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Is |n| < 1 ?

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Is |n| < 1 ?  [#permalink]

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New post 20 Oct 2014, 06:52
1
22
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

57% (02:30) correct 43% (02:31) wrong based on 364 sessions

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Is |n| < 1 ?  [#permalink]

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New post 20 Oct 2014, 22:47
2
6
Bunuel wrote:

Tough and Tricky questions: Absolute Values.



Is |n| < 1 ?

(1) n^x - n < 0
(2) x^(-1) = -2



We need to find whether -1<n<1

St 1 says n^x-n <0 or n (n^x-1 -1)<0 We don't know the value of x so not sufficient


St 2 says 1/x=-2 or x=-1/2...So important question what is this got to do with -1<n<1 : not sufficient

Combining we see that \(\sqrt{n}-n<0\)

So we have \(\sqrt{n}(1-\sqrt{n})<0\)

We will have 2 cases

Case 1 \(\sqrt{n}>0\) 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
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Re: Is |n| < 1 ?  [#permalink]

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New post 28 Oct 2014, 11:03
If x=-0.5, Then why doesnt n^x become "1/sq.root n"
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Re: Is |n| < 1 ?  [#permalink]

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New post 28 Oct 2014, 12:25
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sunaimshadmani wrote:
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

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Re: Is |n| < 1 ?  [#permalink]

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New post 12 Nov 2018, 00:24
Rephrase----> is -1<n<1 ?

1. Nothing about x Insuff
2. Nothing about n Insuff
C Manipulate 1----> n(n^(x-1) - 1) < 0 PUT X = -0.5 ----> 1/sqrt(n)-n < 0 or n.sqrt(n)>1 -----> n > 1. So the answer is NO Sufficient.

Analysis of n.sqrt(n)>1
1. n is positive as GMAT considers sqrt expression to be positive unless WE are reducing the equation.
2. n is greater than 1 as we know that when 0<n<1 The increased exponent decreases the magnitude of positive expression.
Hence by 1 and 2 n>1
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Re: Is |n| < 1 ?  [#permalink]

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New post 12 Nov 2018, 15:04
1
Bunuel wrote:

Is |n| < 1 ?

(1) n^x - n < 0
(2) x^(-1) = -2

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


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


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


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

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


The correct answer is therefore (C), indeed.


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

Regards,
Fabio.
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Re: Is |n| < 1 ?  [#permalink]

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Re: Is |n| < 1 ?   [#permalink] 28 Nov 2019, 15:10
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