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x/|x|<x. which of the following must be true about x ?

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x/|x|<x. which of the following must be true about x ? [#permalink] New post 06 Feb 2005, 08:16
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 May 2014, 04:16, edited 2 times in total.
OA added.
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Re: PS - Inequality [#permalink] New post 01 Sep 2010, 06:32
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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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 [#permalink] New post 06 Feb 2005, 21:32
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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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 [#permalink] New post 07 Feb 2005, 23:11
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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] New post 02 Sep 2010, 02:30
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nonameee wrote:
Bunuel, thanks for your reply. But I have to disagree with you because:

The answer can't be B, since let x=1/2. We get:
(1/2)/(1/2) < (1/2)
1< 1/2
Contradiction.

What do you think about that?


hemanthp wrote:
Yup. IMO - A.

It fails in all other cases.
A. x>1
B. x>-1 => fails for any value between 0 and 1.
C. |x|<1 => fails for any value between 0 and 1.
D. |x|=1 => obviously fails.
E. |x|^2>1 => Fails for negative number less than -1. Take -4. -4/4 < -4 => FALSE.

What is the OA?

Thanks.


hemanthp wrote:
B makes no sense. Either the poster posted the question wrong or the choices wrong (the order probably).


OA for this question is B and it's not wrong.

Consider following:
If x=5, then which of the following must be true about x:
A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x>-10

Answer is E (x>-10), because as x=5 then it's more than -10.

Or:
If -1<x<10, then which of the following must be true about x:
A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x<120

Again answer is E, because ANY x from -1<x<10 will be less than 120 so it's always true about the number from this range to say that it's less than 120.

The same with original question:

If -1<x<0 or x>1, then 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

As -1<x<0 or x>1 then ANY x from these ranges would satisfy x>-1. So B is always true.

x could be for example -1/2, -3/4, or 10 but no matter what x actually is it's IN ANY CASE more than -1. So we can say about x that it's more than -1.

On the other hand A says that x>1, which is not always true as x could be -1/2 and -1/2 is not more than 1.

Hope it's clear.
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Re: PS - Inequality [#permalink] New post 26 Oct 2010, 08:44
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nades09 wrote:
Hi Bunnuel,
Can it be solved as-

x/|x|<x

Hence 1/|x|<1

1/x<1 OR -1/x<1
when 1/x<1
then 1<x = x>1

when -1/x<1
then -1<x = x>-1

The two possible outcomes are x>1 or x>-1
The more restrictive is x>-1
Hence B

Thanks
Neelam


No, that't not correct.

First of all, when you are writing 1/|x|<1 from x/|x|<x you are reducing (dividing) inequality by x:

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

Consider a simple inequality 4>3 and some variable x.

Now, you can't multiply (or divide) both parts of this inequality by x and write: 4x>3x, because if x=1>0 then yes 4*1>3*1 but if x=-1 then 4*(-1)=-4<3*(-1)=-3. Similarly, you can not divide an inequality by x not knowing its sign.

Next, inequality \frac{1}{|x|}<1 holds true in the following ranges: x<-1 and x>1 (and not: x>1 or x>-1, which by the way simply means x>1).

Hope it's clear.
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Re: PS - Inequality [#permalink] New post 24 Sep 2011, 14:20
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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

Answer is A. I hope the following explanation permanently sets A as the right answer for this problem. I see that some people tried simplifying the formula first, and some people performed a common mistake while doing so. Lets look at what many people tried to do.

Step 1: multiply |x| to both sides, resulting in x<x*|x| (this step is correct as |x| will always be positive and thus multiplying |x| to both sides will not cause ambiguity in the direction of the inequality)
Step 2: divide x to both sides, resulting in 1<|x| (this step is incorrect as x could be positive or negative, and such ambiguity restricts such operation. very common mistake that should be avoided at all costs as GMAT question makers will use this against you)

So.... lets go back to Step 1, x<x*|x|
From here, you can further simplify the equation to 0<x(|x|-1) as many people above did, but simplifying the equation this will only create more brain damage than simplify the problem.

The best course of action after Step 1 is to plug in numbers (or even plug in numbers straight to the original equation without performing Step 1).

-2 does not work because -2<-2*|-2| = -2<-4, which is not true
-1 does not work because -1<-1*|-1| = -1<-1, which is not true
-1/2 works because -1/2<-1/2*|-1/2| = -1/2<-1/4, which is true
0 does not work because 0<0*|0| = 0<0, which is not true
1/2 does not work because 1/2<1/2*|1/2| = 1/2<1/4, which is not true
1 does not work because 1<1*|1| = 1<1, which is not true
2 works because 2 < 2*|2| = 2<4, which is true
3 works because 3<3|3 = 3<9, which is true

So we have a situation in which 0>x>-1, and x>1.

Now here is where many people are getting REALLY confused.
Many people say that the right answer is B because, yes, x>-1, BUT it has certain restrictions. The statement x>-1 should also include 0 and 1/2 as the right solutions, but the original equation fails when these numbers are plugged in, making x>1 the only right answer for the equation. Yes, I understand that the answer A disregards the portion where 0>x>-1, but who cares, the question is asking "which of the following must be true about x" and not "what are the solutions for x"
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Re: x/|x|<x. which of the following must be true about x ? [#permalink] New post 13 Aug 2012, 09:07
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stne wrote:
I also think answer should be A, if expert can confirm solution , it would be wonderful


OA for this question is B.

Check this:
x-x-x-which-of-the-following-must-be-true-about-x-13943-40.html#p772618
x-x-x-which-of-the-following-must-be-true-about-x-13943-40.html#p773277
x-x-x-which-of-the-following-must-be-true-about-x-13943-40.html#p807569
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 [#permalink] New post 06 Feb 2005, 16:35
"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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 [#permalink] New post 06 Feb 2005, 16:49
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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 [#permalink] New post 06 Feb 2005, 16:53
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] New post 06 Feb 2005, 16:57
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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 [#permalink] New post 06 Feb 2005, 17:09
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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 [#permalink] New post 06 Feb 2005, 21:45
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, 21:50, edited 1 time in total.
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 [#permalink] New post 06 Feb 2005, 21:50
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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 [#permalink] New post 06 Feb 2005, 23:10
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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 [#permalink] New post 07 Feb 2005, 07:26
(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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 [#permalink] New post 07 Feb 2005, 07:38
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. :-D
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 [#permalink] New post 07 Feb 2005, 07:55
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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 [#permalink] New post 07 Feb 2005, 08:53
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.
  [#permalink] 07 Feb 2005, 08:53
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