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Is |n| < 1 ?

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Joined: 26 Feb 2014
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Is |n| < 1 ?  [#permalink]

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New post Updated on: 27 Feb 2014, 05:45
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Difficulty:

  55% (hard)

Question Stats:

63% (02:15) correct 38% (02:15) wrong based on 176 sessions

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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.

Originally posted by faceharshit on 26 Feb 2014, 09:52.
Last edited by Bunuel on 27 Feb 2014, 05:45, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Is |n| < 1 ?  [#permalink]

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New post 27 Feb 2014, 06:01
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4
faceharshit wrote:
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.


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.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 3 and 7. Thank you.
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Re: Inequality - Data Sufficiency Problem  [#permalink]

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New post 26 Feb 2014, 14:38
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faceharshit wrote:
Hi friends,
Need your help to approach below Inequality problem :-

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.

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:
http://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 :-)
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Magoosh Test Prep


Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
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Re: Is |n| < 1 ?  [#permalink]

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New post 02 Aug 2019, 10:38
Bunuel wrote:
faceharshit wrote:
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.


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.

I am unable to understand the highlighted part. Could you please explain how did we arrive at n>1?
Apologies but i do find inequality a little difficult.

Thanks so much
GMAT Club Bot
Re: Is |n| < 1 ?   [#permalink] 02 Aug 2019, 10:38
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