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Expert replies

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 7:03 am

If the length of the largest straight rod that can be put inside a cube is 10 m, then the surface area of the cube cannot be more than
a) 100 m2
b) 200 m2
c) 400 m2
d) 600 m2
e) Cannot be determined

Last edited by shashank.ism on Mon Feb 08, 2010 7:19 am, edited 1 time in total.

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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 7:09 am

200m^2

d(diag of cube)=10
d^2=(asqrt2)^2+a^2(a=side of cube)
--->3a^2=100
--->surface area of cube=6a^2=2*100=200
hence B

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 7:15 am

shashank.ism wrote:If the length of the largest straight rod that can be put inside a cuboid is 10 m, then the surface area of the cuboid cannot be more than
a) 100 m2
b) 200 m2
c) 400 m2
d) 600 m2
e) Cannot be determined

The straight rod have to be along the diagonal of the cuboid.[Though shashank,I think it will be cube not a cuboid otherwise we have to consider a lot of cases]

10=sqrt(x^2 + x^2 + x^2)
x=10/sqrt(3)
So, surface area will be [spoiler]6*(x^2) = 6x100/3 = 200[/spoiler]

HENCE, b.

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