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In, a Hemisphere igloo, an Eskimo’s head just touches the ro

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In, a Hemisphere igloo, an Eskimo’s head just touches the ro  [#permalink]

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New post Updated on: 26 Mar 2014, 07:28
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In, a Hemisphere igloo, an Eskimo’s head just touches the roof when he stands erect at the centre of the floor, but his son can play over an area of 9856 square units without stooping. If the Eskimo’s height is 65 units, what is his son’s height?

A. 25 units
B. 33 units
C. 35 units
D. 37 units
E. Insufficient data

Originally posted by lnarayanan on 06 Jun 2010, 19:41.
Last edited by Bunuel on 26 Mar 2014, 07:28, edited 2 times in total.
Renamed the topic and edited the question.
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In, a Hemisphere igloo, an Eskimo’s head just touches the ro  [#permalink]

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New post 18 Nov 2010, 06:32
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lnarayanan wrote:
In, a Hemisphere igloo, an Eskimo’s head just touches the roof when he stands erect at the centre of the floor, but his son can play over an area of 9856 square units without stooping. If the Eskimo’s height is 65 units, what is his son’s height?

A) 25 units,
B) 33 units,
C) 35 units,
D) 37 units,
E) Insufficient data


Look at the diagram below:

Image

Now, the RADIUS of the igloo equals to the hight of the Eskimo, so \(R=65\). As the child can play over an area of 9,856 square units then the radius of this playing are is: \(playing \ area=\pi{r^2}=9,856\) --> \(r^2=\frac{9,856}{\pi}\) --> \(r\approx{56}\). Thus the child's height will be \(H=\sqrt{R^2-r^2}=\sqrt{65^2-56^2}=33\).

Answer: B.

Attachment:
AngleSemicircle.gif
AngleSemicircle.gif [ 3.75 KiB | Viewed 10245 times ]

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Re: PS - Polygons  [#permalink]

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New post 20 Nov 2010, 15:31
Bunuel wrote:
lnarayanan wrote:
In,a Hemisphere igloo,an Eskimo’s head just touches the roof when he stands erect at the centre of the floor, but his son can play over an area of 9856 square units without stooping.If the Eskimo’s height is 65 units,What is his son’s height?
A) 25 units , B) 33 units , C) 35 units , D) 37 units ,E) Insufficient data


Look at the diagram below:
Attachment:
AngleSemicircle.gif
Now, the RADIUS of the igloo equals to the hight of the Eskimo, so \(R=65\). As the child can play over an area of 9,856 square units then the radius of this playing are is: \(playing \ area=\pi{r^2}=9,856\) --> \(r^2=\frac{9,856}{\pi}\) --> \(r\approx{56}\). Thus the child's height will be \(H=\sqrt{R^2-r^2}=\sqrt{65^2-56^2}=33\).

Answer: B.



Hi Bunnel ,
Got 2 doubts here
1) Aint the area should be \(playing \ area=\pi{r^2}/2\) since its an hemisphere ?
2) While we are just concentrating on the right half should the area that of the quarter ?
M confused or may be i'm thinking in the wrong direction , Please explain .
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Re: PS - Polygons  [#permalink]

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New post 21 Nov 2010, 01:52
girisshhh84 wrote:
Bunuel wrote:
lnarayanan wrote:
In,a Hemisphere igloo,an Eskimo’s head just touches the roof when he stands erect at the centre of the floor, but his son can play over an area of 9856 square units without stooping.If the Eskimo’s height is 65 units,What is his son’s height?
A) 25 units , B) 33 units , C) 35 units , D) 37 units ,E) Insufficient data


Look at the diagram below:
Attachment:
AngleSemicircle.gif
Now, the RADIUS of the igloo equals to the hight of the Eskimo, so \(R=65\). As the child can play over an area of 9,856 square units then the radius of this playing are is: \(playing \ area=\pi{r^2}=9,856\) --> \(r^2=\frac{9,856}{\pi}\) --> \(r\approx{56}\). Thus the child's height will be \(H=\sqrt{R^2-r^2}=\sqrt{65^2-56^2}=33\).

Answer: B.



Hi Bunnel ,
Got 2 doubts here
1) Aint the area should be \(playing \ area=\pi{r^2}/2\) since its an hemisphere ?
2) While we are just concentrating on the right half should the area that of the quarter ?
M confused or may be i'm thinking in the wrong direction , Please explain .


I think you are just confused with the diagram:

Hemisphere is half of a sphere and the diagram gives the cross section of it. But the base of a hemisphere (the base of an igloo) is still a circle, so the playing area of the child is a circle limited by his height.
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Re: PS - Polygons  [#permalink]

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New post 21 Nov 2010, 14:15
Surface area of sphere is - 4 * pi (r)^2

Shoulden't area of hemisphere be -2 * pi (r)^2 ????
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Re: PS - Polygons  [#permalink]

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New post 26 Mar 2014, 06:12
Bunuel wrote:
lnarayanan wrote:
In,a Hemisphere igloo,an Eskimo’s head just touches the roof when he stands erect at the centre of the floor, but his son can play over an area of 9856 square units without stooping.If the Eskimo’s height is 65 units,What is his son’s height?
A) 25 units , B) 33 units , C) 35 units , D) 37 units ,E) Insufficient data


Look at the diagram below:
Attachment:
AngleSemicircle.gif
Now, the RADIUS of the igloo equals to the hight of the Eskimo, so \(R=65\). As the child can play over an area of 9,856 square units then the radius of this playing are is: \(playing \ area=\pi{r^2}=9,856\) --> \(r^2=\frac{9,856}{\pi}\) --> \(r\approx{56}\). Thus the child's height will be \(H=\sqrt{R^2-r^2}=\sqrt{65^2-56^2}=33\).

Answer: B.



Hi Bunuel,
you can remove the approx sign.

9856/pi = 9856*7/22 = 56 exactly. :)
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Re: PS - Polygons  [#permalink]

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New post 26 Mar 2014, 07:38
seabhi wrote:
Bunuel wrote:
lnarayanan wrote:
In,a Hemisphere igloo,an Eskimo’s head just touches the roof when he stands erect at the centre of the floor, but his son can play over an area of 9856 square units without stooping.If the Eskimo’s height is 65 units,What is his son’s height?
A) 25 units , B) 33 units , C) 35 units , D) 37 units ,E) Insufficient data


Look at the diagram below:
Attachment:
AngleSemicircle.gif
Now, the RADIUS of the igloo equals to the hight of the Eskimo, so \(R=65\). As the child can play over an area of 9,856 square units then the radius of this playing are is: \(playing \ area=\pi{r^2}=9,856\) --> \(r^2=\frac{9,856}{\pi}\) --> \(r\approx{56}\). Thus the child's height will be \(H=\sqrt{R^2-r^2}=\sqrt{65^2-56^2}=33\).

Answer: B.



Hi Bunuel,
you can remove the approx sign.

9856/pi = 9856*7/22 = 56 exactly. :)


\(\pi=3.141592653589793238462643383279502884...\) (it goes on forever) is an irrational number, it cannot be represented as the ratio of two integers.

\(\frac{22}{7}=3.1428...\) is only an approximate value of \(\pi\).

P.S. In that sense this is not a good quality question.
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Re: In, a Hemisphere igloo, an Eskimo’s head just touches the ro  [#permalink]

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New post 14 May 2014, 21:37
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9856 = 7*64*22. 22/7 is what we call as pi (at least an approximation of pi). So 9856/pi = 7*64*22/(22/7) = 7*7*64 = 3136. 3136 is 56^2. Easier than approximation.
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Re: PS - Polygons  [#permalink]

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New post 17 May 2014, 11:03
Bunuel wrote:
lnarayanan wrote:
In,a Hemisphere igloo,an Eskimo’s head just touches the roof when he stands erect at the centre of the floor, but his son can play over an area of 9856 square units without stooping.If the Eskimo’s height is 65 units,What is his son’s height?
A) 25 units , B) 33 units , C) 35 units , D) 37 units ,E) Insufficient data


Look at the diagram below:
Attachment:
AngleSemicircle.gif
Now, the RADIUS of the igloo equals to the hight of the Eskimo, so \(R=65\). As the child can play over an area of 9,856 square units then the radius of this playing are is: \(playing \ area=\pi{r^2}=9,856\) --> \(r^2=\frac{9,856}{\pi}\) --> \(r\approx{56}\). Thus the child's height will be \(H=\sqrt{R^2-r^2}=\sqrt{65^2-56^2}=33\).

Answer: B.


I had the solution to this problem in under a minute or so but couldn't actually compute the answer. Are we really supposed to be able to solve \(\sqrt{\frac{9,856}{\pi}}\) without a calculator? That seems like a stretch to me, but maybe I'm missing something... Is assuming \(\pi \approx \frac{22}{7}\) a standard assumption for this exam?
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Re: PS - Polygons  [#permalink]

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New post 18 May 2014, 01:01
GordonFreeman wrote:
Bunuel wrote:
lnarayanan wrote:
In,a Hemisphere igloo,an Eskimo’s head just touches the roof when he stands erect at the centre of the floor, but his son can play over an area of 9856 square units without stooping.If the Eskimo’s height is 65 units,What is his son’s height?
A) 25 units , B) 33 units , C) 35 units , D) 37 units ,E) Insufficient data


Look at the diagram below:
Attachment:
AngleSemicircle.gif
Now, the RADIUS of the igloo equals to the hight of the Eskimo, so \(R=65\). As the child can play over an area of 9,856 square units then the radius of this playing are is: \(playing \ area=\pi{r^2}=9,856\) --> \(r^2=\frac{9,856}{\pi}\) --> \(r\approx{56}\). Thus the child's height will be \(H=\sqrt{R^2-r^2}=\sqrt{65^2-56^2}=33\).

Answer: B.


I had the solution to this problem in under a minute or so but couldn't actually compute the answer. Are we really supposed to be able to solve \(\sqrt{\frac{9,856}{\pi}}\) without a calculator? That seems like a stretch to me, but maybe I'm missing something... Is assuming \(\pi \approx \frac{22}{7}\) a standard assumption for this exam?


As I've written above this is not a proper GMAT question because we need to approximate \(\pi\) to get the answer, while the question does not ask about approximate height. GMAT would never do that.

As for \(\pi \approx \frac{22}{7}\): this is a good/standard approximation for some problems asking for an approximate answer.
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Re: PS - Polygons  [#permalink]

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New post 18 May 2014, 17:39
Bunuel wrote:
I had the solution to this problem in under a minute or so but couldn't actually compute the answer. Are we really supposed to be able to solve \(\sqrt{\frac{9,856}{\pi}}\) without a calculator? That seems like a stretch to me, but maybe I'm missing something... Is assuming \(\pi \approx \frac{22}{7}\) a standard assumption for this exam?


As I've written above this is not a proper GMAT question because we need to approximate \(\pi\) to get the answer, while the question does not ask about approximate height. GMAT would never do that.

As for \(\pi \approx \frac{22}{7}\): this is a good/standard approximation for some problems asking for an approximate answer.[/quote]

Are we expected to know the square root of 3,136 is 56 off the top of our heads as well? I'm just trying to get a sense for what I need to memorize.
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Re: PS - Polygons  [#permalink]

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New post 18 May 2014, 23:30
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GordonFreeman wrote:
Bunuel wrote:
I had the solution to this problem in under a minute or so but couldn't actually compute the answer. Are we really supposed to be able to solve \(\sqrt{\frac{9,856}{\pi}}\) without a calculator? That seems like a stretch to me, but maybe I'm missing something... Is assuming \(\pi \approx \frac{22}{7}\) a standard assumption for this exam?


As I've written above this is not a proper GMAT question because we need to approximate \(\pi\) to get the answer, while the question does not ask about approximate height. GMAT would never do that.

As for \(\pi \approx \frac{22}{7}\): this is a good/standard approximation for some problems asking for an approximate answer.

Are we expected to know the square root of 3,136 is 56 off the top of our heads as well? I'm just trying to get a sense for what I need to memorize.


No. There is very little memorization that is expected from you. But what is expected is that you will reduce the calculations you need to do using reasoning.

If such a question does come in GMAT, the number will be and easier than 9856. Also, you can easily solve with 9856 too.

\(r^2 = 9856/pi = 9856*7/22\)

r must be an integer otherwise this calculation will become far too cumbersome for GMAT. So 9856 will be completely divisible by 22.
Also, 9856 must have 7 as a factor since perfect squares have powers of prime factors in pairs.
So let's try to split 9856 into factors. We already know that it must have 7 as a factor and 11 as a factor (to be divisible by 22)

\(9856 = 7*1408 = 7*11*128 = 7*11*2^7\) (you must know that 2^7 = 128)

\(r^2 = \frac{7*11*2^7}{2*11} = 7*2^3 = 56\)

Again, \(H = \sqrt{65^2 - 56^2} = \sqrt{(65+56)(65 - 56)} = \sqrt{121*9} = 11*3 = 33\)
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Re: In, a Hemisphere igloo, an Eskimo’s head just touches the ro  [#permalink]

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Re: In, a Hemisphere igloo, an Eskimo’s head just touches the ro   [#permalink] 22 Apr 2019, 21:18
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