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The dimensions of a rectangular solid are 4 inches, 5 inches [#permalink]
20 Aug 2010, 23:20

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A

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Difficulty:

95% (hard)

Question Stats:

31% (03:53) correct
69% (02:50) wrong based on 117 sessions

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?

Re: hard - geometry ! [#permalink]
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?

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

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.

Answer: A.

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

It is definately not a Gmat Question _________________

I will give a Fight till the End

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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

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.

Answer: A.

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

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.

Answer: A.

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 --> \(d=\sqrt{(4^2+4^2)+4^2}\).

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

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.

Answer: A.

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

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.

Answer: A.

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.

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. _________________

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.

Answer: A.

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? : /

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