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What is the value of |x|? (1) x = -|x| (2) x^2 = 4 If [#permalink]
 16 Oct 2005, 01:44
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What is the value of |x|?
(1) x = -|x|
(2) x^2 = 4
If x<>0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x
_________________
Cheers, Rahul.
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For your first question:
What is the value of |x|?
(1) x = -|x|
(2) x^2 = 4
From (1), we can have any negative number or 0.So Statement 1 is not sufficient.
From (2), we know x = +2 or -2. So again, it's not sufficient.
Using both, we know x = -2 so answer is C.
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For the second question, what does '<>' mean? I seen it before, but I can't recall it... very sorry 
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Re: DS - Absolute Value [#permalink]
16 Oct 2005, 07:12
rahulraao wrote: What is the value of |x|?
(1) x = -|x|
(2) x^2 = 4 B because statement 2 is enough to answer the question. from i, x or lxl could be anything. from ii, lxl is 2. so suff. rahulraao wrote: If x<>0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x
for this, D.
from i, x is less than 1 but more than 0 because x^2 cannot be -ve and if x is less than 1, then it must be in fraction. so lxl<1.
from ii, also x is less than 1 but more than 0 because if lxl<1/x, then x must be in fraction.
explain more, if needed....
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ywilfred wrote: For the second question, what does '<>' mean? I seen it before, but I can't recall it... very sorry this means x is not equal to o. x could be +ve or -ve.
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Re: DS - Absolute Value [#permalink]
16 Oct 2005, 07:26
rahulraao wrote: What is the value of |x|?
(1) x = -|x|
(2) x^2 = 4
For this One C as Ywilfred explained
If x<>0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x
For this one I would say A...
Last edited by FN on 16 Oct 2005, 07:38, edited 1 time in total.
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HIMALAYA wrote: ywilfred wrote: For the second question, what does '<>' mean? I seen it before, but I can't recall it... very sorry this means x is not equal to o. x could be +ve or -ve. Well, I prefer to use the programming notation '!='  . Anyway, if x != 0, is |x| <1 ?
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
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ywilfred wrote: 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..... 
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HIMALAYA wrote: ywilfred wrote: 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!
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Re: DS - Absolute Value [#permalink]
16 Oct 2005, 09:37
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
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Re: DS - Absolute Value [#permalink]
16 Oct 2005, 11:42
rahulraao wrote: What is the value of |x|?
(1) x = -|x|
(2) x^2 = 4
If x<>0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x
OK let me take a knock at it from a slightly rested frame of mind
1)
1)
X= -|x|
well, X can be 0=-0, x can be -1= -1 so Insuff
(2) x^2=4 well X=+-(2), but |x|=2 B it is...
2)
x^2<1
well then X is clearly fraction,
|x|<1/x
well we have two conditions when X is positive
x<1/x, then mutliply x by both sides (cause X <> 0)
x^2<1 so good |x|<1
no condition X is negative
-x>1/-x (note the sign changed cause of the negative sign)
multiply -x by both sides
-x*-x<1 or x^2<1
So |X| < 1 sufficient...
D it is...
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Re: DS - Absolute Value
[#permalink]
16 Oct 2005, 11:42
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