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(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
Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink]
30 Nov 2014, 05:48
Bunuel wrote:
smitakokne wrote:
Is m+z > 0
1. m-3z > 0 2. 4z-m > 0
OA : C
Need help in underdstanding how we arrive at C.
(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. _________________
Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink]
23 Dec 2015, 22:28
Expert's post
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. _________________
Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink]
24 Dec 2015, 08:48
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, 08:48
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