Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

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

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

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.

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

Are you interested in applying to business school? If you are seeking advice about the admissions process, such as how to select your targeted schools, then send your questions...