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

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^(-1/2)] - 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

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

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

\(\left| n \right|\,\,\mathop < \limits^? \,\,1\)


\(\left( 1 \right)\,\,{n^x} < n\,\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {n;x} \right) = \left( {2; -1} \right)\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {n;x} \right) = \left( {{1 \over 2};2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)


\(\left( 2 \right)\,\,{1 \over x} = - 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x = - {1 \over 2}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{INSUFF}}.\,\,\,\,\left( {{\rm{trivially}}} \right)\)


\(\left( {1 + 2} \right)\,\,\,\,{1 \over {\sqrt n }} < n\,\,\,\left\{ \matrix{\\
\,\,\,\mathop \Rightarrow \limits^{{\rm{implicitly}}} \,\,\,\,n > 0 \hfill \cr \\
\,\,\,\,\mathop \Rightarrow \limits^{\sqrt n \,\, > \,\,0} \,\,\,n\sqrt n > 1\, \hfill \cr} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{SUFF}}.\)

\(\left( * \right)\,\,\,\left\{ \matrix{\\
\,n > 0 \hfill \cr \\
\,\left| n \right| < 1 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,0 < n < 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \matrix{\\
\,0 < \sqrt n < 1 \hfill \cr \\
\,0 < n < 1 \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,0 < n\sqrt n < 1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{impossible}}\)


The correct answer is therefore (C), indeed.


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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.

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