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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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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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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oss198 wrote:
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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COLLECTION OF QUESTIONS:
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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
Expert's post
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.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0   [#permalink] 07 Sep 2014, 07:56
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