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Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0
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Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0
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Originally posted by Lolaergasheva on 11 Feb 2011, 02:57.
Last edited by Bunuel on 11 Jun 2019, 04:52, edited 2 times in total.
Edited the question.




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Re: Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0
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11 Feb 2011, 07:44
Moved to DS subforum. Lolaergasheva wrote: Is x+1/x3<0 ?
1) 1 < x < 1 2) x^24 < 0 Lolaergasheva, please format the questions properly:Question should read: Is (x+1)/(x3)<0 ?Is \(\frac{x+1}{x3}<0\)? > 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 \(\frac{x+1}{x3}=\frac{positive}{positive}>0\) > 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^24 < 0 > \(2<x<2\). Not sufficient. Answer: A. Check for more about the approach used here: everythingislessthanzero108884.html?hilit=extreme#p868863, here: inequalitiestrick91482.html and here: xyplane71492.html?hilit=solving%20quadratic#p841486And again:
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Re: Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0
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11 Feb 2011, 05:30
Let p=(x+1)/(x3)<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^24 < 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
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10 Jan 2012, 14:50
This one's a thing of beauty. NOTE: No property mentioned about x, so can't crossmultiply the inequality. RULE1: If positive, crossmutiply sign does not change RULE2: If negative, crossmutiply 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
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26 Sep 2013, 18:42
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
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27 Sep 2013, 01:34
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^24<0 (2<x<2) not x^24=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 nonnegative 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
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08 May 2014, 00:15
Bunuel wrote:
Thus the question basically becomes: is \(1<x<3\)?
(1) 1 < x < 1. Sufficient. (2) x^24 < 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
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08 May 2014, 00:40
aniketb wrote: Bunuel wrote:
Thus the question basically becomes: is \(1<x<3\)?
(1) 1 < x < 1. Sufficient. (2) x^24 < 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 ifxisanintegerwhatisthevalueofx1x24x94661.html
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Re: Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0
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16 Jul 2014, 23:02
Hello Bunuel,
if in stmt (1) 1<x1 couldn't x be = 0 and hence answer E?



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Re: Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0
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17 Jul 2014, 08:18
unceldolan wrote: Hello Bunuel,
if in stmt (1) 1<x1 couldn't x be = 0 and hence answer E? Why? If x = 0, then (x + 1)/(x  3) = 1/3 < 0.
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Re: Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0
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04 Oct 2014, 20:09
Interesting question. Below is my solution. (x+1)/(x3) < 0 Y if x+1 > 0 and (x3) < 0
A
(1) 1 < x < 1 0 < x+1 < 2 4 < x3 < 2 Signs of both x+1 and x3 will be opposite, so sufficient.
(2) x^2 < 4 x < 2 2 < x < 2 1 < x+1 < 3 5 < x3 < 1 (x+1)/(x3) does not have a consistent sign, so insufficient.



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Re: Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0
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04 Aug 2015, 06:47
Bunuel wrote: Is (x+1)/(x3)<0 ?
Is \(\frac{x+1}{x3}<0\)? > 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 \(\frac{x+1}{x3}=\frac{positive}{positive}>0\) > 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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Re: Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0
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04 Aug 2015, 07:40
samdighe wrote: Bunuel wrote: Is (x+1)/(x3)<0 ?
Is \(\frac{x+1}{x3}<0\)? > 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 \(\frac{x+1}{x3}=\frac{positive}{positive}>0\) > 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 x3 > 0 > x<1 and x>3 Not possible as no intersection of the ranges. 2.) a>0 and b <0> x+1 > 0 and x3 < 0 > x<1 and x>3 > 1<x<3What 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, \(x^24<0\) > 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: inequalitiestrick91482.htmlinequalityandabsolutevaluequestionsfrommycollection86939.html



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Re: Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0
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05 Aug 2015, 02:02
Is (x + 1)/(x  3) < 0 ? *do (x+1) and (x3) have opposite signs? (1) 1 < x < 1 *When x is within this range (x+1) and (x3) will have opposite signs. Sufficient. (2) x^2  4 < 0 *(x+2)(x2) <0 These two factors should also have opposite signs >(x+2) > 0 and (x2)<0 > x>2 and x<2, or 2 < x < 2. When x is 1, for example, (x+1) and (x3) have opposite signs. When x =  1.5, for example, (x+1) and (x3) are both negative. Not sufficient. *A



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Re: Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0
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26 Mar 2016, 21:05
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
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03 Jul 2016, 00:02
Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0 Can i use componendo dividendo here ?? (just checking as Im not sure) (x + 1 + x 3 )/ (x + 1  x + 3) < 0 (using a/b ===> (a + b)/(a  b) (2x  2)/(4) < 0 x/2  1/2 < 0 x/2 < 1/2 (adding 1/2 on both sides) x < 1 (multiplying by 2 on both sides) now take a look at the answer choices 1) 1 < x < 1 =====> clearly within our given value of x < 1 > hence ok 2) x^2  4 < 0 ===> x^2 < 4 x < 2 or x > 2 so 2 > x > 2, if x is less than 1 than ok but if 1 < x < 2 then not ok Hence ans is A) Please help me out here? Need to verify this method before I applying it elsewhere If its wrong, why is it wrong? Is it because we cannot apply componendo  dividendo to inequalities ?
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Re: Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0
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13 Aug 2016, 05:53
Bunuel wrote: Moved to DS subforum. Lolaergasheva wrote: Is x+1/x3<0 ?
1) 1 < x < 1 2) x^24 < 0 Lolaergasheva, please format the questions properly:Question should read: Is (x+1)/(x3)<0 ?Is \(\frac{x+1}{x3}<0\)? > 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 \(\frac{x+1}{x3}=\frac{positive}{positive}>0\) > 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^24 < 0 > \(2<x<2\). Not sufficient. Answer: A. Check for more about the approach used here: everythingislessthanzero108884.html?hilit=extreme#p868863, here: inequalitiestrick91482.html and here: xyplane71492.html?hilit=solving%20quadratic#p841486And again: Hello Bunuel, Just confirming something  1. When a negative number is moved and multiplied or divided on the other side of an inequality equation, only then will the inequality sign reverse. 2. The inequality sign does not reverse when a number of any sign is moved and multiplied or divided on the other side of the inequality TO A NEGATIVE number. Please confirm if I am right in my understanding. Thanks.



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Re: Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0
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14 Aug 2016, 00:06
TheLordCommander wrote: Bunuel wrote: Moved to DS subforum. Lolaergasheva wrote: Is x+1/x3<0 ?
1) 1 < x < 1 2) x^24 < 0 Lolaergasheva, please format the questions properly:Question should read: Is (x+1)/(x3)<0 ?Is \(\frac{x+1}{x3}<0\)? > 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 \(\frac{x+1}{x3}=\frac{positive}{positive}>0\) > 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^24 < 0 > \(2<x<2\). Not sufficient. Answer: A. Check for more about the approach used here: everythingislessthanzero108884.html?hilit=extreme#p868863, here: inequalitiestrick91482.html and here: xyplane71492.html?hilit=solving%20quadratic#p841486And again: Hello Bunuel, Just confirming something  1. When a negative number is moved and multiplied or divided on the other side of an inequality equation, only then will the inequality sign reverse. 2. The inequality sign does not reverse when a number of any sign is moved and multiplied or divided on the other side of the inequality TO A NEGATIVE number. Please confirm if I am right in my understanding. Thanks. Adding or subtracting the same number on both sides of the inequality doesn't change the sign of the inequality. Multiplying or dividing a ve number reverses the sign of the inequality. By these statements I also mean moving the number(from LHS to RHS). Let me show you how : x 2 > 0 We can say adding 2 on both sides is equivalent to moving  2 to RHS. or x2+2 > +2 or x>2
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Re: Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0
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29 Jan 2017, 05:31
is (x+1)/(x3) < 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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