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Is ab > ab? [#permalink]
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27 Oct 2013, 15:46
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47% (01:36) correct 53% (01:20) wrong based on 294 sessions
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Is ab > ab? 1) a > 1/b 2) a + b > 0
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Re: Is ab > ab? [#permalink]
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Updated on: 27 Oct 2013, 16:54
Is ab > ab?
> Is ab + ab > 0 ?
Notice that this is true only when the product ab is positive. (If ab is negative, ab + ab = 0, which is not greater than 0. If ab is equal to zero, ab + ab = 0, which is not greater than 0.)
Thus, what we really want to know...
Is ab > 0 ?
1) a > 1/b If a = 2 and b = 1, the answer is YES. If a = 2 and b = 1, the answer is NO. > Insufficient.
2) a + b > 0 This can be shown to be insufficient in the same manner; indeed, with the very same numbers. If a = 2 and b = 1, the answer is YES. If a = 2 and b = 1, the answer is NO. > Insufficient.
Thus, even combined the statements are insufficient. Answer E.
Originally posted by Reinfrank2011 on 27 Oct 2013, 16:36.
Last edited by Reinfrank2011 on 27 Oct 2013, 16:54, edited 1 time in total.



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Re: Is ab > ab? [#permalink]
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27 Oct 2013, 16:44
Awesome, thanks for that explanation, it makes a lot more sense now



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Re: Is ab > ab? [#permalink]
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Updated on: 31 Oct 2013, 10:20
Reinfrank2011 wrote: Is ab > ab?
> Is ab + ab > 0 ?
Notice that this is true only when the product ab is positive. (If ab is negative, ab + ab = 0, which is not greater than 0. If ab is equal to zero, ab + ab = 0, which is not greater than 0.)
Thus, what we really want to know...
Is ab > 0 ?
1) a > 1/b If a = 2 and b = 1, the answer is YES. If a = 2 and b = 1, the answer is NO. > Insufficient.
2) a + b > 0 This can be shown to be insufficient in the same manner; indeed, with the very same numbers. If a = 2 and b = 1, the answer is YES. If a = 2 and b = 1, the answer is NO. > Insufficient.
Thus, even combined the statements are insufficient. Answer E. Based on statement 1 however, can't you convert the equation a > 1/ b to a* b>1? Thanks in advance.
Originally posted by TheKraken on 31 Oct 2013, 09:33.
Last edited by TheKraken on 31 Oct 2013, 10:20, edited 1 time in total.



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Re: Is ab > ab? [#permalink]
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31 Oct 2013, 10:13
but if b < 0 then that inequality won't hold as the inequality sign will flip



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Re: Is ab > ab? [#permalink]
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Updated on: 31 Oct 2013, 10:31
SleeB wrote: but if b < 0 then that inequality won't hold as the inequality sign will flip The sign flip is only upon multiplying and dividing by a negative number. Therefore I think OA should be A. Is ab > ab? 1) a > 1/b ** ab>1 both AB has to positive or AB is two negative (3 negatives = Negative). Therefore, AB 2) a + b > 0 Not sufficient as we don't know what values A or B are.
Originally posted by TheKraken on 31 Oct 2013, 10:21.
Last edited by TheKraken on 31 Oct 2013, 10:31, edited 1 time in total.



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Re: Is ab > ab? [#permalink]
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31 Oct 2013, 10:29



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Re: Is ab > ab? [#permalink]
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31 Oct 2013, 10:33
Bunuel wrote: TheKraken wrote: SleeB wrote: Is ab > ab?
1) a > 1/b
2) a + b > 0
but if b < 0 then that inequality won't hold as the inequality sign will flip The sign flip is only upon multiplying and dividing by a negative number. That's the point. Do you know whether b is negative or positive? Do you know whether you should flip the sign of the inequality when multiplying by b? Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign.So you CANNOT multiply a>1/b by b since you don't know the sign of b: if b>0, then you'll have ab>1 but if b<0, then you'll have ab<1 (flip the sign when multiplying by negative value). Hope it's clear. Thanks, Bunuel! The helps a lot.



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Re: Is ab > ab? [#permalink]
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28 May 2014, 12:51
I think that tough part is to be able to rephrase the question asked
Is ab > ab = ab+ab > 0
Is ab<0?
We need to know if a,b have different signs
1) a>1/b, Insufficient
2) a>b, Again could be either positive or negative
Both together, Still insufficient
Answer: E



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Re: Is ab > ab? [#permalink]
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29 May 2014, 05:04
jlgdr wrote: I think that tough part is to be able to rephrase the question asked It is almost always easier to evaluate the behavior of an inequality or equation when everything is moved to one side of the arrow or equal sign. When in doubt, get it all on one side, especially when you're dealing with inequalities in DS. There are exceptions, of course, but as a rule of thumb it is useful. Posted from my mobile device



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Is ab > ab? [#permalink]
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27 Aug 2014, 11:17
SleeB wrote: Is ab > ab?
1) a > 1/b
2) a + b > 0 What is the flaw in my approach ? Experts kindly assist. The question is transformed to (ab)^2+ab>0 => ab ( ab+1) > 0 Hence i need to find if ab is +ve or ab > 0
case 1: Now with a> 1/b ...By cross multiplying, this is transformed to ab > 1 (Cross Multiply) Hence this is sufficient case 2: a+b > 0 means ab is necessarily > 0.Hence this is sufficient
Hence i selected option D, which is not the answer.
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Re: Is ab > ab? [#permalink]
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27 Aug 2014, 11:20



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Is ab > ab? [#permalink]
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27 Aug 2014, 14:16
Bunuel wrote: dshuvendu wrote: SleeB wrote: Is ab > ab?
1) a > 1/b
2) a + b > 0 What is the flaw in my approach ? Experts kindly assist. The question is transformed to (ab)^2+ab>0 => ab ( ab+1) > 0 Hence i need to find if ab is +ve or ab > 0
case 1: Now with a> 1/b ...By cross multiplying, this is transformed to ab > 1 (Cross Multiply) Hence this is sufficient case 2: a+b > 0 means ab is necessarily > 0.Hence this is sufficient
Hence i selected option D, which is not the answer. How did you get (ab)^2+ab>0 ? I have committed a crime now.I missed the square root. I am in a deadlock after the correction, cannot proceed with the transformation As always thanks for the poke .
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Is ab > ab? [#permalink]
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11 Apr 2017, 08:00
When we combine both equations :
a>(1/b) > a(1/b) > 0 (1)
a+b>0 (2)
subracting 1 and 2
a(1/b)>0  a+ b>0 ___________ b(1/b)>0 (1+b^2)/(b) >0
(1+b^2)/(b)<0 (3)
From equation 3 we can say that b is negative since numerator is +ve
hence b<0
in order to equation 1 and 2 to be valid ; a should be positive and greater than b
hence we can say a>0
so a>0 and b<0 ; now we know a and b have opposite signs. Hence C should be sufficient.
Please correct where I am going wrong....










