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# Is ix-yl>lxl-lyl ? 1)y<x 2)x*y<0

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Manager
Joined: 13 Aug 2005
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Is ix-yl>lxl-lyl ? 1)y<x 2)x*y<0 [#permalink]  04 May 2006, 15:31
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Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions
Is ix-yl>lxl-lyl ?
1)y<x
2)x*y<0
VP
Joined: 29 Dec 2005
Posts: 1349
Followers: 7

Kudos [?]: 30 [0], given: 0

Re: DS absolute value [#permalink]  04 May 2006, 17:00
Natalya Khimich wrote:
Is lx-yl > lxl - lyl ?
1) y<x
2) x*y<0

from i, x>y means x or y, both, could be +ve or -ve. if x and y both are +ve, lx-yl = lxl - lyl otherwise not.

from ii, either x or y is +ve and the other one is -ve. so suff because no matter the value of each, lx-yl is always greater than lxl - lyl.

SO B.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5078
Location: Singapore
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St1: y<x
If x = 3, y = -2, then |x-y| = 5, |x| - |y| = 1 |x-y| > |x| - |y|
If x = 2, y = 1, then |x-y| = 1, |x|-|y| = 1 |x-y| = |x|-|y|

Insufficient.

St2: xy < 0

Means either:

(x,y) = (negative, positive) or (positve, negative), or (fraction, fraction)

If x = -3, y = -10, |x-y|=7, |x|-|y|=-7 ; |x-y|>|x|-|y|
If x = -10,y = -3, |x-y| = 7, |x|-|y|=7 ; |x-y| > |x|-|y|

Insufficient.

Using St1 and St2:

If x = -3, y = -10, |x-y|=7, |x|-|y|=-7 ; |x-y|>|x|-|y|
If x = 100, y = -10, |x-y| = 110, |x|-|y|=90; |x-y| >|x|-|y|
If x = 3, y = -7, |x-y|10, |x|-|y|=-4; |x-y| > |x|-|y|
If x = 1/2, y = 1/4, |x-y| = 1/2, |x|-|y| = 1/2

Insufficient.

Ans E
VP
Joined: 29 Dec 2005
Posts: 1349
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Kudos [?]: 30 [0], given: 0

ywilfred wrote:
St2: xy < 0 Means either:

(x,y) = (negative, positive) or (positve, negative), or (fraction, fraction)

If x = -3, y = -10, |x-y|=7, |x|-|y|=-7 ; |x-y|>|x|-|y|
If x = -10,y = -3, |x-y| = 7, |x|-|y|=7 ; |x-y| > |x|-|y|
Insufficient.

x and y both cann't be -ves or +ves.
Manager
Joined: 23 Mar 2006
Posts: 76
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Kudos [?]: 1 [0], given: 0

B it is, though it is tempting to mark C at first glance.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5078
Location: Singapore
Followers: 22

Kudos [?]: 184 [0], given: 0

Professor wrote:
ywilfred wrote:
St2: xy < 0 Means either:

(x,y) = (negative, positive) or (positve, negative), or (fraction, fraction)

If x = -3, y = -10, |x-y|=7, |x|-|y|=-7 ; |x-y|>|x|-|y|
If x = -10,y = -3, |x-y| = 7, |x|-|y|=7 ; |x-y| > |x|-|y|
Insufficient.

x and y both cann't be -ves or +ves.

ah yes !! that was a bad mistake. thanks for pointing it out.
Manager
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Re: DS absolute value [#permalink]  11 Jun 2006, 13:50
Professor wrote:
Natalya Khimich wrote:
Is lx-yl > lxl - lyl ?
1) y<x
2) x*y<0

from i, x>y means x or y, both, could be +ve or -ve. if x and y both are +ve, lx-yl = lxl - lyl otherwise not.

from ii, either x or y is +ve and the other one is -ve. so suff because no matter the value of each, lx-yl is always greater than lxl - lyl.

SO B.

so if both x and y are negative would "lx-yl = lxl - lyl" be true?
thanks
Re: DS absolute value   [#permalink] 11 Jun 2006, 13:50
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