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:

In a XY Plan, Y is above the line -x/2. I prefer to solve this DS with the XY plan.

Stat1 y > -1. This brings us to y above the line y=-1. The area covered could be either above -x/2 (when x>0) or below -x/2 (wehn x<10). (Fig1)

INSUFF.

Stat2 (2^x) - 4 >0
<=> 2^x > 4 = 2^2
<=> x > 2

In the XY plan, it does mean that the concerned area is at right of the line x=2. One more time, their is 2 case, y could be either above the line -x/2 (when y > 0) or below the line -x/2 (the point (3,-100)). (Fig2)

INSUFF.

Both (1) and (2): x > 2 and y > -1 represents an area with a not attainable vertice at (-1;2). By drawing the line -x/2 and this point, we observe that this point is on the line. Thus, the whole area of points such that x>2 and y>-1 is above the line -x/2. (Fig3)

(1) y>-1, clearly insufficient as no info about \(x\).

(2) (2^x)-4>0 --> \(2^x>2^2\) --> \(x>2\) --> also insufficient as no info about \(y\).

(1)+(2) \(y>-1\), or \(2y>-2\) and \(x>2\) --> add this inequalities (remember, you can only add inequalities when their signs are in the same direction and you can only apply subtraction when their signs are in the opposite directions) --> \(2y+x>-2+2\) --> \(x+2y>0\). Sufficient.

(1) y>-1, clearly insufficient as no info about \(x\).

(2) (2^x)-4>0 --> \(2^x>2^2\) --> \(x>2\) --> also insufficient as no info about \(y\).

(1)+(2) \(y>-1\), or \(2y>-2\) and \(x>2\) --> add this inequalities (remember, you can only add inequalities when their signs are in the same direction and you can only apply subtraction when their signs are in the opposite directions) --> \(2y+x>-2+2\) --> \(x+2y>0\). Sufficient.

Answer: C.

I chose C.. thats correct.. bt with different approach

bt Bunuel I didnt get this highlighted thing?? How did u do that?
_________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

(1) y>-1, clearly insufficient as no info about \(x\).

(2) (2^x)-4>0 --> \(2^x>2^2\) --> \(x>2\) --> also insufficient as no info about \(y\).

(1)+(2) \(y>-1\), or \(2y>-2\) and \(x>2\) --> add this inequalities (remember, you can only add inequalities when their signs are in the same direction and you can only apply subtraction when their signs are in the opposite directions) --> \(2y+x>-2+2\) --> \(x+2y>0\). Sufficient.

Answer: C.

I chose C.. thats correct.. bt with different approach

bt Bunuel I didnt get this highlighted thing?? How did u do that?

\(2y+x>-2+2\) --> re-arrange the left hand side as x+2y. As for the right hand side: -2 + 2 = 0. So, we get \(x+2y>0\).

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.
_________________

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is -x < 2y ?

(1) y > -1 (2) (2^x) - 4 > 0

In the original condition, there are 2 variables(x,y), which should match with the number of equations. So you need 2 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2), they become y>-1, 2y>-2 and 2^x>4=2^2 --> x>2, -x<-2, which is -x<-2<2y --> -x<2y. So it is yes and sufficient. Therefore, the answer is C.

-> For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
_________________

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...