Is |n|

Question

Is  |n| < 1  ?

(1)  n^x - n < 0

(2)  x^{(-1)} = -2

My sol:
 1. n (n^(x-1) - 1) < 0
so either n is less than zero OR n^(x-1) - 1 < 0

2. x = -1/2

Clueless on how to analyze above and conclude some answer.

mikemcgarry · Answer

Dear faceharshit , I'm happy to respond. :-)

In writing these questions, I would urge you to pay more attention to mathematical grouping symbols. You can read more about this concept here: https://magoosh.com/gmat/2013/gmat-quant ... g-symbols/ In this question, clearly statement #1 is insufficient by itself, because we know nothing about x. Clearly, statement #2 is insufficient by itself because we know nothing about n. Clearly, we have to combine them to figure anything out. From #2, you are correct, we get x = -1/2. Plug this into #1 - n < 0 Notice, first of all, that we could not make any sensible statement if n were negative, because the square root of a negative is outside the real number system. The fact that this is an ordinary sensible statement automatically precludes negatives. It also precludes n - 0, because 0^(-1/2) is undefined. Add n to both sides. n^(-1/2) < n Since we know that n^(-1/2) is positive, divide both sides by that. 1 < n * n^(+1/2) = n^(3/2) If the 3/2 power of a number is greater than one, then that number must be greater than one. We can answer an affirmative and definitive "no" to the prompt question. Because we can arrive at an answer, that means the combined statements must be sufficient. Answer = (C) Does all this make sense? Mike :-)

Bunuel · Answer

Is |n| < 1 ?

Notice that the question basically asks whether \(-1<n<1\).

(1) n^x - n < 0. Well, this one is clearly insufficient: if \(n=2\) and \(x=0\), the answer is NO but if \(n=\frac{1}{2}\) and \(x=2\), the answer is YES. Not sufficient.

(2) x^(-1) = -2 --> \(\frac{1}{x}=-2\) --> \(x=-\frac{1}{2}\). Not sufficient.

(1)+(2) When we combine we have: \(n^{-\frac{1}{2}} - n < 0\) --> \(\frac{1}{\sqrt{n}}-n<0\).

From \(\frac{1}{\sqrt{n}}\), we can get that n must be a positive number: \(n>0\) (the expression under the square root must be non-negative, also since it's in the denominator it must be greater than 0).

\(\frac{1}{\sqrt{n}}-n<0\);

\(1<\sqrt{n}*n\).

If \(0<n\leq{1}\), then obviously \(1\geq{\sqrt{n}*n}\), thus \(n>1\). We have a NO answer to the question. Sufficient.

Answer: C.

Hope it's clear.

WoundedTiger · Answer

We need to find whether -10 and 1-\sqrt{n}<0 or \sqrt{n}>1 or n^2>1 or n>1 or n<-1 Case 2 \sqrt{n} <0 but it is not possible as the lowest possible value of \sqrt{n}=0 so we need not consider this case.. Thus ans is C

sunaimshadmani · Answer

If x=-0.5, Then why doesnt n^x become "1/sq.root n"

WoundedTiger · Answer

Thanks for pointing out 
If x=-1/2 then the question becomes 1-n^3/2 <0 or n^3/2>1

n>1^2/3 or n>1

Sufficient

We will only have to consider one case as sqrt n is greater than 0

Ans is C

Posted from my mobile device

fskilnik · Answer

\left| n \right|\,\,\mathop < \limits^? \,\,1 \left( 1 \right)\,\,{n^x} < n\,\,\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {n;x} \right) = \left( {2; -1} \right)\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {n;x} \right) = \left( {{1 \over 2};2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right. \left( 2 \right)\,\,{1 \over x} = - 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x = - {1 \over 2}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{INSUFF}}.\,\,\,\,\left( {{\rm{trivially}}} \right) \left( {1 + 2} \right)\,\,\,\,{1 \over {\sqrt n }} < n\,\,\,\left\{ \matrix{ \,\,\,\mathop \Rightarrow \limits^{{\rm{implicitly}}} \,\,\,\,n > 0 \hfill \cr \,\,\,\,\mathop \Rightarrow \limits^{\sqrt n \,\, > \,\,0} \,\,\,n\sqrt n > 1\, \hfill \cr} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{SUFF}}. \left( * \right)\,\,\,\left\{ \matrix{ \,n > 0 \hfill \cr \,\left| n \right| < 1 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,0 < n < 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \matrix{ \,0 < \sqrt n < 1 \hfill \cr \,0 < n < 1 \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,0 < n\sqrt n < 1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{impossible}} The correct answer is therefore (C), indeed. This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.

swatib28 · Answer

Bunuel

Can you please check my below explanation and suggests where am i going wrong ?
Below while working on C choice , we assume that N is +ve but range for N is coming to be negative as well and satisfying the statement..

Whats going wrong here? Please suggest.

bumpbot · Answer

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Is |n| < 1 ? (1) n^x - n < 0 (2) x^(-1) = -2

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