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M09-19

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Is \(|x - 5| \gt 4\)?

Is \(|x-5| \gt 4\)? Opening the absolute value we get the question: is \(x \lt 1\) or \(x \gt 9\)?

(1) \(x^2-4 \lt 0\) and thus \(x^2 \lt 4\) therefore \(|x| \lt 2\) and we can deduce that \(-2 < x < 2\), so we can have an YES answer as well as a NO answer (consider \(x=0\) and \(x=1.5\)). Not sufficient.

(2) \(x^2-1 \lt 0\) thus \(x^2 \lt 1\) therefore \(|x| \lt 1\) and finally, \(-1 \lt x \lt 1\). Sufficient.


Answer: B
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Re: M09-19 [#permalink]

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New post 12 Oct 2014, 10:36
Bunuel wrote:
Official Solution:


Is \(|x-5| \gt 4\)? Opening the absolute value we get the question: is \(x \lt 1\) or \(x \gt 9\)?

(1) \(x^2-4 \gt 0\) and thus \(x^2 \gt 4\) therefore \(|x| \gt 2\) and we can deduct that \(x \lt -2\) or \(x \gt 2\), so we can have an YES answer as well as a NO answer (consider \(x=-5\) and \(x=5\)). Not sufficient.

(2) \(x^2-1 \lt 0\) thus \(x^2 \lt 1\) therefore \(|x| \lt 1\) and finally, \(-1 \lt x \lt 1\). Sufficient.


Answer: B


Hi Bunuel,
I have a doubt in inequality with absolute value questions where we are asked to find if the given statement satisfies the two required inequalities for instance in the above question:
we are required to find if \(x<1 or x >9\), so as per the Second statement we get \(-1<x<1\) so is this sufficient to answer the question?

Because \(x>9\) or not is still left unattended.
SO what exactly is the approach in such questions?

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Re: M09-19 [#permalink]

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New post 12 Oct 2014, 10:48
earnit wrote:
Bunuel wrote:
Official Solution:


Is \(|x-5| \gt 4\)? Opening the absolute value we get the question: is \(x \lt 1\) or \(x \gt 9\)?

(1) \(x^2-4 \gt 0\) and thus \(x^2 \gt 4\) therefore \(|x| \gt 2\) and we can deduct that \(x \lt -2\) or \(x \gt 2\), so we can have an YES answer as well as a NO answer (consider \(x=-5\) and \(x=5\)). Not sufficient.

(2) \(x^2-1 \lt 0\) thus \(x^2 \lt 1\) therefore \(|x| \lt 1\) and finally, \(-1 \lt x \lt 1\). Sufficient.


Answer: B


Hi Bunuel,
I have a doubt in inequality with absolute value questions where we are asked to find if the given statement satisfies the two required inequalities for instance in the above question:
we are required to find if \(x<1 or x >9\), so as per the Second statement we get \(-1<x<1\) so is this sufficient to answer the question?

Because \(x>9\) or not is still left unattended.
SO what exactly is the approach in such questions?

Regards


The question asks whether x < 1 or x > 9 (those are two ranges which make |x - 5| > 4 to hold true).

(2) says that -1 < x < 1, so x IS less than 1, and therefore |x - 5| > 4 to holds true.
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Re: M09-19 [#permalink]

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New post 12 Oct 2014, 10:54
Bunuel wrote:
earnit wrote:
Bunuel wrote:
Official Solution:


Is \(|x-5| \gt 4\)? Opening the absolute value we get the question: is \(x \lt 1\) or \(x \gt 9\)?

(1) \(x^2-4 \gt 0\) and thus \(x^2 \gt 4\) therefore \(|x| \gt 2\) and we can deduct that \(x \lt -2\) or \(x \gt 2\), so we can have an YES answer as well as a NO answer (consider \(x=-5\) and \(x=5\)). Not sufficient.

(2) \(x^2-1 \lt 0\) thus \(x^2 \lt 1\) therefore \(|x| \lt 1\) and finally, \(-1 \lt x \lt 1\). Sufficient.


Answer: B


Hi Bunuel,
I have a doubt in inequality with absolute value questions where we are asked to find if the given statement satisfies the two required inequalities for instance in the above question:
we are required to find if \(x<1 or x >9\), so as per the Second statement we get \(-1<x<1\) so is this sufficient to answer the question?

Because \(x>9\) or not is still left unattended.
SO what exactly is the approach in such questions?

Regards


The question asks whether x < 1 or x > 9 (those are two ranges which make |x - 5| > 4 to hold true).

(2) says that -1 < x < 1, so x IS less than 1, and therefore |x - 5| > 4 to holds true.


So as basically if either of the range is satisfied by the given statements conclusively, then we do not have to worry about the other range ie x>9 ?
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Re: M09-19 [#permalink]

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Re: M09-19 [#permalink]

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New post 12 Oct 2014, 11:08
Got it. Thank you.

If possible, could we find similar questions with two ranges like above question and two statements that proves their sufficiency?
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Re: M09-19 [#permalink]

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New post 24 Mar 2016, 05:27
While the Algebra approach is perfectly fine, I was just wondering whether a Plug-In approach (picking up numbers & then matching with the inequalities) could be used in such cases or not. Is it going to be more confusing or more time confusing than solving the three given inequalities?
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Re: M09-19 [#permalink]

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New post 05 Apr 2016, 05:25
Bunuel wrote:
Official Solution:


Is \(|x-5| \gt 4\)? Opening the absolute value we get the question: is \(x \lt 1\) or \(x \gt 9\)?

(1) \(x^2-4 \gt 0\) and thus \(x^2 \gt 4\) therefore \(|x| \gt 2\) and we can deduct that \(x \lt -2\) or \(x \gt 2\), so we can have an YES answer as well as a NO answer (consider \(x=-5\) and \(x=5\)). Not sufficient.

(2) \(x^2-1 \lt 0\) thus \(x^2 \lt 1\) therefore \(|x| \lt 1\) and finally, \(-1 \lt x \lt 1\). Sufficient.


Answer: B

Dear Bunuel,
I solved this question using algebra and got the question right.
But there is a discrepancy in the result of x between your solution and mine.
Please help me understand the difference

X^2-4>0 can be written as (x-2)(x+2)>0
therefore x>2 or X>-2,
But if I write mod X>2
then X>2
or -X >2 ==> X<-2

How can the same equation have two different solutions ?
Thanks
Teja
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Re: M09-19 [#permalink]

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New post 05 Apr 2016, 05:40
evst wrote:
Bunuel wrote:
Official Solution:


Is \(|x-5| \gt 4\)? Opening the absolute value we get the question: is \(x \lt 1\) or \(x \gt 9\)?

(1) \(x^2-4 \gt 0\) and thus \(x^2 \gt 4\) therefore \(|x| \gt 2\) and we can deduct that \(x \lt -2\) or \(x \gt 2\), so we can have an YES answer as well as a NO answer (consider \(x=-5\) and \(x=5\)). Not sufficient.

(2) \(x^2-1 \lt 0\) thus \(x^2 \lt 1\) therefore \(|x| \lt 1\) and finally, \(-1 \lt x \lt 1\). Sufficient.


Answer: B

Dear Bunuel,
I solved this question using algebra and got the question right.
But there is a discrepancy in the result of x between your solution and mine.
Please help me understand the difference

X^2-4>0 can be written as (x-2)(x+2)>0
therefore x>2 or X>-2,But if I write mod X>2
then X>2
or -X >2 ==> X<-2

How can the same equation have two different solutions ?
Thanks
Teja


How did you get x>2 or x>-2? It does not make sense.
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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: M09-19 [#permalink]

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New post 20 May 2016, 07:54
Bunuel wrote:
Official Solution:


Is \(|x-5| \gt 4\)? Opening the absolute value we get the question: is \(x \lt 1\) or \(x \gt 9\)?



Can you explain "opening the absolute value"? I get what you mean in the end result as I can see if you add 5 to the other side you get 9 and if you subtract 5 you get -1 (how'd it become a positive 1?). But how did you know which one to flip the sign on?

I imagine this is much easier done than said.
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Re: M09-19 [#permalink]

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redfield wrote:
Bunuel wrote:
Official Solution:


Is \(|x-5| \gt 4\)? Opening the absolute value we get the question: is \(x \lt 1\) or \(x \gt 9\)?



Can you explain "opening the absolute value"? I get what you mean in the end result as I can see if you add 5 to the other side you get 9 and if you subtract 5 you get -1 (how'd it become a positive 1?). But how did you know which one to flip the sign on?

I imagine this is much easier done than said.


hi,
there are two ways ..
1) open the MOD..
\(|x-5| \gt 4\)..
first time let x-5 be +ive ... so \(x-5>4.....................x>9...\)
second time take x-5 to be -ive... so \(-(x-5)>4...\) Multiply both sides by '-', the > sign will change........\(.....x-5<4..... x<1.\)
2) critical method...
here it will amount to same as above
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Re: M09-19 [#permalink]

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New post 20 May 2016, 09:11
chetan2u wrote:
redfield wrote:
Bunuel wrote:
Official Solution:


Is \(|x-5| \gt 4\)? Opening the absolute value we get the question: is \(x \lt 1\) or \(x \gt 9\)?



Can you explain "opening the absolute value"? I get what you mean in the end result as I can see if you add 5 to the other side you get 9 and if you subtract 5 you get -1 (how'd it become a positive 1?). But how did you know which one to flip the sign on?

I imagine this is much easier done than said.


hi,
there are two ways ..
1) open the MOD..
\(|x-5| \gt 4\)..
first time let x-5 be +ive ... so \(x-5>4.....................x>9...\)
second time take x-5 to be -ive... so \(-(x-5)>4...\) Multiply both sides by '-', the > sign will change........\(.....x-5<4..... x<1.\)
2) critical method...
here it will amount to same as above


For more check the links below:

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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Re: M09-19 [#permalink]

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I was of the impression that in DS questions the two statements provided cannot contradict each other. Is that true? If yes, aren't the two statements contradicting each other in this question?
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New post 30 Dec 2016, 01:02
CircleOfLife wrote:
I was of the impression that in DS questions the two statements provided cannot contradict each other. Is that true? If yes, aren't the two statements contradicting each other in this question?


Yes, you are absolutely correct. Edited the question. Thank you.
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Collection of Questions:
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Re: M09-19 [#permalink]

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New post 31 May 2017, 01:09
Bunuel I'm not sure I understand - if for statement 2 I plug in x=0,
we have on one hand x-5=0-5=-5 > 4? No.
But |x-5| also means -(x-5), correct? Then if I plug in x=0 here, it becomes -(0-5)=5>4? Yes. We get 2 different answers with x=0 if we go with statement 2. Where am I going wrong? Thanks.
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bianalyst wrote:
Bunuel I'm not sure I understand - if for statement 2 I plug in x=0,
we have on one hand x-5=0-5=-5 > 4? No.
But |x-5| also means -(x-5), correct? Then if I plug in x=0 here, it becomes -(0-5)=5>4? Yes. We get 2 different answers with x=0 if we go with statement 2. Where am I going wrong? Thanks.


First of all an absolute value is ALWAYS non-negative, again it CANNOT be negative.

So, |x - 5|, no matter what x is will be positive or 0. It'll be 0 if x = 5 and positive for ANY other values of x.

If x = 0, the |x - 5| = |0 - 5| = |-5| = 5.

Hope it's clear.
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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: M09-19 [#permalink]

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New post 03 Jun 2017, 08:46
For problem-related to mod, keep in mind if |a|>|b| than a^2>b^2
X2-10x+9>0 which will be rewritten as (x-9)(x-1)>0
Solution set for this equation is (-∞,1) U (9,∞)

Statement 1 says (x-2)(x+2)<0 which means (-2<x<2), and within this solution set stem will be positive as well as negative, NS

Statement 2 says (x-1)(x+1)<0 which means (-1<x<1), and within this solution set stem only be positive only. Sufficient

Hence B is the answer. Hope I am right and it's clear.
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M09-19 [#permalink]

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New post 14 Nov 2017, 00:55
Hi Bunuel,

Many thanks for this great question. I got the solution, the way you showed it. Only today, while I was attempting another question from the GMAT Club tests, I stumbled upon the sign method for solving inequalities. I was just wondering if you or someone else could kindly explain how to solve THIS question using the sign method. Many thanks!
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M09-19   [#permalink] 14 Nov 2017, 00:55

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