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 500,000 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:

What is wrong with the following line of thought for condition 1

1/(x+1) < 1

1 < (x+1)

1 -1 < x+1-1

0 < x Hence x>0.

We can not multiple \(\frac{1}{x+1}< 1\) by \(x+1\) since we don't know whether this expression (\(x+1\)) is positive or negative, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity).

Is x positive?

(1) 1/(x+1) < 1. If \(x=10\) then the answer is YES, but if \(x=-10\) then the answer is NO. Not sufficient.

Or if you want to solve this inequality then: \(\frac{1}{x+1}< 1\) --> \(1-\frac{1}{x+1}>0\) --> \(\frac{x}{x+1}>0\) --> \(x>0\) or \(x<-1\), hence \(x\) could be positive as well as negative. Not sufficient.

(2) (x-1) is a perfect square. Given: \(x-1=\{perfect \ square\}\) --> \(x=\{perfect \ square\}+1\). Now, since perfect square is more than or equal to zero, then \(x=\{perfect \ square\}+1=non-negative+1=positive\). Sufficient.

What is wrong with the following line of thought for condition 1

1/(x+1) < 1

1 < (x+1)

1 -1 < x+1-1

0 < x Hence x>0.

We can not multiple \(\frac{1}{x+1}< 1\) by \(x+1\) since we don't know whether this expression (\(x+1\)) is positive or negative, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity).

Is x positive?

(1) 1/(x+1) < 1. If \(x=10\) then the answer is YES, but if \(x=-10\) then the answer is NO. Not sufficient.

Or if you want to solve this inequality then: \(\frac{1}{x+1}< 1\) --> \(1-\frac{1}{x+1}>0\) --> \(\frac{x}{x+1}>0\) --> \(x>0\) or \(x<-1\), hence \(x\) could be positive as well as negative. Not sufficient.

(2) (x-1) is a perfect square. Given: \(x-1=\{perfect \ square\}\) --> \(x=\{perfect \ square\}+1\). Now, since perfect square is more than or equal to zero, then \(x=\{perfect \ square\}+1=non-negative+1=positive\). Sufficient.

Answer: B.

Hope it's clear.

Bunuel,

How did you arrive at x<-1 in the explanation below?

"\frac{x}{x+1}>0 --> x>0 or x<-1, hence x could be positive as well as negative. Not sufficient."

What is wrong with the following line of thought for condition 1

1/(x+1) < 1

1 < (x+1)

1 -1 < x+1-1

0 < x Hence x>0.

We can not multiple \(\frac{1}{x+1}< 1\) by \(x+1\) since we don't know whether this expression (\(x+1\)) is positive or negative, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity).

Is x positive?

(1) 1/(x+1) < 1. If \(x=10\) then the answer is YES, but if \(x=-10\) then the answer is NO. Not sufficient.

Or if you want to solve this inequality then: \(\frac{1}{x+1}< 1\) --> \(1-\frac{1}{x+1}>0\) --> \(\frac{x}{x+1}>0\) --> \(x>0\) or \(x<-1\), hence \(x\) could be positive as well as negative. Not sufficient.

(2) (x-1) is a perfect square. Given: \(x-1=\{perfect \ square\}\) --> \(x=\{perfect \ square\}+1\). Now, since perfect square is more than or equal to zero, then \(x=\{perfect \ square\}+1=non-negative+1=positive\). Sufficient.

Answer: B.

Hope it's clear.

Bunuel,

How did you arrive at x<-1 in the explanation below?

"\frac{x}{x+1}>0 --> x>0 or x<-1, hence x could be positive as well as negative. Not sufficient."

We have \(\frac{x}{x+1}>0\).

The roots are -1, and 0 (equate the expressions to zero to get the roots and list them in ascending order), this gives us 3 ranges: \(x<-1\), \(-1<x<0\), and \(x>0\).

Now, test some extreme value: for example if \(x\) is very large number then all two terms (x and x+1) will be positive which gives the positive result for the whole expression, so when \(x>0\) the expression is positive. Now the trick: as in the 3rd range expression is positive then in 2rd it'll be negative and finally in 1st it'll be positive again: + - +. So, the ranges when the expression is positive are: \(x<-1\) (1st range) and \(x>0\) (3rd range).

Got it so the condition is met for all numbers less than -1 or greater than 0, but any number between 0 and -1 (Fraction) will result in a negative number.

But even without thinking about the possibility of a fraction, Statement 1 is insufficient because x>0 or x<-1 right?

I know this is a silly, but I didn't realize I could find the roots by doing this:

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...