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If x<>0, is |x| <1? (1) x^2<1 (2) |x| < 1/x From (1) x^2 <1, x is either a positive fraction or a negaive fraction. In both cases, the modulus will be less than 1. So Statement 1 is sufficient.
From (2), |x| < 1/x, two cases: Case 1: x < 1/x x^2 < 1 x^2 - 1<0 (x+1)(x-1) <0 -1<x<1 <-- so always less than 1
Case 2: -x < 1/x -x^2 < 1 x^2 > 1 x^2 - 1 > 0 (x+1)(x-1) > 0 x < -1 or x > 1 <-- not always less than 1. More than 2 solutions using statement 2, so not sufficient. Ans: A
If x<>0, is |x| <1?
ok, here from i, it is clear that x^2 is positive and fraction is the only positive less than 1 and sqrt of any positive fraction is also positive fraction. so i is sufficient to say that x<1 and so does lxl.
from ii, if |x| < 1/x, then x cannot be -ve. because if x is negative, then 1/x cannot be >lxl because lxl is always positive. so sufficient.....
If x<>0, is |x| <1? (1) x^2<1 (2) |x| < 1/x From (1) x^2 <1, x is either a positive fraction or a negaive fraction. In both cases, the modulus will be less than 1. So Statement 1 is sufficient.
From (2), |x| < 1/x, two cases: Case 1: x < 1/x x^2 < 1 x^2 - 1<0 (x+1)(x-1) <0 -1<x<1 <-- so always less than 1
Case 2: -x < 1/x -x^2 < 1 x^2 > 1 x^2 - 1 > 0 (x+1)(x-1) > 0 x < -1 or x > 1 <-- not always less than 1. More than 2 solutions using statement 2, so not sufficient. Ans: A
If x<>0, is |x| <1?
ok, here from i, it is clear that x^2 is positive and fraction is the only positive less than 1 and sqrt of any positive fraction is also positive fraction. so i is sufficient to say that x<1 and so does lxl.
from ii, if |x| < 1/x, then x cannot be -ve. because if x is negative, then 1/x cannot be >lxl because lxl is always positive. so sufficient.....
Ha!! Yes, I miss the negative part out. Sorry!! I've been out of touch with GMAT maths for quite a while!
Re: DS - Absolute Value [#permalink]
16 Oct 2005, 08:37
rahulraao wrote:
What is the value of |x|?
(1) x = -|x| (2) x^2 = 4
B
A) Insuff. x can be any negative number. B) x=2 or -2. Since |2| or |-2| are both = 2, B is suff.
rahulraao wrote:
If x<>0, is |x| <1?
(1) x^2<1 (2) |x| < 1/x
D
A) Suff. since x^2<1, x should be -1 < x < 1. |x| is always less than 1
B) Suff. x has to be a fraction.
If x= -1/2
1/2 < -2 which is false. So x must be positive
If x=1/2
1/2 < 2 true.
If x = 3 or -3
3 < 1/3 or -1/3 Not true.
From above, we know x is a positive fraction (<1) So |x| <1