What is the OA?

OA??

theCodeToGMAT wrote:What is the OA?

dddanny2006 wrote:OA??

theCodeToGMAT wrote:What is the OA?

OA is an acronym for Official Answer..

I am asking what is the OA for the question you posted.

Is x ≥ 0?

1) x^2 = 9x

2) |x| = -x

theCodeToGMAT wrote:

dddanny2006 wrote:OA??

theCodeToGMAT wrote:What is the OA?

OA is an acronym for Official Answer..

I am asking what is the OA for the question you posted.

Is x ≥ 0?

1) x^2 = 9x

2) |x| = -x

Rahul,
I guess OA should be A what you say?

Yep, i also got [spoiler]{A}
[/spoiler]
To find: x ≥ 0

Statement 1: x^2 = 9x
x^2 - 9x = 0
x(x-9) = 0
Either x = 0 or x = 9
SUFFICIENT

Statement 2:
|x| = -x
this is possible when: x < 0 or x = 0
INSUFFICIENT

Answer [spoiler]{A}[/spoiler]

In Statement 2: |x| = -x
Lets take Three possibilities of X
First: Negative
x = -1
|-1| = -(-1)
1= 1
YES
Second: Zero
X = 0
|0| = -(0)
0 =0
Yes
Third : Positive
x = 1
|1| = -(1)
1 != -1
NO
Hence statement2 is insufficient.

This is my method.Please tell me if Im right??

Statement 1 is sufficient.

When it comes to statement 2,
|x| = -x as a result we have x=-x and x=-(-x)=x

We have 2 equations x=x and x=-x


In the equation x=x we test 0,1,-1

When we test 0 and 1 we get a YES.We get a No when we test -1 because -1 is not greater than or equal to 0.We have a Yes and No so insufficient.For the condition x=-x only 0 can be tested because no number is equal to a negative of itself.

Is my technique correct?

Also please read the following article as it has confused me alot.

https://www.manhattangmat.com/strategy-s ... -value.cfm

If you go by this then the statement 2 gives us 2 conditions

x=-x when x>0 and x=x when x<0

In condition x=-x when x>0 if we substitute 1 we get 1!=1.Does this condition become a NO or does it become invalid.

x=x when x<0 since x has to be less than 0 -1=-1 or -2=-2 here -2 doesnt satisfy the condition x>=0 in the question stem and thus its a No.

So seems like both A and D can give us the answer,though both statements contradict each other and it should happen on the GMAT.Only worry for me is the concept.

Thanks and get back to me.

Uva@90 wrote:In Statement 2: |x| = -x
Lets take Three possibilities of X
First: Negative
x = -1
|-1| = -(-1)
1= 1
YES
Second: Zero
X = 0
|0| = -(0)
0 =0
Yes
Third : Positive
x = 1
|1| = -(1)
1 != -1
NO
Hence statement2 is insufficient.

Hi dddanny2006,

The technique you are using is a valid one. I also use the same method when there are more than one term in between modulus.

For example:
|x-1| = 5
Case 1: (x-1) = 5 ==> x = 6
or,
Case 2: -(x-1) = 5 ==> x = -4

But my answer is D instead of A.In Statement 2,I have one condition that cant be tested x=-x when x>0
Im sure you agree with me that x=-x when x>0 cant be tested as no number is equal to its negative barring 0,but 0 cant be tested as x>0 according to the Manhattan article.The other condition is x=x when x<0,this yields a No because any number less than 0 does'nt satifsy the question stem x>=0.So the 2 sub conditions of statement 2 result in a _______(I dont know what can call this equation, a No or what,because it cant be tested at all) and a No.Please work this problem out based on my technique,as Im confused.

Please also note that |X|=-X produces 2 equations only

One is X=-X when X>0 and other is X=-(-X)=X when X<0

So 0 should'nt be tested here as we will break the X<0 and X>0 conditions that arose from removing the modulus signs.

Thanks

theCodeToGMAT wrote:Hi dddanny2006,

The technique you are using is a valid one. I also use the same method when there are more than one term in between modulus.

For example:
|x-1| = 5
Case 1: (x-1) = 5 ==> x = 6
or,
Case 2: -(x-1) = 5 ==> x = -4

Okay dddanny2006.. i see your point here.. Indeed a point of confusion!!!

Anyways, I will tell you how I solve the modulus question..

I simply break the modulus without considering whether i considered x>0 or x<0

If i Say so, then I would write statement 2 as:

Statement 2:

|x| = -x

Case 1:
x = -x
This is ONLY possible when "x" is "0"
Or,
Case 2:
-x = -x ==> x=x
This can mean that x can have any negative or positive value i.e. ....,-2,-1,0,1,2,3,....
So, we cannot say whether x is ALWAYS greater than or equal to "0" as we have negative values which do satisfy the equation.
INSUFFICIENT

Also, I have done many questions following this approach and have never faced such ambiguousy

Well I dont think Im fixating.Its just what the article said.Two conditions each when x>0 and x<0.If I should ignore those 2 conditions,then Im conceptually wrong.Its like Im not folowing the rules.Please explain the same in depth Ceilidh.Thanks.I get the conceptual thinking factor that x=-x is possible only in the case of 0.But why is there an x>0 and x<0 in that article then?

ceilidh.erickson wrote:

dddanny2006 wrote:One is X=-X when X>0 and other is X=-(-X)=X when X<0

So 0 should'nt be tested here as we will break the X<0 and X>0 conditions that arose from removing the modulus signs.

I think you're fixating too much on the algebra, rather than thinking conceptually. Ask yourself - what kinds of numbers behave this way? Negatives and 0.[/quote]

It needs to be mentioned in that article.It has really misguided me,and it put me under alot of stress.Could you please share your inputs on this problem too,Ceilidh?

Heres the link--https://www.beatthegmat.com/absolute-val ... tml#701734

ceilidh.erickson wrote:

dddanny2006 wrote:One is X=-X when X>0 and other is X=-(-X)=X when X<0

So 0 should'nt be tested here as we will break the X<0 and X>0 conditions that arose from removing the modulus signs.

I understand your confusion with the algebra here. Really, the modulus should be "when x>=0" and "when x <= 0"

I think you're fixating too much on the algebra, though. It can be more helpful to simply think conceptually. Ask yourself - what kinds of numbers behave this way? Negatives and 0.