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Is \(M \lt 0\)? 1. \(M = M\) 2. \(M^2 = 9\) Source: GMAT Club Tests  hardest GMAT questions Answer explaination: From statement (1) we have that should be negative or zero. Not enough for the strict condition. From statement (2) can be 3 or 3. Not enough again. If we combine two statements then .Could somebody please explain how can we ever have modulus of any number as negative ? I beleive modulus of a positive number , negative number or zero is positive.... So isnt statement one completely false ? From statement 2 we can infer that M can be either negative or positive .... but cannot answer the question... What should be the answer ?



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Re: m01  Q17  Wrong answer [#permalink]
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25 Sep 2008, 13:19
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Wait. Let's see. S1 states M = M Now we plug in 3 as you suggested: (3) = (3) > 3 = 3. Now we plug in a positive number 3: (3) = (3) > 3 = 3. This is why we know that S1 tells us the \(M \le 0\). Do you agree? Saw the post made but still decided to proceed with mine.
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Re: m01  Q17  Wrong answer [#permalink]
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25 Sep 2008, 13:18
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elmagnifico wrote: dzyubam wrote: Hi. First, modulus of zero is neither positive nor negative. Second, try plugging a negative number in S1. You're right, the M part is always positive, but the lefthand part isn't. That is why we need both S1 and S2 to answer the question. From S2 we know that M is either 3 or 3. From S1 we know that M is either negative or zero. combining the two Statements we get M=3. Answer C.
Hope this helps. how is S1 possible to be negative. i understand that modulus of zero will always be zero, but the first statement doesn't allow for negative result. for example lets plug 3 in S1. 3=3 , thus 3=3 but this is not true! so S1 tells me that M=0, what am i missing here? you took M=3 (+VE) thats not possible M must be ve here.. M=3 (3) = (3) > 3=3
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Re: m01  Q17  Wrong answer [#permalink]
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18 Sep 2008, 06:35
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Hi. First, modulus of zero is neither positive nor negative. Second, try plugging a negative number in S1. You're right, the M part is always positive, but the lefthand part isn't. That is why we need both S1 and S2 to answer the question. From S2 we know that M is either 3 or 3. From S1 we know that M is either negative or zero. combining the two Statements we get M=3. Answer C. Hope this helps.
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Re: m01  Q17  Wrong answer [#permalink]
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06 Nov 2008, 15:47
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S1 is insufficient by itself. Consider M = 3 and M = 0. We have a strict inequality in the question, so we'll need S2. mbaobsessed wrote: I still didn't get this.
Isn't 3 = 3 invalid ?? we cannot even apply the values as such to arrive at the answer.
For me only answer that fits value of M is ve values from S1.
M = M this can be true only if M is ve right ?? hence sufficient ?

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Re: m01  Q17  Wrong answer [#permalink]
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25 Sep 2008, 14:28
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dzyubam wrote: Wait. Let's see. S1 states M = M Now we plug in 3 as you suggested: (3) = (3) > 3 = 3. Now we plug in a positive number 3: (3) = (3) > 3 = 3. This is why we know that S1 tells us the \(M \le 0\). Do you agree?
Saw the post made but still decided to proceed with mine. thanks for your post. both, you and x2suresh deserve a tap on the shoulder. (and kudos:)



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Re: m01  Q17  Wrong answer [#permalink]
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06 Nov 2008, 22:16
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Also look at stmt1 this way.
M is an absolute value and hence >=0.
Left side of the equation is M.
Thus, we are comparing M to some value that is >=0. This is possible only in two cases Case1: when M = 0 Case2, when m<0.
Hence, statement 1 gives M <=0 and is not sufficient to answer the question.



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Re: m01 Q17 [#permalink]
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02 Dec 2010, 06:24
itiskavikatha wrote: Is \(M \lt 0\)?
1. \(M = M\) 2. \(M^2 = 9\) This question was also posted in DS subforum. Below is my solution from there: Is \(m<0\)?(1) \(m=m\) > first of all \(m=m\), (for example: \(3=3=3\)), so we have \(m=m\), as RHS is absolute value which is always nonnegative, then LHS, \({m}\) must also be nonnegative > \(m\geq{0}\) > \(m\leq{0}\), so \(m\) could be either negative or zero. Not sufficient. (2) \(m^2=9\) > \(m=3=positive\) or \(m=3=negative\). Not sufficient. (1)+(2) Intersection of the values from (1) and (2) is \(m=3=negative\), hence answer to the question "is \(m<\)0" is YES. Sufficient. Answer: C. Hope it's clear.
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Re: m01  Q17  Wrong answer [#permalink]
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25 Sep 2008, 13:10
dzyubam wrote: Hi. First, modulus of zero is neither positive nor negative. Second, try plugging a negative number in S1. You're right, the M part is always positive, but the lefthand part isn't. That is why we need both S1 and S2 to answer the question. From S2 we know that M is either 3 or 3. From S1 we know that M is either negative or zero. combining the two Statements we get M=3. Answer C.
Hope this helps. how is S1 possible to be negative. i understand that modulus of zero will always be zero, but the first statement doesn't allow for negative result. for example lets plug 3 in S1. 3=3 , thus 3=3 but this is not true! so S1 tells me that M=0, what am i missing here?



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Re: m01  Q17  Wrong answer [#permalink]
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06 Nov 2008, 14:45
I still didn't get this.
Isn't 3 = 3 invalid ?? we cannot even apply the values as such to arrive at the answer.
For me only answer that fits value of M is ve values from S1.
M = M this can be true only if M is ve right ?? hence sufficient ?




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Re: m01 Q17 [#permalink]
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06 Jul 2010, 06:09
Initially i thought answer to be A but,looking at the explanation by both dzyubam and x2suresh, C seems to be correct answer. Kudos to both of you.
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Re: m01 Q17 [#permalink]
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06 Jul 2010, 13:44
nice explanations thanks. I also thought A at first but you need the information from B



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Re: m01 Q17 [#permalink]
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08 Jul 2010, 02:07
From S1 we can say M>=0 From S2 we can say M=+3,3 so Ans:M=3
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Re: m01 Q17 [#permalink]
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02 Dec 2010, 05:37
The correct answer is C. (1) is not sufficient because M = M means M = 0 or M < 0. For example: 0 = 0 : correct (1) = (1) : correct M > 0 is impossible because if M > 0, then M < 0 but M is nonnegative.
(2) is not sufficient because M = 3 or M = 3.
(1) & (2) together: sufficient because M = 3, so the answer to the question is YES.



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Re: m01 Q17 [#permalink]
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08 Jul 2011, 05:59
I initially considered S1 as enough. Apparently not. I didn't consider the case when M = 0. I always make this mistake!
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Re: m01 Q17 [#permalink]
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11 Jul 2011, 09:04
Beware! This is one of the tricks gmat usually pulls on test takers.
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Re: m01 Q17 [#permalink]
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14 Jan 2012, 22:16
Was in trap  forgot to check for zero. Rest Bunnel explains as it



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Re: m01 Q17 [#permalink]
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14 Jul 2012, 08:13
man...silly mistake! chose A as i overlooked M=0 scenario. C is correct.



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Re: m01 Q17 [#permalink]
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12 Jul 2013, 07:04
Post going through all DS questions from OG, I have noticed that most of the DS questions are like this. Mostly, the tricky ones can be classified in the following buckets: (its not an extensive list)
1) Post deduction and simplification ,both equations are same or one of the equation is an information set that was already present in the question itself 2) Missing out the possibility of 0 as both being positive and negative 3) 45 variables being removed by use of two equations only (GMAT gives questions having variables more than the number of equations, but the even those equations are linked in such a manner that even lesser number of equations are sufficient) 4) square root of a number (if given by GMAT) only considers positive solution only











