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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]

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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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Re: Leak of water from the cone [#permalink]

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07 Oct 2009, 01:53
(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]

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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?
Attachment:

Untitled.png [ 17.59 KiB | Viewed 6178 times ]
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.

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Re: Leak of water from the cone [#permalink]

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07 Oct 2009, 06:14
... misread the second statement.. sorry bout that..

it is A.
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Re: Leak of water from the cone [#permalink]

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30 Jun 2011, 23:03
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]

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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: Leak of water from the cone [#permalink]

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09 Sep 2011, 22:30
Thanks Bunuel and testprepdublin.
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Re: Leak of water from the cone [#permalink]

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01 Oct 2011, 07:19
Bunuel,

You have taken the property of entire cone (i mean r and h ratio) and applied to the partial cone.

Is this because the cone is right circular cone?

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Re: A right circular cone, twice as tall as it is wide at its [#permalink]

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27 Jul 2012, 23:16
Answer must be (E) as no statements individually can give either the value of radius 'r' or height 'h'

Time required (in minutes) to empty the cone = (60/2)* pie * r^2*h
h= 4r

1) Now the original question states that the cone is "partially filled" & not full filled. Thus statement 1 won't suffice.
2) Water level is 4 cm below the full height - Thus not sufficient.

1+2) hr= 3h + 4, which is still insufficient .

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Re: A right circular cone, twice as tall as it is wide at its [#permalink]

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28 Jul 2012, 02:19
fameatop wrote:
Answer must be (E) as no statements individually can give either the value of radius 'r' or height 'h'

Time required (in minutes) to empty the cone = (60/2)* pie * r^2*h
h= 4r

1) Now the original question states that the cone is "partially filled" & not full filled. Thus statement 1 won't suffice.
2) Water level is 4 cm below the full height - Thus not sufficient.

1+2) hr= 3h + 4, which is still insufficient .

OA for this question is A, not E, Check this: a-right-circular-cone-twice-as-tall-as-it-is-wide-at-its-84927.html#p636504
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Re: A right circular cone, twice as tall as it is wide at its [#permalink]

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19 Aug 2012, 01:28
hi Bunuel,
how did you get to the ratio:
R/H=r/h=1/4 ??
thanks!!
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Re: A right circular cone, twice as tall as it is wide at its [#permalink]

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22 Jul 2013, 06:39
From 100 hardest questions.
Bumping for review and further discussion.
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Re: A right circular cone, twice as tall as it is wide at its [#permalink]

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23 Jul 2013, 06:05
kartik222 wrote:
hi Bunuel,
how did you get to the ratio:
R/H=r/h=1/4 ??
thanks!!

kartik,

similar triangle property...

R/H = r/h

since 2R=H

r/h = 1/2

so h = 6.
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Re: Leak of water from the cone [#permalink]

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01 Nov 2013, 03:13
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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?

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=9pi=pi*r^2 --> r=3 --> R/H=r/h=1/4 --> h=12 --> V=1/3*pi*r^2*h=36*pi

Leak rate 2 cubic centimeters per hour --> 36pi/2

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.

Hello Bunuel,

For the second statement , we have the depth of the cone as 4 cms which implies that the Width of the water in the cone at that point is 2cm: Implying the radius to be 1 cm and ; from this the volume of water in the cone can be derived and also the leak rate.

Am i missing something over here ?
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Re: A right circular cone, twice as tall as it is wide at its [#permalink]

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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]

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16 Nov 2013, 10:44
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]

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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.

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]

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03 Dec 2013, 01:39
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.

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: Leak of water from the cone [#permalink]

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06 Dec 2013, 14:34
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.

Would you please remind me why you divided by 3 in volume calc?
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Re: A right circular cone, twice as tall as it is wide at its [#permalink]

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07 Dec 2013, 00:51
kartik222 wrote:
hi Bunuel,
how did you get to the ratio:
R/H=r/h=1/4 ??
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 ???
Re: A right circular cone, twice as tall as it is wide at its   [#permalink] 07 Dec 2013, 00:51

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