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M23-17

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M23-17 [#permalink]

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New post 16 Sep 2014, 01:19
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

63% (01:03) correct 37% (00:58) wrong based on 140 sessions

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Re M23-17 [#permalink]

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New post 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.


Answer: B
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Re: M23-17 [#permalink]

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New post 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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Re: M23-17 [#permalink]

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New post 19 Aug 2016, 02:56
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Re: M23-17 [#permalink]

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New post 21 Aug 2016, 14:09
How all x, y and z numbers can have same value??? they are different negative numbers, right?
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Re: M23-17 [#permalink]

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New post 22 Aug 2016, 01:33
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Re: M23-17 [#permalink]

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New post 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 .
ANSWER:B
Re: M23-17   [#permalink] 08 Mar 2018, 06:38
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M23-17

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