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Is (x + 1)/(x  3) < 0 ? [#permalink]
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11 Feb 2011, 02:57
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Is (x + 1)/(x  3) < 0 ? (1) 1 < x < 1 (2) x^2  4 < 0
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Last edited by Bunuel on 10 May 2013, 01:23, edited 1 time in total.
Edited the question.



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Re: data suff inequalities [#permalink]
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11 Feb 2011, 05:30
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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: data suff inequalities [#permalink]
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11 Feb 2011, 07:44
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Re: data suff inequalities [#permalink]
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17 Mar 2011, 22:19
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.
A



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Re: data suff inequalities [#permalink]
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04 Aug 2011, 00:24
picking numbers here was short and precise:
S2start 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.



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Re: data suff inequalities [#permalink]
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07 Aug 2011, 03:59
Bunuel, thanks for the explanation. +1+1
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Re: Is x+1/x3<0 ? 1) 1 < x < 1 2) x^24 < 0 [#permalink]
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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 ? [#permalink]
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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 ? [#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^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: data suff inequalities [#permalink]
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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: data suff inequalities [#permalink]
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08 May 2014, 00:40
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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: data suff inequalities [#permalink]
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08 May 2014, 05:09
WoundedTiger wrote: 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.htmlThanks woundedtiger, Its quite helpful. Its only one region then,until some exception comes..



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Re: Is (x + 1)/(x  3) < 0 ? [#permalink]
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10 Jul 2014, 01:21
Hi Bunuel,
i have a doubt . In your first explanation above , you have written  frac{x+1}{x3}<0? > roots are 1 and 3.
How are roots 1 & 3 for this inequality .
Thanks in advance .
Kunal .



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Re: Is (x + 1)/(x  3) < 0 ? [#permalink]
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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 ? [#permalink]
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17 Jul 2014, 08:18



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Is (x + 1)/(x  3) < 0 ? [#permalink]
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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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Is (x + 1)/(x  3) < 0 ? [#permalink]
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03 Feb 2015, 18:13
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: Dear Bunuel, in the above question you say that and I know that you got this result from x+1<0 so X<1. However , in X>3 does you got it from that X3>0 as aresult X > 3. My question is, does always the numerator for take the same sign that in the question for example as above questin ( < ) and the denominator take the opposite sign ( > )that in the question . If the answer is not so how you got X>3?
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Re: Is (x + 1)/(x  3) < 0 ? [#permalink]
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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 ? [#permalink]
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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 ? [#permalink]
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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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