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x/|x|<x. which of the following must be true about x ? [#permalink]
 06 Feb 2005, 09:16
Question Stats:
21% (01:57) correct
78% (01:12) wrong based on 37 sessions
x/|x|<x. which of the following must be true about x ? A. x>1 B. x>-1 C. |x|<1 D. |x|=1 E. |x|^2>1 this is a question of the GMATCLUB collection 2. i got a different solution and i just dont know why.
Last edited by Bunuel on 13 Aug 2012, 10:02, edited 1 time in total.
OA added.
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"B".
x/|x|<x-------x<x|x|.....for +ve x...|x| > 1...for -ve x .....|x| < 1....that means x can be -1<x<0....or x > 1....in any case it has to be > -1
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banerjeea_98 wrote: "B".
x/|x|<x-------x<x|x|.....for +ve x...|x| > 1...for -ve x .....|x| < 1....that means x can be -1<x<0....or x > 1....in any case it has to be > -1
lets suppose x be 0.5, which is greater than -1. 0.5/l0.5l=1 which is not less than 0.5.
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MA wrote: banerjeea_98 wrote: "B".
x/|x|<x-------x<x|x|.....for +ve x...|x| > 1...for -ve x .....|x| < 1....that means x can be -1<x<0....or x > 1....in any case it has to be > -1 lets suppose x be 0.5 which is more than -1. 0.5/l0.5l=1 which is not less than 0.5. ya but in no case x can be < -1.....it has to be > -1....but not bet 0 and 1.....x > 1 is wrong.....as the eqn also satisfies when x = -1/2
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Re: PS - Inequality [#permalink]
06 Feb 2005, 17:57
christoph wrote: x/|x|<x. which of the following must be true about x ?
a) x>1
b) x>-1
c) |x|<1
d) |x|=1
e) |x|^2>1
this is a question of the GMATCLUB collection 2. i got a different solution and i just dont know why.
The answer should be A.
x/|x| <x
=> dividing both sides by x
=> 1/|x|<1
thus either x>1 or x<-1 only then 1/|x| will be less than 1.
Thus A.
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MA wrote: banerjeea_98 wrote: MA wrote: banerjeea_98 wrote: "B".
x/|x|<x-------x<x|x|.....for +ve x...|x| > 1...for -ve x .....|x| < 1....that means x can be -1<x<0....or x > 1....in any case it has to be > -1 lets suppose x be 0.5 which is more than -1. 0.5/l0.5l=1 which is not less than 0.5. ya but in no case x can be < -1.....it has to be > -1....but not bet 0 and 1.....x > 1 is wrong.....as the eqn also satisfies when x = -1/2 baner, unclear to me.... pls be specific. Let's imagine a number line with points -1, 0 and 1.....as u can see x can't be <-1...e.g. x = -2......-1 < -2....will not satisfy....
Now let's see -1<x<0......x = -1/2.....-1 < -1/2.....satisfies
Now for 0<x<1.....x = 1/2.....1 < 1/2.....will not satisfy....
For x > 1....x = 2.....1 < 2....satisfies
As u see the soln lies in the following categories:
-1<x<0
x > 1
In all case x > -1 even tho it can't be 0<x<1.....altho I think the ques is lil weird.
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x/|x|<x
if x>0, then 1/x<1 ie, x>1
if x<0, then 1/(-x)>1, ie, x>-1
So combine these two, we know that x must be greater than -1.
baner is right, (B).
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HongHu wrote: x/|x|<x
if x>0, then 1/x<1 ie, x>1
if x<0, then 1/(-x)>1, ie, x>-1
So combine these two, we know that x must be greater than -1.
baner is right, (B).
i think we need AkamaiBrah.
if so, then why 0<x>1 does not fit in the inequality x/|x|<x? i mean why all the values for x>-1 donot satisfy the inequality x/|x|<x? pls do explain........................
Last edited by MA on 06 Feb 2005, 22:50, edited 1 time in total.
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It doesn't say that all x greater than -1 would satisfy the equation. We get two range of x that satisfies the equation (-1,0) and (1,infinite). All that the answer says is that every x in our solution ranges are greater than -1, which is true.
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I believe we're debating solution B, which says x>-1 must be always true.
I can pick positive 1 as an integer that is greater than -1, but does not satisfy x/|x|<x.
If x = +1,
x/|x| = 1 which is not < 1. (1=1)
Now, if there's just one value of x > -1 that does not satisfy the inequality, then we can't say x>-1 is always true.
If you look at my previous post, I've not said I'm solving for x, i'm just stating values of x that hold true for the inequality. I came up with x<-1, and x>1.
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(A) cannot be correct, people.
Let me give you another example. Lets say we have two sets of integers.
S1=(0), S2=(2,3,4)
Now we know that x belongs to S1 or S2. Ask what must be true for x:
(A) x>1
(B) x>-1
(A) is obviously wrong
Because x could be an element of S1 or S2. If x belongs to S1, then it would be 0, and it is NOT greater than 1.
(B) is, however, correct
No matter which set x belongs, x is greater than -1.
You cannot say that well what about x=1? It satisfies x>-1, but is not in either S1 or S2. The mistake you are having here, is instead of A->B, you are trying to say B->A.
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HongHu wrote: (A) cannot be correct, people.
Let me give you another example. Lets say we have two sets of integers.
S1=(0), S2=(2,3,4)
Now we know that x belongs to S1 or S2. Ask what must be true for x:
(A) x>1
(B) x>-1
(A) is obviously wrong
Because x could be an element of S1 or S2. If x belongs to S1, then it would be 0, and it is NOT greater than 1.
(B) is, however, correct
No matter which set x belongs, x is greater than -1.
You cannot say that well what about x=1? It satisfies x>-1, but is not in either S1 or S2. The mistake you are having here, is instead of A->B, you are trying to say B->A.
Glad to see that I am not the only one in minority here, I am sticking with my ans "B" till we get the OA. 
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MA wrote: HongHu wrote: Can you give me one example where x is not greater than -1 in your solution set? do refer my postings: MA wrote: banerjeea_98 wrote: "B".
x/|x|<x-------x<x|x|.....for +ve x...|x| > 1...for -ve x .....|x| < 1....that means x can be -1<x<0....or x > 1....in any case it has to be > -1 lets suppose x be 0.5, which is greater than -1. 0.5/l0.5l=1 which is not less than 0.5. if x=0.5, x is not greater than x/lxl. 0.5/l0.5l = 1, which is greater than 0.5. You need to first find a point in your solution set, and then test if it is greater than -1. x=0.5 is not in your solution set.
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My example is completely parallel to the question, only in the integer and limited form.
Look:
HongHu wrote: (A) cannot be correct, people.
Let me give you another example. Lets say we have two sets of integers.
S1=(0), S2=(2,3,4)
Now we know that x belongs to S1 or S2. Ask what must be true for x:
(A) x>1
(B) x>-1
Compare it with:
S1=(-1,0), S2=(1,infinity)
We know x belongs to S1 or S2. What must be true for x?
You get the same idea.
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greenandwise wrote: A must be the answer because it works in every case and therefore MUST be true.
One value of x could be x=-1/2
x/|x|=-1<-1/2
However this x is not greater than 1. So it is not always true that x>1.
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Re: PS - Inequality [#permalink]
07 Feb 2005, 20:14
DLMD wrote:
if X <0, then X < X* (-X) ---------> X< -X^2 ------> 0 < -X^2-X ---------> 0 > X^2 + X ---------> X <0 or X <-1, so X <-1
The direction of an inequality sign would have to be changed when you multiply a negative number on both side.
For example,
-x<1
=>
x>-1
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Urgh, I just can't talk any sense into you guys.  Read my examples!
A leads to B doesn't mean B leads to A!!! And if B doesn't lead to A it doesn't mean A doesn't lead to B! It's basic logic people!!!
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Re: PS - Inequality [#permalink]
08 Feb 2005, 10:31
DLMD wrote:
if X <0, then X < X* (-X) ---------> X< -X^2 ------> 0 < -X^2-X ---------> 0 > X^2 + X ---------> X <0 or X <-1, so X <-1
Question is x/|x|<x
if x<0, you multiply (-x) on both sides and change the direction of the inequality:
x > x*(-x)
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Re: PS - Inequality [#permalink]
01 Sep 2010, 04:16
Quote: x/|x|<x. which of the following must be true about x ?
a) x>1
b) x>-1
c) |x|<1
d) |x|=1
e) |x|^2>1 Bunuel, I think there's a mistake in the question or in the answer choices: Here's my solution: 1) x<0: -x/x < x -1<x<0 2) x>=0 x/x<x x>1 The solution of the inequality is then: -1<x<0 union x>1 The answer can't be B, since let x=1/2. We get: (1/2)/(1/2) < (1/2) 1< 1/2 Contradiction. I think the answer should be A since it satisfies all the scenarios. Can you please clarify?
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Re: PS - Inequality [#permalink]
01 Sep 2010, 07:32
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x/|x|<x, which of the following must be true about x ? A. x>1 B. x>-1 C. |x|<1 D. |x|=1 E. |x|^2>1 We are given that \frac{x}{|x|}<x (this is a true inequality), so first of all we should find the ranges of x for which this inequality holds true. \frac{x}{|x|}< x multiply both sides of inequality by |x| (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too) --> x<x|x| --> x(|x|-1)>0: Either x>0 and |x|-1>0, so x>1 or x<-1 --> x>1; Or x<0 and |x|-1<0, so -1<x<1 --> -1<x<0. So we have that: -1<x<0 or x>1. Note x is ONLY from these ranges. Option B says: x>-1 --> ANY x from above two ranges would be more than -1, so B is always true. Answer: B. nonameee wrote: Quote: x/|x|<x. which of the following must be true about x ?
a) x>1
b) x>-1
c) |x|<1
d) |x|=1
e) |x|^2>1 Bunuel, I think there's a mistake in the question or in the answer choices: Here's my solution: 1) x<0: -x/x < x -1<x<0 2) x>=0 x/x<x x>1 The solution of the inequality is then: -1<x<0 union x>1 The answer can't be B, since let x=1/2. We get: (1/2)/(1/2) < (1/2) 1< 1/2 Contradiction. I think the answer should be A since it satisfies all the scenarios. Can you please clarify? The options are not supposed to be the solutions of inequality \frac{x}{|x|}<x. Hope it's clear.
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Re: PS - Inequality
[#permalink]
01 Sep 2010, 07:32
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