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# M14-13

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:53
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50% (01:57) correct 50% (01:30) wrong based on 111 sessions

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

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

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

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16 Sep 2014, 00:53
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Official Solution:

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

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

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

(1)+(2) Intersection of the ranges from (1) and (2) is $$x \lt -1$$ and $$x \gt 1$$. Sufficient.

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06 Dec 2014, 18:34
why do we disregard X< -1/2 for the first solution and X> 1/2 for the second solution? I chose E as although X>1 and X<-1 can be derived from both 1 and 2 respectively, the range of 1/2 < x< -1/2 from the information does not align to solution as x can =0
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07 Dec 2014, 05:51
rsamant wrote:
why do we disregard X< -1/2 for the first solution and X> 1/2 for the second solution? I chose E as although X>1 and X<-1 can be derived from both 1 and 2 respectively, the range of 1/2 < x< -1/2 from the information does not align to solution as x can =0

I don't understand what you mean at all...
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05 May 2015, 08:09
I did by wavy curve method.....Efficient
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08 Jun 2015, 06:32
Bunuel wrote:
rsamant wrote:
why do we disregard X< -1/2 for the first solution and X> 1/2 for the second solution? I chose E as although X>1 and X<-1 can be derived from both 1 and 2 respectively, the range of 1/2 < x< -1/2 from the information does not align to solution as x can =0

I don't understand what you mean at all...

He means why do we have to consider only extreme ranges in range intersection you explained...while considering (1) and (2), why do we eliminate x>1/2 and x<-1/2... because if we donot do so we get a range possible between x<-1 and x <-1/2 and also x>1/2 and x > 1 so these in bet ranges are confusing.
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19 Jun 2015, 11:08
Bunuel wrote:
Official Solution:

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

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

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

(1)+(2) Intersection of the ranges from (1) and (2) is $$x \lt -1$$ and $$x \gt 1$$. Sufficient.

Hello Bunuel,
Is there any other method to solve this quickly ?? In a test environment sometimes it's hard to think like this and get answer timely.

Thanks
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08 Jan 2016, 12:08
I have some confusion about this question. From statement 1 taken x<-1 but not X>1/2 & from statement 2 taken only X<-1/2 but not X>1. Why cannot be considered X<-1/2 & X>1/2? would anyone like to clarify my confusion. Thank you.
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21 Feb 2016, 02:49
Hi Bunuel,

I know this has to do something with inequalities changing signs when you go beyond extremes. But I needed a refresher on that. In particular I am stuck with how did you come up with this bit:

Quote:
">" sign means that the given inequality holds true for: x<−1x<−1and x>12x>12. xx could still equal 1, so not sufficient.

Is there any post which I can refer to solidify my concept of finding valid extreme values from Roots of the equation?

Vaibhav.

Ok after doing a bit of looking around, I found this:

http://gmatclub.com/forum/inequalities-trick-91482.html#p804990

This serves the purpose.

Bunuel, is there any other slick/intuitive way of doing this such that one saves time without drawing the line graph and plotting roots on it during the exam?
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21 Feb 2016, 09:31
vabhs192003 wrote:
Hi Bunuel,

I know this has to do something with inequalities changing signs when you go beyond extremes. But I needed a refresher on that. In particular I am stuck with how did you come up with this bit:

Quote:
">" sign means that the given inequality holds true for: x<−1x<−1and x>12x>12. xx could still equal 1, so not sufficient.

Is there any post which I can refer to solidify my concept of finding valid extreme values from Roots of the equation?

Vaibhav.

Ok after doing a bit of looking around, I found this:

http://gmatclub.com/forum/inequalities-trick-91482.html#p804990

This serves the purpose.

Bunuel, is there any other slick/intuitive way of doing this such that one saves time without drawing the line graph and plotting roots on it during the exam?

Solving Quadratic Inequalities - Graphic Approach
Inequality tips

DS Inequalities Problems
PS Inequalities Problems

700+ Inequalities problems
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05 Mar 2016, 15:35
1
Bunuel wrote:
Official Solution:

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

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

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

(1)+(2) Intersection of the ranges from (1) and (2) is $$x \lt -1$$ and $$x \gt 1$$. Sufficient.

Hi Experts! I'm confused:

In statement (1) when you get X<-1 and X>1/2, doesn't this show that |X| is not >1?

When you plot the quadratic equation (2X-1)(X+1)>0 ---> 2X^2+X-1 >0, you get the range X<-1 and X>1/2, thus X obviously has a value at 1 -- meaning that X is not greater than 1. Wouldn't this make (1) sufficient?

Similarly in statement (2), the range is X<-1/2 and X>1, thus again X obviously has a value at -1 -- meaning that X is not less than -1. Again wouldn't this make (2) sufficient?

Thank you!
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23 Nov 2016, 07:28
I found that the most efficient way, for me, was to plug in.

The question asks us to determine if $$x>1$$ or $$x<-1$$, so we just need to test 1,0, and -1 in both statements.

Statement 1 can be satisfied with 1 or any positive integer so it is not sufficient. It is worth noting that it can not be satisfied with -1 or 0.

Statement 2 can be satisfied with -1 or any negative integer, so it is not sufficient. It is with noting that it can not be satisfied with 0 nor with 1.

Combined, we see from stmt 1 that it can not be -1 or 0, and from stmt 2 that It can not be 1 or 0 so x must be either greater than 1 or less than -1.

I tried the curve approach but I did it in over 3 minutes so in this case it's not efficient for me.

Hope this helps.
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25 Dec 2016, 10:47
happyface101 wrote:
Bunuel wrote:
Official Solution:

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

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

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

(1)+(2) Intersection of the ranges from (1) and (2) is $$x \lt -1$$ and $$x \gt 1$$. Sufficient.

Hi Experts! I'm confused:

In statement (1) when you get X<-1 and X>1/2, doesn't this show that |X| is not >1?

When you plot the quadratic equation (2X-1)(X+1)>0 ---> 2X^2+X-1 >0, you get the range X<-1 and X>1/2, thus X obviously has a value at 1 -- meaning that X is not greater than 1. Wouldn't this make (1) sufficient?

Similarly in statement (2), the range is X<-1/2 and X>1, thus again X obviously has a value at -1 -- meaning that X is not less than -1. Again wouldn't this make (2) sufficient?

Thank you!

S1: 2x^2+x-1>0 =>2x^2+x>1=>x(2x+1)>1=>x>1 or x>0....how it could be -1>x>1/2???
thnx
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25 Dec 2016, 10:53
gupta87 wrote:
happyface101 wrote:
Bunuel wrote:
Official Solution:

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

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

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

(1)+(2) Intersection of the ranges from (1) and (2) is $$x \lt -1$$ and $$x \gt 1$$. Sufficient.

Hi Experts! I'm confused:

In statement (1) when you get X<-1 and X>1/2, doesn't this show that |X| is not >1?

When you plot the quadratic equation (2X-1)(X+1)>0 ---> 2X^2+X-1 >0, you get the range X<-1 and X>1/2, thus X obviously has a value at 1 -- meaning that X is not greater than 1. Wouldn't this make (1) sufficient?

Similarly in statement (2), the range is X<-1/2 and X>1, thus again X obviously has a value at -1 -- meaning that X is not less than -1. Again wouldn't this make (2) sufficient?

Thank you!

S1: 2x^2+x-1>0 =>2x^2+x>1=>x(2x+1)>1=>x>1 or x>0....how it could be -1>x>1/2???
thnx

This is totally wrong. You don't solve quadratic inequality this way. How/why did you conclude that x>1 or x>0 from x(2x+1)>1? By the way x>1 or x>0 does not make sense at all.

Solving Quadratic Inequalities - Graphic Approach
Inequality tips

DS Inequalities Problems
PS Inequalities Problems

700+ Inequalities problems

Hope it helps.
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25 Dec 2016, 11:54
understood. but from S1: 2x^2+x-1>0 => (x+1)(2x-1)>0 => x>-1 0r x>1/2...how do u arrive at x<-1 ??
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26 Dec 2016, 10:51
gupta87 wrote:
understood. but from S1: 2x^2+x-1>0 => (x+1)(2x-1)>0 => x>-1 0r x>1/2...how do u arrive at x<-1 ??

The red part does not makes sense. What does it means x>-1 or x>1/2? What can be x in this case? (x+1)(2x-1)>0 holds true for x<-1 and x>1/2. You should realy follow and study the links from my previous post.
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17 Dec 2017, 04:09
Bunuel wrote:
Official Solution:

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

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

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

(1)+(2) Intersection of the ranges from (1) and (2) is $$x \lt -1$$ and $$x \gt 1$$. Sufficient.

Hi Bunuel,

My approach is a bit different from yours but uses the same concept. But I'm getting an incorrect answer. Could you please advise which step I'm following wrong?

As per question Stem -

x >1
X< -1

Choice A ( I didn't multiply by -1 the way you did)

I got equation - -1< X <1/2 - So answer is No for the value of X

Choice B ( I didn't multiply by -1 the way you did)

Equation : -1/2 < X < 1 - So answer is No for the value of X

Hence I marked D in GMAT Club test which is incorrect. Could you please advise how we can solve without multiplying by - 1? And in which step I'm wrong?
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17 Dec 2017, 05:00
2
NeverGiveUp- Arpit wrote:

Hi Bunuel,

My approach is a bit different from yours but uses the same concept. But I'm getting an incorrect answer. Could you please advise which step I'm following wrong?

As per question Stem -

x >1
X< -1

Choice A ( I didn't multiply by -1 the way you did)

I got equation - -1< X <1/2 - So answer is No for the value of X

Choice B ( I didn't multiply by -1 the way you did)

Equation :-1/2 < X < 1 - So answer is No for the value of X

Hence I marked D in GMAT Club test which is incorrect. Could you please advise how we can solve without multiplying by - 1? And in which step I'm wrong?

Hi @NeverGiveUp- Arpit

The highlighted portion is incorrect. even if you are not multiplying with -1, then you need to realize that co-efficient of x will be negative, so on a number line range will start from negative values. Hence when you plot the roots of the quadratic equation on a number line then the range of x will be in the negative regions because the inequality is negative. Refer below image for clarity -

Attachment:
inequality.jpg

Similarly for statement B

basically as per your range 0 is a possible solution. so when you put x=0 in statement 1, you will get 1<0, which is not possible
>> !!!

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17 Dec 2017, 05:10
niks18 wrote:
NeverGiveUp- Arpit wrote:

Hi Bunuel,

My approach is a bit different from yours but uses the same concept. But I'm getting an incorrect answer. Could you please advise which step I'm following wrong?

As per question Stem -

x >1
X< -1

Choice A ( I didn't multiply by -1 the way you did)

I got equation - -1< X <1/2 - So answer is No for the value of X

Choice B ( I didn't multiply by -1 the way you did)

Equation :-1/2 < X < 1 - So answer is No for the value of X

Hence I marked D in GMAT Club test which is incorrect. Could you please advise how we can solve without multiplying by - 1? And in which step I'm wrong?

Hi @NeverGiveUp- Arpit

The highlighted portion is incorrect. even if you are not multiplying with -1, then you need to realize that co-efficient of x will be negative, so on a number line range will start from negative values. Hence when you plot the roots of the quadratic equation on a number line then the range of x will be in the negative regions because the inequality is negative. Refer below image for clarity -

Attachment:
inequality.jpg

Similarly for statement B

basically as per your range 0 is a possible solution. so when you put x=0 in statement 1, you will get 1<0, which is not possible

Makes sense. Cheers mate.
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08 Feb 2018, 02:09
Hello all,

how can we just rewrite (1-2x)(1+x) as (2x-1)(x+1) ? How did we turn around the signs for the first braket?

Best
Re: M14-13 &nbs [#permalink] 08 Feb 2018, 02:09

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