Got this from practice test 1. Can somebody help?
Is m+z > 0?
(1) m-3z > 0
(2) 4z-m > 0
Thanks a lot.
Is m+z > 0?
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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
if m & z can be positive or negative, then i guess answer is E
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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?
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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Sum the 2 inequalities and we get z > 0.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.
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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I'm curious to see your logic for concluding that Z> 0 when combining both the statemets.
wilderness wrote:Sum the 2 inequalities and we get z > 0.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.
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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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.
i.e. m-3z + 4z - m > 0
which give z > 0
Hope it helps.
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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
but same is not true about substraction and also if sign of one of the inequality is different from the other
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
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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ildude02 has it correct (see a few posts above).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.
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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@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
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
the question asks, IS m+z>0? .... i.e. IS m+z +ve?
this can only happen if either
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
Ashish
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