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M24-30

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M24-30 [#permalink]

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New post 16 Sep 2014, 01:22
1
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A
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D
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  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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Re M24-30 [#permalink]

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New post 16 Sep 2014, 01:22
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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 \(x \gt 1\).


Answer: B
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Re: M24-30 [#permalink]

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New post 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 \(x \gt 1\).


Answer: B


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
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Re: M24-30 [#permalink]

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New post 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 \(x \gt 1\).


Answer: B


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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Re: M24-30 [#permalink]

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New post 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.

Finally, B is the answer.
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Re: M24-30 [#permalink]

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New post 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.
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Re: M24-30 [#permalink]

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New post 06 Aug 2016, 11:06
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Re: M24-30 [#permalink]

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New post 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. \(|x|^2 \gt 1\)
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Re: M24-30 [#permalink]

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New post 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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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

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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M24-30 [#permalink]

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New post 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
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Re: M24-30 [#permalink]

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New post 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
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Re: M24-30 [#permalink]

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New post 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.
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Re: M24-30 [#permalink]

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New post 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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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

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M24-30   [#permalink] 30 Jun 2018, 00:35
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