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# Is |x - y|>|x - z|? (1) |y|>|z| (2) x < 0

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Is |x - y|>|x - z|? (1) |y|>|z| (2) x < 0 [#permalink]  01 Nov 2009, 07:42
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Is |x - y|>|x - z|?

(1) |y|>|z|
(2) x < 0
[Reveal] Spoiler: OA

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Re: Is |x - y|>|x - z|? [#permalink]  01 Nov 2009, 08:24
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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]  02 Sep 2010, 01:02
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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]  02 Sep 2010, 01:17
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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|? (1) |y|>|z| (2) x < 0 [#permalink]  23 Apr 2012, 05:49
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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]  01 Nov 2009, 23:52
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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: Inequality DS [#permalink]  23 Aug 2010, 08:34
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Merging similar topics.
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Re: Is |x - y|>|x - z|? [#permalink]  25 Aug 2010, 07:14
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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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Re: Is |x - y|>|x - z|? [#permalink]  06 Sep 2010, 16:25
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Thank you to bunuel again! Super nice explanation!
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Re: Is |x - y|>|x - z|? (1) |y|>|z| (2) x < 0 [#permalink]  23 Apr 2012, 12:28
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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|? (1) |y|>|z| (2) x < 0 [#permalink]  25 Apr 2012, 19:11
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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 -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|? (1) |y|>|z| (2) x < 0 [#permalink]  22 May 2012, 19:20
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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|? (1) |y|>|z| (2) x < 0 [#permalink]  24 Jun 2012, 11:39
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kuttingchai wrote:
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

First, remember what do we use the absolute value for ? We use it to express the distance between two real numbers on the number line.
So, |x - y| means the distance between x and y, and |x| = |x - 0| means the distance between x and 0. Also, |x + 3| = |x - (-3)| expresses the distance between x and -3. Obviously, |x - y| = |y - x| (it is the same distance).

So, the question can be rephrased as "is the distance between x and y greater that the distance between x and z ?" or "is x closer to z than to y?"

(1) |y| > |z| means that the distance from y to 0 is greater than the distance from z to 0. This in fact is not important in this case. But certainly y and z are distinct. Regardless of wheather y > z or y < z (both cases are possible, try to draw the number line and visualize the points), we can place x closer to either y or z. So, (1) is not sufficient.

(2) Is evidently not sufficient. Take a point x on the number line at the left of 0, then you can place y and z as you please, each one can be closer or farther from x. Also, now you can consider y = z, in which case the two distances are equal.

Evidently, considering (1) and (2) together won't help either. For example, put x at the left of 0, y and z on either side of x such that y < x and z > x, both negative. You can play with each distance and put either y or z closer to x.

Therefore, answer, E.
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Re: Modules problem [#permalink]  28 Mar 2013, 01:20
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Can anyone solve the following problem in a completely algebraic way? I mean, without plugging any numbers.
Is |x-z| > |x-y| ?

1. |z| > |y|
2. 0 > x

$$|x-z| > |x-y| = (x-z)^2>(x-y)^2 = x^2+z^2-2xz>x^2+y^2-2xy$$
$$2xy-2xz>y^2-z^2$$
$$2x(y-z)>(y+z)(y-z)$$

1. |z| > |y|
$$z^2>y^2$$
$$z^2-y^2>0=y^2-z^2<0=(y-z)(y+z)<0$$
So one between (y-z) (y+z) is negative, the other is positive, but we cannot tell which one is +ve or -ve.
$$2x(y-z)>(y+z)(y-z)$$
the second part is -ve ((y-z)(y+z)<0) we cannot say anything about 2x(y-z)
Not sufficient

2. 0 > x
x<0 so the first term is -ve (2x) but we cannot say anything about the other part (...(y-z)>(y+z)(y-z))

(1)+(2)
Still not sufficient, let me explain. Here are the combinations ( remeber that one between (y-z) (y+z) is negative and the other positive)
$$2x(y-z)>(y+z)(y-z)$$
Case one (y-z)-ve: 2x<0 (y-z)<0 (y+z)>0
-veNumber*-veNumber>+veNumber*-veNumber
+ve>-ve always true.
Case two (y+z)-ve: 2x<0 (y-z)>0 (y+z)<0
-veNumber*+veNumber>-veNumber*+veNumber
-veNumber>-veNumber we cannot say if this is true, since we have no numerical value

E

Do I deserve a Kudos for this?
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Re: Is |x - y|>|x - z|? [#permalink]  01 Nov 2009, 08:08
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]  25 Aug 2010, 07:28
Expert's post
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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Re: Is |x - y|>|x - z|? (1) |y|>|z| (2) x < 0 [#permalink]  23 Apr 2012, 06:36
Expert's post
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|? (1) |y|>|z| (2) x < 0 [#permalink]  25 Apr 2012, 20:38
Expert's post
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|? (1) |y|>|z| (2) x < 0 [#permalink]  26 Apr 2012, 16:06
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|? (1) |y|>|z| (2) x < 0 [#permalink]  22 May 2012, 17:06
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
Re: Is |x - y|>|x - z|? (1) |y|>|z| (2) x < 0   [#permalink] 22 May 2012, 17:06

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