If x != 0 , is |x| < 1 ?
1) x|x| < x
2) |x| > x

I understand what is explained in the 5/27/2010 session, but my only confusion is that when Ron said that this question cannot be completely solved with algebra. Please see my description below and let me know whether we can use the below mentioned approach.

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Question stem |x| < 1 can be re-written as : -1 < x < 1 (exception x != 0)

Statement 1 :
x|x| < x
x|x| - x < 0
x(|x| -1) < 0
a) x > 0 and |x| - 1 < 0 or |x| < 1 (since in this case x >0)
Since x > o then this inequality gives us 0 < x < 1

b) x < 0 and |x| - 1 > 0 or |x| > 1 (since in this case x <0)
Since x < 0 then this inequality gives us x < -1
Thus Statement 1 is INSUFFICIENT

Statement 2:
|x| > x
As this statement is true it tells us that ‘x’ is always negative , which can be re-written as x < 0
This statement is also INSUFFICIENT as it does not tell us about whether x lies between -1 and 1.

Statement 1 and Statement 2 together:
Statement 1 gives us:
----- 0 < x <1 & x < -1 or we can say x lies between the set [-infinity, -1) (0, 1)
Statement 2 gives us:
----- x < 0 or we can say x lies in the set [-infinity, 0)
Therefore, the only set that satisfies these 2 equations is x < -1 or we can say [-infinity, -1)
Thus, both statement together are SUFFICIENT to confirm answer as NO.

Of course you can use an approach like that, but yours is not a completely algebraic approach, so Ron was right ultimately.. :)

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Tim Sanders
Manhattan GMAT Instructor