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:

OA is C but It seems A also satisfies this condition...

I calculated it as below; AB/7 > 1/14 multiplying both sides by 14 2AB > 1 as A=B 2A^2 >1 and 2 B^2 > 1 Option (C) is true as per this logic While analyzing other options this is what I got... 2A^2>1 A^2>1/2 A>1/sqrt(2) as A=B, B >1/sqrt(2)

so A+B > 1/sqrt(2) + 1/sqrt(2) A+B > 2/sqrt(2) A+B > sqrt(2) This is also going to be greater than 1

This problem is from MGMAT Algebra Strategy Guide:

If AB/7 > 1/14 and A=B While of the following must be greater than 1 ? a) A+B b) 1-A c) 2A^2 d) A^2-1/2 e) A

OA is C but It seems A also satisfies this condition...

I calculated it as below; AB/7 > 1/14 multiplying both sides by 14 2AB > 1 as A=B 2A^2 >1 and 2 B^2 > 1 Option (C) is true as per this logic While analyzing other options this is what I got... 2A^2>1 A^2>1/2 A>1/sqrt(2) as A=B, B >1/sqrt(2)

so A+B > 1/sqrt(2) + 1/sqrt(2) A+B > 2/sqrt(2) A+B > sqrt(2) This is also going to be greater than 1

If AB/7 > 1/14 and A = B, which of the following must be greater than 1?

A. A+B B. 1-A C. 2A^2 D. A^2 - 1/2 E. A

First of all notice that the question asks which of the following MUST be true, not COULD be true.

\(\frac{AB}{7} > \frac{1}{14}\) --> \(2AB>1\). Since A = B, then \(2A^2>1\).

Answer: C.

As for option A: notice that from \(A^2>\frac{1}{2}\) it follows that A, and therefore B, since A = B, can be negative numbers, for example, -1, and in this case A + B = -2 < 1. The problem with your solution is that \(A^2>\frac{1}{2}\) means that \(A>\frac{1}{\sqrt{2}}\) OR \(A<-\frac{1}{\sqrt{2}}\), not only \(A>\frac{1}{\sqrt{2}}\).

Thanks Bunuel, Your explanation is awesome as usual.

But I am a bit confused about below part.

Bunuel wrote:

The problem with your solution is that \(A^2>\frac{1}{2}\) means that \(A>\frac{1}{\sqrt{2}}\) OR \(A<-\frac{1}{\sqrt{2}}\), not only \(A>\frac{1}{\sqrt{2}}\). .

As per my understanding \(A^2>\frac{1}{2}\) means that \(A>+\frac{1}{\sqrt{2}}\) OR \(A>-\frac{1}{\sqrt{2}}\). Could you please elaborate why \(A<-\frac{1}{\sqrt{2}}\) ?

Thanks Bunuel, Your explanation is awesome as usual.

But I am a bit confused about below part.

Bunuel wrote:

The problem with your solution is that \(A^2>\frac{1}{2}\) means that \(A>\frac{1}{\sqrt{2}}\) OR \(A<-\frac{1}{\sqrt{2}}\), not only \(A>\frac{1}{\sqrt{2}}\). .

As per my understanding \(A^2>\frac{1}{2}\) means that \(A>+\frac{1}{\sqrt{2}}\) OR \(A>-\frac{1}{\sqrt{2}}\). Could you please elaborate why \(A<-\frac{1}{\sqrt{2}}\) ?

Let me ask you a question: does \(x^2>4\) mean \(x>2\) or \(x>-2\)? What does \(x>2\) or \(x>-2\) even mean?

\(x^2>4\) --> \(|x|>2\) --> \(x>2\) or \(x<-2\).

Go through the links below to brush up fundamentals on inequalities:

Let me ask you a question: does \(x^2>4\) mean \(x>2\) or \(x>-2\)? What does \(x>2\) or \(x>-2\) even mean?

\(x^2>4\) --> \(|x|>2\) --> \(x>2\) or \(x<-2\).

This conclusion is based on below logic. \(x^2=25\) has TWO solutions, +5 and -5 so I thought \(x^2>4\) would also have two solutions +2 & -2, this is how I got \(x>2\) or \(x>-2\) Could you please let me know why this is not valid ?

I am asking very basic question but I know only after getting my basics clear, I will be able to get good score.... Thanks for bearing with me and my stupid questions .

Let me ask you a question: does \(x^2>4\) mean \(x>2\) or \(x>-2\)? What does \(x>2\) or \(x>-2\) even mean?

\(x^2>4\) --> \(|x|>2\) --> \(x>2\) or \(x<-2\).

This conclusion is based on below logic. \(x^2=25\) has TWO solutions, +5 and -5 so I thought \(x^2>4\) would also have two solutions +2 & -2, this is how I got \(x>2\) or \(x>-2\) Could you please let me know why this is not valid ?

I am asking very basic question but I know only after getting my basics clear, I will be able to get good score.... Thanks for bearing with me and my stupid questions .

Guess you did not follow the links I proposed...

Again, what does x is more than 2 or x is more than -2 mean? What are the possible values of x in this case? For example, can x be 1, since it's more than -2?

\(x^2>4\) indeed has two ranges, which are \(x>2\) or \(x<-2\). Please follow the links in my previous post for more. _________________

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...