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M01-17

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(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 non-negative, then LHS, \({-m}\) must also be non-negative: \(-m\geq 0\). 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: M01-17 [#permalink]

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New post 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 non-negative, then LHS, \({-m}\) must also be non-negative: \(-m\geq 0\). 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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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 non-negative, then LHS, \({-m}\) must also be non-negative: \(-m\geq 0\). 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 M01-17 [#permalink]

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New post 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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New post 01 Feb 2015, 04:36
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|.
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Re: M01-17 [#permalink]

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New post 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: M01-17 [#permalink]

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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: M01-17 [#permalink]

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New post 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<0
math-absolute-value-modulus-86462.html

However 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: M01-17 [#permalink]

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New post 27 Apr 2015, 01:39
Randude wrote:
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<0
math-absolute-value-modulus-86462.html

However 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


Since |0| = 0, then you can include equal sign in either of cases.
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Re M01-17 [#permalink]

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New post 17 Jul 2016, 14:01
I think this is a high-quality question and I agree with explanation.

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Re: M01-17 [#permalink]

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New post 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: M01-17 [#permalink]

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New post 05 Aug 2016, 04:34
varunnair wrote:
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.


HOW did you get this?

From (1): \(m \leq 0\), so \(m\) could be either negative or zero.

From (2): \(m=3=\text{positive}\) or \(m=-3=\text{negative}\).

(1)+(2) \(m=-3=negative\).
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Re: M01-17 [#permalink]

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New post 05 Aug 2016, 10:17
You're right Bunuel. Error on my part. Thanks for the help.

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Re M01-17 [#permalink]

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New post 22 Aug 2016, 06:47
I think this is a high-quality question and I agree with explanation. Good question, need to consider every possible scenario

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M01-17 [#permalink]

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New post 11 May 2017, 16:10
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 non-negative, then LHS, \({-m}\) must also be non-negative: \(-m\geq 0\). 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: M01-17 [#permalink]

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New post 11 Oct 2017, 04:24
Bunuel

As −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: M01-17 [#permalink]

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New post 11 Oct 2017, 04:30
Prashant10692 wrote:
Bunuel

As −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.


First of all notice that any negative number as well as 0 satisfy −m=|−m|.

Next, no, |-m| does not equal to m^2. I think you were thinking about \(\sqrt{m^2}=|m|\).
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Re: M01-17   [#permalink] 11 Oct 2017, 04:30
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