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Is m+z > 0?

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ksutthi Newbie | Next Rank: 10 Posts Default Avatar
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Is m+z > 0?

Post Sun Jul 06, 2008 1:21 am
Got this from practice test 1. Can somebody help?

Is m+z > 0?

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

Thanks a lot.

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szapiszapo Junior | Next Rank: 30 Posts Default Avatar
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Post Sun Jul 06, 2008 3:08 am
does the wording states that m & z can be either positive or negative?

if m & z can be positive or negative, then i guess answer is E

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ksutthi Newbie | Next Rank: 10 Posts Default Avatar
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Post Sun Jul 06, 2008 4:12 am
m and z can be anything. The question is just like that.

The answer is (C). But I was wondering how I can derive that answer.

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ildude02 Master | Next Rank: 500 Posts Default Avatar
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Post Sun Jul 06, 2008 8:35 am
I solved it by plugging in numbers and it took me a while. I wonder if there is an easier way to solve it and I bet there is. Ian, Stuart or anyone, please share your thoughts with the easier approach . This questions format seems to be common and I would appreciate anyone's input on solving this question.

Now that I thought about it more, this is what I came up with when combining both the statements,

3z < m < 4z; for m and z +ve values, m > -z will always be true;
This equation cannot be solved when we consider negative values for Z. You could see that with Z as -ve value, we get,
-3Z < M < -4Z. There cannot be a value to satify this equation since something greater then -3z will always be greater then -4Z. So, that leaves us wth Z always taking a +ve value and whne Z is +ve, M is always +ve as well. That would mean, M > -Z is always TRUE .

Just wanted to make sure is this a valid assumption for not considering negative values for Z?



Last edited by ildude02 on Sun Jul 06, 2008 8:49 am; edited 2 times in total

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wilderness Master | Next Rank: 500 Posts Default Avatar
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Post Sun Jul 06, 2008 8:44 am
ksutthi wrote:
Got this from practice test 1. Can somebody help?

Is m+z > 0?

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

Thanks a lot.
Sum the 2 inequalities and we get z > 0.
If z>0 then they only way (1) is possible is if m is also positive.
Hence both numbers are positive and thus m+z > 0

So C is correct.

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ildude02 Master | Next Rank: 500 Posts Default Avatar
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Post Sun Jul 06, 2008 8:54 am
I'm curious to see your logic for concluding that Z> 0 when combining both the statemets.

wilderness wrote:
ksutthi wrote:
Got this from practice test 1. Can somebody help?

Is m+z > 0?

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

Thanks a lot.
Sum the 2 inequalities and we get z > 0.
If z>0 then they only way (1) is possible is if m is also positive.
Hence both numbers are positive and thus m+z > 0

So C is correct.

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wilderness Master | Next Rank: 500 Posts Default Avatar
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Post Sun Jul 06, 2008 2:03 pm
If both (1) and (2) are separately greater than 0 then their sum must also be greater than 0.
i.e. m-3z + 4z - m > 0
which give z > 0

Hope it helps.

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ildude02 Master | Next Rank: 500 Posts Default Avatar
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Post Sun Jul 06, 2008 5:06 pm
Thanks.

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aj5105 Legendary Member Default Avatar
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Post Tue Jun 23, 2009 2:52 am
Combine (1) + (2)

z>0. This implies m too is >0.

(C)

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rah_pandey Master | Next Rank: 500 Posts Default Avatar
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Post Tue Jun 23, 2009 4:01 am
Please note a>0 and b>0 =>a+b>0

but same is not true about substraction and also if sign of one of the inequality is different from the other

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Domnu Master | Next Rank: 500 Posts
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Post Thu Jun 25, 2009 2:14 pm
I would just graph it with x = m, y = z and look at the regions.

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El Nino Newbie | Next Rank: 10 Posts Default Avatar
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Post Tue Sep 22, 2009 8:14 am
There are values of m and z fulfilling both conditions, some of which result in m+z>0, and others which result in m+z<0.

E.g. m=-2, z=-1 --> m+z<0
m=2, z=-1 --> m+z<0

Similarly, using the graph method suggested, there are values m+z can be positive or negative.

So, I think the statements are insufficient, answer E.

However, the GMAT software and contributors to this site have answer C.

Can someone pls explain??

Thanks

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Post Tue Sep 22, 2009 8:25 am
ksutthi wrote:
Got this from practice test 1. Can somebody help?

Is m+z > 0?

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

Thanks a lot.
ildude02 has it correct (see a few posts above).

When we combine (1) and (2) we get 3z < m < 4z

Now some people see this and conclude that the value of m lies between 3z and 4z (true). They also conclude that z (and thus m) can be either positive or negative (false).

The only time that 3z<4z is when z is positive. When z is negative, then 3z>4z.
So, in addition to telling us that the value of m is between 3z and 4z, the inequality 3z < m < 4z also tells us that z must be positive.
If z is positive, then m must also be positive, in which case m+z must be positive.

So, the answer is C.

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sashish007 Senior | Next Rank: 100 Posts
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Post Thu Dec 16, 2010 9:10 am
@brent, great explanation. but i was a little apprehensive in combining the two at the beginning and thought of the following solution:

the question asks, IS m+z>0? .... i.e. IS m+z +ve?

this can only happen if either
[i] both m & z are +ve OR
[ii] m and z have dissimilar signs with the +ve no. greater in magnitude than the -ve no.
...... [A]

[1] m>3z
m-3z>0
2 variables - 1 equation - NO soln - INS
(since, no. of equations should equal the no. of variables to solve for the variables!)

[2] 4z-m>0
Again, 2 variables - 1 equation - NO soln - INS

COMBINE 1, 2 i.e. Add 1 & 2,
m-3z>0
-m+4z>0 ... ensure that the direction of inequalities are same
---------
z>0 ... z is +ve;
thus, as per [A] m is +ve or -ve

BUT, m > 3z
so, if z is +ve, then m HAS to be +ve
THUS, m+z IS +ve (>0) hence, SUFF (C)

Ashish

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ankurmit Legendary Member
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Post Sun Dec 19, 2010 8:28 am
Grt man!!

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