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 500,000 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:

The weight of a hollow sphere is directly dependent on its [#permalink]

Show Tags

12 Nov 2012, 16:16

1

This post received KUDOS

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

78% (02:10) correct
22% (01:27) wrong based on 163 sessions

HideShow timer Statistics

The weight of a hollow sphere is directly dependent on its surface area. The surface area of a sphere is 4π·R^2, where R is the radius of the sphere. If a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams, a hollow sphere of radius 0.3 cm made of the same metal would weigh how many grams?

If the large sphere is L and the small one is S, then the surface area of S = 4π·0.15^2, and the surface area of L = 4π·0.3^2. Thus the surface area of L is larger than that of S by a factor of 0.3^2/0.15^2 = (0.3/0.15^)2 = 2^2 = 4.

Could someone explain to me the reasoning behind putting the .3 over the .15 in the proportion part?

I think I understand the reason why they throw away the 4π, is it because they are both equal therefore not necessary in solving the problem? (because we are dealing with proportion in weight)

Re: The weight of a hollow sphere is directly dependent on its s [#permalink]

Show Tags

12 Nov 2012, 16:29

This is how I worked it out. Since the question states that Weight is dependent (proportional) to SA, therefore, weight = k (constant) X SA, then find k, use the same formula and you will obtain the answer.

The weight of a hollow sphere is directly dependent on its surface area. The surface area of a sphere is 4π·R^2, where R is the radius of the sphere. If a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams, a hollow sphere of radius 0.3 cm made of the same metal would weigh how many grams?

If the large sphere is L and the small one is S, then the surface area of S = 4π·0.15^2, and the surface area of L = 4π·0.3^2. Thus the surface area of L is larger than that of S by a factor of 0.3^2/0.15^2 = (0.3/0.15^)2 = 2^2 = 4.

Could someone explain to me the reasoning behind putting the .3 over the .15 in the proportion part?

I think I understand the reason why they throw away the 4π, is it because they are both equal therefore not necessary in solving the problem? (because we are dealing with proportion in weight)

The surface area of a sphere of radius 0.15 cm is \(4\pi{r^2}=4\pi{*0.15^2}\); The surface area of a sphere of radius 0.3=2*0.15 cm is \(4\pi{R^2}=4\pi{*2^2*0.15^2}=4*(4\pi{*0.15^2})\).

Since the surface are of the bigger sphere is 4 times the surface area of the smaller sphere and since the weight of a sphere is directly proportional to surface area, then the weight of the bigger sphere is 4 times the weight of the smaller sphere, so 4*8=32 grams.

Re: The weight of a hollow sphere is directly dependent on its [#permalink]

Show Tags

13 Nov 2012, 11:55

2

This post received KUDOS

When its directly proportional, it's all depends on the power of the variable that we are manipulating. if A = pi r^2 then area increases by the square of the multiple by which you increase the radius. if r becomes 2r then the area becomes 4A and if r becomes 1/2 r then area becomes 1/4A. Same thing can be applied if you are manipulating multiple variables.

For e.g. SA = 2 pi r^2 h r has powe of two and h has power of 1 if r increase to 2r and h increases to 2h, SA will increases by 2^2 * 2^1 = 8 times.

For e.g. V = 4/3 pi r^3 now the volume increase by the power of 3 of the mulitplier of r. if r becomes 2r volume goes up by 2^3 = 8 times.

Re: The weight of a hollow sphere is directly dependent on its [#permalink]

Show Tags

30 Sep 2013, 04:51

anon1 wrote:

The weight of a hollow sphere is directly dependent on its surface area. The surface area of a sphere is 4π·R^2, where R is the radius of the sphere. If a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams, a hollow sphere of radius 0.3 cm made of the same metal would weigh how many grams?

If the large sphere is L and the small one is S, then the surface area of S = 4π·0.15^2, and the surface area of L = 4π·0.3^2. Thus the surface area of L is larger than that of S by a factor of 0.3^2/0.15^2 = (0.3/0.15^)2 = 2^2 = 4.

Could someone explain to me the reasoning behind putting the .3 over the .15 in the proportion part?

I think I understand the reason why they throw away the 4π, is it because they are both equal therefore not necessary in solving the problem? (because we are dealing with proportion in weight)

Thinking simply: Weight is propotional to Surface Area .....(1) info given

Weight1 = (some constant) x Surface Area 1 (4π·R1^2)....(2)

Weight2 = (some constant) x Surface Area 2 (4π·R2^2)....(3)

Dividing equation (3) by equation (2) we get

Weight2 /8 gms = (0.30)^2 / (.15)^2

Weight2 = 8 x 4 = 32 gms
_________________

Thanks & Regards Yodee

‘A good plan violently executed now is better than a perfect plan executed next week.’ - General Georg S. Patton

Re: The weight of a hollow sphere is directly dependent on its [#permalink]

Show Tags

04 Aug 2015, 11:47

anon1 wrote:

The weight of a hollow sphere is directly dependent on its surface area. The surface area of a sphere is 4π·R^2, where R is the radius of the sphere. If a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams, a hollow sphere of radius 0.3 cm made of the same metal would weigh how many grams?

If the large sphere is L and the small one is S, then the surface area of S = 4π·0.15^2, and the surface area of L = 4π·0.3^2. Thus the surface area of L is larger than that of S by a factor of 0.3^2/0.15^2 = (0.3/0.15^)2 = 2^2 = 4.

Could someone explain to me the reasoning behind putting the .3 over the .15 in the proportion part?

I think I understand the reason why they throw away the 4π, is it because they are both equal therefore not necessary in solving the problem? (because we are dealing with proportion in weight)

Plug in easy values. Instead of 0.15 use 1 for the small radius and instead of 0.3 use 2 for the small radius. This makes the calculation much easier and it is still correct because the proportions are the same.

So 4pi * 1 = 4pi, and 4pi*4 = 16pi. Therefore the weight of the large sphere is 4 times more than the small one.
_________________

Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.

gmatclubot

Re: The weight of a hollow sphere is directly dependent on its
[#permalink]
04 Aug 2015, 11:47

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...