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Re: Another DS question....probably not hard...I just don't know [#permalink]
07 Jul 2008, 12:01

1

This post received KUDOS

jallenmorris wrote:

Q. Is a^2*b > 0? 1) |a|=b 2) ab<0

1. Insufficient: because a can be either 0 or any other number (+ve or -ve) 2. Insufficient: because ab < 0 gives us that one of them is -ve but we don't know which.

Combined: Sufficient: because from 1 & 2, b can't be -ve so a should be -ve and so a^2*b is > 0 It is C.

Re: Another DS question....probably not hard...I just don't know [#permalink]
07 Jul 2008, 12:23

maratikus wrote:

jallenmorris wrote:

Q. Is a^2*b > 0? 1) |a|=b 2) ab<0

1: insufficient a = b = 1 or a = b = 0 (b >=0) 2: insufficient a = 1, b = -1 or a = -1, b = 1 (a, b are not equal to 0)

1&2: since b >= 0 and b is not equal to zero -> b > 0, a is not equal to 0 -> a^2*b > 0 -> C

With respect to 1 & 2...You're saying neither A nor B can = 0 because anything * 0 = 0 and we're told in 2) that ab <0 (which is not zero). Then with 1, b = |a| and the absolute value of A cannot be nagative. So we know from the 2 statements that B is not 0, and it is not negative. That makes a be negative and not zero (not zero from #2).

Thanks. That helps me out a bunch. +1 for you. _________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Re: Another DS question....probably not hard...I just don't know [#permalink]
07 Jul 2008, 13:51

Q. Is a^2*b > 0? 1) |a|=b 2) ab<0

From 1)

b must be possitive. a's value doesn't matter because a^2 is anyways going to be positive. So far 1) is suff but ... we dont know if a is a non zero or not. hence 1) is not suff.

From 2)

either a or b has to be negative. And neither a or b is zero. if a is -ve and b is +ve then a^2*b > 0 if a is +ve and b is -ve then a^2*b < 0

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