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Re M0117 [#permalink]
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16 Sep 2014, 00:15
Official Solution: (1) \(m=m\). First of all \(m=m\), (for example: \(3=3=3\)), so we have \(m=m\), as the right hand side of the equation (RHS) is absolute value which is always nonnegative, then LHS, \({m}\) must also be nonnegative: \(m\geq 0\). Rewrite as: \(m \leq 0\), so \(m\) could be either negative or zero. Not sufficient. (2) \(m^2=9\). Either \(m=3=\text{positive}\) or \(m=3=\text{negative}\). Not sufficient. (1)+(2) Intersection of the values from (1) and (2) is \(m=3=negative\), hence answer to the question "is \(m \lt 0\)" is YES. Sufficient. 1 Answer: C
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Re: M0117 [#permalink]
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28 Nov 2014, 07:59
Bunuel wrote: Official Solution:
(1) \(m=m\). First of all \(m=m\), (for example: \(3=3=3\)), so we have \(m=m\), as the right hand side of the equation (RHS) is absolute value which is always nonnegative, then LHS, \({m}\) must also be nonnegative: \(m\geq 0\). Rewrite as: \(m \leq 0\), so \(m\) could be either negative or zero. Not sufficient. (2) \(m^2=9\). Either \(m=3=\text{positive}\) or \(m=3=\text{negative}\). Not sufficient. (1)+(2) Intersection of the values from (1) and (2) is \(m=3=negative\), hence answer to the question "is \(m \lt 0\)" is YES. Sufficient. 1
Answer: C But 0 is neither +ve nor ve. So why is it m>=0 and not m>0??



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Re: M0117 [#permalink]
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28 Nov 2014, 08:28
bilbo wrote: Bunuel wrote: Official Solution:
(1) \(m=m\). First of all \(m=m\), (for example: \(3=3=3\)), so we have \(m=m\), as the right hand side of the equation (RHS) is absolute value which is always nonnegative, then LHS, \({m}\) must also be nonnegative: \(m\geq 0\). Rewrite as: \(m \leq 0\), so \(m\) could be either negative or zero. Not sufficient. (2) \(m^2=9\). Either \(m=3=\text{positive}\) or \(m=3=\text{negative}\). Not sufficient. (1)+(2) Intersection of the values from (1) and (2) is \(m=3=negative\), hence answer to the question "is \(m \lt 0\)" is YES. Sufficient. 1
Answer: C But 0 is neither +ve nor ve. So why is it m>=0 and not m>0?? Because \(m=m\) holds for 0 too: 0 = 0 = 0.
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Re M0117 [#permalink]
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31 Jan 2015, 15:31
I think this question is good and helpful. Had a different take. If an integer is prefixed by the  (neg) sign, it cannot be zero by definition. Hence 1 should be sufficient to answer this. Can you clarify my thought process.



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Re: M0117 [#permalink]
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Re: M0117 [#permalink]
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01 Feb 2015, 08:12
Bunuel wrote: padhy02 wrote: I think this question is good and helpful. Had a different take. If an integer is prefixed by the  (neg) sign, it cannot be zero by definition. Hence 1 should be sufficient to answer this. Can you clarify my thought process. That's not correct at all. There is nothing wrong in writing 0=0. Hi Bunuel, there still seems to be some confusion in my head. The article on Wikipedia on 'Sign in Mathematics' made me more confused. Hope such ambiguous/debatable concepts are not tested in the GMAT.



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Re: M0117 [#permalink]
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12 Feb 2015, 09:18
padhy02 wrote: Bunuel wrote: padhy02 wrote: I think this question is good and helpful. Had a different take. If an integer is prefixed by the  (neg) sign, it cannot be zero by definition. Hence 1 should be sufficient to answer this. Can you clarify my thought process. That's not correct at all. There is nothing wrong in writing 0=0. Hi Bunuel, there still seems to be some confusion in my head. The article on Wikipedia on 'Sign in Mathematics' made me more confused. Hope such ambiguous/debatable concepts are not tested in the GMAT. You have to be a little bit careful about the logic you're following on this question. From the outset, the only trustworthy information you have is that m is a real number and you're asked to determine if it's less than 0. The logical path you started going down is the false belief that, because of statement (1), m has to be either a negative or positive real number. This is incorrect. What Statement 1 is really saying is, "If you multiply m by 1 it will equal the absolute value of the m multiplied by 1." Which, as Bunuel said, holds true for 0 as well as any negative real number. Hope this helps you. This is not an ambiguous/debatable concept and I strongly recommend practicing similar questions types as you will likely see multiple on a GMAT exam.



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Re: M0117 [#permalink]
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26 Apr 2015, 12:03
Hi Bunuel, I have a doubt. In the absolute value article you have stated the following: x equals x if x>=0 or x if x<0mathabsolutevaluemodulus86462.htmlHowever in this question you have mentioned : m≤0, so m could be either negative or zero Should the first bold be instead  x if x≤0



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Re: M0117 [#permalink]
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27 Apr 2015, 01:39



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Re M0117 [#permalink]
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17 Jul 2016, 14:01
I think this is a highquality question and I agree with explanation.



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Re: M0117 [#permalink]
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05 Aug 2016, 00:15
Hi Bunuel, both (1) and (2) give us the option that m could be (+) or (), then why do we conclude that the intersection is only the negative value?
Thanks for your help.



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Re: M0117 [#permalink]
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05 Aug 2016, 04:34



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Re: M0117 [#permalink]
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05 Aug 2016, 10:17
You're right Bunuel. Error on my part. Thanks for the help.



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Re M0117 [#permalink]
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22 Aug 2016, 06:47
I think this is a highquality question and I agree with explanation. Good question, need to consider every possible scenario



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Bunuel wrote: (1) \(m=m\). First of all \(m=m\), (for example: \(3=3=3\)), so we have \(m=m\), as the right hand side of the equation (RHS) is absolute value which is always nonnegative, then LHS, \({m}\) must also be nonnegative: \(m\geq 0\). Rewrite as: \(m \leq 0\), so \(m\) could be either negative or zero. Not sufficient.
Too me a long time to understand this. Plugging in values makes it clear. For m =  m  we can find: m = 1 1 =  1 1 = 1 Doesn't make sense. m cannot be 1. m = 0 0 =  0 0 = 0. Make sense. m = 1 1 =  1 1 = 1. Makes sense. So for statement 1, m can be 1 or 0. Insufficient.



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Re: M0117 [#permalink]
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11 Oct 2017, 04:24
BunuelAs −m=−m can't we write m as m^2 ? In that case m = m^2 m^2 + m= 0 m(m+1)=0 m could be either 0 or 1. Please let me know the gaps in my understanding. Thanks in advance Bunuel.
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Re: M0117 [#permalink]
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11 Oct 2017, 04:30



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Re: M0117 [#permalink]
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08 Jan 2018, 21:42
Hey Bunuel, When x < 0, I x I = x Similarly I m I = m means that m < 0. Therefore Statement 1 becomes sufficient. I understood how you got m <= 0 but why can't we apply the above logic of Absolute Values to this question?



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Re: M0117 [#permalink]
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