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A right circular cone, twice as tall as it is wide at its

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A right circular cone, twice as tall as it is wide at its [#permalink] New post   07 Oct 2009, 00:31
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A right circular cone, twice as tall as it is wide at its greatest width, is pointing straight down. The cone is partially filled with water, which is dripping out of a tiny hole in the cone's tip at a rate of 2 cubic centimeters per hour. If the water were to continue to drip out at this rate, how much longer would it take for the cone to empty, assuming that no water is added to the cone and that there is no loss of water from the cone by any other means?

(1) The top surface of the water in the cone is currently \(9\pi\) square centimeters in area.
(2) The top surface of the water in the cone currently is exactly 4 centimeters below the cone's top, measuring vertically.
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A right circular cone, twice as tall as it is wide at its [#permalink] New post   07 Oct 2009, 05:13
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jax91 wrote:
(1) The top surface of the water in the cone is currently 9pi square centimeters in area. -- sufficient on its own (we can get D)
(2) The top surface of the water in the cone currently is exactly 4 centimeters below the cone's top, measuring vertically. -- sufficient on its own ( we have H)

As the ratio of diameter:hieght of any sub-cone formed in this cone will be 1:2

So we need either D or H to get the volume.

IMO D.


Disagree.

A right circular cone, twice as tall as it is wide at its greatest width, is pointing straight down. The cone is partially filled with water, which is dripping out of a tiny hole in the cone's tip at a rate of 2 cubic centimeters per hour. If the water were to continue to drip out at this rate, how much longer would it take for the cone to empty, assuming that no water is added to the cone and that there is no loss of water from the cone by any other means?



H - height of cone, R radius of cone.
h - height of water in cone, r radius of top surface of water in cone.


(1) The top surface of the water in the cone is currently \(9\pi\) square centimeters in area. Top surface area of water = \(9\pi=\pi*r^2\) --> \(r=3\) --> \(\frac{R}{H}=\frac{r}{h}=\frac{1}{4}\) --> \(h=12\) --> \(V=\frac{1}{3}*\pi*r^2*h=36*\pi\) cubic centimeters.

Leak rate 2 cubic centimeters per hour --> \(time=\frac{36\pi}{2}\) hours.

Sufficient.


(2) The top surface of the water in the cone currently is exactly 4 centimeters below the cone's top, measuring vertically. Not sufficient we know that H=h+4, but h can be any value and thus the Volume can be any.


Answer: A.

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Re: Leak of water from the cone [#permalink] New post   07 Oct 2009, 01:53
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(1) The top surface of the water in the cone is currently 9pi square centimeters in area. -- sufficient on its own (we can get D)
(2) The top surface of the water in the cone currently is exactly 4 centimeters below the cone's top, measuring vertically. -- sufficient on its own ( we have H)

As the ratio of diameter:hieght of any sub-cone formed in this cone will be 1:2

So we need either D or H to get the volume.

IMO D.
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Re: Leak of water from the cone [#permalink] New post   30 Jun 2011, 23:03
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It cant be A cause the question says, e cone is partially filled with water.

A can be true only when the cone is fully filled.
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Re: Leak of water from the cone [#permalink] New post   01 Jul 2011, 04:06
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svikram wrote:
It cant be A cause the question says, e cone is partially filled with water.

A can be true only when the cone is fully filled.


If we know the ratio for height:width for any volume in a cone, then that ratio applies to all volumes. This rule is due to the fact that the angles in the cone stay constant when the volume changes. Statement 1 gives us the area at the top of the water. This allows us to find the water volume, using the height:width ratio provided, and subsequently the rate of leaking. A is the answer.
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Re: A right circular cone, twice as tall as it is wide at its [#permalink] New post   19 Aug 2012, 01:28
hi Bunuel,
how did you get to the ratio:
R/H=r/h=1/4 ??
Can you please elaborate?
thanks!!
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Re: A right circular cone, twice as tall as it is wide at its [#permalink] New post   23 Jul 2013, 06:05
kartik222 wrote:
hi Bunuel,
how did you get to the ratio:
R/H=r/h=1/4 ??
Can you please elaborate?
thanks!!



kartik,

similar triangle property...

R/H = r/h

since 2R=H

r/h = 1/2

so h = 6.
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Re: A right circular cone, twice as tall as it is wide at its [#permalink] New post   15 Nov 2013, 16:02
Hi,

have a small question: it is stated that "The cone is partially filled with water, which is dripping out of a tiny hole in the cone's tip at a rate of 2 cubic centimeters per hour"

It has been said that : the leak rate 2 cubic centimeters per hour --> 36pi/2. Meaning that it will take 18pi hours to fill the cone.

But, the cone could be half full. or 3/4 full. We don't know!

Therefore, the result will be different since they ask : "how much longer would it take for the cone to empty, assuming that no water is added to the cone and that there is no loss of water from the cone by any other means"

It could be 18pi/2 or 18pi/4!

Where did i miss something? Plz explain!

Thanks!
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Re: A right circular cone, twice as tall as it is wide at its [#permalink] New post   16 Nov 2013, 10:44
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Paris75 wrote:
Hi,

have a small question: it is stated that "The cone is partially filled with water, which is dripping out of a tiny hole in the cone's tip at a rate of 2 cubic centimeters per hour"

It has been said that : the leak rate 2 cubic centimeters per hour --> 36pi/2. Meaning that it will take 18pi hours to fill the cone.

But, the cone could be half full. or 3/4 full. We don't know!

Therefore, the result will be different since they ask : "how much longer would it take for the cone to empty, assuming that no water is added to the cone and that there is no loss of water from the cone by any other means"

It could be 18pi/2 or 18pi/4!

Where did i miss something? Plz explain!

Thanks!


\(36\pi\) is the volume of water in the cone not the total volume of the cone.
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Re: Leak of water from the cone [#permalink] New post   02 Dec 2013, 13:04
Bunuel wrote:
jax91 wrote:
(1) The top surface of the water in the cone is currently 9pi square centimeters in area. -- sufficient on its own (we can get D)
(2) The top surface of the water in the cone currently is exactly 4 centimeters below the cone's top, measuring vertically. -- sufficient on its own ( we have H)

As the ratio of diameter:hieght of any sub-cone formed in this cone will be 1:2

So we need either D or H to get the volume.

IMO D.


Disagree.

A right circular cone, twice as tall as it is wide at its greatest width, is pointing straight down. The cone is partially filled with water, which is dripping out of a tiny hole in the cone's tip at a rate of 2 cubic centimeters per hour. If the water were to continue to drip out at this rate, how much longer would it take for the cone to empty, assuming that no water is added to the cone and that there is no loss of water from the cone by any other means?
Attachment:
Untitled.png
H - height of cone, R radius of cone.
h - height of water in cone, r radius of top surface of water in cone.

(1) The top surface of the water in the cone is currently \(9\pi\) square centimeters in area. Top surface area of water = \(9\pi=\pi*r^2\) --> \(r=3\) --> \(\frac{R}{H}=\frac{r}{h}=\frac{1}{4}\) --> \(h=12\) --> \(V=\frac{1}{3}*\pi*r^2*h=36*\pi\) cubic centimeters.

Leak rate 2 cubic centimeters per hour --> \(time=\frac{36\pi}{2}\) hours.

Sufficient.

(2) The top surface of the water in the cone currently is exactly 4 centimeters below the cone's top, measuring vertically. Not sufficient we know that H=h+4, but h can be any value and thus the Volume can be any.

Answer: A.




The top surface of the water in the cone is currently \(9\pi\) square centimeters in area. Top surface area of water = \(9\pi=\pi*r^2\) --> \(r=3\) --> \(\frac{R}{H}=\frac{r}{h}=\frac{1}{4}\)

after you solved for r=3, how did you get

\(\frac{R}{H}=\frac{r}{h}=\frac{1}{4}\)

Aren't R, H, and h, all unknown?
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Re: Leak of water from the cone [#permalink] New post   03 Dec 2013, 01:39
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AccipiterQ wrote:
Bunuel wrote:
jax91 wrote:
(1) The top surface of the water in the cone is currently 9pi square centimeters in area. -- sufficient on its own (we can get D)
(2) The top surface of the water in the cone currently is exactly 4 centimeters below the cone's top, measuring vertically. -- sufficient on its own ( we have H)

As the ratio of diameter:hieght of any sub-cone formed in this cone will be 1:2

So we need either D or H to get the volume.

IMO D.


Disagree.

A right circular cone, twice as tall as it is wide at its greatest width, is pointing straight down. The cone is partially filled with water, which is dripping out of a tiny hole in the cone's tip at a rate of 2 cubic centimeters per hour. If the water were to continue to drip out at this rate, how much longer would it take for the cone to empty, assuming that no water is added to the cone and that there is no loss of water from the cone by any other means?
Attachment:
Untitled.png
H - height of cone, R radius of cone.
h - height of water in cone, r radius of top surface of water in cone.

(1) The top surface of the water in the cone is currently \(9\pi\) square centimeters in area. Top surface area of water = \(9\pi=\pi*r^2\) --> \(r=3\) --> \(\frac{R}{H}=\frac{r}{h}=\frac{1}{4}\) --> \(h=12\) --> \(V=\frac{1}{3}*\pi*r^2*h=36*\pi\) cubic centimeters.

Leak rate 2 cubic centimeters per hour --> \(time=\frac{36\pi}{2}\) hours.

Sufficient.

(2) The top surface of the water in the cone currently is exactly 4 centimeters below the cone's top, measuring vertically. Not sufficient we know that H=h+4, but h can be any value and thus the Volume can be any.

Answer: A.




The top surface of the water in the cone is currently \(9\pi\) square centimeters in area. Top surface area of water = \(9\pi=\pi*r^2\) --> \(r=3\) --> \(\frac{R}{H}=\frac{r}{h}=\frac{1}{4}\)

after you solved for r=3, how did you get

\(\frac{R}{H}=\frac{r}{h}=\frac{1}{4}\)

Aren't R, H, and h, all unknown?


We are given that the cone, twice as tall as it is wide, which means that H = 2D --> H = 4R --> R/H = 1/4. Because of similar triangles, the same applies to r and h.

Hope it's clear.
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Re: A right circular cone, twice as tall as it is wide at its [#permalink] New post   07 Dec 2013, 00:51
maaadhu wrote:
kartik222 wrote:
hi Bunuel,
how did you get to the ratio:
R/H=r/h=1/4 ??
Can you please elaborate?
thanks!!



kartik,

similar triangle property...

R/H = r/h

since 2R=H

r/h = 1/2

so h = 6.


I really find these problems very tough. Is there any reference to these problems ???
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A right circular cone, twice as tall as it is wide at its [#permalink] New post   10 Jul 2018, 06:07
maaadhu wrote:
kartik222 wrote:
hi Bunuel,
how did you get to the ratio:
R/H=r/h=1/4 ??
Can you please elaborate?
thanks!!



kartik,

similar triangle property...

R/H = r/h

since 2R=H

r/h = 1/2

so h = 6.


hi

how can you deduce that 2R = H

the cone is twice as tall as it is wide at its greatest width, so

H = 2 x 2R = 4R, which implies

R / 4R = r / h = 1/4, and since, r = 3, h= 12

Isn't that ?

thanks
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Re: A right circular cone, twice as tall as it is wide at its [#permalink] New post   04 Aug 2019, 09:55
erictwendell wrote:
maaadhu wrote:
kartik222 wrote:
hi Bunuel,
how did you get to the ratio:
R/H=r/h=1/4 ??
Can you please elaborate?
thanks!!



kartik,

similar triangle property...

R/H = r/h

since 2R=H

r/h = 1/2

so h = 6.


I really find these problems very tough. Is there any reference to these problems ???


I recommend you to study the basics of 3D Geometries chapter

You can find the link of this from the GMAT Club Math book by bb & Bunuel here https://gmatclub.com/forum/math-3-d-geo ... ml#p792331
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Re: A right circular cone, twice as tall as it is wide at its [#permalink] New post   20 Jan 2023, 07:19
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