Forum Home > GMAT > Quantitative > Problem Solving (PS)
Events & Promotions
|
|
GMAT Club Daily Prep
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Dec 1310:00 AM EST
-12:00 PM EST
Have you ever wondered how to score a PERFECT 805 on the GMAT? Meet Julia, a banking professional who used the Target Test Prep course to achieve this incredible feat. Julia's story is nothing short of an inspiration.
Dec 1410:00 AM EST
-12:00 PM EST
Think a 100% GMAT Verbal score is out of your reach? Target Test Prep will make you think again! Our course uses techniques such as topical study and spaced repetition to maximize knowledge retention and make studying simple and fun.
Dec 1411:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.- Dec 15
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Dec 1608:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!
nusmavrik
Joined: 26 Nov 2009
Last visit: 03 Apr 2022
Posts: 468
Given Kudos: 36
Status:Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Location: Singapore
Concentration: General Management, Finance
Schools: Chicago Booth - Class of 2015
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
37% (03:02) correct
63%
(02:55)
wrong
based on 73
sessions
History
Date
Time
Result
Not Attempted Yet
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 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
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
21 Aug 2010, 01:12
Kudos
Bookmarks
nusmavrik
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
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
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 !!!
21 Aug 2010, 03:11
Kudos
Bookmarks
nusmavrik
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 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 (\(\pi\), \(\sqrt{3}\)).
