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If x ? 0, is x^2/|x| < 1? (1) x < 1 (2) x >

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New post 15 Aug 2006, 21:13
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If x ≠ 0, is x^2/|x| < 1?

(1) x < 1
(2) x > −1


A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


Any takerss???
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New post 15 Aug 2006, 21:21
C

If x ≠ 0, is x^2/|x| < 1?

(1) x < 1
(2) x > −1


If x> 0 then X*X/ X <1
If x< 0 then X*X / -X ( - because mod of a negative number is that number into -1) = X*X/-X <1 => -X< 1=> X> -1

thus C

This is definitely a tricky question; my first inclination would be to mark A.
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New post 15 Aug 2006, 21:28
Dont you think.


since for x<0 |x| will be -x


so x^2 /|x| for x < 0 .. will be x^2 /-x= -x

which will be less than 1

x^2/|x| < 1

so from A we can make out the answer
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New post 15 Aug 2006, 21:38
Check your ans for x = -2, x = -1/2 & x = 1/2 !


Achilless wrote:
Dont you think.


since for x<0 |x| will be -x


so x^2 /|x| for x < 0 .. will be x^2 /-x= -x

which will be less than 1

x^2/|x| < 1

so from A we can make out the answer
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New post 16 Aug 2006, 00:29
C...


The only way that this equation could be less than 1 is if it were between -1 & 1

(eg decimal)
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New post 16 Aug 2006, 00:47
C it is ... the equation holds true when -1<x<1
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New post 16 Aug 2006, 00:59
We don't have to worry about the -ve values. The equality will behave same for +ve and -ve values. We just need to worry about the fractions less than 1 and greater than 0 because x^2 is less than x for |x|<1.

So this is a straight C.
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

  [#permalink] 16 Aug 2006, 00:59
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