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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.

Another easy way to solve this problem... 1) -1<x<1 Plug in 1/2 and -1/2 in original statement, you will see that it is always negative. 2)-2<x<2 -3/2 gives you a positive value...hence not sufficient.

S2-start with 1.5 (the answer is yes) and -1.5 (the answer is no). SO NS S1- 0.5 AND -0.5 the answer is yes in both.though, picking numbers in is risky.

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

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10 Jan 2012, 14:50

1

This post received 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
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I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

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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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.

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?

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

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

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.

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
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"The fool didn't know it was impossible, so he did it."

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.

Is (x + 1)/(x - 3) < 0 ? *do (x+1) and (x-3) have opposite signs? (1) -1 < x < 1 *When x is within this range (x+1) and (x-3) will have opposite signs. Sufficient. (2) x^2 - 4 < 0 *(x+2)(x-2) <0 These two factors should also have opposite signs ->(x+2) > 0 and (x-2)<0 -> x>-2 and x<2, or -2 < x < 2. When x is 1, for example, (x+1) and (x-3) have opposite signs. When x = - 1.5, for example, (x+1) and (x-3) are both negative. Not sufficient. *A
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