Is |x - y| |x - z|? (1) |y| |z| (2) x

Question

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

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

Bunuel · Accepted Answer

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-------- --y----z---0------------ -----------0---z-----y-- -------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.

Hussain15 · Answer

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.

krazo · Answer

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

Bunuel · Answer

Unless otherwise specified, variables could represent the same number.

hemanthp · Answer

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.

gmatjon · Answer

Thank you to bunuel again! Super nice explanation!

shikhar · Answer

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....

Bunuel · Answer

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.

aalba005 · Answer

Great explanation, I never thought of thinking of the problem in terms of positioning. Great strategy.

pinchharmonic · Answer

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

Bunuel · Answer

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<0z>0) --------z--0--------y- (y>0>z)

kuttingchai · Answer

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

EvaJager · Answer

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.

Zarrolou · Answer

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?

SrinathVangala · Answer

+1 Bunuel.....Nice explanation..I went to the extent of taking examples for each case and solving it..This graphical approach is awesome!!!!

madn800 · Answer

Buneul, I think that you have made a mistake in interpreting |y|>|z|
====>|y|/|z|>1
====>either both are -ve or both are +ve.

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

Or

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

Bunuel · Answer

No mistake there.

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

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

MathRevolution · Answer

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.

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

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

In the original condition there are 3 variables (x,y,z) and since we need to match the number of variables and equations, we need 3 equations. Since there is 1 each in 1) and 2), we need 1 more equation and thus E isl likely the answer and it turns out that E actually is the answer.

In a number line, absolute number depicts the distance between two points. Looking back at the question,
Is |x - y| > |x - z|?

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

asks if the distance from y to 0 is greater than the distance from z to 0. Essentially is asks if, assuming x is negative, the distance from x to y is greater then the distance from x to z. if x=-1, y=5, z=-2, then the answer is yes, but if y=-5, z=2, x=-4 then the answer is no and therefore it is not sufficient. Therefore the answer is E. However, keep in mind that these direct substitutions are not the key methods but comparing the number of variables and equations in the original condition are.

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