Is x^n/x^n+1 > x^n+1/x^n? (1) x < 0 (2) n < 0 : GMAT Data Sufficiency (DS)
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Is x^n/x^n+1 > x^n+1/x^n? (1) x < 0 (2) n < 0

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Is x^n/x^n+1 > x^n+1/x^n? (1) x < 0 (2) n < 0 [#permalink]

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New post 27 Sep 2009, 21:29
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Is x^n/x^(n+1)>x^(n+1)/x^n?

(1) x < 0

(2) n < 0
[Reveal] Spoiler: OA

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Last edited by Bunuel on 26 Jan 2012, 10:31, edited 1 time in total.
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New post 27 Sep 2009, 23:38
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The expression can be rephrased as : Is 1/x > x ?
stmt 1. The expression is true for x < 0 except x=-1. But there is no restriction on x. So insuff.
stmt 2. Value of n does not matter.

combining , still the same question, can x be -1 ( and also be < 0 )?
Hence E.
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New post 01 Oct 2009, 03:25
can you please explain it in detail?

Economist wrote:
The expression can be rephrased as : Is 1/x > x ?
stmt 1. The expression is true for x < 0 except x=-1. But there is no restriction on x. So insuff.
stmt 2. Value of n does not matter.

combining , still the same question, can x be -1 ( and also be < 0 )?
Hence E.
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New post 01 Oct 2009, 05:28
x^(n+1) = x^n.x

Now, while reducing the given inequality we need to take care of the inequality sign. ONLY if a -ve value is multiplied/divided on BOTH sides of the equation we need to reverse the inequality.

But, in this case we are not multiplying or dividing each side. We are just canceling out the same factor(-ve or +ve) from each side. So the inequality will remain the same.

In other words, x^n/x^n = 1, regardless of the sign of x^n. So, basically we are just multiplying each side by 1 :)
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New post 23 Oct 2009, 15:26
Economist wrote:
The expression can be rephrased as : Is 1/x > x ?
stmt 1. The expression is true for x < 0 except x=-1. But there is no restriction on x. So insuff.
stmt 2. Value of n does not matter.

combining , still the same question, can x be -1 ( and also be < 0 )?
Hence E.


Is it only for x= -1 or for the range -1<x<0?
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Re: Is x^n/x^n+1 > x^n+1/x^n? (1) x < 0 (2) n < 0 [#permalink]

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New post 26 Jan 2012, 06:17
+1 E

When we simplify the original inequality, using exponents theory, we get 1/x > x.

So both statements are insufficient.
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Re: Is x^n/x^n+1 > x^n+1/x^n? (1) x < 0 (2) n < 0 [#permalink]

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Ranges in above solutions are not correct.

Is \(\frac{x^n}{x^{n+1}}>\frac{x^{n+1}}{x^n}\)?

First of all realistic GMAT question would mention that \(x\neq{0}\).

Anyway: is \(\frac{x^n}{x^{n+1}}>\frac{x^{n+1}}{x^n}\)? --> is \(x^{n-n-1}>x^{n+1-n}\)? --> is \(\frac{1}{x}>x\)? --> is \(\frac{1}{x}-x>0\)? --> is \(\frac{1-x^2}{x}>0\)? --> is \(\frac{(1-x)(1+x)}{x}>0\)? So the question basically asks is \(x<-1\) or \(0<x<1\). (For more on this check: i-have-been-trying-to-understand-inequalities-by-reading-110917.html or range-for-variable-x-in-a-given-inequality-109468.html) Also noitce that the value of n is irrelevant to answer the question.

(1) x < 0. Not sufficient.

(2) n < 0. Not sufficient.

(1)+(2) Still can not answer the question. Not sufficient.

Answer: E.
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Re: Is x^n/x^n+1 > x^n+1/x^n? (1) x < 0 (2) n < 0 [#permalink]

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Bunuel wrote:
Ranges in above solutions are not correct.

Is \(\frac{x^n}{x^{n+1}}>\frac{x^{n+1}}{x^n}\)?

First of all realistic GMAT question would mention that \(x\neq{0}\).

Anyway: is \(\frac{x^n}{x^{n+1}}>\frac{x^{n+1}}{x^n}\)? --> is \(x^{n-n-1}<x^{n+1-n}\)? --> is \(\frac{1}{x}<x\)? --> is \(x-\frac{1}{x}>0\)? --> is \(\frac{x^2-1}{x}>0\)? --> is \(\frac{(x-1)(x+1)}{x}>0\)? So the question basically asks is \(-1<x<0\) or \(x>1\). (For more on this check: i-have-been-trying-to-understand-inequalities-by-reading-110917.html or range-for-variable-x-in-a-given-inequality-109468.html) Also noitce that the value of n is irrelevant to answer the question.

(1) x < 0. Not sufficient.

(2) n < 0. Not sufficient.

(1)+(2) Still can not answer the question. Not sufficient.

Answer: E.


Why have we flipped the inequality sign here? \(\frac{x^n}{x^{n+1}}>\frac{x^{n+1}}{x^n}\)? --> is \(x^{n-n-1}<x^{n+1-n}\)?
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Re: Is x^n/x^n+1 > x^n+1/x^n? (1) x < 0 (2) n < 0 [#permalink]

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New post 15 Sep 2013, 08:28
emailmkarthik wrote:
Bunuel wrote:
Ranges in above solutions are not correct.

Is \(\frac{x^n}{x^{n+1}}>\frac{x^{n+1}}{x^n}\)?

First of all realistic GMAT question would mention that \(x\neq{0}\).

Anyway: is \(\frac{x^n}{x^{n+1}}>\frac{x^{n+1}}{x^n}\)? --> is \(x^{n-n-1}<x^{n+1-n}\)? --> is \(\frac{1}{x}<x\)? --> is \(x-\frac{1}{x}>0\)? --> is \(\frac{x^2-1}{x}>0\)? --> is \(\frac{(x-1)(x+1)}{x}>0\)? So the question basically asks is \(-1<x<0\) or \(x>1\). (For more on this check: i-have-been-trying-to-understand-inequalities-by-reading-110917.html or range-for-variable-x-in-a-given-inequality-109468.html) Also noitce that the value of n is irrelevant to answer the question.

(1) x < 0. Not sufficient.

(2) n < 0. Not sufficient.

(1)+(2) Still can not answer the question. Not sufficient.

Answer: E.


Why have we flipped the inequality sign here? \(\frac{x^n}{x^{n+1}}>\frac{x^{n+1}}{x^n}\)? --> is \(x^{n-n-1}<x^{n+1-n}\)?


It was a typo edited. Thank you. +1.
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Re: Is x^n/x^n+1 > x^n+1/x^n? (1) x < 0 (2) n < 0 [#permalink]

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New post 01 Jan 2014, 11:41
tejal777 wrote:
Is x^n/x^(n+1)>x^(n+1)/x^n?

(1) x < 0

(2) n < 0


The correct answer is E

From question stem we can simplify and get is 1/x > x?

1/x - x > 0 --> Using key points is x<-1 or 0<x<1?

Statement 1

x<0 not sufficient

Statement 2

I don't care about 'n' at this point

Statements 1 and 2 together

I still don't have enough info

Hence E

Hope it helps
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Re: Is x^n/x^n+1 > x^n+1/x^n? (1) x < 0 (2) n < 0 [#permalink]

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New post 03 Jan 2014, 02:57
tejal777 wrote:
Is x^n/x^(n+1)>x^(n+1)/x^n?

(1) x < 0

(2) n < 0


Sol: Given question can be rephrased as is 1/x>x ------> (1-x^2)/x>0

Given x<0, let x=-2 then the above equation holds true but if x= -1/2 then the equation doesn't hold true
(1-1/4)/-1/2 ------>3/4/-1/2---->-6/4 or -3/2 which is less than 0
Ans Statement A is not sufficient

A and D ruled out

St2: given n<0 -----> There is no use for it. Hence B ruled out
With both statements we still don't have anything new.

Hence Ans E
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Re: Is x^n/x^n+1 > x^n+1/x^n? (1) x < 0 (2) n < 0 [#permalink]

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Is x^n/x^n+1 > x^n+1/x^n? (1) x < 0 (2) n < 0 [#permalink]

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New post 04 Sep 2015, 03:49
Forget conventional ways of solving math questions. In DS, Variable approach is

the easiest and quickest way to find the answer without actually solving the

problem. Remember equal number of variables and equations ensures a solution.



Is x^n/x^(n+1)>x^(n+1)/x^n?

(1) x < 0

(2) n < 0

==>transforming the original condition and the question by variable approach method, we have x^n/x^(n+1)>x^(n+1)/x^n? --> 1/x>x?. multiplying x^2 to both sides (since multiplying square values maintain the direction of the inequality sign), x>x^3? , x^3-x>0?, x(x-1)(x+1)>0? gives us -1<x<0 or 1<x?. Using both 1) and 2), the range of que doesn't include the range of con. Therefore E is the answer.
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Is x^n/x^n+1 > x^n+1/x^n? (1) x < 0 (2) n < 0   [#permalink] 04 Sep 2015, 03:49
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