ds: abs

Question

Is |x-y| > |x| - |y| ? 1, y

Any easy approch, within 2 min??

gmat_crack · Answer

You convert this into simple form and try to get the values > 0 or < 0 z = |X-Y| -|X| +|Y| St 1. Not suff. becaause we don't know range of X and Y St2. Not suff. becayse we don't know From St 1 and two case 1 X>Y and X> 0 and Y <0 Z >0 case 2 X > Y and X< 0 and Y >0 Z < 0 Hence E

vipin7um · Answer

|x-y| > |x| - |y| Lets take condition 1. 1, y= |x| - |y|. The equality condition holds when x=y. By suggesting y |x-y| != |x| - |y| HTH...

macca · Answer

hmm...didn't quite get this....so what's your answer??
what did you mean with "again note that |x-y| >= |x| - |y|." when the
question stem says |x-y| > |x| - |y|

SunShine · Answer

E.

One can also pick numbers for x and y, and feed into the satatements, 

Stat1 insufficient
Stat2 does not give much info

combined together, not sufficient

gmatacer · Answer

For these type of questions, I plugin value of x and y

x +ve and Y -ve, say x=3 and y = -2 (this satisfies above equation) and x>y
Now, x -ve and Y -ve, say x=-2 and y=-1(satisfies above equation) and x<y.
So, we can't say from option 1

Drills down to BCE

for same set values(above) for 1t set, x=3 and y=-2. xy(-6)<0 and for second set x=-2 and y=-1, xy(3)>0
Sp, we can't say from option 2 that xy is always less than 0

so, answer is E

macca · Answer

Anybody else

lastchance · Answer

B

1) If x and y both are positive/negative LHS = RHS. 
If y is negative and x is positive, LHS > RHS
Not sufficient

2) xy < 0 means either x or y is negative.
If x is -ve & y positive; LHS > RHS
if y is -ve and x positive; Still LHS > RHS
Sufficent

cheti · Answer

|x-y| > |X|-|Y| |X-Y| is always positive no matter what X and Y are. +ve > |X|-|Y| is the question. A. Since Y

gidimba · Answer

I picked E. I plugged in numbers and it didnt pan out either way.

kripalkavi · Answer

I'm going with C. OA please...

vipin7um · Answer

My answer is D. Since using either condition we can remove the equality part of the inequality.

- Vipin

Professor · Answer

using trail and error approach:

from i, if x = -1 and y = -3
|x-y| = |-1+3| = 2, and |x| - |y| = -2 so true..

if x = 3 and y = 1 
|x-y| = |3-1| = 2, and |3| - |1| = 2 so not true.


from ii, x = -1 and y = 3
|x-y| = |-1-3| = 4, and |x| - |y| = -2 so true..

if x = 3 and y = -1
|x-y| = |3+1| = 4, and |x| - |y| = |3| - |-1| = 2 so true.

B is sufficient.....

gmat_crack · Answer

I revisited the problem I think its B.

willget800 · Answer

B...

A is not suff, try {x,y} ==> {2,1} and {2,-3} 
B is suff, try {x,y} ==> {-1,1} and {1,-1}

chiragr · Answer

Think about Number line....

A is not sufficient as |X-Y| is always positive but |X|-|Y| can be negative also depending on who is far away from origin. 

and y<x will not tell us which point has higher absolute value, (x,y) can be (10, -20) and (10, -2) in first canse |X|-|Y| is negative and in second case its positive. 

B is sufficient as it says x and y are of different sign that means |x-y|, will show distance between this two points.

While |x|-|Y| will show delta between individual distance which has to be smaller than |x-y|

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