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Another DS from GMATPrep -- Absolute Value Inequalities [#permalink]
 10 May 2006, 15:31
Question Stats:
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I don't really understand how to approach the inquality questions on the GMAT... any help on this question (and in general) from GMATPrep would be greatly appreciated...
Is lx-yl > lxl - lyl ?
(1) y < x
(2) xy < 0
Answer is B, but I don't understand it. Thanks for the help.
Marcus
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For (1) y < x
a. if y > 0 and x > 0
then |x-y| = |x| - |y|
e.g. x = 5, y = 1. Evaluating, we get LHS = 4 and RHS = 4
b. if y < 0 and x > 0
then |x-y| > |x| - |y|
e.g. x = 5 and y = (-1). Evaluating, we get LHS = 6 and RHS = 4.
c. if y < 0 and x < 0
then |x-y| > |x| - |y|
e.g. x = (-1) and y = (-5). Evaluating, we get LHS = 4 and RHS = -4
So we cannot say for sure if the statement |x-y| > |x| - |y| will hold true given only condition (1).
For (2) xy < 0
a. y > x
if y > 0 and x < 0
then |x-y| > |x| - |y|
e.g. Taking y = 5 and x = -1 and evaluating, we get LHS = 6 and RHS = (-4).
b. y < x
if y < 0 and x > 0
then |x-y| > |x| - |y|
e.g. y = -5; x = 1. Evaluating, we get LHS = 6 and RHS = -4.
Hence for (2) we always get LHS > RHS i.e. |x-y| > |x| - |y|. Answer is (B).
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Actually I disagree with B.
It says xy < 0. Nothing in the question stem says x and y are both integers.
xy < 0 --> x and y are either:
(x = positive, y = negative)
If x = 3, y = -1, |x-y|=4, |x|-|y| = 2, |x-y| > |x|-|y|
(x = negative, y = negative)
If x = -2, y = 3, |x-y|=5, |x|-|y| = -1, |x-y| > |x|-|y|
(x and y are both positive fractions)
If x = 1/2, y = 1/4, |x-y|=1/2, |x|-|y|=1/4, |x-y| > |x|-|y|
(x and y are both negative fractions)
If x = -1/2, y = -1/4, |x-y| = 1/4, |x|-|y| = 1/4, |x-y| = |x|-|y|
In the last case, xy < 0, but |x-y| = |x|-|y|. So B should be insufficient.
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ywilfred wrote: (x and y are both positive fractions)
If x = 1/2, y = 1/4, |x-y|=1/2, |x|-|y|=1/4, |x-y| > |x|-|y|
Wilfred, How is this condition upholding statement 2 where xy<0?
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pesquadero wrote: ywilfred wrote: (x and y are both positive fractions)
If x = 1/2, y = 1/4, |x-y|=1/2, |x|-|y|=1/4, |x-y| > |x|-|y|
Wilfred, How is this condition upholding statement 2 where xy<0? If x = 1/2,, y = 1/4, xy = 1/8 which is less than 0.
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ywilfred wrote: pesquadero wrote: ywilfred wrote: (x and y are both positive fractions)
If x = 1/2, y = 1/4, |x-y|=1/2, |x|-|y|=1/4, |x-y| > |x|-|y|
Wilfred, How is this condition upholding statement 2 where xy<0? If x = 1/2,, y = 1/4, xy = 1/8 which is less than 0. Not meaning to be rude but xy=1/8 is not less than 0. In fact it is greater than 0. It is between 0 and +1.
If xy= -1/8 then, yes it would be less than 0.
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GMAT Club Legend
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pesquadero wrote: ywilfred wrote: pesquadero wrote: ywilfred wrote: (x and y are both positive fractions)
If x = 1/2, y = 1/4, |x-y|=1/2, |x|-|y|=1/4, |x-y| > |x|-|y|
Wilfred, How is this condition upholding statement 2 where xy<0? If x = 1/2,, y = 1/4, xy = 1/8 which is less than 0. Not meaning to be rude but xy=1/8 is not less than 0. In fact it is greater than 0. It is between 0 and +1. If xy= -1/8 then, yes it would be less than 0. haha.. i must be thinking too much. What a mistake ! 
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Re: Another DS from GMATPrep -- Absolute Value Inequalities [#permalink]
10 May 2006, 20:17
lackeym77 wrote: Is lx-yl > lxl - lyl ?
(1) y < x
(2) xy < 0
was discussed recently: http://www.gmatclub.com/phpbb/viewtopic ... ht=#197015
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Re: Another DS from GMATPrep -- Absolute Value Inequalities
[#permalink]
10 May 2006, 20:17
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