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Geometry 4

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Manager
Joined: 19 Aug 2009
Posts: 79

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26 Sep 2009, 03:23
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Q.
If V is the volume of a cuboid of dimension x, y, z and A is its surface, then A/V will be equal to
1. x^2*y^2*z^2
2. ½( 1/xy + 1/ xz + 1/ yz)
3. ½ ( 1/x +1/y+1/z)

OA- 3
Senior Manager
Joined: 23 Jun 2009
Posts: 361
Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago

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26 Sep 2009, 06:34
1
KUDOS
Volume of it is equal to V= x*y*z
Area of it is equal to the sum of 6 regions on its surface. There are 2 surfaces with dimensions x and y, two with x and z and two with y and z.
So the area = A= 2*x*y+2*x*z+2*y*z
A/V= 2(xy+xz+yz)/xyz= 2(1/x+1/y+1/z).
It is like option c but not the same. Please control OA
Intern
Joined: 27 Aug 2009
Posts: 46

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26 Sep 2009, 07:44
1
KUDOS
virtualanimosity wrote:
Q.
If V is the volume of a cuboid of dimension x, y, z and A is its surface, then A/V will be equal to
1. x^2*y^2*z^2
2. ½( 1/xy + 1/ xz + 1/ yz)
3. ½ ( 1/x +1/y+1/z)

OA- 3

V = xyz
A = 2 (xy + yz + zx)

A/V= 2(xy+yz+zx) / xyz = 2(1/x + 1/y + 1/z)

If question were what is V/A, then
V/A = xyz / [2(xy+yz+zx)] = 1 / [2 (1/x + 1/y + 1/z)]

In the given answer choices, if u put a extra bracket in the answer choice 3, then that will match with V/A.
_________________

Salaries are low in recession. So, working for kudos now.

Manager
Joined: 27 Oct 2008
Posts: 185

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26 Sep 2009, 11:21
1
KUDOS
Area = 2xy + 2yz + 2xz
Volume = xyz
Thus A/V = 2((xy + yz + xz)/xyz)
which is the same as
A/V = 2(1/x + 1/y + 1/z)
Manager
Joined: 18 Jul 2009
Posts: 169
Location: India
Schools: South Asian B-schools

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26 Sep 2009, 11:31
1
KUDOS
area = 2(xy+yz+xz)
volume = xyz

A/V = 2(xy+yz+xz) / xyz = 2 (1/x+1/y+1/z)
_________________

Bhushan S.
If you like my post....Consider it for Kudos

Manager
Joined: 12 Oct 2009
Posts: 115

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13 Oct 2009, 23:25
virtualanimosity wrote:
Q.
If V is the volume of a cuboid of dimension x, y, z and A is its surface, then A/V will be equal to
1. x^2*y^2*z^2
2. ½( 1/xy + 1/ xz + 1/ yz)
3. ½ ( 1/x +1/y+1/z)

OA- 3

OA cannot be 3. A/V will be 2( 1/x +1/y+1/z)

What is the source of this question?
Re: Geometry 4   [#permalink] 13 Oct 2009, 23:25
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