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m09 q22

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Re: m09 q22 [#permalink]  26 Nov 2009, 06:43
Thank you, gmattokyo. I've corrected it. It seems like this question had too many revisions . +1.
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Re: m09 q22 [#permalink]  28 May 2010, 17:06
Guys, I do understand how to do it through plugging in numbers, can someone elaborate the method, where you use algebraic method. It would be great help!
Thank you,
I am considering two options x>0 and x<0
1) x>0
then cross multiplying we get x<x^2 then we find out that x is >1

2) x<0
then cross multiplying we get that x>-x^2 then we get that x is between -1 and 0
thus, x is between -1 and 0 and x>1

Is my reasoning correct?
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Re: m09 q22 [#permalink]  30 May 2010, 23:43
Yes, your reasoning is correct!
mirzohidjon wrote:
Guys, I do understand how to do it through plugging in numbers, can someone elaborate the method, where you use algebraic method. It would be great help!
Thank you,
I am considering two options x>0 and x<0
1) x>0
then cross multiplying we get x<x^2 then we find out that x is >1

2) x<0
then cross multiplying we get that x>-x^2 then we get that x is between -1 and 0
thus, x is between -1 and 0 and x>1

Is my reasoning correct?

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Re: m09 q22 [#permalink]  23 Nov 2010, 06:07
I am unclear on the solution when X < -1.

Why is |X| = - X

I dont get that. Isnt |X| always positive.

Could any one elaborate on this step.
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Re: m09 q22 [#permalink]  23 Nov 2010, 06:35
Expert's post
If $$\frac{x}{|x|} \lt x$$ , which of the following must be true about $$x$$? ($$x \ne 0$$)
A. $$x\gt 2$$
B. $$x \in (-1,0) \cup (1,\infty)$$
C. $$|x| \lt 1$$
D. $$|x| = 1$$
E. $$|x|^2 \gt 1$$

Absolute value properties:
Absolute value is always non-negative: $$|x|\geq{0}$$ (not positive but non-negative, meaning that absolute value can equal to zero), so:
When $$x\leq{0}$$ then $$|x|=-x$$ (note that in this case $$|x|=-negative=positive$$);
When $$x\geq{0}$$ then $$|x|=x$$.

For more check: math-absolute-value-modulus-86462.html

Back to the original question:

$$\frac{x}{|x|}< x$$
Two cases:
A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. But remember that $$x<0$$, so $$-1<x<0$$

B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$.

So the given inequality holds true in two ranges $$-1<x<0$$ and $$x>1$$.

Hope it helps.
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Re: m09 q22 [#permalink]  23 Nov 2010, 06:49
B.
Substitute numbers. I chose 3, -3, -1/3 & 1/3
for 3 & -1/3 the inequality holds good.
Instead of 3 we can have any number between 1 & infinity.
Similarly instead of -1/3 we can have any number between 0 & -1

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Re: m09 q22 [#permalink]  23 Nov 2010, 22:53
Hi all,
i disagree with most of you. My Answer is A.
Explanation:-
x/|x| <x
Case 1: if x<0
If x<0, say a no -3.
Then, -3/|-3|<-3
 -3/3<-3
 -1<-3 which is not true.
Case 2: x=0
Not applicable, as per the given condition.
Case 3: if x>0
Take a no say 1(x>0)
Then, 1/|1|<1
 1/1<1
 1<1 which is not true.
Take another no say 2(x>0)
Then, 2/|2|<2
 2/2<2
 1<2 which is true.
So , for a Question like which of the following must be true about ? ( )

The answer is X>2, which is Option A.
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Re: M09, Q22 [#permalink]  23 Nov 2010, 22:55
Dauren wrote:
What if X=0.5?

In this case X>-1, but the inquality is violated.

Hi Dauren,
Here is my solution. My ANS is A.
Explanation:-
x/|x| <x
Case 1: if x<0
If x<0, say a no -3.
Then, -3/|-3|<-3
 -3/3<-3
 -1<-3 which is not true.
Case 2: x=0
Not applicable, as per the given condition.
Case 3: if x>0
Take a no say 1(x>0)
Then, 1/|1|<1
 1/1<1
 1<1 which is not true.
Take another no say 2(x>0)
Then, 2/|2|<2
 2/2<2
 1<2 which is true.
So , for a Question like which of the following must be true about ?

The answer is X>2, which is Option A.

!!!Give me Kudos if you like my post.!!!!
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Re: m09 q22 [#permalink]  24 Nov 2010, 01:26
Expert's post
321kumarsushant wrote:
Hi all,
i disagree with most of you. My Answer is A.
Explanation:-
x/|x| <x
Case 1: if x<0
If x<0, say a no -3.
Then, -3/|-3|<-3
 -3/3<-3
 -1<-3 which is not true.
Case 2: x=0
Not applicable, as per the given condition.
Case 3: if x>0
Take a no say 1(x>0)
Then, 1/|1|<1
 1/1<1
 1<1 which is not true.
Take another no say 2(x>0)
Then, 2/|2|<2
 2/2<2
 1<2 which is true.
So , for a Question like which of the following must be true about ? ( )

The answer is X>2, which is Option A.

Number plugging is not the best method to solve this question.

OA for this question is B, not A: the given inequality holds true in two ranges $$-1<x<0$$ and $$x>1$$ (see solution in my previous post). You can try values from this ranges to check. So x>2 is not always true.
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Re: m09 q22 [#permalink]  24 Nov 2010, 01:42
Bunuel wrote:
321kumarsushant wrote:
Hi all,
i disagree with most of you. My Answer is A.
Explanation:-
x/|x| <x
Case 1: if x<0
If x<0, say a no -3.
Then, -3/|-3|<-3
 -3/3<-3
 -1<-3 which is not true.
Case 2: x=0
Not applicable, as per the given condition.
Case 3: if x>0
Take a no say 1(x>0)
Then, 1/|1|<1
 1/1<1
 1<1 which is not true.
Take another no say 2(x>0)
Then, 2/|2|<2
 2/2<2
 1<2 which is true.
So , for a Question like which of the following must be true about ? ( )

The answer is X>2, which is Option A.

Number plugging is not the best method to solve this question.

OA for this question is B, not A: the given inequality holds true in two ranges $$-1<x<0$$ and $$x>1$$ (see solution in my previous post). You can try values from this ranges to check. So x>2 is not always true.

x>2 will alwayz be true as it is a part of (-1,0)& (1,infinity)
anyway, i realized my mistake.
the answer will be B.
thnx buddy!
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Re: m09 q22 [#permalink]  24 Nov 2010, 01:48
Expert's post
321kumarsushant wrote:
Bunuel wrote:
321kumarsushant wrote:
Hi all,
i disagree with most of you. My Answer is A.
Explanation:-
x/|x| <x
Case 1: if x<0
If x<0, say a no -3.
Then, -3/|-3|<-3
 -3/3<-3
 -1<-3 which is not true.
Case 2: x=0
Not applicable, as per the given condition.
Case 3: if x>0
Take a no say 1(x>0)
Then, 1/|1|<1
 1/1<1
 1<1 which is not true.
Take another no say 2(x>0)
Then, 2/|2|<2
 2/2<2
 1<2 which is true.
So , for a Question like which of the following must be true about ? ( )

The answer is X>2, which is Option A.

Number plugging is not the best method to solve this question.

OA for this question is B, not A: the given inequality holds true in two ranges $$-1<x<0$$ and $$x>1$$ (see solution in my previous post). You can try values from this ranges to check. So x>2 is not always true.

x>2 will alwayz be true as it is a part of (-1,0)& (1,infinity)
anyway, i realized my mistake.
the answer will be B.
thnx buddy!

No, that's not correct.

Given inequality holds true for $$-1<x<0$$ and $$x>1$$, so if $$x$$ is in the range $$-1<x<0$$ (for example if $$x=-0.5$$) or in the range $$1<x\leq{2}$$ (for example if $$x=1.5$$) then $$x>2$$ won't be true.

Hope it's clear.
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Re: m09 q22 [#permalink]  04 Jun 2011, 02:29
I don't understand Bunuel's following explanation :
Two cases:
A. X < 0 --> X/-X < X --> . But remember that X < 0, so -1 < X < 0

B. X > 0 --> X/X --> 1 < X .

In case A when X < 0 --> Why isn't it like this -X/-X < -X Since all the 3 variables are the same X.

What am I missing here ? HELP !
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Re: m09 q22 [#permalink]  04 Jun 2011, 04:31
So many changes!! I lost track

Although x>2 is correct, option B is more correct as it encompasses the values in option A as well.
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Re: m09 q22 [#permalink]  06 Dec 2011, 03:58
Very good question
My attempt:
given x/|x|<X and x is not equal to 0
option 1:if x > 2 let say x=3 then 3/3 = 1 which is < than 3 hence true. same can be stated about X= 2 hence must not be true.

option 2 : if -1<x<0 and 1<x<infinity. lets assume X is -0.5 then -1 < -.05 true. x=2 then 1< 2 true. this statement must be true for all values of X as per inequality given
answer must be B
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Re: m09 q22 [#permalink]  05 Apr 2012, 06:58
Bunuel wrote:
If $$\frac{x}{|x|} \lt x$$ , which of the following must be true about $$x$$? ($$x \ne 0$$)
A. $$x\gt 2$$
B. $$x \in (-1,0) \cup (1,\infty)$$
C. $$|x| \lt 1$$
D. $$|x| = 1$$
E. $$|x|^2 \gt 1$$

Absolute value properties:
Absolute value is always non-negative: $$|x|\geq{0}$$ (not positive but non-negative, meaning that absolute value can equal to zero), so:
When $$x\leq{0}$$ then $$|x|=-x$$ (note that in this case $$|x|=-negative=positive$$);
When $$x\geq{0}$$ then $$|x|=x$$.

For more check: math-absolute-value-modulus-86462.html

Back to the original question:

$$\frac{x}{|x|}< x$$
Two cases:
A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. But remember that $$x<0$$, so $$-1<x<0$$

B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$.

So the given inequality holds true in two ranges $$-1<x<0$$ and $$x>1$$.

Hope it helps.

I am still confused. I guess I am missing some point.

if |x| > 3 , we solve it as -x< 3 and x> 3 , when we take the negative x we change the inequality sign.

But in the solution it has not been changed. Why? I have gone through math-absolute-value-modulus-86462.html but still confused.

I had come up with solution as E

A. $$x<0$$ --> $$\frac{x}{-x}>x$$ --> $$-1>x$$.

B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$.

So X^2 > 1.
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Re: m09 q22 [#permalink]  05 Apr 2012, 08:42
Expert's post
AmarSharma wrote:
Bunuel wrote:
If $$\frac{x}{|x|} \lt x$$ , which of the following must be true about $$x$$? ($$x \ne 0$$)
A. $$x\gt 2$$
B. $$x \in (-1,0) \cup (1,\infty)$$
C. $$|x| \lt 1$$
D. $$|x| = 1$$
E. $$|x|^2 \gt 1$$

Absolute value properties:
Absolute value is always non-negative: $$|x|\geq{0}$$ (not positive but non-negative, meaning that absolute value can equal to zero), so:
When $$x\leq{0}$$ then $$|x|=-x$$ (note that in this case $$|x|=-negative=positive$$);
When $$x\geq{0}$$ then $$|x|=x$$.

For more check: math-absolute-value-modulus-86462.html

Back to the original question:

$$\frac{x}{|x|}< x$$
Two cases:
A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. But remember that $$x<0$$, so $$-1<x<0$$

B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$.

So the given inequality holds true in two ranges $$-1<x<0$$ and $$x>1$$.

Hope it helps.

I am still confused. I guess I am missing some point.

if |x| > 3 , we solve it as -x< 3 and x> 3 , when we take the negative x we change the inequality sign.

But in the solution it has not been changed. Why? I have gone through math-absolute-value-modulus-86462.html but still confused.

I had come up with solution as E

A. $$x<0$$ --> $$\frac{x}{-x}>x$$ --> $$-1>x$$.

B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$.

So X^2 > 1.

You should read the solution more carefully.

A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$ (not $$-1>x$$ as you've written). But remember that we are considering the range $$x<0$$, so $$-1<x<0$$.

You could check that $$-1>x$$ is not correct by plugging some number, say $$x=-10$$ then $$\frac{x}{|x|}=-1>-10=x$$.

Similar questions to practice:

Hope it's clear.
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Re: m09 q22 [#permalink]  05 Apr 2012, 09:01

this post clarified all the doubts.

x/|x|<x when x<0 we get, x/x> -x [ for example |x| < 3 -> x > -3 when x is negative ]
-> 1 > -x -> -1< x. Voila , the answer.

Took me some time but thanks.
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Re: m09 q22 [#permalink]  04 Dec 2012, 04:10
Expert's post
Bunuel I got this question but the answer weren't the same as you have posted here. They had the original choices as posted in page 1.
I got the correct solution but since the answer choices given were incorrect, I chose the wrong one.

Please do review this question on GC CATS.
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Re: m09 q22 [#permalink]  04 Dec 2012, 04:19
Expert's post
Marcab wrote:
Bunuel I got this question but the answer weren't the same as you have posted here. They had the original choices as posted in page 1.
I got the correct solution but since the answer choices given were incorrect, I chose the wrong one.

Please do review this question on GC CATS.

The correct question is here: if-x-0-and-x-x-x-which-of-the-following-must-be-true-143572.html
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Re: m09 q22 [#permalink]  06 Dec 2013, 19:54
I was a bit confused between A and B, but when I put the value of x = -0.5, then went with B.
IMO B.
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Re: m09 q22   [#permalink] 06 Dec 2013, 19:54

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