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If the volume of a cube is 1 cubic centimeter, then the distance from

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If the volume of a cube is 1 cubic centimeter, then the distance from [#permalink]

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New post 30 Nov 2017, 22:55
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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
[Reveal] Spoiler: OA

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Re: If the volume of a cube is 1 cubic centimeter, then the distance from [#permalink]

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New post 30 Nov 2017, 23:30
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


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If the volume of a cube is 1 cubic centimeter, then the distance from [#permalink]

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New post 01 Dec 2017, 07:47
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Bunuel wrote:
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

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.

Last edited by genxer123 on 01 Dec 2017, 07:53, edited 1 time in total.

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Re: If the volume of a cube is 1 cubic centimeter, then the distance from [#permalink]

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New post 01 Dec 2017, 07:53
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Expert's post
genxer123 wrote:
Bunuel wrote:
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

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:

\(L^2 + W^2+ H^2 = D^2\)

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 is not following BBCode (?). There is a long string of code that is not formatted.


Looks fine to me. Did you try reloading couple of times?
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 135673 [1], given: 12706

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If the volume of a cube is 1 cubic centimeter, then the distance from [#permalink]

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New post 01 Dec 2017, 12:49
Bunuel wrote:
genxer123 wrote:

P.S. Bunuel , I believe your signature is not following BBCode (?). There is a long string of code that is not formatted.


Looks fine to me. Did you try reloading couple of times?

Bunuel, just switched browsers. No idea what is wrong with the other, but this one works. Thanks!

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Re: If the volume of a cube is 1 cubic centimeter, then the distance from [#permalink]

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New post 02 Dec 2017, 07:28
Cube:
Length of diagonal between two vertices on the same plane= side*√2

And,Length of diagonal between two vertices in the different plane= side*√3.
This longer diagnol also passes through center point of cube at midway.

So, for a cube with volume 1 cubic centimeter, Each side = 1 cm.

So, Long Diagonal= √3 cm and midpoint is √3/2 cm from each vertex.

Hence, Ans D

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Kudos [?]: 19 [0], given: 20

Re: If the volume of a cube is 1 cubic centimeter, then the distance from   [#permalink] 02 Dec 2017, 07:28
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