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

Show Tags

21 Aug 2010, 00:20

3

This post received KUDOS

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

31% (03:53) correct
69% (02:48) wrong based on 118 sessions

HideShow timer Statistics

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?

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

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed." - Bernard Edmonds

A person who is afraid of Failure can never succeed -- Amneet Padda

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]

Show Tags

02 Jun 2013, 05:22

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, 22:56, 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? : /

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...

This highly influential bestseller was first published over 25 years ago. I had wanted to read this book for a long time and I finally got around to it...