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M14-13

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New post 18 Jun 2018, 06:19
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New post 22 Apr 2019, 05:08
Hi experts,

On solving the two equations, I am receiving the folowing values of x:
1) x>1/2, x> -1
2) x>-1/2 , x>1

can you please explain how are we getting x<-1 from the first equation and x<-1/2 from the second equation

thanks
Nikita maheshwari
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New post 22 Apr 2019, 09:02
nikitam wrote:
Hi experts,

On solving the two equations, I am receiving the folowing values of x:
1) x>1/2, x> -1
2) x>-1/2 , x>1

can you please explain how are we getting x<-1 from the first equation and x<-1/2 from the second equation

thanks
Nikita maheshwari


I think the solution is elaborated couple of times in the thread. For more on inequalities, please check below links:

9. Inequalities



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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New post 27 Apr 2019, 00:07
I don't agree with the explanation. Sir each option leads to two different sets of answers how can we derive a particular set of inequality.
In first option, either both can be negative or both can be positive for the solution to be greater than zero. This leads to two differrent sets of replies
Similarly second option also does the same
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New post 27 Apr 2019, 03:32
ESHANGOEL wrote:
I don't agree with the explanation. Sir each option leads to two different sets of answers how can we derive a particular set of inequality.
In first option, either both can be negative or both can be positive for the solution to be greater than zero. This leads to two differrent sets of replies
Similarly second option also does the same


The question is 100% correct. Please re-read the thread carefully and follow the links provided to brush up fundamentals on inequalities.
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M14-13  [#permalink]

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New post Updated on: 30 Apr 2019, 23:40
Bunuel , thanks for the explanation.

I was able to get the same answers during the test I did, but I was really confused by the two ranges given.

I was wondering if you can clarify 2 points for me please.

Point 1: The question asks is |x|>1 ? so I thought we had to prove whether x>0 and -x>0 (translating to...) --> x<1
As in, we need to ensure both x>1 and x<1 holds true

Is my interpretation correct?

Point 2:
I'm not sure how you came to a conclusion based on the "intersection" as there are two potential ranges

x>1/2 and x<-1 (from statement 1); and
x>-1/2 and x>1 (from statement 2)

Why did you take x<-1 and x>1 as sufficient to rule out the other range (x>1/2 and x>-1/2)?
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Originally posted by dcummins on 30 Apr 2019, 21:35.
Last edited by dcummins on 30 Apr 2019, 23:40, edited 1 time in total.
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New post 30 Apr 2019, 21:47
dcummins wrote:
Bunuel , thanks for the explanation.

I was able to get the same answers during the test I did, but I was really confused by the two ranges given.

I was wondering if you can clarify 2 points for me please.

Point 1: The question asks is |x|>0 ? so I thought we had to prove whether x>0 and -x>0 (translating to...) --> x<0
As in, we need to ensure both x>0 and x<0 holds true

Is my interpretation correct?

Point 2:
I'm not sure how you came to a conclusion based on the "intersection" as there are two potential ranges

x>1/2 and x<-1 (from statement 1); and
x>-1/2 and x>1 (from statement 2)

Why did you take x<-1 and x>1 as sufficient to rule out the other range (x>1/2 and x>-1/2)?


1. The question asks whether |x| > 1, not whether |x| > 0.

|x| > 1 means x < -1 or x > 1.
While |x| > 0, holds true for ANY value but 0.

2. From (1): \(x \lt -1\)and \(x \gt \frac{1}{2}\)
-----------(-1)--------------------------------(1/2)------------------------[/size]

From (2): \(x \lt - \frac{1}{2}\) and \(x \gt 1\).
------------------------(-1/2)------------------------------------(1)------------------------

Ranges which satisfy BOTH are:
-----------(-1)----------------------------------------------------(1)------------------------
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Re: M14-13  [#permalink]

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New post 01 May 2019, 00:03
Bunuel wrote:
dcummins wrote:
Bunuel , thanks for the explanation.

I was able to get the same answers during the test I did, but I was really confused by the two ranges given.

I was wondering if you can clarify 2 points for me please.

Point 1: The question asks is |x|>0 ? so I thought we had to prove whether x>0 and -x>0 (translating to...) --> x<0
As in, we need to ensure both x>0 and x<0 holds true

Is my interpretation correct?

Point 2:
I'm not sure how you came to a conclusion based on the "intersection" as there are two potential ranges

x>1/2 and x<-1 (from statement 1); and
x>-1/2 and x>1 (from statement 2)

Why did you take x<-1 and x>1 as sufficient to rule out the other range (x>1/2 and x>-1/2)?


1. The question asks whether |x| > 1, not whether |x| > 0.

|x| > 1 means x < -1 or x > 1.
While |x| > 0, holds true for ANY value but 0.

2. From (1): \(x \lt -1\)and \(x \gt \frac{1}{2}\)
-----------(-1)--------------------------------(1/2)------------------------[/size]

From (2): \(x \lt - \frac{1}{2}\) and \(x \gt 1\).
------------------------(-1/2)------------------------------------(1)------------------------

Ranges which satisfy BOTH are:
-----------(-1)----------------------------------------------------(1)------------------------



Thanks mate.

My point 1 was a typo (testing all morning). I meant to ask what you stated, so I amended my post. But my question is still unanswered.

If we are asked is |x|> 1 then we interpret that as 'is x>1 and is x<-1' ?

Sure, so it looks like I failed to flip the sign for x<-1/2 and so the overlap that includes our desired range is given by the combination of statements, yeah?
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Re: M14-13   [#permalink] 01 May 2019, 00:03

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