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gmatnmein2010: Master | Next Rank: 500 Posts; Posts: 185; Joined: Thu Feb 04, 2010 3:06 am; Thanked: 6 times

by gmatnmein2010 » Mon Feb 08, 2010 5:35 am

3. A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?
(A 10( sqrt3- 1)
(B) 5
(C) 10( sqrt2 - 1)
(D) 5( sqrt3 - 1)
(E) 5( sqrt2 - 1)
kindly show me the steps for this

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Source: Beat The GMAT — Problem Solving | Mon Feb 08, 2010 5:35 am

shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pmFollowed by: 2 members; Location: Gandhinagar; Thanked: 41 times

by shashank.ism » Mon Feb 08, 2010 6:01 am

gmatnmein2010 wrote:3. A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?
(A 10( sqrt3- 1)
(B) 5
(C) 10( sqrt2 - 1)
(D) 5( sqrt3 - 1)
(E) 5( sqrt2 - 1)
kindly show me the steps for this

radius of sphere = 10/2 =5
distance from centre to vertices = 10(sqrt. 2)/2 = 5(sqrt. 2)

so shortest possible distance from one of the vertices of the cube to the surface of the sphere = [spoiler]5( sqrt2 - 1)
Ans is E[/spoiler]

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thephoenix: Legendary Member; Posts: 1560; Joined: Tue Nov 17, 2009 2:38 amFollowed by: 5 members; Thanked: 137 times

by thephoenix » Mon Feb 08, 2010 6:44 am

Shortest distance=(diagonal of cube-diameter of sphere)/2=(10*3^1/2-10)/2=5(3^1/2-1)
imo d

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harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members; Location: India; Thanked: 64 times; GMAT Score:760

by harsh.champ » Mon Feb 08, 2010 8:32 am

shashank.ism wrote:
gmatnmein2010 wrote:3. A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?
(A 10( sqrt3- 1)
(B) 5
(C) 10( sqrt2 - 1)
(D) 5( sqrt3 - 1)
(E) 5( sqrt2 - 1)
kindly show me the steps for this
radius of sphere = 10/2 =5
distance from centre to vertices = 10(sqrt. 2)/2 = 5(sqrt. 2)

so shortest possible distance from one of the vertices of the cube to the surface of the sphere = [spoiler]5( sqrt2 - 1)
Ans is E[/spoiler]

Hey shashank,
I think you made a mistake over there.Since we are considering a 3-D object(cube) not a 2-D object(square).
So,length of the diagonal will be 10sqrt(3),not 10sqrt(2).
Besides this,your method of solving is correct.
Hence,the answer will be D not E.

The E option was a trap by the question-setter.

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ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Mon Feb 08, 2010 10:26 am

gmatnmein2010 wrote:3. A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?
(A 10( sqrt3- 1)
(B) 5
(C) 10( sqrt2 - 1)
(D) 5( sqrt3 - 1)
(E) 5( sqrt2 - 1)
kindly show me the steps for this

Square length of the diagonal of a cube joining totally opposite sides = (Sqrt(2)*a)^2 + a^2 = 3a^2
Length of the diagonal = a*sqrt(3)

Distance from the center of the sphere to the vertex = 1/2 full body diagonal = a/2 sqrt(3) = 5 sqrt(3)
Distance from the center of the sphere to the surface of the sphere = 1/2 *a = 5

Distance between vertex and surface of the sphere [spoiler]= 5 sqrt(3)-5 = 5( sqrt3 - 1) =(D)[/spoiler]

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