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If x?0, is |x| <1? (1) x^2<1 (2) |x| < 1/x [#permalink]
 09 Mar 2005, 00:20
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If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x
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Intern
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I will go with D!
i) is sufficient
ii) is also sufficient as it only narrow down the value of X to a positive number.
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SVP
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A)
1) x has to be a fraction => regardless of the sign x is less than 1 => suff
1) |x|x<1 => x can be fraction or x is neg; in both cases the stem is satisfied but |x| is > 1 or <1 or = 1 => insuff
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SVP
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christoph wrote: A)
1) x has to be a fraction => regardless of the sign x is less than 1 => suff
1) |x|x<1 => x can be fraction or x is neg; in both cases the stem is satisfied but |x| is > 1 or <1 or = 1 => insuff
u can't multiply the inequality with a the variable, it changes the inequality...try x = -1/2 in statement 2 as see if that satisifies
Ans shud be "D".
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Director
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-1<x<1
and x not equal to 0
there fore it has to be a fraction..
and has to be <1..
a sufficient
b sufficient..
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SVP
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banerjeea_98 wrote: christoph wrote: A)
1) x has to be a fraction => regardless of the sign x is less than 1 => suff
1) |x|x<1 => x can be fraction or x is neg; in both cases the stem is satisfied but |x| is > 1 or <1 or = 1 => insuff u can't multiply the inequality with a the variable, it changes the inequality...try x = -1/2 in statement 2 as see if that satisifies Ans shud be "D". ...because i dont know whether the variable is neg or pos ! thx baner...
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Senior Manager
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the argument doesnt mention <= or >= it only says < or >
one more for D
in statement 1 only the Sq of a really small no. is gonna be <1 (on both sides of 0)
statement 2: if it were negative there is no way this holds, as a lxl cannot be < (any negative number)
now comes the question ...does this make statement 2 insufficient or sufficient.
I THINK it makes it sufficient cause by math rules it implies that x cannot be negative
therefore we can only look at the possibility of X being positive.
from that angle it is sufficient.
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Gr8. I agree with you all.
Thanx guys.
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GMAT Club Legend
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From 1.
-1<x<1
So (1) is sufficient since we can answer yes/no to |x| < 1
From 2.
Positive x case:
x < 1/x
x^2<1 (sufficient)
Negative x case:
x < -1/-x
x < 1/x
x^2 < 1 (sufficient)
Ans: D
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Manager
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From each of the statements x is between 0 and 1. Thus , yes |x|<1.
D.
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