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

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27 Sep 2009, 22:29
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Is $$\frac{x^n}{x^{n+1}}>\frac{x^{n+1}}{x^n}$$?

(1) x < 0

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

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Last edited by Bunuel on 19 Apr 2017, 11:54, edited 2 times in total.
Edited the question
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Re: Is x^n/x^(n+1) > x^(n+1)/x^n? [#permalink]

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28 Sep 2009, 00:38
2
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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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Re: Is x^n/x^(n+1) > x^(n+1)/x^n? [#permalink]

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01 Oct 2009, 04:25
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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Re: Is x^n/x^(n+1) > x^(n+1)/x^n? [#permalink]

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01 Oct 2009, 06:28
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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Re: Is x^n/x^(n+1) > x^(n+1)/x^n? [#permalink]

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23 Oct 2009, 16:26
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? [#permalink]

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26 Jan 2012, 07:17
+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? [#permalink]

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26 Jan 2012, 11:31
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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.

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Re: Is x^n/x^(n+1) > x^(n+1)/x^n? [#permalink]

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14 Sep 2013, 04:24
1
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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.

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? [#permalink]

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15 Sep 2013, 09:28
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.

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? [#permalink]

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01 Jan 2014, 12:41
tejal777 wrote:
Is x^n/x^(n+1)>x^(n+1)/x^n?

(1) x < 0

(2) n < 0

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? [#permalink]

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03 Jan 2014, 03:57
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? [#permalink]

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Re: Is x^n/x^(n+1) > x^(n+1)/x^n? [#permalink]

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04 Sep 2015, 04:49
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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Re: Is x^n/x^(n+1) > x^(n+1)/x^n? [#permalink]

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29 Mar 2017, 03:23
Is
/frac{x^n}{x^n+1} > /frac{x^n+1}{x^n}

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

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29 Mar 2017, 03:37
techiesam wrote:
Is
/frac{x^n}{x^n+1} > /frac{x^n+1}{x^n}

1) x<0
2) n<0

Merging topics. Please follow the rules when posting a question: https://gmatclub.com/forum/rules-for-po ... 33935.html Pay attention to rule 3.

Also, you might find Writing Mathematical Formulas on the Forum post helpful: https://gmatclub.com/forum/rules-for-po ... l#p1096628
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Re: Is x^n/x^(n+1) > x^(n+1)/x^n? [#permalink]

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21 Apr 2017, 03:44
Is x^n/x^(n+1) > x^(n+1)/x^n?

(1) x < 0

(2) n < 0

x^n/x^(n+1) > x^(n+1)/x^n ==> x^2n > x^2(n+1)
ie X^2 < 1 , therefor the range of X is(-1, 1)

So statement 1 & 2 are not enough to get the sollution
so Ans> E
Re: Is x^n/x^(n+1) > x^(n+1)/x^n?   [#permalink] 21 Apr 2017, 03:44
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