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

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

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20 Jun 2017, 06:32
Marcab wrote:
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) $$-1<x<0$$

We can simplify the given inequality:

x/|x| < x

x < (x)|x|

(x)|x| > x

If x is positive, we can divide both sides by x and obtain |x| > 1.

If x is negative, we can also divide both sides by x, but we have to switch the inequality sign, so we have |x| < 1.

We see that if x is positive, |x| > 1, which is choice C, and if x is negative, |x| < 1, which is choice D. However, since we don’t know whether x is positive or negative, both choice C and choice D “can be true,” not “must be true.”

Let’s analyze further. If x is positive, |x| = x. So, |x| > 1 means x > 1, which is choice A. If x is negative, |x| = -x. So, |x| < 1 means -x < 1 or x > -1. However, because x is negative, we have -1 < x < 0, which is choice E. Again, since we don’t know whether x is positive or negative, both choice A and choice E “can be true,” not “must be true.”

This leaves choice B as the correct answer. In fact, it’s the correct choice because the inequality x > -1 includes both x > 1 and -1 < x < 0. So, regardless of whether x is positive or negative, we can say x > -1.

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

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10 Aug 2017, 21:25
Bunuel wrote:
Marcab wrote:
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 $$-1$$.

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.

P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Please pay attention to the rules #3 and 6. Thank you.

If in case I multiply |x| both sides then inequality will not change and then if i approach like this then how the inequality should be solved :-

X<X|X|
x|x|-x>0
x(|x|-1)>0
now either x>0 or |x|>1,x<0 |x|<1

how to proceed further to solve it to get the range as per the question.
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Joined: 02 Sep 2009
Posts: 50572
Re: If x ≠ 0 and x/|x| < x, which of the following must be true?  [#permalink]

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10 Aug 2017, 22:43
himanshukamra2711 wrote:
Bunuel wrote:
Marcab wrote:
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 $$-1$$.

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.

P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Please pay attention to the rules #3 and 6. Thank you.

If in case I multiply |x| both sides then inequality will not change and then if i approach like this then how the inequality should be solved :-

X<X|X|
x|x|-x>0
x(|x|-1)>0
now either x>0 or |x|>1,x<0 |x|<1

how to proceed further to solve it to get the range as per the question.

$$x(|x|-1)>0$$

Case 1: $$x > 0$$ and $$|x| > 1$$ ($$x < -1$$ or $$x > 1$$) --> $$x > 1$$.

Case 2: $$x < 0$$ and $$|x| < 1$$ ($$-1 < x < 1$$) --> $$-1 < x < 0$$.

Finally, $$-1 < x < 0$$ or $$x > 1$$.
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Re: If x ≠ 0 and x/|x| < x, which of the following must be true?  [#permalink]

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27 Jan 2018, 23:30
given: x/|x| < x
multiply by |x| on both sides (as we know the sign of |x| => safe to multiply by |x|)
x < x|x|
x - x|x| < 0
x(1-|x|) < 0

if x < 0, then |x| has to be less than 1 to hold above inequality => -1 < x < 0
if x > 0, then |x| has to be greater than 1 to hold above inequality => |x| > 1 => x > 1

so range of x : -1 < x < 0, x > 1 => option B covers this range
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Joined: 31 Aug 2016
Posts: 45
Re: If x ≠ 0 and x/|x| < x, which of the following must be true?  [#permalink]

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20 May 2018, 03:12
VeritasPrepKarishma wrote:
Quote:
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

I wrote a post on this beautiful question sometime back:
http://www.veritasprep.com/blog/2012/07 ... -and-sets/

Hello Karishma, I understand everything from this Q. I just have one question.

You can agree that we can say that: x<|x|x since |x|>0 and then reach the same solution by writing x(|x|-1)>0.

I have learned to not divide x/x<x --> x>1 since I lose choices.

In which situation do I deduce that x>1 and in which situation I must write that x^2-x>0 .. and continue from there?
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Joined: 16 Oct 2010
Posts: 8527
Location: Pune, India
Re: If x ≠ 0 and x/|x| < x, which of the following must be true?  [#permalink]

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20 May 2018, 19:26
standyonda wrote:
VeritasPrepKarishma wrote:
Quote:
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

I wrote a post on this beautiful question sometime back:
http://www.veritasprep.com/blog/2012/07 ... -and-sets/

Hello Karishma, I understand everything from this Q. I just have one question.

You can agree that we can say that: x<|x|x since |x|>0 and then reach the same solution by writing x(|x|-1)>0.

I have learned to not divide x/x<x --> x>1 since I lose choices.

In which situation do I deduce that x>1 and in which situation I must write that x^2-x>0 .. and continue from there?

Yes, you can deduce
$$x * (|x|-1)>0$$

For the product of the two factors to be positive, either both are positive or both are negative.

EITHER
x > 0 and |x| - 1 > 0 (which is |x| > 1 ---> x > 1 or x < -1)
So x > 1

OR
x< 0 and |x| - 1 < 0 (which is |x| < 1 --> -1 < x < 1 )
So -1 < x < 0

So any value that x can take will either lie in -1 to 0 to will be greater than 1. So every value will be greater than -1 for sure.

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

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20 May 2018, 21:57
VeritasPrepKarishma

My question is about the solution that was provided were you deduce that x/x=1 instead of moving x on the other side of the inequality. In algebra generally if you do that you lose solutions. Why do you use it like that here? And how do you know that you will not lose a x=0 or something else... ?
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Posts: 8527
Location: Pune, India
Re: If x ≠ 0 and x/|x| < x, which of the following must be true?  [#permalink]

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21 May 2018, 00:14
standyonda wrote:
VeritasPrepKarishma

My question is about the solution that was provided were you deduce that x/x=1 instead of moving x on the other side of the inequality. In algebra generally if you do that you lose solutions. Why do you use it like that here? And how do you know that you will not lose a x=0 or something else... ?

Can you please quote where I have done that. I wouldn't cancel until and unless I am looking for a partial solution.
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Re: If x ≠ 0 and x/|x| < x, which of the following must be true? &nbs [#permalink] 21 May 2018, 00:14

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