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 1511:00 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!- Dec 17
09:00 AM PST
-10:00 AM PST
Join Manhattan Prep instructor Whitney Garner for a fun—and thorough—review of logic-based (non-math) problems, with a particular emphasis on Data Sufficiency and Two-Parts. - Dec 18
10:00 AM PST
-11:00 AM PST
Here is the essential guide to securing scholarships as an MBA student! In this video, we explore the various types of scholarships available, including need-based and merit-based options.
24 Mar 2020, 09:49
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
83% (01:05) correct
17%
(01:09)
wrong
based on 46
sessions
History
Date
Time
Result
Not Attempted Yet
If the volume of one cube is 8 times greater than that of another, what is the ratio of the area of one square face of the larger cube to that of the smaller cube?
(A) 16:1
(B) 12:1
(C) 8:1
(D) 4:1
(E) 2:1
(A) 16:1
(B) 12:1
(C) 8:1
(D) 4:1
(E) 2:1
25 Mar 2020, 01:54
Kudos
Bookmarks
SajjadAhmad
If the volume of one cube is 8 times greater than that of another, what is the ratio of the area of one square face of the larger cube to that of the smaller cube?
(A) 16:1
(B) 12:1
(C) 8:1
(D) 4:1
(E) 2:1
(A) 16:1
(B) 12:1
(C) 8:1
(D) 4:1
(E) 2:1
Solution:
Let x be the side of the smaller cube.
- • The volume of the smaller cube = \(x^3\)
• The volume of the bigger cube = \(8*x^3\)
- o Side of the bigger cube =\(∛(8*x^3) = 2x\)
• Area of one face of larger cube = \(2x*2x=4x^2\)
• Required ratio = \(\frac{4x^2}{x^2}=\frac{4}{1}=4: 1\)
31 Mar 2020, 23:07
Kudos
Bookmarks
This is a very simple question on the concept of Mensuration.
Let the sides of the two cubes be A inches and B inches respectively. The question says that the volume of one cube, say cube A, is 8 times greater than that of another, say cube B.
So, \(A^3\) = 8 * \(B^3\). Reorganising this equation, we have,
\(\frac{A^3 }{ B^3}\) = \(\frac{8}{1}\). Taking cube root on both sides, we have,
\(\frac{A }{ B}\) = \(\frac{2}{1}\).
This means, A = 2B i.e. the side of cube A is double the side of cube B.
The area of one square face of cube B = B^2 and the area of one square face of cube A = \((2B)^2\) = 4\(B^2\).
Since A is the larger cube, required ratio = \(\frac{4B^2 }{ B^2}\) = \(\frac{4}{1}\) or 4:1.
The correct answer option is D.
Hope that helps!
Let the sides of the two cubes be A inches and B inches respectively. The question says that the volume of one cube, say cube A, is 8 times greater than that of another, say cube B.
So, \(A^3\) = 8 * \(B^3\). Reorganising this equation, we have,
\(\frac{A^3 }{ B^3}\) = \(\frac{8}{1}\). Taking cube root on both sides, we have,
\(\frac{A }{ B}\) = \(\frac{2}{1}\).
This means, A = 2B i.e. the side of cube A is double the side of cube B.
The area of one square face of cube B = B^2 and the area of one square face of cube A = \((2B)^2\) = 4\(B^2\).
Since A is the larger cube, required ratio = \(\frac{4B^2 }{ B^2}\) = \(\frac{4}{1}\) or 4:1.
The correct answer option is D.
Hope that helps!
