Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In, a Hemisphere igloo, an Eskimo’s head just touches the ro [#permalink]
06 Jun 2010, 18:41

1

This post received KUDOS

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

85% (hard)

Question Stats:

47% (03:29) correct
53% (02:00) wrong based on 137 sessions

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

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 [ 3.75 KiB | Viewed 3923 times ]

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 \(H=\sqrt{R^2-r^2}=\sqrt{65^2-56^2}=33\).

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 . _________________

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. _________________

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 \(H=\sqrt{R^2-r^2}=\sqrt{65^2-56^2}=33\).

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. _________________

Re: In, a Hemisphere igloo, an Eskimo’s head just touches the ro [#permalink]
14 May 2014, 20:37

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.

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?

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. _________________

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.

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)

Re: In, a Hemisphere igloo, an Eskimo’s head just touches the ro [#permalink]
25 Jun 2015, 01:24

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

The Stanford interview is an alumni-run interview. You give Stanford your current address and they reach out to alumni in your area to find one that can interview you...

Originally, I was supposed to have an in-person interview for Yale in New Haven, CT. However, as I mentioned in my last post about how to prepare for b-school interviews...

Interested in applying for an MBA? In the fourth and final part of our live QA series with guest expert Chioma Isiadinso, co-founder of consultancy Expartus and former admissions...