GMAT Prep DS Q.

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italian7745
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xcusemeplz2009
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Sat Nov 07, 2009 7:27 am

IMO C statement1) m-3z>0 => m>3z not suff. s2) 4z-m<0=>m>4z => not suff again s2) can be written as m-4z>0 now sub 1 and 2 m-3z-m+4z>0 z>0 since m-3z>0 and z>0=>m>0 hence M+Z>0 using bth suff. Hence C _________________ It does not matter how many times you get knocked down , but how many times you get up

palvarez
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Sat Nov 07, 2009 5:35 pm

italian7745 wrote: Is m + z >0 1) m-3z>0 2) 4z-m<0 1. m > 3z, or m +z > 4z we need to know whether z is +ve. 2. m > 4z or m + z > 5z. Same thing, we need to know whether z is +ve combining together m > 3z & m > 4z case1: m > 3z > 4z, in this case, z is negative. Insufficient, since m + z is greater than a negative number, meaning that m+z can be +ve or -ve. case 2: m > 4z > 3z, z is +ve, m +z is +ve. E is the answer

palvarez
Really wants to Beat The GMAT!

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Sat Nov 07, 2009 5:39 pm

xcusemeplz2009 wrote: IMO C statement1) m-3z>0 => m>3z not suff. s2) 4z-m<0=>m>4z => not suff again s2) can be written as m-4z>0 now sub 1 and 2 m-3z-m+4z>0 z>0 Subtraction doesn't preserve inequalities. a > c b > d can we prove that a - b > c - d ? Nope. a = c + k, k > 0 b = d + l, l > 0 a - b = c - d + (k-l) a -b > c - d only when k > l, which we don't know. However, a + b > c + d since a + b = c + d + (k +l) k +l > 0, therefore a + b > c + d

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