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Is |x - y| > |x - z|? [#permalink ]

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01 Nov 2009, 08:42
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Re: Is |x - y| > |x - z|? [#permalink ]

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01 Nov 2009, 09:08

C need to know which is more extreme and what value X is which both statements combined tell you.

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

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01 Nov 2009, 09:24
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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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02 Nov 2009, 00:52
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Bunuel!! I think you have made the record of fastest century of kudos!!

Congrats!!

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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23 Aug 2010, 09:34

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

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25 Aug 2010, 08:14
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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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25 Aug 2010, 08:28

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

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

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02 Sep 2010, 02:17
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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! Thanks.

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

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06 Sep 2010, 17:25
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Thank you to bunuel again! Super nice explanation!

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

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23 Apr 2012, 06:49
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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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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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Re: Is |x - y| > |x - z|? [#permalink ]

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23 Apr 2012, 13:28
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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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25 Apr 2012, 20:11
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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 -y2x < z + y b.) x-y > x -z -y > -zy < 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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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 -y2x < z + y b.) x-y > x -z -y > -zy < 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 < 0y+ 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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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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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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22 May 2012, 20:20
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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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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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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

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