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# M24-30

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Math Expert
Joined: 02 Sep 2009
Posts: 47015

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16 Sep 2014, 01:22
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Difficulty:

35% (medium)

Question Stats:

61% (01:07) correct 39% (01:00) wrong based on 143 sessions

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If $$\frac{x}{|x|} \lt x$$ which of the following must be true about $$x$$?

A. $$x \gt 1$$
B. $$-1 \lt x \lt 0$$ or $$x \gt 1$$
C. $$|x| \lt 1$$
D. $$|x| = 1$$
E. $$|x|^2 \gt 1$$

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Math Expert
Joined: 02 Sep 2009
Posts: 47015

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16 Sep 2014, 01:22
1
1
Official Solution:

If $$\frac{x}{|x|} \lt x$$ which of the following must be true about $$x$$?

A. $$x \gt 1$$
B. $$-1 \lt x \lt 0$$ or $$x \gt 1$$
C. $$|x| \lt 1$$
D. $$|x| = 1$$
E. $$|x|^2 \gt 1$$

Let's see in which ranges $$\frac{x}{|x|} \lt x$$ holds true.

A. If $$x \lt 0$$, then $$|x|=-x$$, so we'll have that $$\frac{x}{-x} \lt x$$ or $$-1 \lt x$$. Since we consider the range when $$x \lt 0$$ then $$-1 \lt x \lt 0$$;

B. If $$x \gt 0$$, then $$|x|=x$$, so we have that $$\frac{x}{x} \lt x$$ or $$1 \lt x$$.

So, given inequality holds true when $$-1 \lt x \lt 0$$ or $$x \gt 1$$.

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Intern
Joined: 24 Jun 2015
Posts: 46

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02 Jul 2015, 10:34
Bunuel wrote:
Official Solution:

If $$\frac{x}{|x|} \lt x$$ which of the following must be true about $$x$$?

A. $$x \gt 1$$
B. $$-1 \lt x \lt 0$$ or $$x \gt 1$$
C. $$|x| \lt 1$$
D. $$|x| = 1$$
E. $$|x|^2 \gt 1$$

Let's see in which ranges $$\frac{x}{|x|} \lt x$$ holds true.

A. If $$x \lt 0$$, then $$|x|=-x$$, so we'll have that $$\frac{x}{-x} \lt x$$ or $$-1 \lt x$$. Since we consider the range when $$x \lt 0$$ then $$-1 \lt x \lt 0$$;

B. If $$x \gt 0$$, then $$|x|=x$$, so we have that $$\frac{x}{x} \lt x$$ or $$1 \lt x$$.

So, given inequality holds true when $$-1 \lt x \lt 0$$ or $$x \gt 1$$.

Hi,

I am confused about this: A. If x<0, then |x|=−x .... I thought that |x| is always positive, why then -x? Could you help me?

Thanks a lot.

Regards.

Luis Navarro
Math Expert
Joined: 02 Sep 2009
Posts: 47015

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02 Jul 2015, 10:38
luisnavarro wrote:
Bunuel wrote:
Official Solution:

If $$\frac{x}{|x|} \lt x$$ which of the following must be true about $$x$$?

A. $$x \gt 1$$
B. $$-1 \lt x \lt 0$$ or $$x \gt 1$$
C. $$|x| \lt 1$$
D. $$|x| = 1$$
E. $$|x|^2 \gt 1$$

Let's see in which ranges $$\frac{x}{|x|} \lt x$$ holds true.

A. If $$x \lt 0$$, then $$|x|=-x$$, so we'll have that $$\frac{x}{-x} \lt x$$ or $$-1 \lt x$$. Since we consider the range when $$x \lt 0$$ then $$-1 \lt x \lt 0$$;

B. If $$x \gt 0$$, then $$|x|=x$$, so we have that $$\frac{x}{x} \lt x$$ or $$1 \lt x$$.

So, given inequality holds true when $$-1 \lt x \lt 0$$ or $$x \gt 1$$.

Hi,

I am confused about this: A. If x<0, then |x|=−x .... I thought that |x| is always positive, why then -x? Could you help me?

Thanks a lot.

Regards.

Luis Navarro

If x is negative, then -x = -negative = positive, so |x| is still positive.

Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: absolute-value-tips-and-hints-175002.html

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

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

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Current Student
Joined: 22 Oct 2015
Posts: 6
GMAT 1: 690 Q49 V36
WE: Investment Banking (Investment Banking)

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07 Mar 2016, 09:33
1
I did this question just by plugging number. It takes a little more time (less than 2 mins anyways), but for me the official solution presented is just too complicated.

A. Let's say x=3/2, than 3/2:3/2<3/2, 1<1.5. This might work.
B. We know x>1 works, so we try -1/2:I1/2I<-1/2, -1<-1/2. Also works, so we eliminate A.
C. IxI<1, x<1 or x>-1, so we test 1/2, 1/2:1/2<1/2, 1<1/2. Incorrect. Eliminate
D. x=+/-1, -1/I1I<-1, -1<-1, Incorrect. Eliminate.
E. x>1, x<-1, we haven't test x<-1, -3/2/I3/2I<-3/2 -1<-3/2. Incorrect. Eliminate.

Current Student
Joined: 03 Apr 2016
Posts: 91
Location: India
Concentration: Operations, General Management
GMAT 1: 720 Q50 V37
WE: Analyst (Computer Software)

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06 Aug 2016, 07:57
why isn't A true? Although A does not cover entire set for x , its indeed true that x> 1.
Math Expert
Joined: 02 Sep 2009
Posts: 47015

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06 Aug 2016, 11:06
shashanksagar wrote:
why isn't A true? Although A does not cover entire set for x , its indeed true that x> 1.

Because A is not always true. For example, when x = -1/2.
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Manager
Joined: 30 Jun 2015
Posts: 60
Location: Malaysia
Schools: Babson '20
GPA: 3.6

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16 Dec 2016, 21:32
Great question. Can you please explain the range of values that $$|x|^2 \gt 1$$ will take?

Will it be x>1 or x<-1 ?

Bunuel wrote:
If $$\frac{x}{|x|} \lt x$$ which of the following must be true about $$x$$?

A. $$x \gt 1$$
B. $$-1 \lt x \lt 0$$ or $$x \gt 1$$
C. $$|x| \lt 1$$
D. $$|x| = 1$$
E. $$|x|^2 \gt 1$$
Math Expert
Joined: 02 Sep 2009
Posts: 47015

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17 Dec 2016, 03:50
cuhmoon wrote:
Great question. Can you please explain the range of values that $$|x|^2 \gt 1$$ will take?

Will it be x>1 or x<-1 ?

Bunuel wrote:
If $$\frac{x}{|x|} \lt x$$ which of the following must be true about $$x$$?

A. $$x \gt 1$$
B. $$-1 \lt x \lt 0$$ or $$x \gt 1$$
C. $$|x| \lt 1$$
D. $$|x| = 1$$
E. $$|x|^2 \gt 1$$

Yes. $$|x|^2 \gt 1$$ means x < -1 or x > 1.
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Intern
Joined: 15 Aug 2013
Posts: 37

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08 Dec 2017, 11:10
Hi Bunuel ,
Could you please explain why are we including zero in this range.
Since we consider the range when x<0 then −1<x<0.

Thanks
Abhinash
Intern
Joined: 15 Aug 2013
Posts: 37

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08 Dec 2017, 11:12
abhinashgc wrote:
Hi Bunuel ,
Could you please explain why are we including zero in this range.
Since we consider the range when x<0 then −1<x<0.

Thanks
Abhinash

Sorry I got it.Thanks
Intern
Joined: 19 Nov 2012
Posts: 20

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30 Jun 2018, 00:19
Hi Bunuel,

Your solution is elegant. However, lets say I tried something like whats given below, then how do I get to the same ans?

$$x / |x| < x$$
x < x |x|
x( |x| - 1) > 0
case 1: x> 0 OR |x| -1 > 0
x > 0 OR x> 1 OR x < -1
case 2: x < 0 OR |x| - 1 < 0
x< 0 OR -1 < x < 1

Thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 47015

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30 Jun 2018, 00:35
sanjay1810 wrote:
Hi Bunuel,

Your solution is elegant. However, lets say I tried something like whats given below, then how do I get to the same ans?

$$x / |x| < x$$
x < x |x|
x( |x| - 1) > 0
case 1: x> 0 OR |x| -1 > 0
x > 0 OR x> 1 OR x < -1
case 2: x < 0 OR |x| - 1 < 0
x< 0 OR -1 < x < 1

Thanks.

x(|x| - 1) > 0

Case 1:
x > 0 AND |x| - 1 > 0. Notice that if x > 0, then |x| = x, so we'd get x > 0 AND x - 1 > 0;
x > 0 AND x > 1
So, in this case we have x > 1.

Case 2:
x < 0 OR |x| - 1 < 0. Notice that if x < 0, then |x| = -x, so we'd get x < 0 AND -x - 1 < 0;
x < 0 AND -1 < x;
So, in this case we have -1 < x < 0.

As you can see we got the same range as in the solution above.
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Re: M24-30   [#permalink] 30 Jun 2018, 00:35
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# M24-30

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