Last visit was: 17 Jul 2024, 19:48 It is currently 17 Jul 2024, 19:48
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0

SORT BY:
Tags:
Show Tags
Hide Tags
Intern
Joined: 04 Feb 2011
Posts: 36
Own Kudos [?]: 848 [174]
Given Kudos: 42
Location: US
Math Expert
Joined: 02 Sep 2009
Posts: 94384
Own Kudos [?]: 641789 [69]
Given Kudos: 85694
Manager
Joined: 10 Mar 2013
Posts: 136
Own Kudos [?]: 497 [30]
Given Kudos: 2412
GMAT 1: 620 Q44 V31
GMAT 2: 610 Q47 V28
GMAT 3: 700 Q49 V36
GMAT 4: 690 Q48 V35
GMAT 5: 750 Q49 V42
GMAT 6: 730 Q50 V39
GPA: 3
General Discussion
Director
Joined: 04 Jan 2011
Status:-=Given to Fly=-
Posts: 756
Own Kudos [?]: 433 [7]
Given Kudos: 78
Location: India
GMAT 1: 650 Q44 V37
GMAT 2: 710 Q48 V40
GMAT 3: 750 Q51 V40
GPA: 3.5
WE:Education (Education)
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
4
Kudos
2
Bookmarks
Let p=(x+1)/(x-3)<0

Statement 1:
-1<x<1
So Range of p is between 0 and -1
Therefore, p is less than 0.
Sufficient.

Statement 2:
x^2-4 < 0
x^2 < 4
Take sq. rt. on both sides:
|x| < 2

=> -2<x<2
1/5<p<-3 [Range]
So we cannot say if p is less than 0
In Sufficient!

Ans A
Manager
Joined: 29 Jul 2011
Posts: 52
Own Kudos [?]: 167 [2]
Given Kudos: 6
Location: United States
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
2
Kudos
This one's a thing of beauty.

NOTE: No property mentioned about x, so can't cross-multiply the inequality.
RULE1: If positive, cross-mutiply sign does not change
RULE2: If negative, cross-mutiply sign changes

1. This means that x is + or - fraction. Substituting gives us that the inequality will hold true for all fractions of x, both + or -. Suff.
2. x^2 - 4 < 0 => x^2 < 4 => x < 2 or x > -2. If x = 1.5, inequality holds true. If x = -1.5, it does not. Insuff.

A
Intern
Joined: 09 Apr 2013
Status:Onward and upward!
Posts: 10
Own Kudos [?]: 25 [0]
Given Kudos: 72
Location: United States
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
This may be a dumb question- but I thought on the GMAT that all square roots of positive integers are considered positive. Hence, B would be sufficient because it is assumed that x could only be positive 2.?? Help. When should I use that rule?
Math Expert
Joined: 02 Sep 2009
Posts: 94384
Own Kudos [?]: 641789 [10]
Given Kudos: 85694
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
3
Kudos
7
Bookmarks
TAL010 wrote:
This may be a dumb question- but I thought on the GMAT that all square roots of positive integers are considered positive. Hence, B would be sufficient because it is assumed that x could only be positive 2.?? Help. When should I use that rule?

First of all notice that we have x^2-4<0 (-2<x<2) not x^2-4=0 (x=-2 or x=2).

Next, when the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root. That is, $$\sqrt{25}=5$$, NOT +5 or -5.

In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only non-negative value on the GMAT.

Hope it helps.
Intern
Joined: 24 Aug 2012
Status:MBA Aspirant
Posts: 30
Own Kudos [?]: 121 [1]
Given Kudos: 20
Location: India
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
1
Kudos
Bunuel wrote:

Thus the question basically becomes: is $$-1<x<3$$?

(1) -1 < x < 1. Sufficient.
(2) x^2-4 < 0 --> $$-2<x<2$$. Not sufficient.

Hi Bunuel thanks for the above explaination and answer.

I have a doubt though, for 2nd statement possible regions are x<-2,-2<x<2 and x>2. and the signs come to +,-,+ so we selected middle one for the answer here. But will it be possible that more than one regions satisy the inequality? If yes then what should we do?
Director
Joined: 25 Apr 2012
Posts: 529
Own Kudos [?]: 2319 [1]
Given Kudos: 740
Location: India
GPA: 3.21
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
1
Kudos
aniketb wrote:
Bunuel wrote:

Thus the question basically becomes: is $$-1<x<3$$?

(1) -1 < x < 1. Sufficient.
(2) x^2-4 < 0 --> $$-2<x<2$$. Not sufficient.

Hi Bunuel thanks for the above explaination and answer.

I have a doubt though, for 2nd statement possible regions are x<-2,-2<x<2 and x>2. and the signs come to +,-,+ so we selected middle one for the answer here. But will it be possible that more than one regions satisy the inequality? If yes then what should we do?

No,There will be no other region satisfying the inequality. You can try and putting in values of the x<-2 and x>2 and see it for yourself.
Consider x=-50 or x=50...There is only 1 range where the inequality holds true.

Check out the solution for a similar question and look at graphical representation to understand how to handle such questions

if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html
Intern
Joined: 16 Jul 2011
Posts: 35
Own Kudos [?]: 9 [1]
Given Kudos: 166
Concentration: Marketing, Real Estate
GMAT 1: 550 Q37 V28
GMAT 2: 610 Q43 V31
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
1
Kudos
Bunuel wrote:

Is (x+1)/(x-3)<0 ?

Is $$\frac{x+1}{x-3}<0$$? --> roots are -1 and 3, so 3 ranges: $$x<-1$$, $$-1<x<3$$ and $$x>3$$ --> check extreme value: if $$x$$ some very large number then $$\frac{x+1}{x-3}=\frac{positive}{positive}>0$$ --> in the 3rd range expression is positive, then in 2nd it'll be negative and in 1st it'll be positive again: + - +. So, the range when the expression is negative is: $$-1<x<3$$.

Thus the question basically becomes: is $$-1<x<3$$?

Bunuel: I did not understand this part. Could you please throw some more light on this? Where did 'roots' come into the picture? And then how did you derive the subsequnt range? Confused
SVP
Joined: 20 Mar 2014
Posts: 2359
Own Kudos [?]: 3650 [7]
Given Kudos: 816
Concentration: Finance, Strategy
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
5
Kudos
2
Bookmarks
samdighe wrote:
Bunuel wrote:

Is (x+1)/(x-3)<0 ?

Is $$\frac{x+1}{x-3}<0$$? --> roots are -1 and 3, so 3 ranges: $$x<-1$$, $$-1<x<3$$ and $$x>3$$ --> check extreme value: if $$x$$ some very large number then $$\frac{x+1}{x-3}=\frac{positive}{positive}>0$$ --> in the 3rd range expression is positive, then in 2nd it'll be negative and in 1st it'll be positive again: + - +. So, the range when the expression is negative is: $$-1<x<3$$.

Thus the question basically becomes: is $$-1<x<3$$?

Bunuel: I did not understand this part. Could you please throw some more light on this? Where did 'roots' come into the picture? And then how did you derive the subsequnt range? Confused

Look at it this way:

You have been given an inequality of the form:

a/b <0

Now this is true ONLY for 2 cases
1.) a<0 and b >0 ---> x+1 < 0 and x-3 > 0 ---> x<-1 and x>3 Not possible as no intersection of the ranges.
2.) a>0 and b <0---> x+1 > 0 and x-3 < 0 ---> x<-1 and x>3 ---> -1<x<3

What it means is that all values of x lying in the range (-1,3) ONLY will satisfy the given inequality or will give "Yes" to the question asked. All other values may or may not give you a "yes".

Per statement 1, -1<x<1 and this lies in the range mentioned above and is thus sufficient with a definite "yes".

Per statement 2, $$x^2-4<0$$ ---> -2<x<2 . Thus this range will give you a "yes" when x = 0.5 but will give you a "no" when x = -1.5. Thus this statement is not sufficient.

A is thus the correct answer.

A few good posts on inequalities are: inequalities-trick-91482.html
inequality-and-absolute-value-questions-from-my-collection-86939.html
Manager
Joined: 24 May 2013
Posts: 56
Own Kudos [?]: 147 [3]
Given Kudos: 99
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
2
Kudos
1
Bookmarks
Is (x + 1)/(x - 3) < 0 ?

plotting the inequality we get the zone -1<x<3 satisfying the above inequality.

(1) -1 < x < 1
as this range is subset of -1<x<3 so the inequality will always be -ve. Sufficient.

(2) x^2 - 4 < 0

-2<x<2
For this range, we can have both +ve and -ve values for the given inequality. Insufficient.

Hence A.
Attachments

Inequality.png [ 3.83 KiB | Viewed 57716 times ]

Director
Joined: 26 Oct 2016
Posts: 506
Own Kudos [?]: 3409 [3]
Given Kudos: 877
Location: United States
Schools: HBS '19
GMAT 1: 770 Q51 V44
GPA: 4
WE:Education (Education)
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
3
Kudos
is (x+1)/(x-3) < 0?

1) -1 < x < 1
2) x^2 - 4 < 0

So, let's see what the question is asking. Is the ratio of 2 quantities negative=> either the numerator OR the denominator must be negative for this to be true but not BOTH.

1) The numerator will be positive because x is greater than -1. The denominator will always be negative since the maximum x can get is very close to 1, but subtracting 3 from it, will result in a negative number. So, we have a negative result. SUFFICIENT.

2) X^2 <4 => -2<X<2. Now, the denominator, again, will be always negative since the MAX x can get is very close to 2. BUT, now the numerator can be positive [for x>-1] or negative [for -2< x<-1]. Hence, we can get a positive or a negative end result. INSUFFICIENT.

Hope this helps.

Hence A.
Director
Joined: 06 Jan 2015
Posts: 732
Own Kudos [?]: 1600 [0]
Given Kudos: 579
Location: India
Concentration: Operations, Finance
GPA: 3.35
WE:Information Technology (Computer Software)
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
Quote:
Is (x + 1)/(x - 3) < 0 ?

(1) -1 < x < 1
(2) x^2 - 4 < 0

HI GMATGuruNY, MentorTutoring ,GMATBusters,

$$(x + 1)/(x - 3) < 0$$ ? ==> Multiple by $$(x-3)$$ on both side ==> $$(x + 1)(x - 3) < 0$$

Does it not mean Either $$x<-1$$ or $$x<3$$?
GMAT Tutor
Joined: 27 Oct 2017
Posts: 1895
Own Kudos [?]: 5788 [3]
Given Kudos: 238
WE:General Management (Education)
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
3
Kudos
Hi

Since, we dont know whether (x-3) is positive or negative , we cannot multiply (x-3) both sides.
it can be done as below.

NandishSS wrote:
Quote:
Is (x + 1)/(x - 3) < 0 ?

(1) -1 < x < 1
(2) x^2 - 4 < 0

HI GMATGuruNY, MentorTutoring ,GMATBusters,

$$(x + 1)/(x - 3) < 0$$ ? ==> Multiple by $$(x-3)$$ on both side ==> $$(x + 1)(x - 3) < 0$$

Does it not mean Either $$x<-1$$ or $$x<3$$?

Attachment:

WhatsApp Image 2020-02-29 at 8.17.44 PM.jpeg [ 99.23 KiB | Viewed 28990 times ]
Volunteer Expert
Joined: 16 May 2019
Posts: 3507
Own Kudos [?]: 6979 [3]
Given Kudos: 500
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
3
Kudos
NandishSS wrote:
Quote:
Is (x + 1)/(x - 3) < 0 ?

(1) -1 < x < 1
(2) x^2 - 4 < 0

HI GMATGuruNY, MentorTutoring ,GMATBusters,

$$(x + 1)/(x - 3) < 0$$ ? ==> Multiple by $$(x-3)$$ on both side ==> $$(x + 1)(x - 3) < 0$$

Does it not mean Either $$x<-1$$ or $$x<3$$?

Hello, NandishSS. Pardon the delay in my response, but today was my busiest day, and this is the first time I have found in which I can adequately reply. I approached the problem a little differently, namely by reinterpreting the question. Rather than manipulate anything in the original inequality, which would lead to problems, I questioned what it would take for the expression to be negative. There were only two ways:

1) negative/positive
2) positive/negative

With this in mind, I jumped straight into the statements, starting with (1). Looking at the lower extreme, if "x" were -0.999 or some similar value, then we would meet condition 2) above:

(-0.999 + 1)/(-0.999 - 3) = positive/negative

What about the other extreme, though? If "x" were, say, 0.999, then we would meet condition 2) again:

(0.999 + 1)/(0.999 - 3) = positive/negative

Consistency leads to a SUFFICIENT answer. Thus, the answer was down to (A) or (D). Looking at statement (2), I considered that whether I moved 4 to the other side of the inequality or left it where it was, "x" had to be between -2 and 2 for the inequality to hold. I repeated the same process as before, ignoring the exact values of the expressions but just paying attention to the general tendency of positive/negative:

(-1.999 + 1)/(-1.999 - 3) = negative/negative

(1.999 + 1)/(1.999 - 3) = positive/negative

Conflicting information is the opposite of what we want, so I wrote (2) off as NOT SUFFICIENT, leaving only (A) as the answer. I hope that helps. If you have further questions, please ask.

- Andrew
GMAT Club Legend
Joined: 03 Oct 2013
Affiliations: CrackVerbal
Posts: 4918
Own Kudos [?]: 7803 [2]
Given Kudos: 220
Location: India
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
2
Kudos
Top Contributor
Is $$\frac{(x+1) }{ (x-3)}$$ < 0?

Multiplying and dividing the LHS by (x-3), we have,

Is $$\frac{(x+1) (x-3) }{ (x-3)^2}$$ < 0?

Since $$(x-3)^2$$ is always non-negative, the question can be rephrased as,
Is (x+1) (x-3) < 0?

The solution for this inequality is -1 < x < 3. Therefore, the question can be rephrased as,
Is -1<x<3??

This is the question we now try to answer using the statements.

From statement I alone, -1<x<1.
Is -1<x<3? A definite yes.

Statement I alone is sufficient to answer the question. Answer options B, C and E can be eliminated. Possible answer options are A or D.

From statement II alone, $$x^2$$ – 4 < 0.
Therefore, (x-2) (x+2) < 0 or -2<x<2.

Is -1<x<3?? No definite answer since x may be in this range or may not be.
Statement II alone is insufficient. Answer option D can be eliminated.

The correct answer option is A.

Hope that helps!
Aravind B T
Manager
Joined: 19 Nov 2020
Posts: 86
Own Kudos [?]: 41 [0]
Given Kudos: 121
Location: India
GMAT 1: 700 Q48 V38
GPA: 3.6
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
Bunuel wrote:
Moved to DS subforum.

Lolaergasheva wrote:
Is x+1/x-3<0 ?

1) -1 < x < 1
2) x^2-4 < 0

Lolaergasheva, please format the questions properly:

Is (x+1)/(x-3)<0 ?

Is $$\frac{x+1}{x-3}<0$$? --> roots are -1 and 3, so 3 ranges: $$x<-1$$, $$-1<x<3$$ and $$x>3$$ --> check extreme value: if $$x$$ some very large number then $$\frac{x+1}{x-3}=\frac{positive}{positive}>0$$ --> in the 3rd range expression is positive, then in 2nd it'll be negative and in 1st it'll be positive again: + - +. So, the range when the expression is negative is: $$-1<x<3$$.

Thus the question basically becomes: is $$-1<x<3$$?

(1) -1 < x < 1. Sufficient.
(2) x^2-4 < 0 --> $$-2<x<2$$. Not sufficient.

Hi Bunuel,

I went through some links for inequalities.. can you please clarify some questions regarding inequalities..
1. When should we add 0 as a range in the graphic method.. because in some post I saw that 0 must always be added to mark ranges in the graph and I was wondering why you have not separated here..
2. When we have something on the numerator and denominator.. do we just directly equate/</> the denominator to zero just as it were the numerator?

And thank you, I now have it in my mind that I should never multiply/ divide an inequality equation by a variable if we do not know its value.
Non-Human User
Joined: 09 Sep 2013
Posts: 34005
Own Kudos [?]: 852 [0]
Given Kudos: 0
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
Moderator:
Math Expert
94384 posts