Is (x + 1)/(x - 3)

Question

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

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

Bunuel · Accepted Answer

Moved to DS subforum. Lolaergasheva, please format the questions properly: Question should read: Is (x+1)/(x-3)<0 ? Is \frac{x+1}{x-3}<0 ? --> roots are -1 and 3, so 3 ranges: x<-1 , -13 --> 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 -2everything-is-less-than-zero-108884.html?hilit=extreme#p868863, here: inequalities-trick-91482.html and here: xy-plane-71492.html?hilit=solving%20quadratic#p841486 And again: Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/ Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/ No posting of PS/DS questions is allowed in the main Math forum.

TooLong150 · Answer

Interesting question. Below is my solution.
(x+1)/(x-3) < 0
Y if x+1 > 0 and (x-3) < 0

A

(1) 
-1 < x < 1
0 < x+1 < 2
-4 < x-3 < -2
Signs of both x+1 and x-3 will be opposite, so sufficient.

(2)
x^2 < 4
|x| < 2
-2 < x < 2
-1 < x+1 < 3
-5 < x-3 < -1
(x+1)/(x-3) does not have a consistent sign, so insufficient.

AmrithS · Answer

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  
So we cannot say if p is less than 0
In Sufficient!

Ans A

shinbhu · Answer

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

TAL010 · Answer

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?

Bunuel · Answer

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 {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.

aniketb · Answer

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?

WoundedTiger · Answer

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

sameersanjeev · Answer

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 :roll:

ENGRTOMBA2018 · Answer

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 -2inequalities-trick-91482.html inequality-and-absolute-value-questions-from-my-collection-86939.html

Vinayprajapati · Answer

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.

anairamitch1804 · Answer

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.

NandishSS · Answer

HI GMATGuruNY , MentorTutoring , GMATBusters , Can you please help me with my understanding? (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 ?

GMATBusters · Answer

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. https://gmatclub.com/forum/download/file.php?id=53133 WhatsApp Image 2020-02-29 at 8.17.44 PM.jpeg

AndrewN · Answer

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

CrackverbalGMAT · Answer

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

Bhu750 · Answer

Lolaergasheva, please format the questions properly: Question should read: Is (x+1)/(x-3)<0 ? Is \frac{x+1}{x-3}<0 ? --> roots are -1 and 3, so 3 ranges: x<-1 , -13 --> 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 -2 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.

bumpbot · Answer

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

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