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ashish.jere
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Post subject: Is |x -y| > |x| + |y|? Posted: Wed Jul 08, 2009 2:28 pm |
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Is |x -y| > |x| + |y|?
(1) y < x
(2) xy < 0
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cfaking
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Post subject: Re: Inequality Posted: Thu Jul 09, 2009 4:03 am |
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Posts: 39 Location: India
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is it D? Cond (1) satisfies for all Numbers
use Number line and lets consider Cond (2)
------------Y1------------O--X1---------------
---------------------X2---O------------Y2----- |X-Y| is basically the distance between Y1 & X1 or between X2 & Y2
and they are always Equal to |x| + |y|
Good Q
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RonPurewal
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Post subject: Re: Inequality Posted: Fri Jul 10, 2009 7:14 am |
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ashish.jere wrote: Is |x -y| > |x| + |y|? let's REPHRASE. you should all be able to give the canonical rephrase of the left-hand side. the expression |x - y| represents the DISTANCE between 'x' and 'y' on the number line.on the right hand side, each INDIVIDUAL absolute value represents the distance between that number and ZERO.so |x| is the distance between 0 and x, and |y| is the distance between 0 and y. once you have these 2 realizations in place, you can combine them: * if x and y are on opposite sides of 0, then adding the distances |x| and |y| will produce the combined distance between x and y. therefore, the above inequality won't be true, since "=" will be the true statement. * if x and y are on the same side of 0, then either one of the distances |x| and |y|, alone, will be greater than the distance between x and y. therefore, their sum will certainly be greater, and so the prompt inequality will be true. * if either x or y (or both) IS 0, then "=" holds, and the inequality is false. therefore, we can rephrase the question: REPHRASE: are x and y both positive or both negative?that rephrase is a LOT of work, but, once you have it in hand, it's really, really easy to use the statements. Quote: (1) y < x this doesn't tell whether x and y have the same sign. insufficient. Quote: (2) xy < 0 this means x and y have OPPOSITE signs, so the answer to the question is NO. SUFFICIENT. ans (b). --
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RonPurewal
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Post subject: Re: Inequality Posted: Fri Jul 10, 2009 7:16 am |
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you can also solve this problem by PLUGGING IN NUMBERS:
statement (1)
try 2 positives: y = 1, x = 2 |x - y| = 1 |x| + |y| = 3 "no"
try one of each sign: y = -1, x = 2 |x - y| = 3 |x| + |y| = 3 "yes"
insufficient.
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statement (2)
first interpret the statement as meaning "x and y have opposite signs"
then plug in all manner of pairs of numbers with opposite signs, to your heart's content, until you are satisfied that the inequality is NEVER satisfied.
definite "no" --> sufficient.
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ashish.jere
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Post subject: Re: Inequality Posted: Mon Jul 13, 2009 12:54 pm |
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sd
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Post subject: Re: Inequality Posted: Mon Jul 13, 2009 10:31 pm |
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I still don't get how the answer is B? I think it should be D.
Ron, please help. I totally understand your explanation for statement 1. I understand that given y < x, irrespective of whatever examples you chose, you always get that |x-y| <= |x| + |y|.
Isn't this enough to answer the question. The question asked is
Is |x-y| > |x| + |y|?
And we are able to answer the above question as NO using just statement 1. So isn't statement 1 sufficient then?
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kamalsinghy
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Post subject: Re: Inequality Posted: Tue Jul 14, 2009 9:10 pm |
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Hi Ron,
I agree with 'sd'. Even I chose D just because of the inequality sign. From the statement 1 itself, in any case, |x-y| > |x|+|y| is not possible.
Thanks, Kamal
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cfaking
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Post subject: Re: Inequality Posted: Wed Jul 15, 2009 12:11 pm |
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Posts: 39 Location: India
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Ron, Great Explanations!!But in the 2nd Example,Q is asking whether |x - y| >|x| + |y| and the ANSWER is NO, no matter what combination of {x,y} you pick hence OA-D RonPurewal wrote: you can also solve this problem by PLUGGING IN NUMBERS:
statement (1)
try 2 positives: y = 1, x = 2 |x - y| = 1 |x| + |y| = 3 "no"
try one of each sign: y = -1, x = 2 |x - y| = 3 |x| + |y| = 3 "yes"
insufficient.
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statement (2)
first interpret the statement as meaning "x and y have opposite signs"
then plug in all manner of pairs of numbers with opposite signs, to your heart's content, until you are satisfied that the inequality is NEVER satisfied.
definite "no" --> sufficient.
_________________ Many of the great achievements of the world were accomplished by tired and discouraged men who kept on working.
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vivekwrites
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Post subject: Re: Inequality Posted: Thu Jul 16, 2009 3:02 pm |
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|x-y| can never be greater than |x|+|y|. We dont need any of the statements. I think the poster copied the problem wrong.
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RonPurewal
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Post subject: Re: Inequality Posted: Tue Jul 21, 2009 5:59 pm |
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yeah. actually, all of you are right.
there must be a typo in this question, since the answer to the question prompt as written is AUTOMATICALLY "no", even with neither of the numbered statements taken as true.
gmat questions NEVER work like that.
to the original poster (or to any other poster familiar with this question): do you have the correct version?
--
actually, one of two things is almost certainly the case here:
(1) ">" is supposed to be "<" (the most likely explanation) -- in fact, i unconsciously read ">" as "<", since that's the only sense in which the question wouldn't be trivial.
(2) this is not actually a GMAT PREP SOFTWARE question at all (remember, those are the only questions meant to be posted in this folder), and is instead from some other, more dodgy source. hmm.
i may have to delete the thread if (2) is the case, which would be somewhat of a shame (as the rephrasing parts are still valuable).
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ai42
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Post subject: Re: Is |x -y| > |x| + |y|? Posted: Sun Jul 26, 2009 3:29 pm |
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I found it. It's a GMATPREP I question. It should be: is |x-y|>|x|-|y|? 1) y<x 2) xy<0
OA - B
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ashish.jere
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Post subject: Re: Is |x -y| > |x| + |y|? Posted: Mon Jul 27, 2009 1:56 pm |
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I am sorry for the typo guys.
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www.mbachase.com
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Post subject: Re: Is |x -y| > |x| + |y|? Posted: Thu Jul 30, 2009 8:08 am |
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[Attention mbachase.com: we appreciate your desire to help GMAT students. Unfortunately, we don't host posts from competitors on our forums. If you'd like to offer free services to your students (and I think that's a great idea!), then you might consider starting your own forums. Good luck! Stacey]
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