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Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
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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
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Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
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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
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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?
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Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
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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.
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Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
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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.

Answer: A.

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?
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Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
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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.

Answer: A.

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
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Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
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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 :roll:
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Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
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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 :roll:


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
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Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
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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.
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Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
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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.
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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,

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 \( x<3 \)?
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Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
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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,

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 \( x<3 \)?


Attachment:
WhatsApp Image 2020-02-29 at 8.17.44 PM.jpeg
WhatsApp Image 2020-02-29 at 8.17.44 PM.jpeg [ 99.23 KiB | Viewed 28990 times ]
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Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
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NandishSS wrote:
Quote:
Is (x + 1)/(x - 3) < 0 ?

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


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 \( x<3 \)?

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
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Re: Is (x + 1)/(x - 3) < 0 ? (1) -1 < x < 1 (2) x^2 - 4 < 0 [#permalink]
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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!
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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:

Question should read:

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

Is \(\frac{x+1}{x-3}<0\)? --> roots are -1 and 3, so 3 ranges: \(x<-1\), \(-1<x<3\) 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.

Answer: A.

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.
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