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:

nice method, but i m afraid there is some mistake in ur explanation.

if y>4. say y=5 now plugging this value in eq2, i.e. option 2

x-2y < -6 ===> x-10<-6 ===> x<4 ===> so x can also be -ve... in this case xy<0

plugging the value of y=5 in eq1, i.e. option 1

x-y > -2 ===> x>y+2 ===> x>7 ===> x=+ve.... in this case xy>0

two answers. hence data insuff. Answer:E

here is my method.. as u know, 1 & 2 are individually insuff.. taking 1&2 together

x-y>-2 ===> x>y+2 x-2y<-6 ===> x<2y-6

so combining both the above equations (y+2)< x < (2y-6)

by plugging values, if y= 2 ==> 4<x<-2 ==> x<-2 & x>4 two cases arise. so, data insuff. Answer: E

Not so. Lagomez's answer is correct.

Is xy>0?

(1) x-y > -2 (2) x-2y <-6

Note that question basically asks whether \(x\) and \(y\) have the same sign.

Statements alone are not sufficient as was shown above.

Now, remember we can subtract inequalities with the signs in opposite direction.

\(x-y-(x-2y)>-2-(-6)\) --> \(y>4\). As \(y>4\) and (from 1) \(x>y-2\), hence \(x>2\) (we can add inequalities when their signs are in the same direction, so \(y+x>4+y-2\) --> \(x>2\)) --> both \(x\) and \(y\) are positive. Sufficient.

Answer: C.

When you got that for y=5 --> x<4, it's correct, but it doesn't mean that x can be negative, as this is not ALL information you have. If you plug the same value y=5 in (1) you'll get --> x>3, which is also correct.

i think ans should be E... agreed y>4.. substitute in eq 1...x>2... substitute in eq 2... x<2y-6...x<8-6...x<2..... combining the two , can we get a value of x, which is both < 2 and>2,, i dont think

When you substitute in equation 2, you take the value of y=4 (which can not be as y>4).

Statements alone are not sufficient as was shown above.

Now, remember we can subtract inequalities with the signs in opposite direction.

x-y-(x-2y)>-2-(-6) --> y>4. As y>4 and (from 1) x>y-2, hence x>2. Sufficient.

Answer: C.

When you got that for y=5 --> x<4, it's correct, but it doesn't mean that x can be negative, as this is not ALL information you have. If you plug the same value y=5 in (1) you'll get --> x>3, which is also correct.

Basically when y=5 --> 3<x<4.

Hope it's clear.

oh yeah... too bad that i forgot to plug the value of y=5 in both the equations... thanx a ton Bunuel... kudos..
_________________

i think ans should be E... agreed y>4.. substitute in eq 1...x>2... substitute in eq 2... x<2y-6...x<8-6...x<2..... combining the two , can we get a value of x, which is both < 2 and>2,, i dont think

When you substitute in equation 2, you take the value of y=4 (which can not be as y>4).

C is correct.

If I substituted x=2 into each of the equations to solve for Y, i got two different answers. Y>4 and Y<4, depending on which equation you substitute it into.

Can't seem to find this in the forum. From the Gprep test:

Is xy>0? 1. x-y > -2 2. x-2y <-6

Can't seem to get my head around this. Appreciate any help.

answer c Statement 1: x = 5 and y = 4 answer is yes x = -1 and y = 0 answer is no

Statement 2: x = -7 and y = 0 answer is no x = -10 and y = -1 answer is yes

Combined: x - y > -2 x - 2y < -6

or -x + y < 2 x - 2y < - 6 -y < -4 y > 4 so you know y is positive

What about x..plug a value of y > 4 in the equation and you'll get x > 0

Correct answer is C...

As pointed by Bunuel, when u plug in y > 4 in both the above equations, you get x range as 3<x<4.... Therefore xy > 0!

You shouldn't consider only x - 2y < -6 for determining the value of x. Both the equations should be considered!

Cheers! JT
_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

i think ans should be E... agreed y>4.. substitute in eq 1...x>2... substitute in eq 2... x<2y-6...x<8-6...x<2..... combining the two , can we get a value of x, which is both < 2 and>2,, i dont think

When you substitute in equation 2, you take the value of y=4 (which can not be as y>4).

C is correct.

If I substituted x=2 into each of the equations to solve for Y, i got two different answers. Y>4 and Y<4, depending on which equation you substitute it into.

I don't quite understand what's your point. Answer to the original question is C.

xy>0 means that the point (x,y) is either in I quadrant or in III.

(1) x-y > -2 the points satisfying this inequality can be in ALL four quadrants, refer to the first graph msunny provided (blue area). Hence insufficient.

(2) x-2y <-6 the points satisfying this inequality can be in I, II or in III quadrants, refer to the second graph msunny provided (blue area). Hence insufficient.

(1)+(2) Points satisfying BOTH inequalities are only in I quadrant: intersection of blue areas. Hence sufficient.

I apologize if my point is unclear, but my question is that, after I calculated X>2 (I solved for X first), I plugged in X=2 into the first equation, and I got Y<4. What I am confused about is that if Y<4, that means Y can be negative; however, x>2, so if Y is negative and X is always positive, then you would be in quadrant IV.

If plugged in X=2 into the second equation, then you get y>4, which is sufficient since X>2 and Y>4, so you will always be in quadrant I.

I apologize if my point is unclear, but my question is that, after I calculated X>2 (I solved for X first), I plugged in X=2 into the first equation, and I got Y<4. What I am confused about is that if Y<4, that means Y can be negative; however, x>2, so if Y is negative and X is always positive, then you would be in quadrant IV.

If plugged in X=2 into the second equation, then you get y>4, which is sufficient since X>2 and Y>4, so you will always be in quadrant I.

Your X value is > 2, do not substitute 2 as X in the equation. Substituting X = 3 would be a right approach. On doing that for first eq, you get: 3-y>-2 i.e. y < 5 --- [1] Substituting X = 3 in equation 2, you get:

3-2y<-6 i.e. y > 4.5....

Therefore if X = 3, then y range is from 4.5 to 5 which means both X and Y are positive. Hence XY > 0.

Hope this is clear..

Thanks, JT
_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

I apologize if my point is unclear, but my question is that, after I calculated X>2 (I solved for X first), I plugged in X=2 into the first equation, and I got Y<4. What I am confused about is that if Y<4, that means Y can be negative; however, x>2, so if Y is negative and X is always positive, then you would be in quadrant IV.

If plugged in X=2 into the second equation, then you get y>4, which is sufficient since X>2 and Y>4, so you will always be in quadrant I.

First of all when you solved system of inequalities and got x>2 and you want to continue by number plugging you should plug correct numbers. You got x>2, hence you should plug x more than 2. Any value more than 2 for x will give you the RANGE for y which is positive.

You'll get only positive range for y, when you test ANY value of x>2, as the solution of this system of inequalities is x>2 and y>4 (refer to the solution in my first post in this topic). And vise-versa: if you try values of y>4 you'll get only positive range for x.

we CAN NOT ADD INEQUALITY, though we can add equality.

a>b and c>d

we can not have a+c>b+d . because there is some redundant

we use this method

y-2<x<2y-6 y-2<2y-6 4<y that means x>2, C is correct.

The red part is not correct.

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

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

[rss2posts title=The MBA Manual title_url=https://mbamanual.com/2016/11/22/mba-vs-mim-guest-post/ sub_title=MBA vs. MiM :3qa61fk6]Hey, guys! We have a great guest post by Abhyank Srinet of MiM-Essay . In a quick post and an...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...