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M01-17

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Director
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Re: M01-17  [#permalink]

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New post 05 Jul 2018, 04:21
1
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]

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New post 20 Jul 2018, 10:45
Hi,

I re read the solution, but unable to understand the flaw in my logic
Please advise
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New post 20 Jul 2018, 10:54
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Re: M01-17  [#permalink]

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New post 20 Jul 2018, 11:03
Hi,

Thanks for the response. It is clear now
So if the question was is m>=0 then statement would be sufficient
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Re: M01-17  [#permalink]

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New post 13 Mar 2019, 12:30
Bunuel wrote:
Is \(m \lt 0\)?


(1) \(-m = |-m|\)

(2) \(m^2 = 9\)


Hi Bunuel, chetan2u

I read solution but not able find my doubt, not sure if I have missed anything.

I think statement 1 is only true when "m" is negative as RHS |-m| is always going to be positive no matter what is value of "m" and LHS as per statement is "-m" which can be only be converted to positive when value of "m" is negative to make it equal to RHS as "- * - = +"

Could you please advise where I am going wrong.
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New post 13 Mar 2019, 21:36
Gmatprep550 wrote:
Bunuel wrote:
Is \(m \lt 0\)?


(1) \(-m = |-m|\)

(2) \(m^2 = 9\)


Hi Bunuel, chetan2u

I read solution but not able find my doubt, not sure if I have missed anything.

I think statement 1 is only true when "m" is negative as RHS |-m| is always going to be positive no matter what is value of "m" and LHS as per statement is "-m" which can be only be converted to positive when value of "m" is negative to make it equal to RHS as "- * - = +"

Could you please advise where I am going wrong.


\(-m=|-m|\) holds for 0 too: -0 = 0 = |-0|.
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Re: M01-17  [#permalink]

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New post 14 Mar 2019, 03:20
Bunuel wrote:
Is \(m \lt 0\)?


(1) \(-m = |-m|\)

(2) \(m^2 = 9\)



#1
to be true , m can be -ve or 0 ;insufficient
#2
m=+/-3
insufficient
from 1 & 2
m= -3
IMO C
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Re: M01-17  [#permalink]

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New post 28 Apr 2019, 18:55
Made a stupid mistake when solving this in a GMATClub test.

Is m<0?

(1)\(−m=|−m|\)

Test m= -1

-(-1) = |-1|
1=1
is m< 0 --> yes

Test m = 0
-0= |-0|
0=0
is m <0 --> No

---Insufficient

(2)\(m^2=9\)

m= + / - 3

Yes or No
Insufficient

(1)+(2)

If m = -3 the equation in (1) is satisfied since
-(-3) = |-3|
3=3

but if m= 3 the equation does not hold true
i.e.
Is -(3) = |3| ? No
-3 is not equal to 3 obviously

Therefore combined it must be true that m < 0
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Re: M01-17   [#permalink] 28 Apr 2019, 18:55

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