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

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Re: Is m+z > 0 ? [#permalink] New post 24 Nov 2013, 08:10
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.

Answer: C.

For graphic approach refer to: is-m-z-0-1-m-3z-0-2-4z-m-75657.html

Hi Bunuel,

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?
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Re: Is m+z > 0 ? [#permalink] New post 24 Nov 2013, 08:13
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ronr34 wrote:
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.

Answer: C.

For graphic approach refer to: is-m-z-0-1-m-3z-0-2-4z-m-75657.html

Hi Bunuel,

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.
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Re: Is m+z > 0 ? [#permalink] New post 30 Nov 2013, 00:34
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.

Answer: C.

For graphic approach refer to: is-m-z-0-1-m-3z-0-2-4z-m-75657.html




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.
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Re: Is m+z > 0 ? [#permalink] New post 30 Nov 2013, 03:07
Expert's post
mohnish104 wrote:
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.

Answer: C.

For graphic approach refer to: is-m-z-0-1-m-3z-0-2-4z-m-75657.html




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.


Please read the whole thread before posting:
is-m-z-0-1-m-3z-0-2-4z-m-106381.html#p1248558
is-m-z-0-1-m-3z-0-2-4z-m-106381-20.html#p1296461
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Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink] New post 23 Jul 2014, 12:15
Is that correct?

1) \(m-3z>0\) <=> \(m>3z\) Insuff
2) \(4z-m>0\) <=> \(m<4z\) Insuff

1+2)
we multiply 1) by 4, \(4m>12z\)
we multiply 2) by 3, \(3m<12z\) <=> \(-3m>-12z\)

As we can add two inequalities \(4m-3m>0\) so \(m>0.\)

from1), \(m>3z\)
from 2), \(m<4z\) <=> \(-m>-4z\)

As we can add two inequalities \(0>-z\) so \(z>0\)

so \(m+z>0\)
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Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink] New post 23 Jul 2014, 17:03
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Is that correct?

1) \(m-3z>0\) <=> \(m>3z\) Insuff
2) \(4z-m>0\) <=> \(m<4z\) Insuff

1+2)
we multiply 1) by 4, \(4m>12z\)
we multiply 2) by 3, \(3m<12z\) <=> \(-3m>-12z\)

As we can add two inequalities \(4m-3m>0\) so \(m>0.\)

from1), \(m>3z\)
from 2), \(m<4z\) <=> \(-m>-4z\)

As we can add two inequalities \(0>-z\) so \(z>0\)

so \(m+z>0\)


Yes, that's correct.

You might find the following post helpful: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1379270 (Tips about inequalities).
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Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink] New post 24 Jul 2014, 09:39
thank you for the link!
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Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink] New post 28 Jul 2014, 00:37
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
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Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink] New post 07 Sep 2014, 07:41
Hi Bunuel, I have a doubt in this question. it is asked whether m+z>0 this implies that is m>-z?

Statement 1 says: m-3Z>0 then m>3z and 3z is definitely greater than -z then isnt m>-z. Please clarify.
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Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink] New post 07 Sep 2014, 07:56
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snehamd1309 wrote:
Hi Bunuel, I have a doubt in this question. it is asked whether m+z>0 this implies that is m>-z?

Statement 1 says: m-3Z>0 then m>3z and 3z is definitely greater than -z then isnt m>-z. Please clarify.


3z is not always greater than -z. It's only true when z is positive, but if z is negative, then 3z < -z.
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Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink] New post 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.

Answer: C.

For graphic approach refer to: is-m-z-0-1-m-3z-0-2-4z-m-75657.html



bunuel,
i have a doubt here...

z > 0, plugging this in statement 1 gives us m > 0

but plugging this is statement 2,
4z > m , if z=1, m=0, 4 > 0, m+z > 0 yes
but when z = 1, m = -1, 4 > -1, but m + z is not greater than 0. NO

in that case, how is C the answer.
in conditions like these, do we plug in either of the statemnts and check.

would really appreciate your help on this.
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Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink] New post 15 Nov 2015, 06:21
from I we get as follows

m | z | m-3z | m+z
---------------------------
4 | 1 | 1 | 5 ---- this case true
-1 | -8 | 23 | -9 ---- this case fails


from II we get as follows

z | m | 4z-m | m+z
---------------------------
4 | 1 | 15 | 5 ---- this case true
-1 | -8 | 8 | -9 ---- this case fails

1+2 also both values give yes and no
so option is E
If i am wrong please correct me
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Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink] New post 15 Nov 2015, 08:22
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ashokingmat wrote:
from I we get as follows

m | z | m-3z | m+z
---------------------------
4 | 1 | 1 | 5 ---- this case true
-1 | -8 | 23 | -9 ---- this case fails


from II we get as follows

z | m | 4z-m | m+z
---------------------------
4 | 1 | 15 | 5 ---- this case true
-1 | -8 | 8 | -9 ---- this case fails

1+2 also both values give yes and no
so option is E
If i am wrong please correct me


The values are reversed in your examples.
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Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink] New post 23 Dec 2015, 17:17
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
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Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink] New post 23 Dec 2015, 17:50
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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] New post 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.
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Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink] New post 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.
Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0   [#permalink] 24 Dec 2015, 08:48

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