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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.

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.

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

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.

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

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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

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03 Sep 2015, 00:24

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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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04 Sep 2015, 04:49

Expert's post

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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