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M09-22

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New post 16 Sep 2014, 00:40
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If \(x \ne 0\) and \(\frac{x}{|x|} \lt x\), which of the following must be true?

A. \(x \gt 1\)
B. \(x \gt -1\)
C. \(|x| \lt 1\)
D. \(|x|>1\)
E. \(-1 \lt x \lt 0\)

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Re M09-22  [#permalink]

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New post 16 Sep 2014, 00:40
Official Solution:

If \(x \ne 0\) and \(\frac{x}{|x|} \lt x\), which of the following must be true?

A. \(x \gt 1\)
B. \(x \gt -1\)
C. \(|x| \lt 1\)
D. \(|x|>1\)
E. \(-1 \lt x \lt 0\)


Notice that we are asked to find which of the options MUST be true, not COULD be true.

Let's see what ranges does \(\frac{x}{|x|} \lt x\) give for \(x\). Two cases:

If \(x \lt 0\) then \(|x|=-x\), hence in this case we would have: \(\frac{x}{-x} \lt x\), which is the same as \(-1 \lt x\). But remember that we consider the range \(x \lt 0\), so \(-1 \lt x \lt 0\);

If \(x \gt 0\) then \(|x|=x\), hence in this case we would have: \(\frac{x}{x} \lt x\), which is the same as \(1 \lt x\).

So, \(\frac{x}{|x|} \lt x\) means that \(-1 \lt x \lt 0\) or \(x \gt 1\).

Only option which is ALWAYS true is B. ANY \(x\) from the range \(-1 \lt x \lt 0\) or \(x \gt 1\) will definitely be more the \(-1\).

As for other options:

A. \(x \gt 1\). Not necessarily true since \(x\) could be -0.5;

C. \(|x| \lt 1\), so \(-1 \lt x \lt 1\). Not necessarily true since \(x\) could be 2;

D. \(|x| \gt 1\), so \(x \lt -1\) or \(x \gt 1\). Not necessarily true since \(x\) could be -0.5;

E. \(-1 \lt x \lt 0\). Not necessarily true since \(x\) could be 2.


Answer: B
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Re: M09-22  [#permalink]

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New post 15 Oct 2014, 00:20
1
Hi Bunuel
B cannot be the answer!

If i put X= 0.5 which is X > -1, Then the
L.H.S = 1
And R.H.S = 0.5.
R.H.S is not greater than L.H.S
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Re: M09-22  [#permalink]

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New post 15 Oct 2014, 00:24
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VarunBhardwaj wrote:
Hi Bunuel
B cannot be the answer!

If i put X= 0.5 which is X > -1, Then the
L.H.S = 1
And R.H.S = 0.5.
R.H.S is not greater than L.H.S


x cannot be 0.5 because it does not satisfy given condition that \(\frac{x}{|x|} \lt x\). Please re-read the solution.
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Re: M09-22  [#permalink]

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New post 15 Oct 2014, 03:40
Apologies.
Thanks for explaining. I completely misunderstood the question stem.
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Re: M09-22  [#permalink]

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New post 08 Jul 2015, 08:13
Bunuel wrote:
VarunBhardwaj wrote:
Hi Bunuel
B cannot be the answer!

If i put X= 0.5 which is X > -1, Then the
L.H.S = 1
And R.H.S = 0.5.
R.H.S is not greater than L.H.S


x cannot be 0.5 because it does not satisfy given condition that \(\frac{x}{|x|} \lt x\). Please re-read the solution.




Hi Bunuel,

I am confused with the MUST be true condition, I understand that MUST be true is like SUFFICIENCY for DS problems, I do not know If I am wrong or right but for instance... If X > -1 then in a DS question the answer is YES for -1 < x < 0 and x > 1, but the answer is NO for 0 < x < 1... then it would not be sufficient... I understand that MUST is like in ALL cases the same answer (In all cases affirm the question stem), but I am confused know, could you help me?

Thanks a lot.

Regards.

Luis Navarro
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Re: M09-22  [#permalink]

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New post 08 Jul 2015, 08:48
luisnavarro wrote:
Bunuel wrote:
VarunBhardwaj wrote:
Hi Bunuel
B cannot be the answer!

If i put X= 0.5 which is X > -1, Then the
L.H.S = 1
And R.H.S = 0.5.
R.H.S is not greater than L.H.S


x cannot be 0.5 because it does not satisfy given condition that \(\frac{x}{|x|} \lt x\). Please re-read the solution.




Hi Bunuel,

I am confused with the MUST be true condition, I understand that MUST be true is like SUFFICIENCY for DS problems, I do not know If I am wrong or right but for instance... If X > -1 then in a DS question the answer is YES for -1 < x < 0 and x > 1, but the answer is NO for 0 < x < 1... then it would not be sufficient... I understand that MUST is like in ALL cases the same answer (In all cases affirm the question stem), but I am confused know, could you help me?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700


Go through the discussion of the same question here: if-x-0-and-x-x-x-which-of-the-following-must-be-true-143572.html

Also, practice must or could be true questions here: search.php?search_id=tag&tag_id=193

Hope it helps.
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Re M09-22  [#permalink]

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New post 17 Nov 2015, 12:29
I think this is a high-quality question.
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M09-22  [#permalink]

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New post 16 Jun 2016, 04:53
valmikee wrote:
I think this the explanation isn't clear enough, please elaborate. This given condition would not hold true if x=1. Hence, I don't think option B is the appropriate choice. Can someone please help clarify?


valmikee: the inequality gives us either \(x >1\) or \(-1<x<0\)

Hence, in any case of satisfied solution of inequality, x will always be greater than -1 --> must be true that: \(x >-1\) (B)
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Re: M09-22  [#permalink]

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New post 21 Jun 2016, 19:57
this question is amazing. im trying to figure out - how does answer E potentially result in the value 2 if x cannot take on two distinctive values? would you mind giving me a simple example with values? thank you
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New post 21 Jun 2016, 20:41
lydennis8 wrote:
this question is amazing. im trying to figure out - how does answer E potentially result in the value 2 if x cannot take on two distinctive values? would you mind giving me a simple example with values? thank you

lydennis8 - maybe some misunderstanding?

I think Bunuel only said that (E) not necessary true since if x = 2 then the inequality is still hold.

Say x= 3 then: \(3/|3|=1 < 3,\) (correct) and so on...

Hope it's clear :)
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Re: M09-22  [#permalink]

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New post 21 Jun 2016, 23:35
Linhbiz thanks for that, misunderstood the wording
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M09-22  [#permalink]

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New post 21 Jul 2016, 05:28
Hello, thanks for interesting question.

how can |x| = -x when x<0? if x <0 then |-x|=x?

I think when x<0, -x/|x|<-x. isn't it?

Please explain the flaws in my logic.

thank u!
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New post 21 Jul 2016, 08:24
lshoshiashvili wrote:
Hello, thanks for interesting question.

how can |x| = -x when x<0? if x <0 then |-x|=x?

I think when x<0, -x/|x|<-x. isn't it?

Please explain the flaws in my logic.

thank u!


This is a property of an absolute value:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

You should brush-up fundamentals on modulus:

Theory on Absolute Values: math-absolute-value-modulus-86462.html
The E-GMAT Question Series on ABSOLUTE VALUE: the-e-gmat-question-series-on-absolute-value-198503.html
Properties of Absolute Values on the GMAT: properties-of-absolute-values-on-the-gmat-191317.html
Absolute Value: Tips and hints: absolute-value-tips-and-hints-175002.html

DS Absolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Absolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Absolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

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New post 07 Aug 2016, 02:55
I think this is a high-quality question and I agree with explanation.
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Re M09-22  [#permalink]

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New post 20 May 2017, 09:13
I think this the explanation isn't clear enough, please elaborate. I understood the algebraic method but I had a doubt. An inequality problem can be solved by plugging in values also right? If we try to plug-in values into option A, the -0.5 value which we arrive by solving algebraically, will go against the statement itself. Is it that we can't solve some inequality problems by plugging in?
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Re: M09-22  [#permalink]

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New post 20 May 2017, 09:37
vil1 wrote:
I think this the explanation isn't clear enough, please elaborate. I understood the algebraic method but I had a doubt. An inequality problem can be solved by plugging in values also right? If we try to plug-in values into option A, the -0.5 value which we arrive by solving algebraically, will go against the statement itself. Is it that we can't solve some inequality problems by plugging in?


I'll try again.

The question asks if \(-1 \lt x \lt 0\) or \(x \gt 1\), then which of the following must be true.

Since \(-1 \lt x \lt 0\) or \(x \gt 1\), then it must be true to say about x that x > -1.

For example, x can be, among many other values, -0.9, -0.89292838, -0.76539, -0,5, ... (because \(-1 \lt x \lt 0\)) as well as x can be 3, \(\pi\), 4.17, \(\sqrt{71}\), ... (because \(x \gt 1\)). Any of them is greater than -1. For ANY possible x, so for ANY x from \(-1 \lt x \lt 0\) and \(x \gt 1\), it will be true to say that x is greater than -1.

Option A, which says that \(x \gt 1\), is NOT always true because if x is from \(-1 \lt x \lt 0\), say if x is -0.14, then \(x \gt 1\) will NOT be true.

Hope it's clear.
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New post 21 Aug 2017, 08:03
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aamir89 wrote:
Bunuel wrote:
Official Solution:

If \(x \ne 0\) and \(\frac{x}{|x|} \lt x\), which of the following must be true?

A. \(x \gt 1\)
B. \(x \gt -1\)
C. \(|x| \lt 1\)
D. \(|x|&gt;1\)
E. \(-1 \lt x \lt 0\)

Answer: B



Bunnel though i totally agree with your solution but in selection of option i do not agree. the question asks which of the following 'must be true"

if we are selecting B i.e., x>-1 then if x=1 the inequality does not hold true, in 0<x<1 also the inequality doesnt hold true.
But in E for every value of x in that interval the inequality holds true. i accept for x>1 also the inequality holds true but for every value in -1<x<0 the inequality holds true where as in B :x>-1 at some places the inequality is true and in some places the inequality does not hold true. B can be selected in case of "could be true" question.


Hi,

This question created a lot of confusion for me in the past until an instructor started with the basic. What does the question mean?

This part 'If \(x \ne 0\) and \(\frac{x}{|x|} \lt x\)' does NOT ask us to solve the question and find range of x but rather this part tell us a Truth or Fact, the other part asks us to state 'What Facts do we know about the solution of the inequality?


So you, me and other had successfully obtained the range. The question becomes if solution of inequality x/|x|<x is −1<x<0 or x>1, what MUST BE TRUE about EVERY X?

I quoted what he said below:
The OA does NOT imply that EVERY value greater than -1 is a valid solution for x/|x|<x.
It implies the reverse:
That every valid solution for x/|x|<x is greater than -1.

P.S.: This question or much similar to it appeared in GMATPrep or question pack 1 .

I hope it helps
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Re M09-22  [#permalink]

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New post 29 Aug 2017, 23:19
I think this is a poor-quality question and I don't agree with the explanation. Hi. Could you pls point me on mistake in my thoughts.
We have two options for X here:
x>1 or 0>x>-1
According to correct answer choice x>-1, x can be 1/2, for example. Let's check it for this solution 1<1/2 - seems not correct.
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Re: M09-22  [#permalink]

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New post 29 Aug 2017, 23:25
StaicyT wrote:
I think this is a poor-quality question and I don't agree with the explanation. Hi. Could you pls point me on mistake in my thoughts.
We have two options for X here:
x>1 or 0>x>-1
According to correct answer choice x>-1, x can be 1/2, for example. Let's check it for this solution 1<1/2 - seems not correct.


This is explained several times on this thread: we are given that -1 < x < 0 or x > 1. Whatever the actual value of x could be (again it could be only -1 < x < 0 or x > 1) it would be correct to say that x is greater than -1.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M09-22 &nbs [#permalink] 29 Aug 2017, 23:25

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