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:

(1) m - 3z > 0. Insufficient on its own. (2) 4z - m > 0. Insufficient on its own.

(1)+(2) Remember we can add inequalities with the sign in the same direction --> \(m-3z+4z-m>0\) --> \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) --> \(m+z=positive+positive>0\). Sufficient.

When I took both statements and in my head made the stipulations, I ended up with 3z<m<4z, but that did not yield the right answer. Can you tell me what I am missing doing this vs. actually adding the equations?

(1) m - 3z > 0. Insufficient on its own. (2) 4z - m > 0. Insufficient on its own.

(1)+(2) Remember we can add inequalities with the sign in the same direction --> \(m-3z+4z-m>0\) --> \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) --> \(m+z=positive+positive>0\). Sufficient.

When I took both statements and in my head made the stipulations, I ended up with 3z<m<4z, but that did not yield the right answer. Can you tell me what I am missing doing this vs. actually adding the equations?

You missed the last step: from 3z<4z it follows that z>0, thus 3z=positive --> (3z=positive)<m --> m=positive --> m+z=positive. _________________

(1) m - 3z > 0. Insufficient on its own. (2) 4z - m > 0. Insufficient on its own.

(1)+(2) Remember we can add inequalities with the sign in the same direction --> \(m-3z+4z-m>0\) --> \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) --> \(m+z=positive+positive>0\). Sufficient.

I have a doubt: 1 & 2 are obviously insuff. Combining 1 & 2 we get z>0. Now according to 1) m-3z>0 (so we get m>0). My doubt starts here 2) says 4z-m>0 so when z>0, m>0 or m<0. For eg z=2 & m=-2, we will still get 4z-m>0 So according to me answer should be E or am I missing something here.

(1) m - 3z > 0. Insufficient on its own. (2) 4z - m > 0. Insufficient on its own.

(1)+(2) Remember we can add inequalities with the sign in the same direction --> \(m-3z+4z-m>0\) --> \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) --> \(m+z=positive+positive>0\). Sufficient.

I have a doubt: 1 & 2 are obviously insuff. Combining 1 & 2 we get z>0. Now according to 1) m-3z>0 (so we get m>0). My doubt starts here 2) says 4z-m>0 so when z>0, m>0 or m<0. For eg z=2 & m=-2, we will still get 4z-m>0 So according to me answer should be E or am I missing something here.

Bunuel my line of reasoning was that the numbers are aligned like 3z<m<4z. Now m falls between 3z and 4z which can only be possible if both m and z are positive ( as the multiplier is positive). Now adding two positive numbers will always be more than 0 . Hence both are sufficient to prove

(1) m - 3z > 0. Insufficient on its own. (2) 4z - m > 0. Insufficient on its own.

(1)+(2) Remember we can add inequalities with the sign in the same direction --> \(m-3z+4z-m>0\) --> \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) --> \(m+z=positive+positive>0\). Sufficient.

Do you think algebraic approach works even if an answer choice is E in such similar questions? Or do we have to resort to graphic approach/number picking then? Any tips on how to attack these questions in general?

Thanks _________________

Please consider giving 'kudos' if you like my post and want to thank

Do you think algebraic approach works even if an answer choice is E in such similar questions? Or do we have to resort to graphic approach/number picking then? Any tips on how to attack these questions in general?

Thanks

IMO, these questions require a combination of algebraic and number plugging in order to arrive at the final answer. Without looking at a particular question, it is very dangerous to define your strategy.

I did not use any graphical method as such but did employ number plugging for proving that statements 1 and 2 are not sufficient on their own and then use algebra to show that when the 2 statements are combined, we get z>0. _________________

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 m+z > 0

(1) m-3z > 0 (2) 4z-m > 0

In the original condition, there are 2 variables(m,z), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. In 1)&2), add the 2 equations, which is m-3z+4z+m>0, z>0. Then m>z>0 is derived from m>3z>z, which is m>0. So, m+z>0 is always 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. _________________

with the first statement, if you apply the numbers -1,0 and 1. You will conclude that m is positive when z is positive. Hence I thought one is sufficient.

gmatclubot

Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0
[#permalink]
24 Dec 2015, 09:48

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...