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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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23 Dec 2015, 18:50
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NoHalfMeasures wrote: Bunuel, Skywalker18, Engr2012, and other experts : 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.



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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23 Dec 2015, 23:28
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) m3z > 0 (2) 4zm > 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 m3z+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.
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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09 May 2016, 21:55
smitakokne wrote: Is m+z > 0
(1) m3z > 0 (2) 4zm > 0 Responding to a pm: Quote: When I look at such question, I know I will figure out something if I consider both statements together. I feel comfortable with applying inequalities rules there. But I find it harder to approach individual statements in such cases. I get lost in figuring out cases to reject (or approve) one statement on its own. At times, I take time and eventually rush onto considering both statements together.
Can you please help me out if I'm lacking some approach or concepts?
There really isn't one suitable method for all inequality questions. The approach depends on the kind of question. Very rarely will I plug in numbers here to find cases where the inequality holds or does not hold. This is how I will do this question: "Is m+z > 0?" Here I think that m and z could both be positive or one of them could be positive with greater absolute value than the other. If both are negative, this doesn't hold. Go on to stmnts. (1) m3z > 0 Here, I will try to segregate the variables to get relation between them. m > 3z The moment I see this, I naturally go to the number line. (z positive): ______________________________ 0 ____z_________________3z___________m____________________ OR (z negative): __3z_________(m)______________z___(m)__ 0 _______________(m)____________________ Both z and m could be positive, both negative or z negative m positive. Not sufficient. (2) 4zm > 0 4z > m (z positive): __________________(m)____________ 0 __(m)_______z_____________________(m)_____4z______ OR (z negative): ______m_______4z____________________z_____ 0 _______________________ Both z and m could be positive, both negative or m negative z positive. Not sufficient. Both together, 3z < m < 4z (z positive): ______________________________ 0 ____z_________________3z_____m_______4z___________ OR (z negative): ___________4z________3z___________________z_____ 0 _______________________ m has to be to the left of 4z but right of 3z. Not possible. So m and z both must be positive. So m + z > 0 Answer (C)
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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09 May 2016, 22:52
Thank you so much Karishma. You're a saviour!
Is it possible to visualise such problems rather than drawing on number line? I think that will be prone to errors.
Besides, I have noticed that problem statements concerning number line involves too many possibilities as you explained in detail. For the same reason, I feel these questions are designed to put the test taker in a situation where it consumes more than required time? My gut will always push me to rush on these questions therefore.
Rephrasing the question is vital I guess. Let me know if the only way through such questions is to draw the number line and analyze?
Cheers!
Posted from my mobile device



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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10 May 2016, 20:43
HarisinghKhedar wrote: Thank you so much Karishma. You're a saviour!
Is it possible to visualise such problems rather than drawing on number line? I think that will be prone to errors.
Besides, I have noticed that problem statements concerning number line involves too many possibilities as you explained in detail. For the same reason, I feel these questions are designed to put the test taker in a situation where it consumes more than required time? My gut will always push me to rush on these questions therefore.
Rephrasing the question is vital I guess. Let me know if the only way through such questions is to draw the number line and analyze?
Cheers!
Posted from my mobile device Most questions can be solved using many different methods. You can use the graphical approach discussed by walker here: http://gmatclub.com/forum/ismz75657.htmlAs for too many possibilities on the number line, it seems that way when I draw it out but here is how I see it in my mind. Stmtn 1: There are only 2 cases: z is positive (or 0) or z is negative. If z is to the right of 0, 3z is further to the right of z and m is further to the right of 3z. m has to be positive. If z is to the left of 0, 3z is further to the left of z but m is anywhere on the right of 3z. m could be negative or positive. That's all I need to know.
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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14 Oct 2016, 05:55
smitakokne wrote: Is m+z > 0
(1) m3z > 0 (2) 4zm > 0 We must determine whether the sum of m and z is greater than zero. Statement One Alone: m – 3z > 0 We can manipulate the inequality in statement one to read: m > 3z However, we still cannot determine whether the sum of m and z is greater than zero. Statement one alone is insufficient to answer the question. We can eliminate answer choices A and D. Statement Two Alone: 4z – m > 0 We can manipulate the inequality in statement two to read: 4z > m However, we still cannot determine whether the sum of m and z is greater than zero. Statement two alone is insufficient to answer the question. We can eliminate answer choice B. Statements One and Two Together: Using our statements together, we can add together our two inequalities. Rewriting statement 2 as –m + 4z > 0, we have m – 3z > 0 + (m +4z > 0) z > 0 Since we know that z > 0 and that m > 3z (from statement one), m must also be greater than zero. Thus, the sum of m and z is greater than zero. Answer: C
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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30 Dec 2017, 22:38
In this solution  We have 4z>m... If z>0, we can't say for sure that m is positive.. However, we can say m>0 from the equation when m>3z.. because if z>0, m has to be >0. What am I missing here? Can anyone help please ScottTargetTestPrep wrote: smitakokne wrote: Is m+z > 0
(1) m3z > 0 (2) 4zm > 0 We must determine whether the sum of m and z is greater than zero. Statement One Alone: m – 3z > 0 We can manipulate the inequality in statement one to read: m > 3z However, we still cannot determine whether the sum of m and z is greater than zero. Statement one alone is insufficient to answer the question. We can eliminate answer choices A and D. Statement Two Alone: 4z – m > 0 We can manipulate the inequality in statement two to read: 4z > m However, we still cannot determine whether the sum of m and z is greater than zero. Statement two alone is insufficient to answer the question. We can eliminate answer choice B. Statements One and Two Together: Using our statements together, we can add together our two inequalities. Rewriting statement 2 as –m + 4z > 0, we have m – 3z > 0 + (m +4z > 0) z > 0 Since we know that z > 0 and that m > 3z (from statement one), m must also be greater than zero. Thus, the sum of m and z is greater than zero. Answer: C



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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30 Dec 2017, 22:46
cuhmoon wrote: In this solution 
We have 4z>m...
If z>0, we can't say for sure that m is positive..
However, we can say m>0 from the equation when m>3z.. because if z>0, m has to be >0.
What am I missing here? Can anyone help please Hi cuhmoonyou have answered your own question. By combining Statement 1 & 2 you get z>0 and from Statement 1 you get m>3z>0. Can you tell what is the exact problem you are facing here?



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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31 Dec 2017, 00:26
by St2) 4z>m.. If z is 0, that does not necessarily imply m>0.. m, may or may not be positive. This is where I got stuck niks18 wrote: cuhmoon wrote: In this solution 
We have 4z>m...
If z>0, we can't say for sure that m is positive..
However, we can say m>0 from the equation when m>3z.. because if z>0, m has to be >0.
What am I missing here? Can anyone help please Hi cuhmoonyou have answered your own question. By combining Statement 1 & 2 you get z>0 and from Statement 1 you get m>3z>0. Can you tell what is the exact problem you are facing here?



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0 [#permalink]
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31 Dec 2017, 00:32
cuhmoon wrote: by St2) 4z>m.. If z is 0, that does not necessarily imply m>0.. m, may or may not be positive. This is where I got stuck niks18 wrote: cuhmoon wrote: In this solution 
We have 4z>m...
If z>0, we can't say for sure that m is positive..
However, we can say m>0 from the equation when m>3z.. because if z>0, m has to be >0.
What am I missing here? Can anyone help please Hi cuhmoonyou have answered your own question. By combining Statement 1 & 2 you get z>0 and from Statement 1 you get m>3z>0. Can you tell what is the exact problem you are facing here? Hi cuhmoonYes for Statement 2 that is a possibility. Hence Statement 2 alone is not sufficient to answer the question. But when you combine both the statements then z>0 and even if z=0 (which is not possible in this case) then from Statement 1: m>0




Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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