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.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
Apr 3001:30 AM EDT
-02:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
May 0306:30 AM PDT
-08:30 AM PDT
Verbal trouble on GMAT? Fix it NOW! Join Sunita Singhvi for a focused webinar on actionable strategies to boost your Verbal score and take your performance to the next level.
30 Nov 2017, 23:55
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
66% (01:08) correct
34%
(01:06)
wrong
based on 102
sessions
History
Date
Time
Result
Not Attempted Yet
If the volume of a cube is 1 cubic centimeter, then the distance from any vertex to the center point inside the cube is
(A) 1/2 cm
(B) √2/2 cm
(C) √2 cm
(D) √3/2 cm
(E) √3 cm
(A) 1/2 cm
(B) √2/2 cm
(C) √2 cm
(D) √3/2 cm
(E) √3 cm
01 Dec 2017, 00:30
Kudos
Bookmarks
D
Side of the cube =1 cm
Distance between the opposite vertices = sqrt(3)
Hence the distance between the vertex to the center of the cube =sqrt(3)/2
Sent from my iPhone using GMAT Club Forum
Side of the cube =1 cm
Distance between the opposite vertices = sqrt(3)
Hence the distance between the vertex to the center of the cube =sqrt(3)/2
Sent from my iPhone using GMAT Club Forum
Bunuel
From one vertex of a cube to the center is half the length of the "space" diagonal that runs from one vertex, through the center, to a vertex on the opposite side.
The length of the space diagonal, D, is found with a variation on the Pythagorean theorem:
\(L^2 + W^2+ H^2 = D^2\)
Find side length(s):
The cube's volume, in cubic centimeters, is
\(s^3 = 1\)
\(\sqrt[3]{s^3} = \sqrt[3]{1}\)
So \(s = 1\)
Length, width, and height = 1
Using space diagonal formula:
\(1^2 + 1^2 + 1^2 = D^2\)
\(3 = D^2\)
\(\sqrt{3} = \sqrt{D^2}\)
\(D = \sqrt{3}\)
We need half that distance:
\(\frac{\sqrt{3}}{2}\)
Answer D
P.S. Bunuel , I believe your signature, or something in your posts, is not following BBCode (?). There is a long string of code that is not formatted.
