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:

Re: Gprep DS: Is xy>o? [#permalink]
25 Jan 2011, 14:34

Dear Bunuel,

cold you please explain this " Now, remember we can subtract inequalities with the signs in opposite direction." I seem to have forgotten basic arithmetic operations... ((

Re: Gprep DS: Is xy>o? [#permalink]
25 Jan 2011, 14:43

Expert's post

0987654312 wrote:

Dear Bunuel,

cold you please explain this " Now, remember we can subtract inequalities with the signs in opposite direction." I seem to have forgotten basic arithmetic operations... ((

Thank youuu

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 \(5>1\) --> \(3-5<4-1\).

Re: Gprep DS: Is xy>o? [#permalink]
25 Jan 2011, 14:53

Bunuel wrote:

0987654312 wrote:

Dear Bunuel,

cold you please explain this " Now, remember we can subtract inequalities with the signs in opposite direction." I seem to have forgotten basic arithmetic operations... ((

Thank youuu

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 \(5>1\) --> \(3-5<4-1\).

As pointed out above, using graphs is extremely quick and efficient in such questions. The only region where x-y> -2 and x-2y<-6 intersect is the first quadrant (as shown in the graph below). If you are uncomfortable with drawing accurate lines quickly, check out 'Bagging the graphs - Parts I, II and III' at http://www.veritasprep.com/blog/category/gmat/quarter-wit-quarter-wisdom/ Part III discusses a question exactly like this.

As pointed out above, using graphs is extremely quick and efficient in such questions. The only region where x-y> -2 and x-2y<-6 intersect is the first quadrant (as shown in the graph below). If you are uncomfortable with drawing accurate lines quickly, check out 'Bagging the graphs - Parts I, II and III' at http://www.veritasprep.com/blog/category/gmat/quarter-wit-quarter-wisdom/ Part III discusses a question exactly like this.

Attachment:

Ques1.jpg

Dear Karishma,

thank you very much for the link provided:very helpful! the approach with the graphs is very efficient! However, just one more comment as a clarification: I understand why the the lines are drawn in this way and i see that they intersect in Q1. However, how can i relate this to the question directily, i .e. what would the right logic be in order to answer the question (is xy>0?)?

As pointed out above, using graphs is extremely quick and efficient in such questions. The only region where x-y> -2 and x-2y<-6 intersect is the first quadrant (as shown in the graph below). If you are uncomfortable with drawing accurate lines quickly, check out 'Bagging the graphs - Parts I, II and III' at http://www.veritasprep.com/blog/category/gmat/quarter-wit-quarter-wisdom/ Part III discusses a question exactly like this.

Attachment:

Ques1.jpg

Dear Karishma,

thank you very much for the link provided:very helpful! the approach with the graphs is very efficient! However, just one more comment as a clarification: I understand why the the lines are drawn in this way and i see that they intersect in Q1. However, how can i relate this to the question directily, i .e. what would the right logic be in order to answer the question (is xy>0?)?

Dear Karishma,

one more question for you. I read the link you suggested and the explanations are brilliant ! I was wandering, can you provide me with a link to this: ( it is taken from the link posted below) "In part I of Graphs, I had also mentioned “Learn how to draw a line from its equation in under ten seconds and you shall solve the related question in under a minute"

As pointed out above, using graphs is extremely quick and efficient in such questions. The only region where x-y> -2 and x-2y<-6 intersect is the first quadrant (as shown in the graph below). If you are uncomfortable with drawing accurate lines quickly, check out 'Bagging the graphs - Parts I, II and III' at http://www.veritasprep.com/blog/category/gmat/quarter-wit-quarter-wisdom/ Part III discusses a question exactly like this.

Attachment:

Ques1.jpg

Dear Karishma,

thank you very much for the link provided:very helpful! the approach with the graphs is very efficient! However, just one more comment as a clarification: I understand why the the lines are drawn in this way and i see that they intersect in Q1. However, how can i relate this to the question directily, i .e. what would the right logic be in order to answer the question (is xy>0?)?

Question: Is xy > 0 xy > 0 when either both x and y are positive or both are negative. If the intersection lies in only first quadrant, both x and y are positive (In I Quadrant, x > 0 and y > 0) If the intersection lies in only third quadrant then both x and y are negative because in III quadrant, x < 0 and y < 0. As long as your points lie in I and III quadrants only, xy will be > 0. If your points lie in II or IV quadrant too, xy can be negative too because either x or y (both not both) is negative in II and IV quadrant.

Since in the graph above, the intersection lies only in I quadrant, it means x and y are both +ve and hence xy > 0. _________________

"In part I of Graphs, I had also mentioned “Learn how to draw a line from its equation in under ten seconds and you shall solve the related question in under a minute"

Re: Is xy>0? (1) x-y>-2 (2) x-2y<-6 [#permalink]
17 Jun 2014, 09:08

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. _________________

Type of Visa: You will be applying for a Non-Immigrant F-1 (Student) US Visa. Applying for a Visa: Create an account on: https://cgifederal.secure.force.com/?language=Englishcountry=India Complete...

I started running back in 2005. I finally conquered what seemed impossible. Not sure when I would be able to do full marathon, but this will do for now...