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Ds  Challenge 2  inequalities [#permalink]
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08 Oct 2006, 19:23
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If X$Y = (x^2  y^2)/ (xy) then what is the value of X$Y ?
1. X + Y = 3
2. X  Y = 2



Director
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A
(x^2  y^2) = (x+y) (xy) = x+y)
__________ __________
(xy) (xY)
Statement 1 is SUFF



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If X$Y = (x^2  y^2)/ (xy) then what is the value of X$Y ?
1. X + Y = 3
2. X  Y = 2
X$Y= (x+y) * (xy)/(xy)
= x + y
A.



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Thats was my answer too , not the OA according Challenge. Any more tries ?



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asaf wrote: If X$Y = (x^2  y^2)/ (xy) then what is the value of X$Y ?
1. X + Y = 3
2. X  Y = 2
X$Y= (x+y) * (xy)/(xy) = x + yA.
What did you wrongly assume in your last step?



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Thanks for the attempt kevin the OA says its C
here is the OE
C is the correct answer.
We can simplify the stem. x$y = (x + y) * (x  y ) / (xy)
We can simplify it further and will end up with x$y = x + y (when x =/=y is not equal to). Therefore, insufficient if x = y.
Statement 2 does not help us much. Insufficient.
Combining the two statements, we get that x+y is 3 and x is not equal to y. It is sufficient as it is right now, but if you would like to solve for x and y, you can. x = 2.5 and y = 0.5;
I am not sure about this OA/OE , can some Math experts validate/invalidate this OE with a better example.



Director
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Again, x=y=0 or in this case even just x=y.



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kevincan wrote: asaf wrote: If X$Y = (x^2  y^2)/ (xy) then what is the value of X$Y ?
1. X + Y = 3
2. X  Y = 2
X$Y= (x+y) * (xy)/(xy) = x + yA.
What did you wrongly assume in your last step?[/quote]
I honestly dont see anything wrong. Please do correct me.



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We cannot assume (xy) as anything  what if it is 0, then we cannot divide, we need it to be a non zero number which B gives.



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It must be C.
Don't fall into the trap that xy can be cancelled in both Numerator and Denominator......
We can do this provided xy !=0.
That's nowhere mentioned in 1.
So it must be C
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Normally, the question will already contain x != y. In addition, the equation exists and is given: so we cannot have x=y. It is not as we are searching the solution and we are transforming an equation.
In my eyes, the answer is more A than C.



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Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.
IT is given that x$y = (x^2y^2)/(xy).
That's it ...............
We know that the above function is difined if x!=y. But how can we consider that the given statment implies this.................
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cicerone wrote: Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.
IT is given that x$y = (x^2y^2)/(xy). That's it ............... We know that the above function is difined if x!=y. But how can we consider that the given statment implies this.................
Well, I would like to say u no
An equation is definied and implies by its existence that x!=y ... In my eyes, it's widely enough



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Fig wrote: cicerone wrote: Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.
IT is given that x$y = (x^2y^2)/(xy). That's it ............... We know that the above function is difined if x!=y. But how can we consider that the given statment implies this................. Well, I would like to say u no An equation is definied and implies by its existence that x!=y ... In my eyes, it's widely enough
Hey fig u mean to say that when x=y the function is not defined.
Now 1 says x+y =3
I will take x=1.5 and y=1.5...............
Then where is the function to calculate with the above values , i mean when they are same the function is not defined.
So i say that 1 is not sufficient
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cicerone wrote: Fig wrote: cicerone wrote: Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.
IT is given that x$y = (x^2y^2)/(xy). That's it ............... We know that the above function is difined if x!=y. But how can we consider that the given statment implies this................. Well, I would like to say u no An equation is definied and implies by its existence that x!=y ... In my eyes, it's widely enough Hey fig u mean to say that when x=y the function is not defined. Now 1 says x+y =3 I will take x=1.5 and y=1.5............... Then where is the function to calculate with the above values , i mean when they are same the function is not defined. So i say that 1 is not sufficient
Yes... To me, we have a system of equations :
(1) x$y = (x^2y^2)/(xy) (implying by existence of the equation x!=y)
(2) x+y =3
So, in my eyes, (1) implies an equation (3) x!=y that implies the couple (x;y) = (1,5;1,5) a not valid possibility
In the real GMAT, the problem will be clarified by the writings of x!=y.



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Fig wrote: cicerone wrote: Fig wrote: cicerone wrote: Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.
IT is given that x$y = (x^2y^2)/(xy). That's it ............... We know that the above function is difined if x!=y. But how can we consider that the given statment implies this................. Well, I would like to say u no An equation is definied and implies by its existence that x!=y ... In my eyes, it's widely enough Hey fig u mean to say that when x=y the function is not defined. Now 1 says x+y =3 I will take x=1.5 and y=1.5............... Then where is the function to calculate with the above values , i mean when they are same the function is not defined. So i say that 1 is not sufficient Yes... To me, we have a system of equations : (1) x$y = (x^2y^2)/(xy) (implying by existence of the equation x!=y) (2) x+y =3 So, in my eyes, (1) implies an equation (3) x!=y that implies the couple (x;y) = (1,5;1,5) a not valid possibility In the real GMAT, the problem will be clarified by the writings of x!=y.
That's what i am precisely saying........... we definitely need that information..........
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Nope... Not definitly
The problem states an equation that needs to avoid certain values of x. Since an equation can exist only for values of x that makes this equation works, impossible values of x are, even if not stated, implicitely banned by the existence of the equation.
In real GMAT, the impossible values of x will be precised next to the equation and not in the statment 1 and 2
Last edited by Fig on 10 Oct 2006, 14:32, edited 2 times in total.



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Great Discussion guys !!!
I guess C makes sense now .
First rule in GMAT DS , never assume what is not stated and it can be stated anywhere.
xy !=0 is not stated in this question and cannot be assumed.
HongHu any takes on this one ?



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I agree with Fig and disagree with the OE. My thumb of rules for GMAT question is that if something appears in the denominator then it is not zero. However, if you don't know whether it is not zero, you can not divide both sides of an equation by it.
For example, if the stem says x/y=1. We can assume that y is not zero. However, if the stem says x=y. We can not say x/y=1 because y may be zero.
By the same token, if the stem says xy/y=1, we can say that x=1 and y<>0. But if the stem says xy=y, then all we can do is xyy=0, or y(x1)=0. In other words there are two solutions to this equation: x=1 and y=anything, or y=0 and x=anything.
What do you all think?
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In the real GMAT, the question will exclude all impossible cases
So... people who are not 100% in trust for this way to see an equation will have finally no more troubles than those who trust.
The rule of x*y=y given by HongHu is the one to remember guys I fairly believe it appears to anyone of us during the GDay










