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Is |x - y| > |x - z|?

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Is |x - y| > |x - z|?

(1) |y| > |z|
(2) x < 0
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Re: Is |x - y| > |x - z|? [#permalink]

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C

need to know which is more extreme and what value X is which both statements combined tell you.
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Hussain15 wrote:
Is |x - y|>|x - z|?

(1) |y|>|z|
(2) x < 0


Good question. I believe the answer is E.

Basically the question asks whether the distance between x and y is greater than the distance between x and z?

(1) |y|>|z|, means y is farther from 0 than z, which gives us the following possible cases:

--y--------0---z-- or --y----z--0-- or --0----z------y or --z---0----------y--

Now depending on the position of x, the distance between x and y may or may not be greater than distance between x and z.

(2) x<0. Clearly insufficient as we know nothing about y and z:

---x----0---


(1)+(2) x<0 and |y|>|z|

--y------------x--0--z-- in this case |x - y|>|x - z|

BUT

--x--y---------z--0----- in this case |x - y|<|x - z|

Two different answers, so insufficient.

Answer: E.
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Re: Is |x - y| > |x - z|? [#permalink]

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Great explanation! I am not sure about the answer, will post OA in a while.
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Re: Is |x - y| > |x - z|? [#permalink]

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In general for these types of questions, if I'm testing values, are the variables different from each other.. eg should I consider the case where x = y, or x = z unless otherwise stated?

Sorry if this is a dumb question.. lol
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Re: Is |x - y| > |x - z|? [#permalink]

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New post 25 Aug 2010, 08:28
krazo wrote:
In general for these types of questions, if I'm testing values, are the variables different from each other.. eg should I consider the case where x = y, or x = z unless otherwise stated?

Sorry if this is a dumb question.. lol


Unless otherwise specified, variables could represent the same number.
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I think we need to develop intuition for these kind of questions... from looking at question i was not able to say the answer but after looking at choices, i started to feel it must be E. After less then a minute of thinking i was sure it has to be E.
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It is E.
We can make out pretty quickly that these are heavily based on values of X, Y, and Z. So it has to be C or E.
You are subtracting Y and Z from a fixed value of X. Having absolute values is not going to help.
Nice post but!
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Thank you to bunuel again! Super nice explanation!
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New Approach ,
Bunuel Please correct me if this approach is wrong ...

to find if |x-y|>|x-z| ?

Squaring both sides (since both sides are +ve)

(x-y)2-(x-z)2 >0
(x-y-x+z)(x-y+x-z) >0
(z-y) (2x-y-z) >0

From 1 --> Red part i negative but cant say anything about green part so insufficient.
From 2 --> know nothing so insufficient

from both also nothing can be concluded so E....
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Re: Is |x - y| > |x - z|? [#permalink]

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New post 23 Apr 2012, 07:36
shikhar wrote:
New Approach ,
Bunuel Please correct me if this approach is wrong ...

to find if |x-y|>|x-z| ?

Squaring both sides (since both sides are +ve)

(x-y)2-(x-z)2 >0
(x-y-x+z)(x-y+x-z) >0
(z-y) (2x-y-z) >0

From 1 --> Red part i negative but cant say anything about green part so insufficient.
From 2 --> know nothing so insufficient

from both also nothing can be concluded so E....


Yes, you can square both parts of the inequality (since both are non-negative) and rewrite the question: is (z-y)(2x-y-z)>0?

But from |y|>|z| you can not say that z-y is negative: if y=2 and z=1 then YES but if y=-2and z=1 then NO.

So, overall, I'd still suggest number line approach.
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Great explanation, I never thought of thinking of the problem in terms of positioning. Great strategy.
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Re: Is |x - y| > |x - z|? [#permalink]

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bunuel, would love your feedback on my method

|x-y| > |x - z|

this applies when both signs are the same, or signs are different.

ie
|x| > |y|

-x > y
x > y


a.)
-(x-y) > x - z
-x + y > x -z
-2x > -z -y
2x < z + y

b.)
x-y > x -z
-y > -z
y < z



stmt 1:

i did the same thing here:
a.)
y > z

b.)
y < -z

or z < y < -z

which shows z is negative and not much about. in any case, x is not even mentioned so this is insufficient

stmt 2:

x < 0

again y and z are not mentioned, insufficient


stmt 1 + stmt 2:

attacking (a) from the original problem stem

y > z ?

according to stmt 1 this is false since z < y < -z.
so (a) from the stem is sufficient.

attacking (b) from the original stem

2x < y + z?


we know z < y < -z, so add a z to each

2z < y + z < 0

y+ z < 0

and x < 0


so you have

negative number * 2 < negative number

and since we don't haev any relationship between what 2x would be or y +z would be, it would be insufficient E
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Re: Is |x - y| > |x - z|? [#permalink]

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New post 25 Apr 2012, 21:38
pinchharmonic wrote:
bunuel, would love your feedback on my method

|x-y| > |x - z|

this applies when both signs are the same, or signs are different.

ie
|x| > |y|

-x > y
x > y


a.)
-(x-y) > x - z
-x + y > x -z
-2x > -z -y
2x < z + y

b.)
x-y > x -z
-y > -z
y < z



stmt 1:

i did the same thing here:
a.)
y > z

b.)
y < -z

or z < y < -z

which shows z is negative and not much about. in any case, x is not even mentioned so this is insufficient

stmt 2:

x < 0

again y and z are not mentioned, insufficient


stmt 1 + stmt 2:

attacking (a) from the original problem stem

y > z ?

according to stmt 1 this is false since z < y < -z.
so (a) from the stem is sufficient.

attacking (b) from the original stem

2x < y + z?


we know z < y < -z, so add a z to each

2z < y + z < 0

y+ z < 0

and x < 0


so you have

negative number * 2 < negative number

and since we don't haev any relationship between what 2x would be or y +z would be, it would be insufficient E


There are several flows above. For example: |y|>|z| does not mean that z is negative. Consider simple example: y=2 and x=1.

|y|>|z| simply means that means y is farther from 0 than z:

---y-------0--z------- (y<0<z)
---y----z--0---------- (y<z<0)
-----------0--z-----y- (y>z>0)
--------z--0--------y- (y>0>z)
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Re: Is |x - y| > |x - z|? [#permalink]

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New post 26 Apr 2012, 17:06
thanks bunuel, great point i'll think of this type of problem in a more conceptual manner next time
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Re: Is |x - y| > |x - z|? [#permalink]

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New post 22 May 2012, 18:06
try assuming values y=3 z=2 ... x can be any no. +ve / -ve (take both cases to decide between A & C) but eventualy both did not give answer so E is correct
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Re: Is |x - y| > |x - z|? [#permalink]

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I used the approach below :

1) |y|>|z| will give 4 different cases , I assumed only two
y=5 , z=2 & y=-5 , z= -2
No information about x , so x can be =+ ve or -ve , let x = 7
x-y=7-5 =2& x-z= 7-2=5

x-y=7-(-5)=12 & x-z = 7-(-2)=9
contradictory results . Insufficient

2)x<0 does not tell anything about y,z so insufficient.

1&2)
Repeating the above values y=5 , z=2 & y=-5 , z= -2
x=-3
x-y= -3+5= 2 , x-z = -3+2 = -1
x-y= -3-5= -8 , x-z= -3-2=-5

Contradictory results , Insufficient.
Hence E.
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Re: Is |x - y| > |x - z|? [#permalink]

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New post 25 May 2012, 19:58
picked E purely based on picking numbers approach. took 114 sec (1.54 min) how much time have you guys taken on this sum on avg?
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Re: Is |x - y| > |x - z|? [#permalink]

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New post 24 Jun 2012, 10:16
Voted for E

Solved it following way

(A) |Y| > |Z|


Case 1 :

---[-X =-4]-------[-Y = -3]------[-Z = -2]------[0]-------[Z = 2]-------[Y=3]-------[X=4]------


Case 2 :

---[-Y = -3]------[-Z = -2]------[-X =-1]-------[0]-------[X=1]-------[Z = 2]-------[Y=3]------

when x>0

|X-Y| = |1-3| = 2 and |X-Z| = |1-2| = 1
therefore |X-Y| > |X-Z|

|X-Y| = |4-3| = 1 and |X-Z| = |4-2| = 2
therefore |X-Y| < |X-Z|

when x<0

|X-Y| = |-1+3| = 2 and |X-Z| = |-1+2| = 1
therefore |X-Y| > |X-Z|

|X-Y| = |-4+3| = 1 and |X-Z| = |-4+2| = 2
therefore |X-Y| < |X-Z|

not sufficient

(B) X<0 dont know about Y and Z, therefore nit sufficient

(C) X < 0 and |Y| > |Z|

when x<0

|X-Y| = |-1+3| = 2 and |X-Z| = |-1+2| = 1
therefore |X-Y| > |X-Z|

|X-Y| = |-4+3| = 1 and |X-Z| = |-4+2| = 2
therefore |X-Y| < |X-Z|
[not sufficent]

therefore E
Re: Is |x - y| > |x - z|?   [#permalink] 24 Jun 2012, 10:16

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