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

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Senior Manager
Joined: 28 Jun 2007
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 [#permalink]

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27 Oct 2007, 13:53
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Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 6 sessions

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

(1) y < x

(2) xy < 0

Kudos [?]: 52 [0], given: 0

Manager
Joined: 19 Aug 2007
Posts: 167

Kudos [?]: 66 [0], given: 0

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27 Oct 2007, 14:14
1) if x > y then if theyre both positive then
|x-y| is the same as |x| - |y| so in this case |x-y| is not greater

but if x is positive and ys i negative, letws say x=3 and y=-2
then |x-y| = 5 which is greater than |x| - |y| = 1
insufficient

2) xy<0 means one is negative and one is positive.
if x is positive and y is negative, then we know that |x-y| is greater.

if x is negative and y is positive, lets say x=-2 and y=3
then |x-y| =5 and |x| - |y| is -1 so again its greater
so this should be sufficient.
can anyone confirm?

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VP
Joined: 08 Jun 2005
Posts: 1144

Kudos [?]: 246 [0], given: 0

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27 Oct 2007, 14:32
this post was edited !

this problem (as all absolute value problems do) ask you about distances on the number line.

You can ask ---> is the distance from x to y is greater then the difference in the distance from x to 0 and from y to 0.

statement 1

the distance from x to zero and y to zero will be greater then the distance of x to y only when x and y don't have the same sign.

insufficient

statement 2

the distance from x to zero and y to zero will be greater then the distance of x to y only when x and y don't have the same sign.

from statemtn 2 we can tell that x,y don't have the same sign !

sufficient.

Last edited by KillerSquirrel on 28 Oct 2007, 01:12, edited 3 times in total.

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SVP
Joined: 29 Aug 2007
Posts: 2472

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27 Oct 2007, 15:53
gluon wrote:
Is |x - y| > |x| - |y|?

(1) y < x
(2) xy < 0

B. if xy < 0, one is -ve and the other is +ve. so in that case, |x - y| > |x| - |y| is always true.

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VP
Joined: 08 Jun 2005
Posts: 1144

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28 Oct 2007, 01:13
GMAT TIGER wrote:
gluon wrote:
Is |x - y| > |x| - |y|?

(1) y < x
(2) xy < 0

B. if xy <0> |x| - |y| is always true.

Perfect !

see my edited post

Kudos [?]: 246 [0], given: 0

Re: DS: absolute values   [#permalink] 28 Oct 2007, 01:13
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