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

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Q. Is a^2*b > 0? 1) |a|=b 2) ab<0 [#permalink]  07 Jul 2008, 11:48
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Q. Is a^2*b > 0?
1) |a|=b
2) ab<0
_________________

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

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Re: Another DS question....probably not hard...I just don't know [#permalink]  07 Jul 2008, 11:59
1
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According to me, the answer is C.

1) |a| = b, which means that b should be positive.
2) says that one of a or b should be negative.

From 1 we know that "b" is positive hence "a" should be negative and we know that a^2*b < 0.
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Re: Another DS question....probably not hard...I just don't know [#permalink]  07 Jul 2008, 12:01
1
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.
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Re: Another DS question....probably not hard...I just don't know [#permalink]  07 Jul 2008, 12:04
1
KUDOS
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
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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.
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J Allen Morris
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Re: Another DS question....probably not hard...I just don't know [#permalink]  07 Jul 2008, 13:30
jallenmorris wrote:
Q. Is a^2*b > 0?
1) |a|=b
2) ab<0

1: a can be 0. Insuff.

2: b could be -. INsuff.

Together, suff b/c a is not 0.
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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

therefore not sufficient

TOGETHER:

a and b not equal to zero.
b is +ve
a is -ve

Sufficient.

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Re: Another DS question....probably not hard...I just don't know [#permalink]  07 Jul 2008, 17:00
jallenmorris wrote:
Q. Is a^2*b > 0?
1) |a|=b
2) ab<0

a^2*b = (a^2)* b ?
or
a^2*b = a^(2* b) ?
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Re: Another DS question....probably not hard...I just don't know [#permalink]  07 Jul 2008, 18:28
jallenmorris wrote:
Q. Is a^2*b > 0?
1) |a|=b
2) ab<0

1. If a^(2b) then answer is C

2. If (a^2)b then answer is E

Please tell us which is the case - 1 or 2
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Re: Another DS question....probably not hard...I just don't know [#permalink]  07 Jul 2008, 19:12
I should not have copied and pasted, I should have used the math functionality provided by the forums.

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

a^2*b = (a^2)* b ?
or
a^2*b = a^(2* b) ?

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Re: Another DS question....probably not hard...I just don't know [#permalink]  07 Jul 2008, 19:57
Well, C is the answer if a and b are integers.
Wouldn't the result be a complex number if a negative a were raised to a decimal power ? So E ????
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Re: Another DS question....probably not hard...I just don't know [#permalink]  07 Jul 2008, 20:34
jallenmorris wrote:
Is a^2*b > 0?
1) |a|=b
2) ab<0

Is (a^2)*b > 0?

edit, C, forgot to consider that a can be 0
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Re: Another DS question....probably not hard...I just don't know [#permalink]  07 Jul 2008, 22:28
jallenmorris wrote:
Q. Is a^2*b > 0?
1) |a|=b
2) ab<0

good one. go for C as a in (1) could be 0.
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Re: Another DS question....probably not hard...I just don't know [#permalink]  08 Jul 2008, 03:56
I've decided that when I get a DS like this, I'm going to consider a few numbers. I'm going to start with the following set:

{-2, -1, -0.5, 0, 0.5, 1, 2}

That should give an good indication as to how the inequality acts with both + and - numbers.
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Re: Another DS question....probably not hard...I just don't know [#permalink]  08 Jul 2008, 07:35
C

1. a>0,b>0 yes
a<0,b> 0 yes
a= b =0 no

insuff

2.a>0 b<0 no
a<0 b>0 yes

insuff

from 1 and 2
a< 0 b >0
Suff

C
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Re: Another DS question....probably not hard...I just don't know [#permalink]  10 Jul 2008, 02:36
x97agarwal wrote:
jallenmorris wrote:
Q. Is a^2*b > 0?
1) |a|=b
2) ab<0

1. If a^(2b) then answer is C

2. If (a^2)b then answer is E

Please tell us which is the case - 1 or 2

IMO for both cases viz. a^(2b) or (a^2)b the answer is C.

the solution:
a != 0 and b != 0 (a&b not equal to 0) from ab<0
& b>0 from |a|=b ---1
=> a<0
=> a^(2b)>0 ---2
& a^2>0 ---3

hence for the case
1. a^(2b) >0 from ---2
2. (a^2)b > 0 from ---1 & 3

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Re: Another DS question....probably not hard...I just don't know [#permalink]  10 Jul 2008, 09:50
assuming we are working with a^(2b) > 0, and the following assumptions

B is not negative, B = |A|
A is negative since AB< 0

use -2, and -.5 for A. and 2,.5 for B respectively.

A = -2: (-2)^(2*2) = 16
A = -.5: (-.5)^(2*.5) = -.5

answers on both side of 0.....get E
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Re: Another DS question....probably not hard...I just don't know [#permalink]  13 Jul 2008, 10:02
To prove, we need:
a) find out if a or b or both are zero?
b) Whether b is +ve or -ve.

1) b = |a| means b is non-negative i.e. either +ve or 0 - Insufficient.

2) ab<0 means a & b are not 0. but either a or b is -ve

Combining: we can say that b is +ve and a is -ve.
C.
Re: Another DS question....probably not hard...I just don't know   [#permalink] 13 Jul 2008, 10:02
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