Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

That's what I am saying. Since x cannot take this value then how can B be answer. How can x>-1 when 0<x<1 is not accepted?

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. \(-1<x<0\)

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.

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

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|>1\)

(E) \(-1<x<0\)

Explanations required for this one. Not convinced at all with the OA.

My range is -1<x<0 and x>1.

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

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

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

If \(x>0\) then \(|x|=x\), hence in this case we would have: \(\frac{x}{x}<x\) --> \(1<x\).

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

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

Answer: B.

As for other options:

A. \(x>1\). Not necessarily true since \(x\) could be -0.5; C. \(|x|<1\) --> \(-1<x<1\). Not necessarily true since \(x\) could be 2; D. \(|x|>1\) --> \(x<-1\) or \(x>1\). Not necessarily true since \(x\) could be -0.5; E. \(-1<x<0\). Not necessarily true since \(x\) could be 2.

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

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|>1\)

(E) \(-1<x<0\)

Hi Karishma Can you pls help me with the answer to the above link. I was able to solve the inequality My answer after solving inequality is -1<x<0 or x>1 So how can be the answer not E The point of elimination for option e in the official explanation is as given below:-How can x be 2 when the range is less than 0...... E. −1<x<0. Not necessarily true since x could be 2.

A 'must be true' question! They are absolutely straight forward if you get the fundamental but they can drive you crazy if you don't.

"My answer after solving inequality is -1<x<0 or x>1" Perfect. That is the range of x for which the inequality works. So tell me, what values can x take? -1/2, -1/3, -2/3, 1.4, 2, 500, 123498 etc... Now the question is "which of the following must be true?"

(A) \(x>1\) Are all these values greater than 1? No.

(B) \(x>-1\) Are all these values greater than -1? Yes. The answer. Note that you dont have to establish that all value greater than -1 should work for the inequality. You only have to establish that all values which work for the inequality must satisfy this condition.

(C) \(|x|<1\) Not true for all values of x.

(D) \(|x|>1\) Not true for all values of x.

(E) \(-1<x<0\) Not true for all values of x. x can take values 1.4, 2, 500 etc

Appreciate if someone could point out where I am going wrong here.

x / |x| < x

Since x is non zero, dividing by x on both sides

1 / |x| < 1

Taking reciprocal,

|x| > 1

Then I just jumped into Choice D. Didn't even look at the others.

Await your valued views.

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know its sign.

So you cannot divide both parts of inequality x / |x| < x by x as you don't know the sign of this unknown: if x>0 you should write 1/|x|<1 BUT if x<0 you should write 1/|x|>1.

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

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|>1\)

(E) \(-1<x<0\)

m09 q22

Explanations required for this one. Not convinced at all with the OA.

My range is -1<x<0 and x>1.

can you please, explain me what does the must be true clause mean in this question ??

0.1 to 0.9999999999 none satisfy this relation and more over they are greater than -1 and when again we have an option called > 1, why do we choose this to be wrng ? do we have any value > 1 but still don't satisfy this question ??

please explain, i did understand from -1 to 0 there are values which accept this relation but accpeting this doesn't mean we can omit from 0 to -9 ...

Im confused, im out of nuts . please help me for this

Question: if \(-1<x<0\) or \(x>1\), then which of the following must be true? Notice that \(-1<x<0\) or \(x>1\) is given to be true: x is either from {-1, 0} or from {1, +infinity}

Re: If x#0 and x/|x|<x, which of the following must be true? [#permalink]

Show Tags

04 Dec 2012, 05:14

If you choose 0.9, then B fails. I just reviewed m09 q22 on the forum. You have changed the answer choices in that thread but not yet on GC CAT.
_________________

Re: If x#0 and x/|x|<x, which of the following must be true? [#permalink]

Show Tags

04 Dec 2012, 19:48

1

This post was BOOKMARKED

\(\frac{x}{|x|}<x\)

Test the values for x.

(1) x = 0 ==> No! Since it is given that x is not 0 (2) x = -1 ==> -1 < -1 ==> No! (3) x = 1 ==> 1 < 1 ==> No! (4) x = -2 ==> -1 < -2 ==> No! (5) x = \(\frac{-1}{4}\) ==> -1 < \(\frac{-1}{4}\) ===> Yes! (5) x = 2 ==> 1 < 2 ==> Yes!

Our range: -1 < x < 0 or x > 1

What must always be true in that range? x > -1 always

Re: If x#0 and x/|x|<x, which of the following must be true? [#permalink]

Show Tags

08 Dec 2012, 19:05

Marcab wrote:

Hey thanks. Its crystal clear now.

Yes - same here.

The word in that explanation that helped me the most is "satify". I think the difficulty of this question is good. The learning moment is also exactley what I needed. The language is what confused me on the first attempt. I think it would be understood by more people if the question had the english rephrased to: "... which of the following statements can be satisfied by all possible values of x".

Having said that, I learnt a lot about absolute values on the number plane trying to get my head around this explanation, so maybe it's helping us learn in the best way possible

Re: If x#0 and x/|x|<x, which of the following must be true? [#permalink]

Show Tags

16 Dec 2012, 21:04

But your dividing my the same number x on both sides, whether its positive or negative, implying both sides will simultaneously be negative together or positive together which doesnt change the sign.

Re: If x#0 and x/|x|<x, which of the following must be true? [#permalink]

Show Tags

18 Dec 2012, 00:48

Ans:

we take two cases as we see there is a modulus sign. The equation becomes x(1-|x|)<0 and then after solving for both cases we get x to be always greater than -1, so the answer is (B).
_________________

Re: GMAT quant DS question from GMAT club tests [#permalink]

Show Tags

01 May 2013, 21:28

rongali wrote:

If x≠0 and x / |x| < x, which of the following must be true

A) x>1 B) x>−1 c) |x|<1 D) |x|>1 E) −1<x<0

why is B an answer, as the equation wont hold true for values between 0<x<1

Mark as a guessHide Answer

I think the answer is absolutely correct. Firstly the question asks for a "MUST be true" option. Both the ranges for x, as rightly calculated above are : x>1 OR -1<x<0. However, none of the options subscribe to MUST be true type.It is so because x could be 2 OR x could be -0.5 However, the option B, where x>-1 will always be true, irrespective of the two ranges. Any value lying in the range -1<x<0 IS always in tandem with x>-1 AND any value for x>1 WILL also have x>-1.
_________________

Re: GMAT quant DS question from GMAT club tests [#permalink]

Show Tags

01 May 2013, 21:38

doe007 wrote:

When x = 1/2, x is GT -1, but x / |x| is NOT < x Thus we cannot say that x / |x| < x for all x > -1. Option B is wrong.

We are not saying that x / |x| < x for all x > -1. That defies the whole purpose of breaking down the given inequality into 2 ranges. All we are saying is that for the 2 given ranges, the value of x MUST BE x>-1. You are taking the value of x=0.5, which in the first place is invalid for the given problem.
_________________

Spending two weeks in States, travelling coast to coast, visiting all the exceptional B-Schools. It reinvigorated the desire to go all in this MBA process. Before visiting Haas (...

Military MBA Acceptance Rate Analysis Transitioning from the military to MBA is a fairly popular path to follow. A little over 4% of MBA applications come from military veterans...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...