N:Dure » Sat Dec 25, 2010 11:30 pm

The longest distance between any two points in a right circular cylinder is 13. The height and the diameter are positive integers where the height exceeds the diameter.

A) 75 TT
B) The volume of the cylinder

a) A is bigger
b) B is bigger
c) Both are equal
d) Insufficient

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Source: Beat The GMAT — Problem Solving | Sat Dec 25, 2010 11:30 pm

N:Dure: Master | Next Rank: 500 Posts; Posts: 181; Joined: Sat Oct 16, 2010 1:57 pm; Thanked: 4 times

by N:Dure » Sat Dec 25, 2010 11:47 pm

Got it! Thanks

It's a 5:12:13 triangle. The base is 5 i.e radius is 2.5, and height is 12

(2.5)^2 TT *12 = 75 TT

C

But it makes you think by "the longest distance" that this is the diameter.

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Sat Dec 25, 2010 11:50 pm

N:Dure wrote:The longest distance between any two points in a right circular cylinder is 13. The height and the diameter are positive integers where the height exceeds the diameter.

A) 75 TT
B) The volume of the cylinder

a) A is bigger
b) B is bigger
c) Both are equal
d) Insufficient

In a right circular cylinder, the distance between any two points will be maximum when they lie on the opposite face and also diametrically opposite to each other. Refer to the image below,

Thus longest distance between any two points in a right circular cylinder = length of the red line = âˆš[(2r)Â² + (h)Â²] = âˆš[(d)Â² + (h)Â²], where d is the diameter and h is the height of the cylinder and h > d.

Now, âˆš[(d)Â² + (h)Â²] = 13
=> (d)Â² + (h)Â² = (13)Â²

Only integral solution of the above equation for d and h are d = 5 and h = 12. As h > d, we cannot take the other possible set, i.e. d = 12 and h = 5.

Therefore, the volume of the cylinder = Ï€rÂ²h = Ï€(5/2)Â²(12) = 75Ï€

Thus the quantity in A and the quantity in B are same.

The correct answer is C.

Anurag Mairal, Ph.D., MBA
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N:Dure: Master | Next Rank: 500 Posts; Posts: 181; Joined: Sat Oct 16, 2010 1:57 pm; Thanked: 4 times

by N:Dure » Sun Dec 26, 2010 12:04 am

Thank you sir for elucidating.

In some other questions, he says the "longest distance bet. 2 points" and by this he means the diameter. So how to differentiate between them? Is this pertinent only to the right cylinder?

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by Anurag@Gurome » Sun Dec 26, 2010 12:10 am

N:Dure wrote:In some other questions, he says the "longest distance bet. 2 points" and by this he means the diameter. So how to differentiate between them? Is this pertinent only to the right cylinder?

Depends upon the shape of the object. For example,

1. For a quadrilateral, the distance between two points is longest when they are diagonally opposite.
2. For a circle, the distance between two points is longest when they are diametrically opposite.
3. For a cube/rectangular parallelepiped, the distance between two points is longest when they are diagonally opposite and the diagonal is the main diagonal, i.e. the one of the points is the bottom-corner point and another is the opposite top-corner point.
4. For a right circular cylinder already explained.

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N:Dure: Master | Next Rank: 500 Posts; Posts: 181; Joined: Sat Oct 16, 2010 1:57 pm; Thanked: 4 times

by N:Dure » Sun Dec 26, 2010 12:27 am

But for a normal cylinder, when he says longest distance, is it the diameter then? or is it the triangle's hypotenuse?

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Sun Dec 26, 2010 12:29 am

N:Dure wrote:But for a normal cylinder, when he says longest distance, is it the diameter then? or is it the triangle's hypotenuse?

For a normal cylinder too it is the same as right circular cylinder, i.e. the hypotenuse of the right-angled triangle.

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