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

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

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04 May 2006, 16:31
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100% (01:01) correct 0% (00:00) wrong based on 5 sessions

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Is ix-yl>lxl-lyl ?
1)y<x
2)x*y<0
VP
Joined: 29 Dec 2005
Posts: 1343
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Kudos [?]: 62 [0], given: 0

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04 May 2006, 18:00
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.
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04 May 2006, 18:45
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
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04 May 2006, 20:13
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
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05 May 2006, 01:36
B it is, though it is tempting to mark C at first glance.
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05 May 2006, 02:26
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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11 Jun 2006, 14:50
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, 14:50
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