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 (>).

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

Show Tags

12 Sep 2010, 00:37

Bunuel wrote:

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

03 May 2014, 20:22

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

27 Jul 2017, 05:11

from 1 alone y can be negative and y can be positive say y=-1 and x=1. 7+2=9. 9>0

say y=1 and x=1 7-2=5. 5>0

1 alone is insufficient. A and D are out.

statement 2 says -y<x. rewrite 0<x+y y can be negative and y can be positive. say y=-1 and x=2 2-1 = 1. 1>0. works

say y=1 and x=1 1+1=2. 2>0. works.

B is out, so we are left with either C or E - 50/50 chance we'd get it right, not bad but combining, we see that both statements say that y can be > or < 0. C is out, and E is the answer.

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

Show Tags

19 Oct 2017, 11:10

Bunuel , could you illustrate the graphical solution for this problem too?

The graph of the first statement passes through the origin and lies to the right of the line in the 1st quadrant and to its left in the third quadrant. If x is positive, only values in the first quadrant hold. Then y>0, is it not? OA a?