Is m+z > 0?

ksutthi · Posted: Sun Jul 06, 2008 2:21 am Post subject: Is m+z > 0?

Got this from practice test 1. Can somebody help?

Is m+z > 0?

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

Thanks a lot.

szapiszapo · Posted: Sun Jul 06, 2008 4:08 am Post subject:

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

ksutthi · Posted: Sun Jul 06, 2008 5:12 am Post subject:

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.

ildude02 · Posted: Sun Jul 06, 2008 9:35 am Post subject:

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?

wilderness · Posted: Sun Jul 06, 2008 9:44 am Post subject: Re: Is m+z > 0?

ildude02 · Posted: Sun Jul 06, 2008 9:54 am Post subject: Re: Is m+z > 0?

I'm curious to see your logic for concluding that Z> 0 when combining both the statemets.

wilderness · Posted: Sun Jul 06, 2008 3:03 pm Post subject:

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.

ildude02 · Posted: Sun Jul 06, 2008 6:06 pm Post subject:

Thanks.