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

Question

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

(1) x < 0

(2) n < 0

Bunuel · Accepted Answer

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.

Economist · Answer

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.

reply2spg · Answer

can you please explain it in detail?

Economist · Answer

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

eresh · Answer

Is it only for x= -1 or for the range -1<x<0?

metallicafan · Answer

+1 E

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

So both statements are insufficient.

HKHR · Answer

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} ?

Bunuel · Answer

It was a typo edited. Thank you. +1.

jlgdr · Answer

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

WoundedTiger · Answer

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

MathRevolution · Answer

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.

techiesam · Answer

Is 
/frac{x^n}{x^n+1} > /frac{x^n+1}{x^n}

1) x<0
2) n<0

Bunuel · Answer

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

sreenu7464 · Answer

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

shashankism · Answer

DS: \frac{x^n}{x^{n+1}}>\frac{x^{n+1}}{x^n} \frac{1}{x}>\frac{x Statement 1 : x<0 -1\frac{x x = -1, 1/x}=\frac{x x<-1, 1/x}< x NOT SUFFICIENT Statement 2: n<0 has no significance NOT SUFFICIENT Answer E

ocelot22 · Answer

First simplify the question stem: x^n/x^n+1 > x^n+1/x^n --> x^-1>x --> 1/x >x ----> is x negative or a fraction or a negative fraction? (1) x<0 Let x=-1 1/-1 >-1 No let x = -3 1/-3 >-3 Yes NS (2) n<0 Question does not depend on n, NS (1) and (2) Since (1) NS and question does not depend on (2) NS OA is E

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