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silivest60
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I'd like some feedback on my solving of this GMATPrep question....I appreciate your comments
Thanks
My solution:
(1) m-3z>0 => there are three cases:
either: (I) m>0 and z>0 and m>3z, in which case answer is YES
or (II) m<0 and z<0 and m<3z, in which case answer is NO
or (III) m>0 and z<0 , in which case answer may be YES or NO
Therefore (1) is insufficient
(2) 4z-m>0 => there are also three cases:
either: (I) m>0 and z>0 and 4z>m, in which case answer is YES
or (II) m<0 and z<0 and 4z<m, in which case answer is NO
or (III) m<0 and z>0 , in which case answer may be YES or NO
Therefore (2) is insufficient
Now, combining (1) and (2) => we only get case (I) m>0 and z>0 and 3z<m<4z, which answers YES
Case (II) m<0 and z<0 does not hold true since statement (1) and (2) contradict each other: (1) says m<3z , and (2) says m>4z
Therefore Choice (C) is the answer.
Thanks
My solution:
(1) m-3z>0 => there are three cases:
either: (I) m>0 and z>0 and m>3z, in which case answer is YES
or (II) m<0 and z<0 and m<3z, in which case answer is NO
or (III) m>0 and z<0 , in which case answer may be YES or NO
Therefore (1) is insufficient
(2) 4z-m>0 => there are also three cases:
either: (I) m>0 and z>0 and 4z>m, in which case answer is YES
or (II) m<0 and z<0 and 4z<m, in which case answer is NO
or (III) m<0 and z>0 , in which case answer may be YES or NO
Therefore (2) is insufficient
Now, combining (1) and (2) => we only get case (I) m>0 and z>0 and 3z<m<4z, which answers YES
Case (II) m<0 and z<0 does not hold true since statement (1) and (2) contradict each other: (1) says m<3z , and (2) says m>4z
Therefore Choice (C) is the answer.
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