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Source: — Data Sufficiency |
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by pemdas » Wed Feb 22, 2012 3:53 pm
dreamv wrote:Is m+z>0?

1) m-3z>0
2) 4z-m>0
the easiest way is to pick values for plug-in
st(1) m=4,z=-4 ---> 4+12>0 and m+z=4-4=0 answer is No; m=4,z=1 ---> 4-3>0 and m+z=4+1>0 answer is Yes, Not Sufficient
st(2) m=-4,z=4 ---> 16+4>0 and m+z=-4+4=0 answer is No; m=-4,z=1 ---> 4+4>0 and m+z=-4+1>0 answer is Yes, Not Sufficient
Combining st(1&2): adding two inequalities (m-3z)+(4z-m)>0 we obtain z>0 and adding 3z>0 to (m-3z)>0 we get m>0, hence m+z>0 Sufficient

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by Anurag@Gurome » Wed Feb 22, 2012 7:52 pm
dreamv wrote:Is m+z>0?

1) m-3z>0
2) 4z-m>0
(1) m - 3z > 0
If m = 1, z = -1, then m + z = 0
If m = 2 and z = -1, then m + z > 0
No definite answer; NOT sufficient.

(2) 4z - m > 0
If m = -1, z = 0, then m + z < 0
If m = 2 and z = 1, then m + z > 0
No definite answer; NOT sufficient.

Combining (1) and (2), add the inequalities, m - 3z > 0 and 4z - m > 0
z > 0, which implies z is always positive. This implies that m is also a positive integer, as from statement 1 we get that m > 3z. This implies that m + z > 0; SUFFICIENT.

The correct answer is C.
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