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# Is a^2*b > 0?

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Is a^2*b > 0? [#permalink]  17 Jun 2007, 20:58
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Is a^2*b > 0?

(1) |a| = b
(2) ab < 0
[Reveal] Spoiler: OA

Last edited by Bunuel on 07 May 2015, 06:37, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: Is a^2*b > 0? [#permalink]  17 Jun 2007, 21:27
stmt 1: |a| = b => b >= 0. INSUFFICIENT as we cannot say if a^2*b >0, it can be equal to 0.

stmt 2: ab<0. INSUFFICIENT.

Either a<0 or b<0. If a<0 a^2*b >0 and if b<0, a^2*b <0.
combining stmt 1 and 2. b>0 and a<0.
Since, a^2 >0, a^2*b > 0. SUFFICIENT.

OA?

Last edited by sumande on 18 Jun 2007, 01:33, edited 1 time in total.
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Re: Is a^2*b > 0? [#permalink]  18 Jun 2007, 00:08
if abs(a)=b then b is always >0
Answer would be A if a and b were not 0
But we never know
So (1) is not suff
(2) is not suff neither

Therefore C
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Re: Is a^2*b > 0? [#permalink]  07 Jul 2008, 11:59
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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: Is a^2*b > 0? [#permalink]  07 Jul 2008, 12:01
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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: Is a^2*b > 0? [#permalink]  07 Jul 2008, 12:04
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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: Is a^2*b > 0? [#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
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings CEO Joined: 29 Mar 2007 Posts: 2589 Followers: 16 Kudos [?]: 238 [0], given: 0 Re: Is a^2*b > 0? [#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. Director Joined: 20 Sep 2006 Posts: 658 Followers: 2 Kudos [?]: 78 [0], given: 7 Re: Is a^2*b > 0? [#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. ANSWER: C VP Joined: 03 Apr 2007 Posts: 1376 Followers: 3 Kudos [?]: 271 [0], given: 10 Re: Is a^2*b > 0? [#permalink] 07 Jul 2008, 17:00 jallenmorris wrote: Q. Is a^2*b > 0? 1) |a|=b 2) ab<0 Allen, please help me understand this : a^2*b = (a^2)* b ? or a^2*b = a^(2* b) ? Senior Manager Joined: 23 May 2006 Posts: 327 Followers: 2 Kudos [?]: 114 [0], given: 0 Re: Is a^2*b > 0? [#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 SVP Joined: 30 Apr 2008 Posts: 1891 Location: Oklahoma City Schools: Hard Knocks Followers: 34 Kudos [?]: 470 [0], given: 32 Re: Is a^2*b > 0? [#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. _________________ ------------------------------------ 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: Is a^2*b > 0? [#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: Is a^2*b > 0? [#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: Is a^2*b > 0? [#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: Is a^2*b > 0? [#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.
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Re: Is a^2*b > 0? [#permalink]  07 May 2015, 12:23
Hello Experts!
Kindly share your views and methods on this question.
Thanks
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Re: Is a^2*b > 0? [#permalink]  07 May 2015, 21:47
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Expert's post
Celestial09 wrote:
Hello Experts!
Kindly share your views and methods on this question.
Thanks
Celestial

Hi Celestial09,

The question is a classic example where students miss out on the concept that $$|x| => 0$$ and not $$|x| > 0$$. Let's analyze the question statement first to see what exactly we need to find.

Analyze the given information in the question
The question asks us if $$a^2*b > 0$$. We know that $$a^2 => 0$$. To answer the question, we need to know for sure if $$a$$ and $$b$$ are not equal to 0 along with the sign of $$b$$. Please note that sign of $$a$$ does not make any difference as $$a^2$$ will never be negative. With this understanding, let's evaluate the statements now.

Analyze statement-I independently
St-I tells us that $$|a| = b$$. Since $$|a| => 0$$, it means that $$b => 0$$. Using this information, $$|a| = b = 0$$ or $$|a| = b > 0$$. Since the statement does not tell us for sure if $$a$$ and $$b$$ are not equal to 0 we can't say if $$a^2 * b > 0$$.

Analyze statement-II independently
St-II tells us that $$ab < 0$$ i.e. $$a$$, $$b$$ have the opposite signs. So,

i. If $$a > 0$$, then $$b < 0$$. This would mean $$a^2 * b < 0$$
ii. If $$a < 0$$, then $$b > 0$$. This would mean that $$a^2 * b > 0$$ as $$a^2$$ is positive irrespective of the sign of $$a$$.

Thus, the statement does not tell us for sure if $$a^2 * b > 0$$.

However, the statement tells us about an important nature of both $$a$$ and $$b$$. Since product of $$a$$ and $$b$$ is not equal to 0, we can say that neither of $$a$$ or $$b$$ is equal to zero. This nature of $$a$$ and $$b$$ would be helpful when we combine the analysis from st-I & II

Combine analysis from st-I & II
St-I tells us that $$|a| = b => 0$$ and st-II tells us that $$b < 0$$ or $$b > 0$$ and $$a$$ not equal to 0.

Combining both the statements, we can say that $$|a| = b > 0$$. Since none of $$a$$ and $$b$$ is equal to $$0$$ and $$b > 0$$, we can safely say that $$a^2 > 0$$ and $$b > 0$$ i.e. $$a^2* b > 0$$.

Hence combining both the statements is sufficient to answer our question.

Hope its clear!

Regards
Harsh
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Re: Is a^2*b > 0? [#permalink]  08 May 2015, 03:01
Expert's post
Is a^2*b > 0?

Since a^2 is a non-negative value (0 or positive), then for a^2*b to be positive a must not be 0 and b must be positive.

(1) |a| = b. Absolute value of a number (|a|) is also a non-negative value (0 or positive), thus this statement implies that $$b\geq{0}$$. Since both a and b can be 0, then this statement is NOT sufficient.

(2) ab < 0. This statement rules out any of the unknowns being 0 but b can be positive as well as negative. Not sufficient.

(1)+(2) Since from (2) $$b\neq{0}$$ then from (1) $$b>0$$. So, we have that neither of a and b is 0 and b is positive, therefore a^2*b > 0. Sufficient.

Hope it's clear.
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Re: Is a^2*b > 0? [#permalink]  08 May 2015, 03:02
Expert's post
Bunuel wrote:
Is a^2*b > 0?

Since a^2 is a non-negative value (0 or positive), then for a^2*b to be positive a must not be 0 and b must be positive.

(1) |a| = b. Absolute value of a number (|a|) is also a non-negative value (0 or positive), thus this statement implies that $$b\geq{0}$$. Since both a and b can be 0, then this statement is NOT sufficient.

(2) ab < 0. This statement rules out any of the unknowns being 0 but b can be positive as well as negative. Not sufficient.

(1)+(2) Since from (2) $$b\neq{0}$$ then from (1) $$b>0$$. So, we have that neither of a and b is 0 and b is positive, therefore a^2*b > 0. Sufficient.

Hope it's clear.

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if-r-s-and-t-are-nonzero-integers-is-r-5-s-3-t-4-negative-165618.html
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Hope it helps.
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Re: Is a^2*b > 0?   [#permalink] 08 May 2015, 03:02
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