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If the volume of the cone is SH/3 where S is the area of

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If the volume of the cone is SH/3 where S is the area of [#permalink]

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If the volume of the cone is SH/3 where S is the area of the base and H the height. AB=BC. What is the ratio of the volume of the cone (with BC as height) to the volume of the rest of the cone (with AB as height). (for graph see attached. Didn't know how to add picture to this post)


1. 1:3
2. 1:4
3. 1:6
4. 1:7
5. 1:8

Thanks for helping. I must have forgotten something about cone characteristics here...
[Reveal] Spoiler: OA

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Re: Geometry [#permalink]

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New post 21 Jan 2012, 05:33
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Splendidgirl666 wrote:
If the volume of the cone is SH/3 where S is the area of the base and H the height. AB=BC. What is the ratio of the volume of the cone (with BC as height) to the volume of the rest of the cone (with AB as height). (for graph see attached. Didn't know how to add picture to this post)

1. 1:3
2. 1:4
3. 1:6
4. 1:7
5. 1:8

Thanks for helping. I must have forgotten something about cone characteristics here...


Note that the wording is not 100% accurate, plus it's highly unlikely you'll need this staff for the GMAT.

Attachment:
untitled.JPG
untitled.JPG [ 13.91 KiB | Viewed 4021 times ]
Anyway, notice that since AB=BC (AC=2*BC) then R=2r (from similar triangles property). Now, the area of the base of a little upper cone (cone BC) will be \(s=\pi*r^2\) and as the area of the base of the bigger cone (cone AC) is \(S=\pi*R^2=\pi*4r^2=4s\) then \(s=\frac{S}{4}\).

Next, \(volume_{BC}=\frac{\frac{S}{4}*\frac{H}{2}}{3}=\frac{SH}{24}\)

\(volume_{AB}=volume_{AC}-volume_{BC}=\frac{SH}{3}-\frac{SH}{24}=\frac{7SH}{24}\);

Ratio --> \(\frac{volume_{BC}}{volume_{AB}}=\frac{\frac{SH}{24}}{\frac{7SH}{24}}=\frac{1}{7}\).

Answer: D.

P.S. You can attach an image file as you attached ppt file.
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Re: If the volume of the cone is SH/3 where S is the area of [#permalink]

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New post 29 May 2013, 02:25
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I think the easiest way to solve this problem is keeping in mind the idea that if any dimension (for example, height h) of similar figures are in ratio 1:x, then their areas (S) will be in ratio 1:x^2, and their volumes will be in ratio 1:x^3

First,
H(bc) - the height BC
H(whole cone) - the height of the whole cone
V(bc) - the volume of the smaller cone with the height BC
V(ab) - the volume of the larger cone with the height AB
V(whole cone) - the volume of the whole cone with the height AC

H(bc):H(whole cone) = 1:2, therefore V(bc):V(whole cone) = 1:8.

Let's V(bc) = x, then V(bc):V(whole cone) = x:8x.

Consequently, subtracting V(bc) from V(whole cone) we can calculate V(ab):
V(ab) = 8x-x=7x
and the ratio of the volume of the top half of the cone (with BC as height) to the volume of the bottom half of the cone (with AB as height) =
x:7x = 1/7

Answer D

Last edited by thewind on 29 May 2013, 02:46, edited 2 times in total.

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Re: Geometry [#permalink]

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New post 29 May 2013, 02:31
Bunuel wrote:
...plus it's highly unlikely you'll need this staff for the GMAT.

BTW, Bunuel how do you differentiate what is out of the GMAT scope?
Sorry if I'm asking an obvious question :)

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Re: If the volume of the cone is SH/3 where S is the area of [#permalink]

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New post 30 Nov 2013, 23:00
Buneul, actually this is a CAT question. However, here goes the solution. As, you said r=R/2 and AB=BC=H/2.

Now, volume of cone with height BC=\(Pi*r^2*BC\)----->1
Volume of frustum of a cone with height AB=\(\frac{Pi*AB*(R^2+r^2+rR)}{3}\)----->2
Substitute values, and do \(\frac{1}{2}\) to get the answer.

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Re: If the volume of the cone is SH/3 where S is the area of [#permalink]

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Re: If the volume of the cone is SH/3 where S is the area of [#permalink]

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New post 27 Feb 2017, 06:21
Hello from the GMAT Club BumpBot!

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Re: If the volume of the cone is SH/3 where S is the area of [#permalink]

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New post 21 Aug 2017, 02:30
Splendidgirl666 wrote:
If the volume of the cone is SH/3 where S is the area of the base and H the height. AB=BC. What is the ratio of the volume of the cone (with BC as height) to the volume of the rest of the cone (with AB as height). (for graph see attached. Didn't know how to add picture to this post)


1. 1:3
2. 1:4
3. 1:6
4. 1:7
5. 1:8

Thanks for helping. I must have forgotten something about cone characteristics here...


Let r be the radius of the cone. S = pi * r^2
So, radius of smaller cone = r/2
Volume of smaller cone = 1/3* pi * (r/2)^2 *H/2 = 1/3*pi*r^2*H/8
(AB= BC = H/2)
Volume of the rest of the cone = 1/3* pi * r^2 - 1/3*pi*r^2*H/8 = 1/3*7/8*pi*r^2*H

Volume of smaller cone/Volume of the rest of the cone = 1/7

Answer D
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Re: If the volume of the cone is SH/3 where S is the area of   [#permalink] 21 Aug 2017, 02:30
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