Contradictory answers

[dddanny2006]

Is x ≥ 0?

1) x^2 = 9x

2) |x| = -x

Going back to the article,Statement 1 is absolutely sufficient.When it comes to statment 2,this is my way of solving it and it contradicts with the answer you've provided.

|x|=-x =>x=+(-x)=-x,this is true when x>0, and x=-(-x),this is true when x<0
Therefore statement 2 gives us 2 conditions x=x and x=-x.Now since x=-x when x>0,we cant test 0 there.Now,we also have x=x when x<0,so -5=-5,-6=-6,these dont satisfy the condition in the question stem and thus its a NO.Thus the answer is D in contradiction to A what the book says.

In which way is my understanding of the question wrong?

I have followed this technique from here--
https://www.manhattangmat.com/strategy-s ... -value.cfm

Absolute value expressions start to become difficult when variable expressions are placed inside the bars. For example, /x/. Upon a cursory examination, the expression /x/ seems like it should be equal to x. Since there is no sign in front of the x, the absolute value bars should be able to be removed without jeopardizing the "guarantee of positive." What this line of reasoning fails to account for, however, is that x itself could be negative! When dealing with absolute value expressions that contain variables, two scenarios must be considered: (1) the scenario whereby the expression inside the bars is positive and (2) the scenario whereby the expression inside the bars is negative.

In this example, for scenario (1) if x > 0, the expression /x/ can simply be represented as x; for scenario (2) if x < 0, the expression /x/ must be represented as (-x). Notice that in the negative scenario, we don't simply remove the absolute value bars. We remove the absolute value bars and negate the entire expression within.

Let's look at a more complicated example: the expression /x - 3/. As always, we must consider both the positive and negative scenarios. When is the expression inside the absolute value bars positive? Not simply when x > 0, but when x - 3 > 0 or when x > 3. Likewise the expression will be negative when x < 3.

To recap, the two scenarios are:
(1) /x - 3/ can be rewritten as x - 3 when x > 3
(2) /x - 3/ can be rewritten as -(x - 3) or 3 - x when x < 3

One more for the road: /3x + y/.
(1) /3x + y/ can be rewritten as 3x + y when 3x + y > 0
(2) /3x + y/ can be rewritten as -(3x + y) when 3x + y < 0

[theCodeToGMAT]

What is the OA?

[dddanny2006]

OA??

What is the OA?

[theCodeToGMAT]

OA??

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

[Uva@90]

OA??

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?

[theCodeToGMAT]

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]

[Uva@90]

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.

[dddanny2006]

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.

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.

[theCodeToGMAT]

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

[dddanny2006]

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

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

[theCodeToGMAT]

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

[ceilidh.erickson]

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.

[dddanny2006]

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?

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]

[dddanny2006]

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

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.