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

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Re: PS - Inequality [#permalink] New post 01 Sep 2010, 23:20
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?
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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, 02:12
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
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Re: PS - Inequality [#permalink] New post 26 Oct 2010, 08:44
Expert's post
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 Jul 2011, 18:12
A graph helps me visualize these types of problems. If you graph \frac{x}{|x|} (red line) and x (blue line), the green areas represent the region in which the inequality \frac{x}{|x|}<x is satisfied.

The question asks what MUST BE TRUE if the inequality holds. In other words, IF \frac{x}{|x|}<x IS SATISFIED, then what must be true of x?

The green areas represent the regions where this inequality holds. What must be true of both green areas?

The x values in both regions must be greater than -1. B is the correct answer.
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Re: PS - Inequality [#permalink] New post 24 Sep 2011, 10:48
praveenvino wrote:
I am bit confused here. i agree the range of x, -1<x<0 or x>1. But doesn't x>-1 covers the range 0<x<1 also?

Yes it does. But question says, if we pick any number in the range -1<x<0(ex: -0.5,-0.25) or x>1(ex: 2,3,4) will that number be greater than -1 (x>-1). We can see that yes that number will be greater than -1.

praveenvino wrote:
So we have that: -1<x<0 or x>1. Note x is ONLY from these ranges.

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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: PS - Inequality [#permalink] New post 24 Sep 2011, 16:36
syh244 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

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"


Your analysis is amazing, but the summing up is not entirely correct.

As you yourself said:
the question is asking "which of the following must be true about x" and not "what are the solutions for x".

When the condition is given:
x/|x|<x

And we get the range for x:
-1<x<0 OR x>1.

It is imperative that "x" MUST BE in that range. So, x just can't be less than equal to -1, OR Anything between 0 AND 1, inclusive, such as "1/2".

If there is a value of x, it must be in the range specified above.

So,
(A) x>1: It is one of the possible ranges. But, we don't know what will x bear. What if x=-0.5. Not necessarily true.
(B) x>-1: No matter which value "x" bears, it will always be >-1. Must be true.
(C) |x|<1: This CAN be true, but what if x = 2. Not necessarily true.
(D) |x|=1: This cannot be true. Both -1 AND +1 are outside the range.
(E) |x|^2>1: x<-1 OR x>1. What if x=-0.5. Not necessarily true.

Ans: "B"
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Re: PS - Inequality [#permalink] New post 24 Sep 2011, 17:05
x/mod(x) < x

when x>=0 , mod(x) = x

=> given expression becomes x/x < x => 1<x => x>1 -------expression 1

when x<0, mod(x) = -x

=> given expression becomes x/(-x) < x => -1<x => x>-1--------expression 2


a. x>1 , need not be true

x <0 can be possible too which is not greater than 1.

b. x>-1 , must be true.

both the expressions 1 and 2 , x>1 and x>-1 means x is greater than -1 for sure.

c. mod(x) <1, need not be true

when x>=0, mod(x)<1 => x<1 ( this is not true from expression 1)

d. mod(x)= 1 , need not be true.

lets say x = -1/2 => mod(x) = 1/2 and this is not equal to 1.
e. mod(x)^2 > 1, need not be true.

lets say x= -1/2 => mod(x)^2 = 1/4 and this is not greater than 1.


Answer is B.
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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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Re: x/|x|<x. which of the following must be true about x ? [#permalink] New post 24 Aug 2012, 14:04
+1 B

Solving the inequality, we have two ranges:
x > 1 or x > -1

Whatever the real value of x is, it is always higher than -1.
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Re: x/|x|<x. which of the following must be true about x ? [#permalink] New post 04 Oct 2012, 06:38
HERE! the key is "must be true" and not could be or should be true. so, answer is B by above explanation of bunuel.
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Re: PS - Inequality [#permalink] New post 27 Dec 2012, 23:55
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Bunuel wrote:

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.


Bunuel,
As you've mentioned that we're to verify the range of x for which the given inequality holds good.
So for x>-1, x can have the value like 1/2. So in that case the inequality doesn't hold good for sure. Aren't we validating the inequality to be true ?Now,'must be true' means it has to satisfy all the possible plug-in values taking one from each of the category i.e. positive fraction and integer and negative fraction and integer as per the given conditions.

Whereas, for x>1 it does satisfy for all the possible values like x=3/2,4 etc. and the inequality holds good.So.how can we ignore the above case where the inequality clearly becomes false ?
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Re: PS - Inequality [#permalink] New post 28 Dec 2012, 04:04
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debayan222 wrote:
Bunuel wrote:

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.


Bunuel,
As you've mentioned that we're to verify the range of x for which the given inequality holds good.
So for x>-1, x can have the value like 1/2. So in that case the inequality doesn't hold good for sure. Aren't we validating the inequality to be true ?Now,'must be true' means it has to satisfy all the possible plug-in values taking one from each of the category i.e. positive fraction and integer and negative fraction and integer as per the given conditions.

Whereas, for x>1 it does satisfy for all the possible values like x=3/2,4 etc. and the inequality holds good.So.how can we ignore the above case where the inequality clearly becomes false ?


I think you don't understand what is given and what is asked.

Given: -1<x<0 and x>1 (that's what x/|x|<x means).

Now, the question asks which of the following MUST be true.

You are saying: "so for x>-1, x can have the value like 1/2." That's not correct: if -1<x<0 and x>1, then how x can be 1/2?
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Re: x/|x|<x. which of the following must be true about x ? [#permalink] New post 12 May 2013, 08:11
The question ( must be true about x) asks about x not the solution of the inequality , thus we solve inequality and we see from answer choices if x always is inside our solution of the inequality

x-/x/*x < 0 , i.e x (1-/x/) <0 holds true in 2 cases

a) x+ve and /x/>1 , i.e. x+ve in the range x<-1 ( this is equivalent to x>1) or x>1 thus in this case x is always >1

b) x-ve and /x/<1 , i.e. x-ve and -1<x<1 ( from /x/<1) but since x is always -ve in this assumption therefore the range becomes -1<x<0

now we have 2 ranges that is x>1 and -1<x<0 now we check each answer choice vs. those ranges , x>-1 is always true ( must be true about x) in those ranges

Thus B is the answer
Hope this helps
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Re: PS - Inequality [#permalink] New post 13 May 2013, 10:28
Vijo wrote:
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.



Wrong , you can not divide x this way, bcos its sign is not known.
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Re: x/|x|<x. which of the following must be true about x ? [#permalink] New post 25 May 2013, 22:47
I think x>1 holds true always but not x>-1
0<x<1 comes under the range x>-1 and it does not satisfy the equation.
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Re: x/|x|<x. which of the following must be true about x ? [#permalink] New post 26 May 2013, 03:00
Expert's post
pradeepkamp wrote:
I think x>1 holds true always but not x>-1
0<x<1 comes under the range x>-1 and it does not satisfy the equation.


That is incorrect.

Go through Bunuel's Solution.

if-x-0-and-x-x-x-which-of-the-following-must-be-true-143572.html
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Re: PS - Inequality [#permalink] New post 24 Mar 2014, 06:40
Bunuel 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

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.



Hi Bunuel,
Even I have the same questions, can you help clear the confusion with X = 1/2.

Thanks.
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Re: PS - Inequality [#permalink] New post 24 Mar 2014, 06:51
Expert's post
seabhi wrote:
Hi Bunuel,
Even I have the same questions, can you help clear the confusion with X = 1/2.

Thanks.


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] 24 Mar 2014, 06:51
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