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

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Intern
Joined: 08 Jun 2009
Posts: 33
Is m + z > 0? (1) m 3z > 0 (2) 4z m > 0 Clear [#permalink]

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16 Jun 2009, 07:28
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Is m + z > 0?

(1) m – 3z > 0
(2) 4z – m > 0

Current Student
Joined: 13 May 2008
Posts: 141
Schools: LBS

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16 Jun 2009, 08:01
Jozu wrote:
Is m + z > 0?

(1) m – 3z > 0
(2) 4z – m > 0

Both together are sufficient but neither alone is enough ... examine (1) which essentially says m > 3z which doesn't suffice the eliminate the possibility that both m & z are negative or positive .. suppose m=-1 and z = -3... which would say -1> -9 or m=10 and z =1 where 10>3 ... so eq1 not enough

same issue with eq2 only vice versa here... but if you combine both equations you can see that only when z is a positive number the two eqs can be satisfied and any positive number for z would create a upper and lower bound for m.. thereby ensuring that both z and m are positive and hence m+z>0...
Manager
Joined: 28 Jan 2004
Posts: 201
Location: India

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21 Jun 2009, 22:34
Stmt. 1
only tells that m>3z not suff. as we dont know the signs of M and Z.
Insuff

Stmt. 2
tell m<4Z. insuff.

Combine..
m-3z + 4z - m > 0+0
z>0 or z is +ve.
If z is +ve M has to be +ve because it is >3Z.

Hence M+Z is +ve or >0.
Director
Joined: 04 Jan 2008
Posts: 893

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22 Jun 2009, 01:47
yes combining both

we get z>0

hence m is also positive

hence C

mdfrahim wrote:
Stmt. 1
only tells that m>3z not suff. as we dont know the signs of M and Z.
Insuff

Stmt. 2
tell m<4Z. insuff.

Combine..
m-3z + 4z - m > 0+0
z>0 or z is +ve.
If z is +ve M has to be +ve because it is >3Z.

Hence M+Z is +ve or >0.

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Senior Manager
Joined: 23 Jun 2009
Posts: 357
Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago

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23 Jun 2009, 06:05
IMO C.

Gathering both inequalities together, we can do it.

It is a problem in GmatPrep1.
Manager
Joined: 16 Apr 2009
Posts: 230
Schools: Ross

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23 Jun 2009, 08:09
IMO, it's C and the explanations are good..
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Manager
Joined: 30 Nov 2008
Posts: 91

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03 Jul 2009, 07:50
Guys,

I got C also, but I substituted z=x and y=m and solved it graphically? Could you do it this way?
Re: inequality   [#permalink] 03 Jul 2009, 07:50
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