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Director
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Is M < 0 ? (1) -M = |-M| (2) M^2 = 9 [#permalink]
 20 Nov 2007, 12:07
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Is M < 0 ?
(1) -M = |-M|
(2) M^2 = 9
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(C) for me 
M < 0 ?
Stat 1
-M = |-M|
implies that:
-M >=0
<=> M =< 0 ... M can be 0.
INSUFF.
Stat 2
M^2 = 9
<=> M=3 or M=-3
INSUFF.
Both 1 and 2
We have:
o M =< 0
AND
o M=3 or M=-3
Implies that M = -3 < 0
SUFF.
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Director
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Re: DS: Is M < 0 ? [#permalink]
20 Nov 2007, 12:32
eyunni wrote: Is M < 0 ?
(1) -M = |-M|
(2) M^2 = 9
A. if -M = l-Ml, then M must be -ve.
2 says m is either +ve or -ve.
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Re: DS: Is M < 0 ? [#permalink]
20 Nov 2007, 12:34
Himalayan wrote: eyunni wrote: Is M < 0 ?
(1) -M = |-M|
(2) M^2 = 9 A. if -M = l-Ml, then M must be -ve. 2 says m is either +ve or -ve. In statement 1, M can be 0, making M out of M < 0
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Director
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Fig,
I have a very basic question here.
I see that m <= 0. The question is: m < 0? Isn't the answer straightaway "NO"? Because m can be 0 or less than 0.
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eyunni wrote: Fig,
I have a very basic question here.
I see that m <= 0. The question is: m < 0? Isn't the answer straightaway "NO"? Because m can be 0 or less than 0.
Well to me, the answers to the question are:
> YES <=> M < 0
> NO <=> M >= 0
> NON ANSWERABLE : any other cases such as M =< 0 or -10 < M < 10
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Director
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Fig wrote: eyunni wrote: Fig,
I have a very basic question here.
I see that m <= 0. The question is: m < 0? Isn't the answer straightaway "NO"? Because m can be 0 or less than 0. Well to me, the answers to the question are: > YES <=> M <0> NO <M>= 0 > NON ANSWERABLE : any other cases such as M =< 0 or -10 < M < 10 woow...Fig, I did not find this point in OG. Is this a potential GMAT type question?
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eyunni wrote: Fig wrote: eyunni wrote: Fig,
I have a very basic question here.
I see that m <= 0. The question is: m < 0? Isn't the answer straightaway "NO"? Because m can be 0 or less than 0. Well to me, the answers to the question are: > YES <=> M <0> NO <M>= 0 > NON ANSWERABLE : any other cases such as M =< 0 or -10 < M < 10 woow...Fig, I did not find this point in OG. Is this a potential GMAT type question? I would like to say pretty usual, in fact. It's a kind of Yes/No question
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Director
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Re: DS: Is M < 0 ? [#permalink]
20 Nov 2007, 14:14
Fig wrote: Himalayan wrote: eyunni wrote: Is M < 0 ?
(1) -M = |-M|
(2) M^2 = 9 A. if -M = l-Ml, then M must be -ve. 2 says m is either +ve or -ve. In statement 1, M can be 0, making M out of M < 0 thats good point fig. thanx. 
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Re: DS: Is M < 0 ? [#permalink]
20 Nov 2007, 19:09
eyunni wrote: Is M < 0 ?
(1) -M = |-M|
(2) M^2 = 9
I get C.
1: M can be 0 or -1 etc...
2: M is -3 or 3
together M must be -3.
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Re: DS: Is M < 0 ? [#permalink]
20 Nov 2007, 19:12
GMATBLACKBELT wrote: eyunni wrote: Is M < 0 ?
(1) -M = |-M|
(2) M^2 = 9 I get C. 1: M can be 0 or -1 etc... 2: M is -3 or 3 together M must be -3. How can M be -1? Wouldn't that mean that statement 1 would be:
--1(ie +1) = |-1|?
Doesn't statement 1 mean that M has to be positive or zero?
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"A" for me here
1) -M = |-M| only when M = 0 --> Suff
2) M = 3 or -3 --> Insuff
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Whatever wrote: "A" for me here
1) -M = |-M| only when M = 0 --> Suff
2) M = 3 or -3 --> Insuff
For statment 1, It's not only M=0 that works. We could try any negative number.
We can pick one example:
o If M = -10, then -M=-(-10) = 10 and |-M| = |-(-10)| = |10| = 10...
So statment 1 alone is not enough
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Fig wrote: Whatever wrote: "A" for me here
1) -M = |-M| only when M = 0 --> Suff
2) M = 3 or -3 --> Insuff For statment 1, It's not only M=0 that works. We could try any negative number. We can pick one example: o If M = -10, then -M=-(-10) = 10 and |-M| = |-(-10)| = |10| = 10... So statment 1 alone is not enough Can someone explain how M can be negative in Statement 1?
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alrussell wrote: Fig wrote: Whatever wrote: "A" for me here
1) -M = |-M| only when M = 0 --> Suff
2) M = 3 or -3 --> Insuff For statment 1, It's not only M=0 that works. We could try any negative number. We can pick one example: o If M = -10, then -M=-(-10) = 10 and |-M| = |-(-10)| = |10| = 10... So statment 1 alone is not enough Can someone explain how M can be negative in Statement 1? I can try
As we have abs ( something ) >= 0, we have:
|-M| >= 0
<=> -M >= 0 as |-M| = -M
<=> M =< 0
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Fig wrote: alrussell wrote: Fig wrote: Whatever wrote: "A" for me here
1) -M = |-M| only when M = 0 --> Suff
2) M = 3 or -3 --> Insuff For statment 1, It's not only M=0 that works. We could try any negative number. We can pick one example: o If M = -10, then -M=-(-10) = 10 and |-M| = |-(-10)| = |10| = 10... So statment 1 alone is not enough Can someone explain how M can be negative in Statement 1? I can try As we have abs ( something ) >= 0, we have: |-M| >= 0 <M>= 0 as |-M| = -M <=> M =< 0 Sorry I am being slow - what I am getting for 1 is -M = |-M| so M = |M| and |M| is never negative... so on that basis I concluded that M has to be positive ... I'm not sure where I have gone wrong?
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alrussell wrote: Fig wrote: alrussell wrote: Fig wrote: Whatever wrote: "A" for me here
1) -M = |-M| only when M = 0 --> Suff
2) M = 3 or -3 --> Insuff For statment 1, It's not only M=0 that works. We could try any negative number. We can pick one example: o If M = -10, then -M=-(-10) = 10 and |-M| = |-(-10)| = |10| = 10... So statment 1 alone is not enough Can someone explain how M can be negative in Statement 1? I can try As we have abs ( something ) >= 0, we have: |-M| >= 0 <M>= 0 as |-M| = -M <=> M =< 0 Sorry I am being slow - what I am getting for 1 is -M = |-M| so M = |M| and |M| is never negative... so on that basis I concluded that M has to be positive ... I'm not sure where I have gone wrong? In bold, it's not correct
We have |M| = |-M|.... but we do not have -M = M
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Fig wrote: Quote: Sorry I am being slow - what I am getting for 1 is -M = |-M| so M = |M| and |M| is never negative... so on that basis I concluded that M has to be positive ... I'm not sure where I have gone wrong? In bold, it's not correct We have |M| = |-M|.... but we do not have -M = M Thanks - so where is the error in my statement below:
Lets take say - M=-3/3 In that case, the Abs is |3|. If -M=|-M| -M=|-3|
If that's the case doesnt -M=-3? or does it equal 3 / -3
Im really lost as to what the rule is here... any chance you could explain in words (in order to help a rather math challeneged lawyer understand the concepts at play?)
THANKS!
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alrussell wrote: Fig wrote: Quote: Sorry I am being slow - what I am getting for 1 is -M = |-M| so M = |M| and |M| is never negative... so on that basis I concluded that M has to be positive ... I'm not sure where I have gone wrong? In bold, it's not correct We have |M| = |-M|.... but we do not have -M = M Thanks - so where is the error in my statement below: Lets take say - M=-3/3 In that case, the Abs is |3|. If -M=|-M| -M=|-3| If that's the case doesnt -M=-3? or does it equal 3 / -3 Im really lost as to what the rule is here... any chance you could explain in words (in order to help a rather math challeneged lawyer understand the concepts at play?) THANKS! I will try ... I'm perhaps less expert in pure word explanations
I think u have a mixture in your reasoning between inferences from abs properties and lines of calculation.
First of all, -M = -3 cannot be chosen because M must be negative ... But, I can take this value and demonstrate that it's not working
So, in your example, -M = -3 implies that M = 3. We will consider seperatly the left side of the original equation and then the right one. The objective is to find the same value, indicating that both sides are equal one another.
Sides:
o At left, -M = -3.
o At right, |-M| = |-3| = 3
To conclude on your example, -3 is not equal to 3 implies that M cannot be 3 or -M cannot be -3 and at the same satisfies the equation -M = |-M|
Now, back to the original equation. Luckily, we have 1 property of abs that is very useful : abs (something) >= 0.
By using the original equation, we know now that -M must be positive or equal to 0 to make M a solution to -M = |-M|. M is, by this way, negative or 0.
Another way to see it is to set Z = -M. That means Z = |Z| implies Z positive or 0. Thus, M, the opposite of Z on a number line, is negative or 0.
I hope that helps
PS : I recommand u to practice some questions from here.... http://www.gmatclub.com/phpbb/viewtopic.php?t=39533
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alrussell wrote: Fig wrote: I will try ... I'm perhaps less expert in pure word explanations I think u have a mixture in your reasoning between inferences from abs properties and lines of calculation. First of all, -M = -3 cannot be chosen because M must be negative ... But, I can take this value and demonstrate that it's not working So, in your example, -M = -3 implies that M = 3. We will consider seperatly the left side of the original equation and then the right one. The objective is to find the same value, indicating that both sides are equal one another. Sides: o At left, -M = -3. o At right, |-M| = |-3| = 3 To conclude on your example, -3 is not equal to 3 implies that M cannot be 3 or -M cannot be -3 and at the same satisfies the equation -M = |-M| Now, back to the original equation. Luckily, we have 1 property of abs that is very useful : abs (something) >= 0. By using the original equation, we know now that -M must be positive or equal to 0 to make M a solution to -M = |-M|. M is, by this way, negative or 0. Another way to see it is to set Z = -M. That means Z = |Z| implies Z positive or 0. Thus, M, the opposite of Z on a number line, is negative or 0. I hope that helps PS : I recommand u to practice some questions from here.... http://www.gmatclub.com/phpbb/viewtopic.php?t=39533Thanks VERY MUCH! You are welcome... I understand that it should be very hard to tackle the Q section in your case ...
Good Luck on it & do not hesitate to post/ask/comment in this section of GMATClub
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