Is |n|

Question

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

Himalayan · Answer

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.

vshaunak@gmail.com · Answer

Think more. OA is not 'E'

ywilfred · Answer

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.

jimmyjamesdonkey · Answer

How would you rate this problem --> Easy/Medium/Hard in terms of what you would find on the GMAT DS?

Andr359 · Answer

(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.

Himalayan · Answer

now its your turn to release the OA.

sumande · Answer

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)

vshaunak@gmail.com · Answer

OA is 'C'. Stmt1: n^x -n < 0 n < 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

wudy · Answer

(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

trahul4 · Answer

I will go with C . 

Agree with vshaunak.

vshaunak@gmail.com · Answer

n can't be -ve as n^-3/2 will be undefined for -ve values of n.

vshaunak@gmail.com · Answer

Apologies.... 
I put the stmt2 wrongly.
The correct stmt2 is  x^-1 = -2 
I have edited the question.

Himalayan · Answer

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Himalayan · Answer

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.

anonymousegmat · Answer

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

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

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