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Re: DS: Manhattan GMAT Inequality [#permalink]
14 Feb 2008, 09:42

I say B.

I. x/|x| < x Let's say X<0, x= -2 => -2/2<-2 false x= -1/2 => -1<-1/2 true if x>0, x=2 => 1<2 true x= 1/2 => 1<1/2 false. So X>1 or -1<x<0. If x=2 => |x| >1, if x=-1/2 => 1/2>-1/2 => |x|<1 not suff

II. |x| > x if x=-2 => 2>-2 true, if x=2 => 2>2 false So X<0 for |x| > x. if x=-2 => |-2|=2>1 false for every x<0 suff.

Re: DS: Manhattan GMAT Inequality [#permalink]
14 Feb 2008, 10:02

moni77 wrote:

I say B.

I. x/|x| < x Let's say X<0, x= -2 => -2/2<-2 false x= -1/2 => -1<-1/2 true if x>0, x=2 => 1<2 true x= 1/2 => 1<1/2 false. So X>1 or -1<x<0. If x=2 => |x| >1, if x=-1/2 => 1/2>-1/2 => |x|<1 not suff

II. |x| > x if x=-2 => 2>-2 true, if x=2 => 2>2 false So X<0 for |x| > x. if x=-2 => |-2|=2>1 false for every x<0 suff.

i think u may be correct.

why cant we cross multiply given that |x| is a nonneg number that wont flip the inequality sign?
_________________

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Re: DS: Manhattan GMAT Inequality [#permalink]
14 Feb 2008, 10:15

bmwhype2 wrote:

moni77 wrote:

I say B.

I. x/|x| < x Let's say X<0, x= -2 => -2/2<-2 false x= -1/2 => -1<-1/2 true if x>0, x=2 => 1<2 true x= 1/2 => 1<1/2 false. So X>1 or -1<x<0. If x=2 => |x| >1, if x=-1/2 => 1/2>-1/2 => |x|<1 not suff

II. |x| > x if x=-2 => 2>-2 true, if x=2 => 2>2 false So X<0 for |x| > x. if x=-2 => |-2|=2>1 false for every x<0 suff.

i think u may be correct.

why cant we cross multiply given that |x| is a nonneg number that wont flip the inequality sign?

You can do that too. x< x |x|, so |x|>1 if x>0 and |x|<1 if x<0. Try -2.-1/2, 2 and 1/2 and see the results.

Re: DS: Manhattan GMAT Inequality [#permalink]
14 Feb 2008, 11:01

This one is tricky because it is easy to lose sight of the original question. The question is basically asking whether x is between -1 and 1.

(1) This tells us either that a) -1 < x, or b) 1 < x. Neither of these scenarios by themselves tell us that x is bounded between -1 and 1. It might fall between these, but it might not.

(2) This statement tells us the x can only be negative. Again, by itself not sufficient.

(1) U (2) Now we know that x is negative, so we are looking at scenario (1)a), that -1 < x. But since x must be negative the more precise range is (-1,0). Together they are sufficient.