If x ? 0, is (x^2/|x|) < 1? (1) x < 1 (2) x > : DS Archive
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# If x ? 0, is (x^2/|x|) < 1? (1) x < 1 (2) x >

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If x ? 0, is (x^2/|x|) < 1? (1) x < 1 (2) x > [#permalink]

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25 Apr 2006, 11:31
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If x ≠ 0, is (x^2/|x|) < 1?
(1) x < 1
(2) x > −1
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25 Apr 2006, 14:29
Condition 1: If x = -1 then the equation = 1 if x<-1 equation is >1 - Sufficient
Condtion 2 : if x= 1 the equation = 1if x > 1 equation > 1 Sufficient
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25 Apr 2006, 14:50

1) if x=.5 then .25/.5 = 1/2 which answers yes. But if x=-2 then 4/2 = 2 which answers no

2) if x=-.5 then .25/.5 = 1/2 which answers yes. But if x=5 then 25/5 = 5 which answers no

Taken together, the value is always less than 1, hence C
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25 Apr 2006, 15:57
x^2/abs(X) < 1 <=> x^2<abs(x) this is true only if 0<X<1 or if -1<x<0
Since the stem already tells us that x is not =0, then we are looking for x>-1 and x<1

1) Insufficient. e.g. with x=-2 => x^2>abs(x) and with x=1/2 => x^2<abs(x)

2) Insufficient: e.g with x=2 => x^2>abs(x) and with x=-1/2 => x^2<abs(x)

Together: Suffiient. If e.g with x=1/2 and with x=-1/2 = we always have: x^2<abs(x)

therefore C
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26 Apr 2006, 10:34
OA is 'C'. I used the same approach as you guys.
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26 Apr 2006, 11:02
C.

I got it by plugging in values. Is there a better quicker way?
26 Apr 2006, 11:02
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