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x<0<y+z.

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Manager
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x<0<y+z. [#permalink] New post 30 Aug 2016, 09:59
00:00

Question Stats:

50% (00:18) correct 50% (01:13) wrong based on 2 sessions
x<0<y+z

z!=0

Quantity A
Quantity B
(y+z)/x
y/z



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
Intern
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Re: x<0<y+z. [#permalink] New post 16 Oct 2016, 08:28
My answer is D.
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GRE 1: Q167 V156
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Re: x<0<y+z. [#permalink] New post 16 Oct 2016, 08:57
Expert's post
Sonalika42 wrote:
x<0<y+z

z!=0

Quantity A
Quantity B
(y+z)/x
y/z



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.



Here we can see that \(z\neq 0\). So z can either be 0 or 1.

Depending on the value of z Quantity B can be finite or infinite (as y cannot be 0 such that y/z can be written in 0/0 form).

Thus option D is the only possibility.
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Re: x<0<y+z. [#permalink] New post 12 Nov 2016, 12:51
Expert's post
Sonalika42 wrote:
x<0<y+z

z!=0

Quantity A
Quantity B
(y+z)/x
y/z



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.


Can you please explain z!=0
Some spaces would help remove any ambiguity (and they're free! :) )
z! (z factorial cannot equal 0), so presumably != is supposed to mean "no equal to"?
In future posts, you might want to write "not equal to" or add a space to show that there are no factorials (e.g., z != ), or use this ≠

Cheers,
Brent
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Re: x<0<y+z. [#permalink] New post 15 Dec 2016, 22:52
My answer was D. In response to the above, I assume that z! = 0 means z x z-1 x z-2 etc. so the only number that makes sense here is that z MUST equal 0. In that case option B above is an illegal number (since you can't have 0 in the denominator). Quantity A will always be negative because x is negative. I am not sure though how z! could equal 1 as per the above response?
Re: x<0<y+z.   [#permalink] 15 Dec 2016, 22:52
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