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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
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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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Math Expert
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Re M23-17  [#permalink]

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New post 16 Sep 2014, 01:19
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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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New post 19 Aug 2016, 02:56
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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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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
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Re: M23-17  [#permalink]

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New post 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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