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:

summing the above two equations, can't we derive 5y > 0 and thus y > 0?

If \(y=0\) and \(x\) is any positive integer both statements hold true and the answer to the question is NO. If \(y=1\) and \(x=2\) again both statements hold true and the answer to the question is YES. Two different answers. Not sufficient.

Answer: E.

As for your question: when we sum \(7x - 2y > 0\) and \(7x + 7y > 0\) we'll get \(14x+5y>0\) and not \(5y > 0\).

As for your question: when we sum \(7x - 2y > 0\) and \(7x + 7y > 0\) we'll get \(14x+5y>0\) and not \(5y > 0\).

Hope it helps.

Thanks. Oh ya,, silly mistake from me. I meant subtracting equation 2 from 1 will lead us to 9y >0 and eventually y > 0.

Am i wrong in performing such subtractions for inequalities?

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\). Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

So we can only add \(7x - 2y > 0\) and \(7x + 7y > 0\) as their signs are in the same direction (>).

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\). Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

So we can only add \(7x - 2y > 0\) and \(7x + 7y > 0\) as their signs are in the same direction (>).

Hope it helps.

Absolutely fantastic. Thanks Bunuel once again. This has indeed cleared one of the flaws in my understanding.

Re: If x and y are integers and x > 0, is y > 0? [#permalink]

Show Tags

15 Oct 2013, 02:29

Hello from the GMAT Club BumpBot!

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.
_________________

Re: If x and y are integers and x > 0, is y > 0? [#permalink]

Show Tags

03 May 2014, 21:22

If x and y are integers and x > 0, is y > 0?

(1) 7x – 2y > 0 (2) -y < x

Bunuel, please have a look at this.

statement1. 7x - 2y > 0 => x > 2y/7---------(1)

y can be negative or positive. Not sufficient

statement2. -y < x => x + y > 0---------(2)

Again, y can be negative or positive. Not Sufficient.

statement 1 & 2 Together.

from eq. (1) & (2), [a value > 2y/7] + y > 0 lets consider least value of x (actually this would be lesser than the least). 2y/7 + y > 0 => 9y/7 > 0 or (9/7)y>0 therefore, y must be > than 0.

Again, y can be negative or positive. Not Sufficient.

statement 1 & 2 Together.

from eq. (1) & (2), [a value > 2y/7] + y > 0 lets consider least value of x (actually this would be lesser than the least). 2y/7 + y > 0 => 9y/7 > 0 or (9/7)y>0 therefore, y must be > than 0.

Answer C.

Why are you considering the least value of x??? Nothing prevents x to be greater than this value and in this case your logic will not hold.
_________________

Re: If x and y are integers and x > 0, is y > 0? [#permalink]

Show Tags

31 Oct 2015, 12:12

Hello from the GMAT Club BumpBot!

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.
_________________

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...