The volume of a sphere with radius r is (4/3)*pi*r^3. A solid sphere

Hi,

the V = \(\frac{4}{3}pi*r^3\) and this volume weighs 40 pounds...

If it is a shell, you can find V of solid part by subtracting inner empty sphere from total..
V of entire sphere \(\frac{4}{3} * pi *(5r)^3\)and V of inner empty space = \(\frac{4}{3} * pi *(2r)^3\)..

so V of solid =\(\frac{4}{3} * pi *(5r)^3 - \frac{4}{3}* pi *(2r)^3........................ = \frac{4}{3}*pi*r^3*(125-8)....................... = \frac{4}{3}pi*r^3*117\)

NOW \(\frac{4}{3}pi*r^3\) weighs 40 pounds,
so \(\frac{4}{3}pi*r^3*117\) will weigh \(40*117 = 4680\)...

E
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A very tricky question
The answer is E
Weight of the sphere with radius r is =40
Now the radius of the outer shell =5r if we consider it as sphere then its weight would be 125 times 40=5000
Now the sphere with inner radius 2r will have 8 times 40 as weight =320
Subtract the weight ans we get the weight of the shell =5000-320=4680

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\(\frac {4}{3} \pi r^3\) = \(40\)

\(\frac{22}{7} r^3\) = \(30\)

\(\frac{11}{7} r^3\) = \(15\)

\(r^3\) = \(15*7/11\)

Now think about the semi circle



So, Volume will be = \(\frac{2}{3}*\frac{22}{7}*( 5r^3 - 2r^3)\)

Or Volume = \(\frac{2}{3}*\frac{22}{7}*( 117r^3)\)

We know r^3 = \(15*7/11\) ; so substitute it ....

Volume = \(\frac{4}{3}*\frac{22}{7}*( 117*15*\frac{7}{11})\) ( Check the highlighted part in the question we require the volume of the complete sphere )

Volume = 4,680

You may ask the question why the question mentions



It's given only to show that the thickness of the spherical ball (which is hollow from inside )

Hence I completely endorse answer option (E)
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The way you explained was really good, thanks a lot.
One small question, how did you consider v=w?

Hi..

It is not that the volume and weight is being equalled.
We are finding a relationship between the two as the weight will depend on the volume
Say you have a cake with a weight of 20 pounds and you cut 10 piece. What will be the weight of each piece. Ofcourse 20/10..
Here we are talking of volume and not pieces but finally that entire cake has a certain volume and by making 10 EQUAL parts, we are cutting the volume in 10 parts.
Otherwise volume*density= weight.
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Hi,

We can use the Density = Mass/Volume Formula. From the Statement Density = 40/\(\frac {4}{3} \pi r^3\)
As the above figure is made up of same material Density will be same. So,

40/\(\frac {4}{3} \pi r^3\) = x/\(\frac {4}{3} \pi (5r)^3\) - \(\frac {4}{3} \pi (2r)^3\)

x = 4680.

VeritasPrepKarishma , Bunuel I'm not getting the concept of this problem. Could you please explain this problem with a simple approach? Thanks.

The volume of a sphere with radius r is \(\frac {4}{3} \pi r^3\). A solid sphere of radius r that is made of a certain material weighs 40 pounds. The figure above shows half of a spherical shell made of the same material. What is the weight, in pounds, of the entire spherical shell if the outer radius is 5r and the inner radius is 2r?

A. 120
B. 360
C. 840
D. 1,080
E. 4,680

We have a sphere with radius of 5r with spherical empty space inside with radius of 2r.

The volume of a sphere with radius 5r is \(\frac {4}{3} \pi (5r)^3=125(\frac {4}{3} \pi r^3)\).

The volume of a sphere with radius 2r is \(\frac {4}{3} \pi (2r)^3=8(\frac {4}{3} \pi r^3)\).

The volume of solid material of the given hollow sphere = the volume of the sphere - the volume of the empty space = \(125(\frac {4}{3} \pi r^3)-8(\frac {4}{3} \pi r^3)=117(\frac {4}{3} \pi r^3)\).

Since solid sphere of radius r having the volume of \(\frac {4}{3} \pi r^3\), that is made of a certain material weighs 40 pounds, then our hollow sphere made of the same material and having the volume of \(117(\frac {4}{3} \pi r^3)\), will weigh \(117*40=4,680\).

Answer: E.

Hope it's clear.
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Bunuel For the volume of the sphere why are we taking "outer radius 5r" and "inner radius 2r" for the volume of the empty space? I can't identify it from the pic which one is the inner or outer radius, which one is solid sphere and spherical shell. Please explain these to me. Thanks.

We have a sphere with radius of 5r with spherical empty space inside with radius of 2r.

Here is a cross section:




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Hello Bunuel the reason why we realised that there is an empty space inside the sphere is because of the use of word 'shell' ?

It is a straightforward question, but the visual does not make it clear that the inner sphere is empty especially when the text of the question doesn't make it apparent. I was stuck on it for a while on GMATPrepExam, then assumed that the question is asking for the difference in volume adjusted weight. It is surprising because in PS questions, the stem is generally very clear and without any ambiguity. For example, even in the same question - formula of a sphere is provided. Most of the times, PS questions don't mind giving redundant information through text that is already there in the pic.

Bunuel VeritasKarishma Have you come across other questions for geometry such as these, in which text is silent on a critical information and one has to rely on visual approach?

The question says that it is a spherical shell. It means it is hollow inside. A sphere is the one which is solid.
Also the question mentions outer and inner radii. So that is a hint that it is hollow inside. This is a GMAT prep question so it means you are expected to know what a spherical shell is. The diagram doesn't tell us much.
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Hi VeritasKarishma Bunuel

Even i had this confusion. Based on the diagram and the shaded portion, it appears as if the sphere with radius has the mass, while the area between outer radius and inner radius is hollow. And this is not clear from the text as well. it could also mean that the radius 5r is ony covering it. In that case, the volume of mass would be by 2r sphere. Fortunately, there was no answer choice corresponding to it.

Hi BrentGMATPrepNow, question ask What is the weight, in pounds, of the entire (total weight) spherical shell (mean empty) if the outer radius is 5r and the inner radius is 2r (presumed solid if outer is empty)? My confusions are regarding correct part (inner/outer) of solid and hollow part (picture might not be accurate), total weight is not addition but deduction? Thanks Brent

The question: What is the weight, in pounds, of the entire spherical shell if the outer radius is 5r and the inner radius is 2r?
In other words, we have an empty spear with radius 2r INSIDE a sphere with radius 5r.
In other words, the sphere has a thickness of 3r.

That's said, I wouldn't worry about this question. It seems like other students are having a hard time interpreting the information as well.
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Ok thanks BrentGMATPrepNow. Don't think this is OG type of question due to the ambiguious wordings.