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Height of a Spherical Vessel?

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by Haaress » Thu Apr 29, 2010 8:48 am
A cylindrical vessel of a certain height and radius can hold 30 liters of water in it when filled to the brim of the vessel. If all the water in the vessel is transferred to a spherical vessel whose height and radius is the same as that of the cylindrical vessel, what percentage of the capacity of the spherical vessel will remain empty after the transfer?


A. 25%


B. 33.33%


C. 50%


D. 0%


E. 16.67%
Last edited by Haaress on Thu Apr 29, 2010 9:28 am, edited 1 time in total.
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by DanaJ » Thu Apr 29, 2010 9:20 am
First off, this is not something that you will likely see on G-day, because you are not required to know the volume of a sphere. Second, there's no such thing as the height of a sphere: the only measurement that characterizes a sphere is its radius.

You should start off with the two formulas for volume:

cylinder: Pi*(r^2)*h

sphere: [4*Pi*(r^3)]/3

I suspect that you are supposed to deduce that h is 2*r, as shown in the picture. If you put a sphere next to a cylinder, I think the logical step is to assume that.

So now you know that h = 2r. Then the formula for the volume of the cylinder is: 2*Pi*(r^3).

As you can see, the volume of the sphere is smaller than the volume of the cylinder. Therefore, there will be no room left in the sphere.

I will reiterate however that this is not something you should expect on test day. I am 100% you will not be required to use the formula for the volume of a sphere, since there is only one problem with a sphere in GMATprep and the formula si readily given to you in the text of the problem.
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by 4GMAT_Mumbai » Thu Apr 29, 2010 9:26 am
Hi,

I am not sure if the volume of the sphere will be more than that of the cylinder, as depicted in the attachment.

If the sphere has the same height as the cylinder, then h = 2 times r.

Volume of cylinder = 2 Pi (r^3)

Volume of sphere = 4 Pi (r^3) / 3.

Thus, the sphere has 66.66% volume of the cylinder. Or it will hold 20 liters of water ... I guess I am missing something here ...
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by Haaress » Thu Apr 29, 2010 9:42 am
Thanks Dana. Having seen it on the 4gmat.com site , I thought of it as an air -tight question and subsequently wasted some time on it. Anyway, I am glad that I have this all cleared up now.
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by sumanr84 » Thu Apr 29, 2010 10:23 pm
4GMAT_Mumbai wrote:Hi,

I am not sure if the volume of the sphere will be more than that of the cylinder, as depicted in the attachment.

If the sphere has the same height as the cylinder, then h = 2 times r.

Volume of cylinder = 2 Pi (r^3)

Volume of sphere = 4 Pi (r^3) / 3.

Thus, the sphere has 66.66% volume of the cylinder. Or it will hold 20 liters of water ... I guess I am missing something here ...
Volume of cylinder = 2 Pi (r^3)..this is wrong formula..Correct one would include height of the cylinder ( Pi * (r^2) * h )
I am on a break !!
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by harshavardhanc » Fri Apr 30, 2010 12:04 am
sumanr84 wrote: If the sphere has the same height as the cylinder, then h = 2 times r.

Volume of cylinder = 2 Pi (r^3)
Volume of cylinder = 2 Pi (r^3)..this is wrong formula..Correct one would include height of the cylinder ( Pi * (r^2) * h )
it's not the formula, which the poster is talking about. It's a deduction from the formula after putting in h=2r.
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Harsha
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