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# M23-17

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Math Expert
Joined: 02 Sep 2009
Posts: 59589

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16 Sep 2014, 01:19
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Difficulty:

35% (medium)

Question Stats:

69% (01:31) correct 31% (01:35) wrong based on 134 sessions

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If $$x$$, $$y$$, and $$z$$ are negative numbers, is $$x \lt y \lt z$$?

(1) $$x+z=2y$$.

(2) $$xz = yz$$.

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Joined: 02 Sep 2009
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16 Sep 2014, 01:19
Official Solution:

(1) $$x+z=2y$$. If $$x=y=z=-1$$, then the answer is NO but if $$x=-3$$, $$y=-2$$ and $$z=-1$$, then the answer is YES. Not sufficient.

(2) $$xz = yz$$. Since $$z \ne 0$$, then we can reduce given equation by $$z$$ and we'll get $$x=y$$, so $$x \lt y \lt z$$ is not true. Sufficient.

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Joined: 06 May 2016
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18 Aug 2016, 12:41
Hi Bunuel,

x+z= 2y

or y = (x+z)/2
or we can say y is average of x and z. So y should be between x and z.

Am I missing anything here?

Kris
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19 Aug 2016, 02:56
gmatravi wrote:
Hi Bunuel,

x+z= 2y

or y = (x+z)/2
or we can say y is average of x and z. So y should be between x and z.

Am I missing anything here?

Kris

The solution above gives an example which shows that this is not necessarily true: x = y = z = -1.
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Joined: 28 May 2014
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Schools: ISB '17, NUS '17

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21 Aug 2016, 14:09
How all x, y and z numbers can have same value??? they are different negative numbers, right?
Math Expert
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22 Aug 2016, 01:33
akashbolster wrote:
How all x, y and z numbers can have same value??? they are different negative numbers, right?

Unless it is explicitly stated otherwise, different variables CAN represent the same number.
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Joined: 21 Jun 2016
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08 Mar 2018, 06:38
1
If x, y, and z are negative numbers, is x<y<z?

(1) x+z=2y
from this equation we can say that y is the mean of x & z .
so there are two possibilities
x<y<z or z<y<x
statement 1 : insufficient . Option A and Option D gone
(2) xz=yz
from this statement we can find z=0 or x=y
Z=0 cant be as we have been told z is a negative intezer
so x=y must be valid . question whether x<y<z fails . Statement 2 is sufficient .
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30 Nov 2019, 02:57
Bunuel

would it make sense to write the first statement as "x + z - 2y < 0"
I somehow only got that the values could be equal when I wrote it in this manner, with the statement before I tried to fit numerous values which of course does not make a lot of sense..
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Re: M23-17   [#permalink] 30 Nov 2019, 02:57
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# M23-17

Moderators: chetan2u, Bunuel