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Given x < y < z; where x, y, z are sides of a cuboid.

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Given x < y < z; where x, y, z are sides of a cuboid. [#permalink] New post 29 Aug 2016, 06:54
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60% (00:58) correct 40% (01:23) wrong based on 5 sessions
Given x < y < z; where x, y, z are sides of a cuboid.


Quantity A
Quantity B
Volume of cuboid with edges x+10, y, z
Volume of cuboid with edges x, y, z+10




A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Nov 2017, 23:50, edited 2 times in total.
Added the OA.
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Re: Given x < y < z [#permalink] New post 29 Aug 2016, 06:54
In my opinion answer should be D.

I don't have the official answer that's the reason i have put the qs here.
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Re: Given x < y < z [#permalink] New post 08 Nov 2017, 07:48
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Hallo there,

in my humble opinion the correct answer here should be A.

Lets evaluate and compare both expressions:
A
(X+10)(Y)(Z)

XYZ + YZ*10

B
(X)(Y)(Z+10)

XYZ + XY*10

Now while only comparing YZ*10 and XY*10 we now A>B because Z>X

I hope my reasoning is not fallacious here.
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Re: Given x < y < z [#permalink] New post 08 Nov 2017, 15:53
Expert's post
[youtube][/youtube]
simon1994 wrote:
Hallo there,

in my humble opinion the correct answer here should be A.

Lets evaluate and compare both expressions:
A
(X+10)(Y)(Z)

XYZ + YZ*10

B
(X)(Y)(Z+10)

XYZ + XY*10

Now while only comparing YZ*10 and XY*10 we now A>B because Z>X

I hope my reasoning is not fallacious here.


Looks absolutely correct!
_________________

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Re: Given x < y < z   [#permalink] 08 Nov 2017, 15:53
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Given x < y < z; where x, y, z are sides of a cuboid.

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