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rahuljones1
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rahuljones1 wrote:If x is not equal to 0, is |x| less than 1?
(1)
x
--- < x
|x|
(2) |x| > x
IMO C
explantion if anwer is correct
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schumi_gmat
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IMO E
Equation reduces to -1<x<1? x not equal to zero
1) x/ lxl < x
x= 3, expression is true 1<3
x=-3, -1<-3 .....false
x=-1/2. -1<-1/2.......true
x=1/2...1 < 1/2 .....false
It is true for -1<x<0 and x>1
Insuff
2) lxl>x
This is true for x<0 but not for 0<x<1.
Insuff.
Together,
x<0 or x>1
Whats OA?
Equation reduces to -1<x<1? x not equal to zero
1) x/ lxl < x
x= 3, expression is true 1<3
x=-3, -1<-3 .....false
x=-1/2. -1<-1/2.......true
x=1/2...1 < 1/2 .....false
It is true for -1<x<0 and x>1
Insuff
2) lxl>x
This is true for x<0 but not for 0<x<1.
Insuff.
Together,
x<0 or x>1
Whats OA?
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scoobydooby
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i get C
1) x/ lxl < x
=>x<x*|x| (|x|>0, given x not =0)
=>x-x*|x|<0
=>x(1-|x|)<0
either x<0, 1-|x|>0=>|x|<1........a or
x>0, 1-|x|<0=>|x|>1.........b
gives both yes and no answers, not sufficient
2) |x| > x
=>|x|-x>0
if x>0, x-x>0=> 0>0 not possible
so x must be negative. given that x is not 0
|x| may be <1 or >1 cant say
not sufficient
combining, we get x<0 from 2).
so |x|<1 from a in statement 1)
hence, C
1) x/ lxl < x
=>x<x*|x| (|x|>0, given x not =0)
=>x-x*|x|<0
=>x(1-|x|)<0
either x<0, 1-|x|>0=>|x|<1........a or
x>0, 1-|x|<0=>|x|>1.........b
gives both yes and no answers, not sufficient
2) |x| > x
=>|x|-x>0
if x>0, x-x>0=> 0>0 not possible
so x must be negative. given that x is not 0
|x| may be <1 or >1 cant say
not sufficient
combining, we get x<0 from 2).
so |x|<1 from a in statement 1)
hence, C
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schumi_gmat
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scoobydooby
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schumi,
from 1) you get -1<x<0 or x>1
and from 2) x<0
when you combine, x>1 is ruled out. you get -1<x<0 which would make |x|<1 so C would be sufficient.
from 1) you get -1<x<0 or x>1
and from 2) x<0
when you combine, x>1 is ruled out. you get -1<x<0 which would make |x|<1 so C would be sufficient.
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schumi_gmat
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Thanks Scooby,
Is my assumption, lxl<1 ===> -1<x<1 correct?
I got it this way.
for x>0, x<1
for x<0, -x<1 i.e. x>-1
Hence -1<x<1
If this is correct, then the conditions does hold true for 0<x<1
Is my assumption, lxl<1 ===> -1<x<1 correct?
I got it this way.
for x>0, x<1
for x<0, -x<1 i.e. x>-1
Hence -1<x<1
If this is correct, then the conditions does hold true for 0<x<1
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Ian Stewart
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Another way to look at the above:rahuljones1 wrote:If x is not equal to 0, is |x| less than 1?
(1) x/ |x| < x
(2) |x| > x
Using only 1), we can multiply on both sides by |x|, since |x| must be positive:
x < |x|*x
If x is positive, dividing by x tells us that |x| > 1. If x is negative, dividing by x (and reversing the inequality, since we're dividing by a negative) tells us that |x| < 1. Insufficient.
Using only 2), if x is positive, then |x| = x, while if x is negative, |x| > x. So S2 simply tells us that x is negative, which is insufficient.
Using 1) + 2) together, we know x is negative, so as we saw when analyzing S1 alone, this tells us that |x| < 1 (indeed, it must be that -1 < x < 0). So C.
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maihuna
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If x is not equal to 0, is |x| less than 1?
(1)
x
--- < x
|x|
(2) |x| > x
=============================
Q is asking for whether : |x|<1 or -1<x<1
1). x<x|x|
if x<0, x<-x2 or x^2+x <0 or x(x+1)<0
as x<0 we know x+1 >0 or x>-1 so it says -1<x<0
here |x| <1
: for x>0 x<x^2 or x(x-1)<0 as x>0 x<1 so in range 0<x<1
so this case is ok
2. |x| > x
if x<0 -x>x or 0>2x or x<0 nothing new
if x>0 x>x or 0>0 not possible
so for sure we know x<0 as oter case is not possible here but less than 0 doesnt mean it cant be less than 1 it can well be -5 or -5000 so not sufficient
Combining 1 & 2 we know for x<0 and so we can take the range -1<x<0
so |x| <1 as it is between 0 and 1
i will go with CCCCCCCCCCCCCCC
(1)
x
--- < x
|x|
(2) |x| > x
=============================
Q is asking for whether : |x|<1 or -1<x<1
1). x<x|x|
if x<0, x<-x2 or x^2+x <0 or x(x+1)<0
as x<0 we know x+1 >0 or x>-1 so it says -1<x<0
here |x| <1
: for x>0 x<x^2 or x(x-1)<0 as x>0 x<1 so in range 0<x<1
so this case is ok
2. |x| > x
if x<0 -x>x or 0>2x or x<0 nothing new
if x>0 x>x or 0>0 not possible
so for sure we know x<0 as oter case is not possible here but less than 0 doesnt mean it cant be less than 1 it can well be -5 or -5000 so not sufficient
Combining 1 & 2 we know for x<0 and so we can take the range -1<x<0
so |x| <1 as it is between 0 and 1
i will go with CCCCCCCCCCCCCCC