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Re: M01-17 [#permalink]
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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Re: M01-17 [#permalink]
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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]
I think this is a high-quality question and I agree with explanation.
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Re: M01-17 [#permalink]
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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Re: M01-17 [#permalink]
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Only thing I would like to highlight once again for this problem is that have a set of numbers to test on each and every DS question. For statement (1) it was very important to test for zero. It does not strike naturally for this statement that m could be zero.

Hence develop a habit to test numbers systematically. Use a standard set of numbers during your practice sessions so that this becomes a habit. After reading the constraints mentioned in question stem and the 2 statements, you could test for (assuming no constraints) -2,\(\frac{-3}{2}\),-1,\(\frac{-1}{2}\),0,\(\frac{1}{2}\),1,\(\frac{3}{2}\) and 2.

Please read my overall understanding on DS questions and few useful tips:-

https://gmatclub.com/forum/data-suffici ... l#p2073589
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Re: M01-17 [#permalink]
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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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re M01-17 [#permalink]
I think this is a high-quality question.
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Re M01-17 [#permalink]
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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