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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]  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.

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]  24 Nov 2013, 08:13
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Expert's post
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.

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]  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.

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]  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.

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.

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]  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]  23 Jul 2014, 17:03
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Expert's post
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.

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Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink]  24 Jul 2014, 09:39
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Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink]  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]  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]  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.
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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.

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.
Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0   [#permalink] 30 Nov 2014, 05:48

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