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

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Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink]  16 Dec 2010, 09:24
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Is m+z > 0

(1) m-3z > 0
(2) 4z-m > 0
[Reveal] Spoiler: OA
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Re: Is m+z > 0 ? [#permalink]  16 Dec 2010, 09:29
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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
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Re: Is m+z > 0 ? [#permalink]  17 Dec 2010, 09:13
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.

Bunuel,

is m+z > 0 the same as m/z > -1 ?

if so, then 1. would be m/z > 3, which is SUFF
and 2. would be m/z < 4, which is INSUFF

so I would have said A

Please correct me if I'm wrong - my inequality skills are a bit rusty
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Re: Is m+z > 0 ? [#permalink]  17 Dec 2010, 09:25
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psirus 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.

Bunuel,

is m+z > 0 the same as m/z > -1 ?

if so, then 1. would be m/z > 3, which is SUFF
and 2. would be m/z < 4, which is INSUFF

so I would have said A

Please correct me if I'm wrong - my inequality skills are a bit rusty

No, it's not correct.

When you are writing m/z>-1 from m+z>0 you are actually dividing both parts of inequality by z: never multiply or reduce (divide) inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

So if we knew that z>0 then m+z>0 --> m/z+1>0 and if we knew that z<0 then m+z>0 --> m/z+1<0.

Hope it's clear.
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Re: Is m+z > 0 ? [#permalink]  17 Dec 2010, 10:32
Thanks Bunuel! That's helps a lot!
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Re: Is m+z > 0 ? [#permalink]  19 Jan 2011, 22:14
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.

Hello Bunnel,

As you said in the above post..
"(1)+(2) Remember we can add inequalities with the sign in the same direction -->"
what should we have different signs...
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Re: Is m+z > 0 ? [#permalink]  20 Jan 2011, 02:19
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jullysabat 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.

Hello Bunnel,

As you said in the above post..
"(1)+(2) Remember we can add inequalities with the sign in the same direction -->"
what should we have different signs...

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.
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Re: DS Inequality question [#permalink]  26 May 2011, 10:10
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Plus in arbitary values to check is m+z is always greater than 0 or there are other possibilities

st-1 ==> m - 3z >0
m =1 and z = -1 ==> m+z = 0
m=2 and z = -1 ==> m+z >0
not sufficient

st-2 ==> 4z-m>0
m = -1 and z = 0.1 ==> m+z < 0
m=2 and z = 1 ==> m+z >0
not sufficient

combining both

m-3z + 4z -m >0
z > 0.... (P)
from 1, m > 3z and from 2, 4z > m
==> 3z < m < 4z... (Q)
from (P) and (Q) m+Z > 0

Answer C.

Last edited by agdimple333 on 26 May 2011, 11:13, edited 1 time in total.
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Re: DS Inequality question [#permalink]  26 May 2011, 11:07
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Wow - Thank you agdimple333. That was quick and clear!

I was combining both to get that 3z < m < 4z but I did not add them to get that z > 0 as well. Now that I think again if m lies between 3z and 4z, then by this equation alone z has to be GT 0 because there is NO z < 0 that will ever satisfy this equation. Thanks again.
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Re: Is m+z > 0 ? [#permalink]  11 Aug 2011, 04:27
Brunel, could you help me to figure out where I am wrong.

From statement 1: M-3Z > 0

Adding 3Z to both sides of inequality

M>3Z

is Z is negative, then 3 times that negative number is definitely less than that number

for example Z = -.25

3Z = -.75 < Z

Z = -1
3Z = - 3 < Z

Zero does not apply because, then both sides would be equal.

Since we its evident that both Z and M are positive Z+M is greater than zero...

Where am I going wrong with my reasoning?
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Re: Is m+z > 0 ? [#permalink]  11 Aug 2011, 07:27
manishgeorge wrote:
Brunel, could you help me to figure out where I am wrong.

From statement 1: M-3Z > 0

Adding 3Z to both sides of inequality

M>3Z

is Z is negative, then 3 times that negative number is definitely less than that number

for example Z = -.25

3Z = -.75 < Z

Z = -1
3Z = - 3 < Z

Zero does not apply because, then both sides would be equal.

Since we its evident that both Z and M are positive Z+M is greater than zero...

Where am I going wrong with my reasoning?

m>3z
m=0; z=-1; m>3z; m+z=-1<0
m=1; z=0; m>3z; m+z=1>0
m=+1; z=-1; m>3z; m+z=0

Thus, knowing that m>3z is not sufficient to find whether m+z>0
Not Sufficient.
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Is m + z > 0? [#permalink]  17 Jan 2012, 19:50
As all the other answers A and B not sufficient.

I line the inequalities up together

3z<m<4z; only way this holds is if all are positive. so C
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Re: prep 1 ds [#permalink]  16 May 2012, 10:23
I'd like you to conform my solution to this problem.
I got this right but not sure if I solved it correctly.

Stm 1) m -3z > 0

m > 3z --- > since both m and n can be negative or positive, it is insufficient.

Stm 2) 4z - m >0

4z > m ------ > again, since both m and n can be negative or positive, it is insufficient.

Together )
m > 3z and m < 4z ---- > 3z < m < 4z ---> if either m or n is negative, this inequality is impossible.
Hence, this stm tells us that both m and n are positive.
In other words, m + n > 0 --------- > Sufficient.
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GMATPrep DS Question: Is m+z>0? (1) m-3z>0 (2) 4z-m>0 [#permalink]  21 Nov 2012, 19:03
Is m+z>0?

(1) m > 3z
(2) m < 4z.

The answer is given as (C) Both statements together.

I don't understand how. Can someone please explain?

I think the answer is (E) Not sufficient with both

(1) tells you that m > 3z (2) tells you that m < 4z. Either of the two cases taken individually is not sufficient. So Rule out (A), (B), (D)

Therefore, 3z<m<4z when you combine the two.

Now z can take both -ve and +ve values. So, m + z can be either -ve or +ve depending on the value of z. Hence, (E).

Please help.

Last edited by Vips0000 on 21 Nov 2012, 19:30, edited 2 times in total.
Edited the question.placed stems properly
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Re: GMATPrep DS Question: Is m+z>0? (1) m-3z>0 (2) 4z-m>0 [#permalink]  21 Nov 2012, 19:34
arvindbhat1887 wrote:
Is m+z>0?

(1) m > 3z
(2) m < 4z.

The answer is given as (C) Both statements together.
I don't understand how. Can someone please explain?
I think the answer is (E) Not sufficient with both
(1) tells you that m > 3z (2) tells you that m < 4z. Either of the two cases taken individually is not sufficient. So Rule out (A), (B), (D)
Therefore, 3z<m<4z when you combine the two.
Now z can take both -ve and +ve values. So, m + z can be either -ve or +ve depending on the value of z. Hence, (E).
Please help.

Stem 1: m>3z
or m-3z >0
doesnt tell us anything about m+z. Not sufficient.

Stem 2: m<4z
or m-4z <0
or 4z-m >0
tells us nothing again. Not sufficient.

combining, we get that 3z <m <4z
also, if we add both equations,
m-3z >0
4z-m >0

we get, z>0
Thus since m>3z
=> m is also >0

therefore m+z >0
Sufficient.

Ans C it is.
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Re: GMATPrep DS Question: Is m+z>0? (1) m-3z>0 (2) 4z-m>0 [#permalink]  21 Nov 2012, 20:06
z must be positive because we know 3z < 4z. Only positive values of z make this statement true.

If m is greater than 3z, it must be positive as well. Therefore, m+z is positive.

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Re: Is m+z > 0 ? [#permalink]  21 Jul 2013, 01:29
Bunuel wrote:

(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 one question when statements are combined, I got till the part z > 0 now I got confused here since in statement 1 I could prove it m>3z ( m is positive) but didn't how to apply z>0 to the second statement >>> 4z - m > 0

So when you reach that stage you can apply it to any of the statement and conclude its sufficient or you have to conclude using both statement separately?
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Re: Is m+z > 0 ? [#permalink]  21 Jul 2013, 01:37
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fozzzy wrote:
Bunuel wrote:

(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 one question when statements are combined, I got till the part z > 0 now I got confused here since in statement 1 I could prove it m>3z ( m is positive) but didn't how to apply z>0 to the second statement >>> 4z - m > 0

So when you reach that stage you can apply it to any of the statement and conclude its sufficient or you have to conclude using both statement separately?

You got that m is positive with (1), so can stop there. If you combine you get that 4z>m>3z>0.
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Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink]  25 Jul 2013, 05:41
smitakokne wrote:
Is m+z > 0

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

1) m-3z>0 insuff
m+z>4z

2) 4z-m>0 insuff
m+z<5z

1)+2) = 4z <m+z< 5z. This inequality holds only if z>0. And if z>0 then m+z>0 hence sufficient
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Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink]  26 Sep 2013, 08:53
I looked at it so:

to rephrase the question: m>(-z) (this is what we are trying to find)

1. m>3z insuff
2. m>4z insuff

combined: 3z<m<4z - suff

is this way good?
am i missing something here?
Re: Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0   [#permalink] 26 Sep 2013, 08:53

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