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# The dimensions of a rectangular solid are 4 inches, 5 inches

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The dimensions of a rectangular solid are 4 inches, 5 inches [#permalink]

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20 Aug 2010, 23:20
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What is the difficulty level - hard, medium or simple?

The dimensions of a rectangular solid are 4 inches, 5 inches, and 8 inches. If a cube, a side of which is equal to one of the dimensions of the rectangular solid, is placed entirely within the sphere just large enough to hold the cube, what the ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube?

(A) 10:17
(B) 2:5
(C) 5:16
(D) 25:7
(E) 32:25
[Reveal] Spoiler: OA

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Last edited by nusmavrik on 21 Aug 2010, 00:59, edited 3 times in total.
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Re: hard - geometry ! [#permalink]

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21 Aug 2010, 00:12
nusmavrik wrote:
This must be 750 level. Please help me crack this in 2 mins or less.

The dimensions of a rectangular solid are 4 inches, 5 inches, and 8 inches. If a cube, a side of which is equal to one of the dimensions of the rectangular solid, is placed entirely within the sphere just large enough to hold the cube, what the ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube?

(A) 2:3
(B) 2:5
(C) 5:16
(D) 25:7
(E) 32:25

[Reveal] Spoiler:
A. 2:3 (approx)

I am pretty sure about the thing marked in red. It should state rectangular solid and not the sphere........if that is the case find the answer below.....

cube with side greater than 4 cannot fit in the cuboid (4,5 & 8)

so cube has volume of 4^3 = 64

volume of cuboid = 4*5*8 = 160

area not occupied by cube = 96

so the answer is 64:96 or 2:3 (its perfect 2:3) not approx as stated in your OA

hope this helps !!!
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Re: geometry ! [#permalink]

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21 Aug 2010, 02:11
nusmavrik wrote:
What is the difficulty level - hard, medium or simple?

The dimensions of a rectangular solid are 4 inches, 5 inches, and 8 inches. If a cube, a side of which is equal to one of the dimensions of the rectangular solid, is placed entirely within the sphere just large enough to hold the cube, what the ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube?

(A) 10:17
(B) 2:5
(C) 5:16
(D) 25:7
(E) 32:25

The cube must have a side of 4 inches to fit in rectangular solid. Now as the cube is inscribed in sphere then the radius of this sphere equals to half of the cube's diagonal --> $$d=\sqrt{4^2+4^2+4^2}=4\sqrt{3}$$, so radius of the sphere is $$r=\frac{d}{2}=2\sqrt{3}$$

Volume of the cube will be $${volume}_{cube}={side}^3=4^3=64$$;
Volume of the sphere will be $${volume}_{sphere}=\frac{4}{3}\pi{r^3}=\frac{4}{3}\pi{24\sqrt{3}}={32\sqrt{3}\pi}$$.

The ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube is $$\frac{64}{32\sqrt{3}\pi-64}$$, and after some calculations you'll get that this ratio is closer to 10/17 than to any other ratios in answer choices.

P.S. I really doubt that GMAT would offer such a problem, as it requires lengthy calculations and bad approximations ($$\pi$$, $$\sqrt{3}$$).
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Re: geometry ! [#permalink]

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21 Aug 2010, 12:50
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It is definately not a Gmat Question
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Re: geometry ! [#permalink]

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24 Aug 2010, 07:56
Bunuel wrote:
nusmavrik wrote:
What is the difficulty level - hard, medium or simple?

The dimensions of a rectangular solid are 4 inches, 5 inches, and 8 inches. If a cube, a side of which is equal to one of the dimensions of the rectangular solid, is placed entirely within the sphere just large enough to hold the cube, what the ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube?

(A) 10:17
(B) 2:5
(C) 5:16
(D) 25:7
(E) 32:25

The cube must have a side of 4 inches to fit in rectangular solid. Now as the cube is inscribed in sphere then the radius of this sphere equals to half of the cube's diagonal --> $$d=\sqrt{4^2+4^2+4^2}=4\sqrt{3}$$, so radius of the sphere is $$r=\frac{d}{2}=2\sqrt{3}$$

Volume of the cube will be $$volume_{cube}={side}^3=4^3=64$$;
Volume of the sphere will be $${volume}_{sphere}=\frac{4}{3}\pi{r^3}=\frac{4}{3}\pi{24\sqrt{3}}={32\sqrt{3}\pi}$$.

The ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube is $$\frac{64}{32\sqrt{3}\pi-64}$$, and after some calculations you'll get that this ratio is closer to 10/17 than to any other ratios in answer choices.

P.S. I really doubt that GMAT would offer such a problem, as it requires lengthy calculations and bad approximations ($$\pi$$, $$\sqrt{3}$$).

thanx nusma for posting sucha question ...however pls explain why do we have
third 4^2 .... in calculation of diagonal .. I am surely missing something
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Re: geometry ! [#permalink]

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24 Aug 2010, 08:12
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gauravnagpal wrote:
Bunuel wrote:
nusmavrik wrote:
What is the difficulty level - hard, medium or simple?

The dimensions of a rectangular solid are 4 inches, 5 inches, and 8 inches. If a cube, a side of which is equal to one of the dimensions of the rectangular solid, is placed entirely within the sphere just large enough to hold the cube, what the ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube?

(A) 10:17
(B) 2:5
(C) 5:16
(D) 25:7
(E) 32:25

The cube must have a side of 4 inches to fit in rectangular solid. Now as the cube is inscribed in sphere then the radius of this sphere equals to half of the cube's diagonal --> $$d=\sqrt{4^2+4^2+4^2}=4\sqrt{3}$$, so radius of the sphere is $$r=\frac{d}{2}=2\sqrt{3}$$

Volume of the cube will be $$volume_{cube}={side}^3=4^3=64$$;
Volume of the sphere will be $${volume}_{sphere}=\frac{4}{3}\pi{r^3}=\frac{4}{3}\pi{24\sqrt{3}}={32\sqrt{3}\pi}$$.

The ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube is $$\frac{64}{32\sqrt{3}\pi-64}$$, and after some calculations you'll get that this ratio is closer to 10/17 than to any other ratios in answer choices.

P.S. I really doubt that GMAT would offer such a problem, as it requires lengthy calculations and bad approximations ($$\pi$$, $$\sqrt{3}$$).

thanx nusma for posting sucha question ...however pls explain why do we have
third 4^2 .... in calculation of diagonal .. I am surely missing something

Because it's not a diagonal of a face, it's a diagonal of a cube.

Diagonal of a face of a cube ($$=\sqrt{4^2+4^2}$$) and the side of a cube ($$=4$$) form right triangle, hypotenuse of this right triangle would be diagonal of a cube --> $$d=\sqrt{(4^2+4^2)+4^2}$$.

Hope it's clear.
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Re: geometry ! [#permalink]

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27 Aug 2010, 23:17
BTW what is source of such difficult problem
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Re: geometry ! [#permalink]

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20 Nov 2010, 21:03
prashantbacchewar wrote:
BTW what is source of such difficult problem

I m not able to understand, why the side of the cube is 4 ? It is not mentioned that the cube is inside the rectangular box
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21 Nov 2010, 00:40
sap wrote:
prashantbacchewar wrote:
BTW what is source of such difficult problem

I m not able to understand, why the side of the cube is 4 ? It is not mentioned that the cube is inside the rectangular box

Yes, the question should mention that the cube can be entirely placed in the rectangular solid.
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Re: geometry ! [#permalink]

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21 Nov 2010, 13:32
If cube is to be fitted in sphere, it will take 10 minutes to fit it within it

if cube is to be fitted in Rectangular solid, only then its possible to solve it within 2 mins.

Any case it is 700 plus level question.

My answer in scene 1 is - 10/17
My answer in scene2 is - 2/3
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Re: The dimensions of a rectangular solid are 4 inches, 5 inches [#permalink]

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02 Jun 2013, 04:22
nusmavrik wrote:
What is the difficulty level - hard, medium or simple?

The dimensions of a rectangular solid are 4 inches, 5 inches, and 8 inches. If a cube, a side of which is equal to one of the dimensions of the rectangular solid, is placed entirely within the sphere just large enough to hold the cube, what the ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube?

(A) 10:17
(B) 2:5
(C) 5:16
(D) 25:7
(E) 32:25

Last edited by cumulonimbus on 29 May 2014, 21:56, edited 1 time in total.
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Re: geometry ! [#permalink]

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29 May 2014, 21:54
Bunuel wrote:
nusmavrik wrote:
What is the difficulty level - hard, medium or simple?

The dimensions of a rectangular solid are 4 inches, 5 inches, and 8 inches. If a cube, a side of which is equal to one of the dimensions of the rectangular solid, is placed entirely within the sphere just large enough to hold the cube, what the ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube?

(A) 10:17
(B) 2:5
(C) 5:16
(D) 25:7
(E) 32:25

The cube must have a side of 4 inches to fit in rectangular solid. Now as the cube is inscribed in sphere then the radius of this sphere equals to half of the cube's diagonal --> $$d=\sqrt{4^2+4^2+4^2}=4\sqrt{3}$$, so radius of the sphere is $$r=\frac{d}{2}=2\sqrt{3}$$

Volume of the cube will be $${volume}_{cube}={side}^3=4^3=64$$;
Volume of the sphere will be $${volume}_{sphere}=\frac{4}{3}\pi{r^3}=\frac{4}{3}\pi{24\sqrt{3}}={32\sqrt{3}\pi}$$.

The ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube is $$\frac{64}{32\sqrt{3}\pi-64}$$, and after some calculations you'll get that this ratio is closer to 10/17 than to any other ratios in answer choices.

P.S. I really doubt that GMAT would offer such a problem, as it requires lengthy calculations and bad approximations ($$\pi$$, $$\sqrt{3}$$).

Hi Bunnel,

Why will the answer be different if I use "a" as the side of the cube.

a^3 gets cancelled, and i get, 1/(4*pi*3^(1/2)) = 7/(88*3^(1/2)-7) = .04
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Re: geometry ! [#permalink]

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30 May 2014, 00:25
cumulonimbus wrote:
Bunuel wrote:
nusmavrik wrote:
What is the difficulty level - hard, medium or simple?

The dimensions of a rectangular solid are 4 inches, 5 inches, and 8 inches. If a cube, a side of which is equal to one of the dimensions of the rectangular solid, is placed entirely within the sphere just large enough to hold the cube, what the ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube?

(A) 10:17
(B) 2:5
(C) 5:16
(D) 25:7
(E) 32:25

The cube must have a side of 4 inches to fit in rectangular solid. Now as the cube is inscribed in sphere then the radius of this sphere equals to half of the cube's diagonal --> $$d=\sqrt{4^2+4^2+4^2}=4\sqrt{3}$$, so radius of the sphere is $$r=\frac{d}{2}=2\sqrt{3}$$

Volume of the cube will be $${volume}_{cube}={side}^3=4^3=64$$;
Volume of the sphere will be $${volume}_{sphere}=\frac{4}{3}\pi{r^3}=\frac{4}{3}\pi{24\sqrt{3}}={32\sqrt{3}\pi}$$.

The ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube is $$\frac{64}{32\sqrt{3}\pi-64}$$, and after some calculations you'll get that this ratio is closer to 10/17 than to any other ratios in answer choices.

P.S. I really doubt that GMAT would offer such a problem, as it requires lengthy calculations and bad approximations ($$\pi$$, $$\sqrt{3}$$).

Hi Bunnel,

Why will the answer be different if I use "a" as the side of the cube.

a^3 gets cancelled, and i get, 1/(4*pi*3^(1/2)) = 7/(88*3^(1/2)-7) = .04

Please, show your work so that it's clear how you get the expression in your post.

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Re: geometry ! [#permalink]

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11 Jun 2014, 09:49
Hi Bunuel,
Y the cube must have a side of 4 inches only to fit in? Please help me understand.
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Re: geometry ! [#permalink]

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12 Jun 2014, 06:57
maggie27 wrote:
Hi Bunuel,
Y the cube must have a side of 4 inches only to fit in? Please help me understand.

Because the dimensions of a rectangular solid in which we should place a cube are 4 inches, 5 inches, and 8 inches. If the cube's side is more than 4 it won't fit in the rectangular solid.
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Re: geometry ! [#permalink]

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15 Jun 2014, 12:39
Bunuel wrote:
nusmavrik wrote:
What is the difficulty level - hard, medium or simple?

The dimensions of a rectangular solid are 4 inches, 5 inches, and 8 inches. If a cube, a side of which is equal to one of the dimensions of the rectangular solid, is placed entirely within the sphere just large enough to hold the cube, what the ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube?

(A) 10:17
(B) 2:5
(C) 5:16
(D) 25:7
(E) 32:25

The cube must have a side of 4 inches to fit in rectangular solid. Now as the cube is inscribed in sphere then the radius of this sphere equals to half of the cube's diagonal --> $$d=\sqrt{4^2+4^2+4^2}=4\sqrt{3}$$, so radius of the sphere is $$r=\frac{d}{2}=2\sqrt{3}$$

Volume of the cube will be $${volume}_{cube}={side}^3=4^3=64$$;
Volume of the sphere will be $${volume}_{sphere}=\frac{4}{3}\pi{r^3}=\frac{4}{3}\pi{24\sqrt{3}}={32\sqrt{3}\pi}$$.

The ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube is $$\frac{64}{32\sqrt{3}\pi-64}$$, and after some calculations you'll get that this ratio is closer to 10/17 than to any other ratios in answer choices.

P.S. I really doubt that GMAT would offer such a problem, as it requires lengthy calculations and bad approximations ($$\pi$$, $$\sqrt{3}$$).

I was stuck. The answer here is 0.49 (on calculation). 2/5 = 0.4 and 10/17=0.59. Not sure which one is closer? : /
Re: geometry !   [#permalink] 15 Jun 2014, 12:39
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