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Given x < y < z; where x, y, z are sides of a cuboid.
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Updated on: 11 Nov 2017, 23:50
Updated on: 11 Nov 2017, 23:50
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73% (01:15) correct
26% (00:52) wrong based on 15 sessions
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Given x < y < z; where x, y, z are sides of a cuboid.
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.
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.
Originally posted by Sonalika42 on 29 Aug 2016, 06:54.
Last edited by Bunuel on 11 Nov 2017, 23:50, edited 2 times in total.
Last edited by Bunuel on 11 Nov 2017, 23:50, edited 2 times in total.
Added the OA.
Re: Given x < y < z
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29 Aug 2016, 06:54
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.
I don't have the official answer that's the reason i have put the qs here.
Re: Given x < y < z
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08 Nov 2017, 07:48
08 Nov 2017, 07:48
2
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.
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.
