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Is a^2*b > 0?
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Updated on: 03 Jun 2015, 05:51
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46% (01:22) correct 54% (01:04) wrong based on 313 sessions
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Is \(a^2*b > 0\)? (1) \(a = b\) (2) \(ab < 0\)
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Originally posted by iamba on 17 Jun 2007, 20:58.
Last edited by Harley1980 on 03 Jun 2015, 05:51, edited 4 times in total.
added formatting



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Re: Is a^2*b > 0?
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Updated on: 18 Jun 2007, 01:33
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.
Therefore, answer: C
OA?
Originally posted by sumande on 17 Jun 2007, 21:27.
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?
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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?
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07 Jul 2008, 11:59
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?
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07 Jul 2008, 12:01
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?
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07 Jul 2008, 12:04
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?
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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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Re: Is a^2*b > 0?
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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: Is a^2*b > 0?
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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



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Re: Is a^2*b > 0?
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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) ?



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Re: Is a^2*b > 0?
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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: Is a^2*b > 0?
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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: Is a^2*b > 0?
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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?
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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 Please correct me if i made a mistake.



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Re: Is a^2*b > 0?
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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?
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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 nonnegative 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.



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Re: Is a^2*b > 0?
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07 May 2015, 12:23
Hello Experts! Kindly share your views and methods on this question. Thanks Celestial



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Re: Is a^2*b > 0?
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07 May 2015, 21:47
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 questionThe 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 statementI independentlyStI 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 statementII independentlyStII 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 stI & II Combine analysis from stI & IIStI tells us that \(a = b => 0\) and stII 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?
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08 May 2015, 03:01



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Re: Is a^2*b > 0?
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