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Re: D0110
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15 Feb 2017, 06:51
Bunuel wrote: Official Solution:
(1) \(\frac{1}{x} \gt  1\). Rewrite as \(\frac{1+x}{x} \gt 0\), two cases: A. \(x \gt 0\) and \(1+x \gt 0\), (\(x \gt 1\)). So, \(x \gt 0\); B. \(x \lt 0\) and \(1+x \lt 0\), (\(x \lt 1\)). So, \(x \lt 1\). So, the given inequality holds true in two ranges: \(x \gt 0\) and \(x \lt 1\), thus \(x\) may or may not be greater than 1. Not sufficient. (2) \(\frac{1}{x^5} \gt \frac{1}{x^3}\). Rewrite as \(\frac{1x^2}{x^5} \gt 0\), two cases: A. \(x \gt 0\) (it's the same as \(x^5 \gt 0\)) and \(1x^2 \gt 0\), (\(1 \lt x \lt 1\)). So, \(0 \lt x \lt 1\); B. \(x \lt 0\) and \(1x^2 \lt 0\), (\(x \lt 1\) or \(x \gt 1\)). So, \(x \lt 1\); We got that given inequality holds true in two ranges: \(0 \lt x \lt 1\) and \(x \lt 1\), ANY \(x\) from this ranges will be less than 1. So the answer to our original question is NO. Sufficient.
Answer: B can we use pluggin method to make it easy ?? where can we use and where we cannot depend on pluggin method ?
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Re: D0110
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06 May 2017, 04:28
(1) 1x>−11x>−1. Rewrite as 1+xx>01+xx>0, two cases:
A. x>0x>0 and 1+x>01+x>0, (x>−1x>−1). So, x>0x>0;
B. x<0x<0 and 1+x<01+x<0, (x<−1x<−1). So, x<−1x<−1.
How can we distribute the numerator and denominator.Is it okay to do it?



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Re: D0110
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06 May 2017, 20:17
Bunuel wrote: Official Solution:
(1) \(\frac{1}{x} \gt  1\). Rewrite as \(\frac{1+x}{x} \gt 0\), two cases: A. \(x \gt 0\) and \(1+x \gt 0\), (\(x \gt 1\)). So, \(x \gt 0\);
I don't understand this. Is there a rule or something that says when a/b > 0 then a>0 and b>0?



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Re: D0110
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07 May 2017, 02:06



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Re: D0110
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08 May 2017, 07:09
Bunuel wrote: Oh ok. I actually already knew that, I just have never seen it written like that in inequality form and could not make the connect. I read all the quant manhattan books already



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Re: D0110
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03 Jun 2017, 22:45
Hi Bunuel, How did you write 1/x1>0 as x+1/x without knowing the sign of x?



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Re: D0110
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04 Jun 2017, 06:31



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Re: D0110
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04 Jun 2017, 11:31
pratyushk1 wrote: I think this is a highquality question and I don't agree with the explanation. Can you please explain S1 (case 1). 1+x>0 and x>0. So it should be x>1 and x>0. How have you written x>0 after that. Hi Let me try. You have written 1+x>0 and x>0 from Bunuel's first explanation to this question. you are correct that 1+x > 0 translates to x > 1. And we already have x > 0. So when we are given that x>1 and x>0 (these are Two different ranges of x given) we have to figure out the common range  that will be our final range for x. You see x>1 includes all numbers greater than 1 till infinity. So this means 1<x<0 and 0<x<infinity But when we say x>0 it includes all numbers from o to infinity. So this means 0<x<infinity So if you look at both the ranges above.. which range is common between both? Its 0<x<infinity.. So when we say that x>1 and x>0, then our conclusion is that x>0 (because that is the only range which is common to both the above ranges).. I think thats why Bunuel wrote (in his very first explanation to this question) that x>0 (You can also think of it this way: Say your boss tells you that you will receive your salary after 10th June and your Super Boss tells you that you will receive your salary after 12th June.. and if you know that both these statements are true  then the only conclusion which can be drawn from these is that your salary will come only after 12th June..from 10th to 12th June is NOT possible.)



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Re: D0110
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02 Aug 2017, 18:51
Iit possible to do this?
\(\frac{1}{x^5} > \frac{1}{x^3}\)become
\(x^3\)>\(x^5\) then \(x^3\)\(x^5\)>0 then \(x^3\)(1\(x^2\))>0



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Re: D0110
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06 Sep 2017, 19:58
Hey, is this approach right:
Just for a:
1/x > 1 ,
if x > 0 ,
a) multiply both sides by x
x(1/x) > x(1)
1 > x
b) Divide by 1 and flip signs,
1 < x
if x < 0 a) multiply both sides by x and flip signs cos x is negative
1 < x
b) divide by 1 and flip signs
1 > x
Same answer as with your process. Not sure if it is right though



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06 Sep 2017, 21:17
jjacosta37 wrote: Hey, is this approach right:
Just for a:
1/x > 1 ,
if x > 0 ,
a) multiply both sides by x
x(1/x) > x(1)
1 > x
b) Divide by 1 and flip signs,
1 < x
if x < 0 a) multiply both sides by x and flip signs cos x is negative
1 < x
b) divide by 1 and flip signs
1 > x
Same answer as with your process. Not sure if it is right though 1/x > 1: 1. If x > 0, then x > 1 but since we are considering the range when x > 0, then x > 0. 2. If x < 0, then x < 1.
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Re: D0110
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24 Sep 2017, 08:21
Hey Bunuel, Nice question, appreciate all your help. Why exactly do we rewrite statement 1 as Rewrite as [1+x][/x]=0? Thanks.
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30 Nov 2017, 04:57
I think this is a poorquality question and I don't agree with the explanation. In the case of option one. whithout solving it, by just considering the inequality, by taking inverse cant we write that x<1?



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30 Nov 2017, 05:40



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17 Dec 2017, 00:04
I think when attempting these kind of inequality questions in timed condition in tests .
We need to figure out an easier approach and the one I could think of apart from Bunnels excellent solution was:
Imagine a number line and divide it into 1,0 and 1
And now we need to look into the statements and start plugging values just as we do for inuquality.



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23 Dec 2017, 16:22
I think this is a highquality question and the explanation isn't clear enough, please elaborate. I think its extremely important that explanations are student friendly, the explanations by expert Bunnel are amazing but often it feels likes those explanations assume that students have all the knowledge. We dont. The jargon of variables without some explanation can leave big gaps to fill. For this question, i felt explanation could have been MUCH MUCH easier. Take it as constructive critique.



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23 Dec 2017, 16:39
Statement 1) \(\frac{1}{x}\)\(>1\)
\(\frac{1}{x + 1}\) \(>0\)
\(\frac{(1+x)}{x}\) \(>0\)
Now if \(x=2\) , then statement 2 becomes \(\frac{(1 + 2)}{2}\) \(= 3/2\)which is greater 1 But if\(x=1/2\) then statement 2 becomes \(\frac{1.5}{.5}\)\(>0\) ( which is true) but less than 1
Therefore statement 1 is insufficient.
Statement 2) \(\frac{1}{x^5}\)> \(\frac{1}{x^3}\)
\(\frac{1}{x^5}\)  \(\frac{1}{x^3}\) >0
\(\frac{(x^3 x^5)}{x^5x^3}\) \(>0\)
\(\frac{x^3[1x^2]}{x^5x^3}\) \(>0\)
\(\frac{(1  x^2)}{x^5}\)\(>0\)
Now Only way expression above is \(>0\) when \(x^2\)is <1 therefore \(1< x < 1\) Which answers the question \(x>1\) ? as NO. Hence Statement 2 is sufficient.



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mtk10 wrote: I think this is a highquality question and the explanation isn't clear enough, please elaborate. I think its extremely important that explanations are student friendly, the explanations by expert Bunnel are amazing but often it feels likes those explanations assume that students have all the knowledge. We dont. The jargon of variables without some explanation can leave big gaps to fill. For this question, i felt explanation could have been MUCH MUCH easier. Take it as constructive critique. Dear friend, It's assumed that a student is not attempting questions, especially harder ones, and especially GMAT Club's hard questions, if the fundamentals are not clear. Solution for this particular questions includes all necessary steps and cannot possibly include underline theory. Still, if the solution is not clear you can always ask specific questions (will be glad to assist) and/or read the whole thread, often there you can find missing gaps. 9. Inequalities For more check Ultimate GMAT Quantitative Megathread
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07 Jan 2018, 04:50
I think this is a highquality question and the explanation isn't clear enough, please elaborate. Hi, How did we go from (x+1)/x>0 to x+1>0 because we cant multiply both sides by x as we dont know whether x is positive or negative. Best Regards.



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07 Jan 2018, 05:06
Furqan90 wrote: I think this is a highquality question and the explanation isn't clear enough, please elaborate. Hi, How did we go from (x+1)/x>0 to x+1>0 because we cant multiply both sides by x as we dont know whether x is positive or negative. Best Regards. For \(\frac{1+x}{x} \gt 0\) we should consider two cases: Case A  both numerator and denominator are positive: \(x \gt 0\) and \(1+x \gt 0\), (\(x \gt 1\)). So, \(x \gt 0\); Case B  both numerator and denominator are negative: \(x \lt 0\) and \(1+x \lt 0\), (\(x \lt 1\)). So, \(x \lt 1\). So, the given inequality holds true in two ranges: \(x \gt 0\) and \(x \lt 1\), thus \(x\) may or may not be greater than 1. Not sufficient. Can you please tell me which step is unclear?
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