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# M14 #13

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Manager
Joined: 12 Feb 2012
Posts: 108
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Kudos [?]: 20 [0], given: 28

Re: M14 #13 [#permalink]  27 Feb 2012, 19:07
Hey Bunuel or anyone,

In statement (1) I found the domain of x by finding the roots and plugging different values within the sections to find if it satified the inequility. But how do we find the domain of x through this other approach???

either {(1-2x) > 0 and (1+x) < 0 } OR { (1-2x) < 0 and (1+x) > 0 }

Hence, x > 1/2 or x < -1 ???

What I arrived at using this approach was:

{ 1/2 > x and x < -1 } OR { 1/2 < x and x>-1} ------->>> { 1/2 > x and x < -1 } OR { 1/2 < x }

but how do we combine this OR statement to arrive ---->>>> x > 1/2 or x < -1 ????

We have two contradictory expressions 1/2 > x or 1/2 < x

Manager
Status: I will not stop until i realise my goal which is my dream too
Joined: 25 Feb 2010
Posts: 235
Schools: Johnson '15
Followers: 2

Kudos [?]: 33 [0], given: 16

Re: M14 #13 [#permalink]  29 Apr 2012, 09:59
dczuchta wrote:
Hello. I did not see this question under the search, although I'm not certain I searched correctly.

If $$x$$ is an integer, is $$|x| \gt 1$$ ?

1. $$(1 - 2x)(1 + x) \lt 0$$
2. $$(1 - x)(1 + 2x) \lt 0$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Answer is C because combined we come conclude that neither 0,-1 and 1 can be X and therefore [x]>1

My question: What is the most effiecient way of answering this question? It it best to just plug in numbers and by doing so, conclude that between S1 and S2 X is neither 0, 1, or -1?

My route was:

1) Simplifying S1 to.... 2X>1 so X>1/2 OR X<-1
2) Simplifying S2 to....X>1 OR 2X<-1 so X>-1/2

Is that incorrect? And if so, how would you combine S1 and S2 and exclude numbers?
Thank you.

1: gives
x>1/2 and x<-1

so we get both YES and NO

2: gives x> 1 and X<-1/2
so both YES and NO

so A B D are knocked off

together we get that the values are between 1/2 and -1/2, hence the ANSWER is NO..hence C
_________________

Regards,
Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

Satyameva Jayate - Truth alone triumphs

Intern
Joined: 15 Apr 2012
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Re: M14 #13 [#permalink]  01 May 2012, 08:29
Bunel,

I have read your post on this topic in detail. But the reason why the inequality sign was flipped (from the original question) for the first part of the expression (1-2x)>0 and not for the second part of the expression (1 + x)<0 is not clear.

Can you please clarify?

thanks.
Senior Manager
Joined: 22 Dec 2011
Posts: 299
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Kudos [?]: 147 [0], given: 32

Re: M14 #13 [#permalink]  23 Sep 2012, 06:57
Bunuel wrote:
This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach": x2-4x-94661.html#p731476

If x is an integer, is |x| > 1?

First of all: is $$|x| > 1$$ means is $$x<-1$$ (-2, -3, -4, ...) or $$x>1$$ (2, 3, 4, ...), so for YES answer $$x$$ can be any integer but -1, 0, and 1.

(1) (1 - 2x)(1 + x) < 0 --> rewrite as $$(2x-1)(x+1)>0$$ (so that the coefficient of x^2 to be positive after expanding): roots are $$x=-1$$ and $$x=\frac{1}{2}$$ --> "$$>$$" sign means that the given inequality holds true for: $$x<-1$$ and $$x>\frac{1}{2}$$. $$x$$ could still equal to 1, so not sufficient.

(2) (1 - x)(1 + 2x) < 0 --> rewrite as $$(x-1)(2x+1)>0$$: roots are $$x=-\frac{1}{2}$$ and $$x=1$$ --> "$$>$$" sign means that the given inequality holds true for: $$x<-\frac{1}{2}$$ and $$x>1$$. $$x$$ could still equal to -1, so not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is $$x<-1$$ and $$x>1$$. Sufficient.

SIMPLY AWESOMEEEEEE!
Manager
Joined: 28 Apr 2013
Posts: 169
Location: India
GPA: 4
WE: Medicine and Health (Health Care)
Followers: 1

Kudos [?]: 43 [0], given: 84

Re: M14 #13 [#permalink]  20 Jan 2014, 05:20
Hi;

Neither condition 1 nor condition 2 suffice. But i guess drawing the possible outcomes of condition 1 and 2 on no. line will get the solution. Answer is C

_________________

Thanks for Posting

LEARN TO ANALYSE

+1 kudos if you like

Intern
Joined: 06 Feb 2014
Posts: 33
Location: United States
GRE 1: 1380 Q780 V600
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WE: Operations (Aerospace and Defense)
Followers: 0

Kudos [?]: 24 [0], given: 54

M14 #13 [#permalink]  17 Aug 2014, 10:41
Bunuel wrote:
This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach": x2-4x-94661.html#p731476

If x is an integer, is |x| > 1?

First of all: is $$|x| > 1$$ means is $$x<-1$$ (-2, -3, -4, ...) or $$x>1$$ (2, 3, 4, ...), so for YES answer $$x$$ can be any integer but -1, 0, and 1.

(1) (1 - 2x)(1 + x) < 0 --> rewrite as $$(2x-1)(x+1)>0$$ (so that the coefficient of x^2 to be positive after expanding): roots are $$x=-1$$ and $$x=\frac{1}{2}$$ --> "$$>$$" sign means that the given inequality holds true for: $$x<-1$$ and $$x>\frac{1}{2}$$. $$x$$ could still equal to 1, so not sufficient.

(2) (1 - x)(1 + 2x) < 0 --> rewrite as $$(x-1)(2x+1)>0$$: roots are $$x=-\frac{1}{2}$$ and $$x=1$$ --> "$$>$$" sign means that the given inequality holds true for: $$x<-\frac{1}{2}$$ and $$x>1$$.$$x$$ could still equal to -1, so not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is $$x<-1$$ and $$x>1$$. Sufficient.

Hi Bunuel,

I understood the question and got the answer right by finding the intersection for inequalities. The approach I normally use for such questions is as follows:

(1 - 2x)(1 + x) < 0

for above statement to be true the 2 expressions should have opposite signs, giving us (x > 1/2 and x > -1) OR (x < 1/2 and x < -1)
Now, finding intersection gives us x > 1/2 OR x < -1.
As a result, x < -1 is fine but x > 1/2 includes 1 as an option - making the statement insufficient.

Now my question is, how did you use the logic below to skip some of the steps above and directly go to the conclusion: x<-1 and x> 1/2. . What's the rule if instead of greater than (>) sign it was less than (<).

If I can understand the logic, I believe it will help save some valuable time.

">" sign means that the given inequality holds true for: x<-1 and x> 1/2.
M14 #13   [#permalink] 17 Aug 2014, 10:41

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# M14 #13

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