moneyman wrote:Is m+z>0 ??
(1) m-3z>0
(2)4z-m>0
Fromt this I deduced that 3z<m<4z> 0 or z > 0
If z > 0, m > 3z i.e. m > 0
So m + z > 0
So go for C. Got me, Maxx?
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camitava
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Correct me If I am wrong
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Amitava
Regards,
Amitava
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codesnooker
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Though I have not started working on inequalities, so it is hard for me to understand the above solution posted by Camitava.
However, there is also another way to solve this problem without having much knowledge about inequalities. Here it goes:-
Statement 1: m - 3z > 0 => m > 3z
Statement 2: 4Z - m > 0 => 4z > m
From both statement as you deduce that 4z > m > 3z
Now lets check the solution with negative values to prove that m + z < 0
Lets take z = -1 (simplest one)
i.e -4 > m > -3
which is incorrect as -4 can't greater than -3.
Now lets check with some negative fractional value.
Lets take -1/12 (divisible by both 3 and 4)
therefore equation becomes, -0.33 > m > -0.25, which again incorrect, as -0.33 cannot be greater than -0.25
So, it means z cannot be negative. So, if z is not negative, then m also cannot be negative.
Therefore adding 2 positive number yields a positive number.
i.e. m + z > 0
There C is the answer
However, there is also another way to solve this problem without having much knowledge about inequalities. Here it goes:-
Statement 1: m - 3z > 0 => m > 3z
Statement 2: 4Z - m > 0 => 4z > m
From both statement as you deduce that 4z > m > 3z
Now lets check the solution with negative values to prove that m + z < 0
Lets take z = -1 (simplest one)
i.e -4 > m > -3
which is incorrect as -4 can't greater than -3.
Now lets check with some negative fractional value.
Lets take -1/12 (divisible by both 3 and 4)
therefore equation becomes, -0.33 > m > -0.25, which again incorrect, as -0.33 cannot be greater than -0.25
So, it means z cannot be negative. So, if z is not negative, then m also cannot be negative.
Therefore adding 2 positive number yields a positive number.
i.e. m + z > 0
There C is the answer
Last edited by codesnooker on Mon Mar 31, 2008 5:27 am, edited 3 times in total.
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codesnooker
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Hi Camitava, Can you explain here how you have taken z > 0 ?camitava wrote:
From this I deduced that 3z<m<4z> 0 or z > 0
I didn't understand this point.
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camitava
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codesnooker wrote:Hi Camitava, Can you explain here how you have taken z > 0 ?camitava wrote:
From this I deduced that 3z<m<4z> 0 or z > 0
I didn't understand this point.
Sorry! I really don;t know how did I mess up while posting my last post.
I meant to say that - add the two equs : m-3z>0
and 4z-m>0.
So m - 3z + 4z - m > 0
or z > 0
From this m > 3z or m > 0
So m + z > 0
Got me, Codesnooker?
Correct me If I am wrong
Regards,
Amitava
Regards,
Amitava
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camitava
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Ohhh! Got it now... Last time I forgot to uncheck the option - Disable HTML in this post or may be I messed up in some other way ...
Correct me If I am wrong
Regards,
Amitava
Regards,
Amitava
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codesnooker
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codesnooker
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Yes HTML code is ruining the answers in case of questions of inequalities. Disable them, before posting answers or questions of inequalities
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Stuart@KaplanGMAT
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Just as a general tip, you can turn html off under your profile and then just unclick the box if you ever decide you need it.
As a quick point on this discussion (if it's even needed anymore), Codesnooker got to:
4z > 3z
which can only be true if z is positive.
And, since m > 3z, this means that m is also positive.
So, since both z and m are positive, it must be true that:
m + z
is also positive.
As a quick point on this discussion (if it's even needed anymore), Codesnooker got to:
From here, we can see that:From both statement as you deduce that 4z > m > 3z
4z > 3z
which can only be true if z is positive.
And, since m > 3z, this means that m is also positive.
So, since both z and m are positive, it must be true that:
m + z
is also positive.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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