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Is |n| < 1 ? (1) (n^x) n < 0 (2) x^ 1 = 2 OA later..

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Director
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Is |n| < 1 ? (1) (n^x) n < 0 (2) x^ 1 = 2 OA later.. [#permalink] New post 08 Jul 2007, 09:46
00:00
A
B
C
D
E

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(N/A)

Question Stats:

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Is |n| < 1 ?

(1) (n^x) – n < 0

(2) x^–1 = –2

OA later...

Last edited by vshaunak@gmail.com on 10 Jul 2007, 09:29, edited 1 time in total.
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Re: DS Modulus [#permalink] New post 08 Jul 2007, 11:37
vshaunak@gmail.com wrote:
Is |n| < 1?
(1) (n^x) – n < 0
(2) x–1 = –2
OA later...


Should be E.

from 1:
n^x – n < 0
n^x < n

if n = 4, and x = 1/2, (n^x) < (n) and |n| is grater than 1
if n = 2, and x = -1, (n^x) < (n) and |n| is grater than 1
if n = 0.25, and x = 2, (n^x) < (n) and |n| is smaller than 1. so nsf.

from 2:
x – 1 = –2
x = – 1. also nsf.

From 1 and 2:

if x = -1 and n = 2, (n^x) < (n) and |n| is grater than 1
if x = -1 and n = -0.5, (n^x) < (n) and |n| is smaller than 1

nsf.
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 [#permalink] New post 09 Jul 2007, 17:08
Think more. OA is not 'E'
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 [#permalink] New post 09 Jul 2007, 18:00
Try st2 first:
x-1 = -2
x = -1 --> nothing about n, insufficient.

try st1:
If x is even, say 2, then:
n^2-n < 0 if n is a positive integer
n^2-n < 0 if n is a positive fraction

If x is odd, say 3, then:
n^3-n < 0 if n is a negative integer
n^3-n < 0 if n is a positive fraction

We don't know status of n, insufficient.

Using both:
1/n - n <0> 0 --> out
If n = positive integer, say 3, then 1/n-n <0> possible
If n = negative fraction, say -1/3, then 1/n-n <0> possible
If n = negative integer, say -3, then 1/n-n > 0 --> out

Two possibilites for n, insufficient.

E for me.
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 [#permalink] New post 09 Jul 2007, 18:03
How would you rate this problem --> Easy/Medium/Hard in terms of what you would find on the GMAT DS?
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Re: DS Modulus [#permalink] New post 09 Jul 2007, 18:09
vshaunak@gmail.com wrote:
Is |n| < 1 ?

(1) (n^x) – n < 0

(2) x–1 = –2



(1) Insuf right away.

(2) Same as (1).

(1&2) x = -1, then: 1/n - n = (1-n^2)/n < 1, which holds for (-1, 0) and (1, oo+), therefore E.

Last edited by Andr359 on 09 Jul 2007, 18:53, edited 3 times in total.
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 [#permalink] New post 09 Jul 2007, 18:17
vshaunak@gmail.com wrote:
Think more. OA is not 'E'



now its your turn to release the OA.
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Re: DS Modulus [#permalink] New post 09 Jul 2007, 21:06
vshaunak@gmail.com wrote:
Is |n| < 1 ?

(1) (n^x) – n < 0

(2) x–1 = –2

OA later...


I go with C.

From both statements together: (1-n^2)/n < 0

If n < -1, both numerator and denominator will be less than 0, and (1-n^2)/n will be greater than zero.

So, n can be within (-1,0) to satisfy the inequality. Hence, |n| < 1. (C)
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 [#permalink] New post 10 Jul 2007, 00:33
OA is 'C'.

Stmt1:
n^x -n < 0
n[n^(x-1) - 1] < 0
n < 0 and n^(x-1) - 1 > 0 or n > 0 and n^(x-1) - 1 < 0

Lets say
n < 0 and n^(x-1) - 1 > 0
n = -4 , x =3 ===> |n| > 1
n=-1/2, x =-1 ====> |n| < 1
So INSUFF

Now n > 0 and n^(x-1) - 1 < 0
n =2 , x =5/4 ====>|n| > 1
n=1/2 , x=5 ====>|n| < 1
So INSUFF

Stmt2: clearly INSUFF

Taken them together:
n> 0 and n^(-3/2) < 1 or n< 0 and n^(-3/2) > 1
for n> 0,
n * sqrt(n) > 1
|n| > 1


For n < 0 NOT POSSIBLE as sqrt(n) requires n to be +ve.

Hence SUFF
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 [#permalink] New post 10 Jul 2007, 04:34
vshaunak@gmail.com wrote:
OA is 'C'.

Stmt1:
n^x -n < 0
n[n^(x-1) - 1] < 0
n <0> 0 or n > 0 and n^(x-1) - 1 < 0

Lets say
n <0> 0
n = -4 , x =3 ===> |n| > 1
n=-1/2, x =-1 ====> |n| <1> 0 and n^(x-1) - 1 <0>|n| > 1
n=1/2 , x=5 ====>|n| <1> 0 and n^(-3/2) < 1 or n<0> 1
for n> 0,
n * sqrt(n) > 1
|n| > 1


For n < 0 NOT POSSIBLE as sqrt(n) requires n to be +ve.

Hence SUFF


(1)n^x<n
(2)x=-1
(1)+(2): n^(-1)<n 1/n<n
case 1:if 0<n; 1<n^2; 1<n
case 2:if n<0; n^2<1; -1<n<0
Answer is E.

The two cases listed below will call answer C into question.
case 1: n=10, x=-1; comply (1) and (2), 1<|n|
case 2: n=-0.5, x=-1; comply (1) and (2), |n|<1
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 [#permalink] New post 10 Jul 2007, 07:55
I will go with C .

Agree with vshaunak.
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 [#permalink] New post 10 Jul 2007, 08:26
wudy wrote:
vshaunak@gmail.com wrote:
OA is 'C'.

Stmt1:
n^x -n < 0
n[n^(x-1) - 1] < 0
n <0> 0 or n > 0 and n^(x-1) - 1 < 0

Lets say
n <0> 0
n = -4 , x =3 ===> |n| > 1
n=-1/2, x =-1 ====> |n| <1> 0 and n^(x-1) - 1 <0>|n| > 1
n=1/2 , x=5 ====>|n| <1> 0 and n^(-3/2) < 1 or n<0> 1
for n> 0,
n * sqrt(n) > 1
|n| > 1


For n < 0 NOT POSSIBLE as sqrt(n) requires n to be +ve.

Hence SUFF


(1)n^x<n
(2)x=-1
(1)+(2): n^(-1)<n 1/n<n
case 1:if 0<n; 1<n^2; 1<n
case 2:if n<0; n^2<1; -1<n<0
Answer is E.

The two cases listed below will call answer C into question.
case 1: n=10, x=-1; comply (1) and (2), 1<|n|
case 2: n=-0.5, x=-1; comply (1) and (2), |n|<1


n can't be -ve as n^-3/2 will be undefined for -ve values of n.
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 [#permalink] New post 10 Jul 2007, 09:28
Apologies....
I put the stmt2 wrongly.
The correct stmt2 is x^-1 = -2
I have edited the question.
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 [#permalink] New post 10 Jul 2007, 10:53
vshaunak@gmail.com wrote:
Apologies....
I put the stmt2 wrongly.
The correct stmt2 is x^-1 = -2
I have edited the question.



:roll: :shock: :shock:
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 [#permalink] New post 10 Jul 2007, 11:13
dahcrap wrote:
vshaunak@gmail.com wrote:
Apologies....
I put the stmt2 wrongly.
The correct stmt2 is x^-1 = -2
I have edited the question.


Even still I would stick with E


seems it makes sense now cuz x = -1/2 and (n^x) <n> 1. if n is 1 or less than 1 but grater than 0, (n^x) < (n) doesnot hold true. so it should be C.
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 [#permalink] New post 10 Jul 2007, 11:29
dahcrap wrote:
vshaunak@gmail.com wrote:
Apologies....
I put the stmt2 wrongly.
The correct stmt2 is x^-1 = -2
I have edited the question.


Even still I would stick with E


I am not sure how to approach this problem... but
does n^-1/2 mean we take the negative square root?

or do we go about it like this:

if n=1/4, (1/4)^-1/2 = 1/sqrt(1/4) = 1/1/2 =2. |2-.25|>1

if n=4 4^-1/2 = 1/sqrt4 = 1/2. |.5-4| > 1


i'd go with C
  [#permalink] 10 Jul 2007, 11:29
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