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# Ds - Challenge 2 - inequalities

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Ds - Challenge 2 - inequalities [#permalink]

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08 Oct 2006, 18:23
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If X$Y = (x^2 - y^2)/ (x-y) then what is the value of X$Y ?

1. X + Y = 3

2. X - Y = 2
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08 Oct 2006, 18:26
A

(x^2 - y^2) = (x+y) (x-y) = x+y)
__________ __________

(x-y) (x-Y)

Statement 1 is SUFF
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08 Oct 2006, 22:02
If X$Y = (x^2 - y^2)/ (x-y) then what is the value of X$Y ?

1. X + Y = 3

2. X - Y = 2

X$Y= (x+y) * (x-y)/(x-y) = x + y A. Senior Manager Joined: 11 Jul 2006 Posts: 382 Location: TX Followers: 1 Kudos [?]: 14 [0], given: 0 ### Show Tags 09 Oct 2006, 13:58 Thats was my answer too , not the OA according Challenge. Any more tries ? GMAT Instructor Joined: 04 Jul 2006 Posts: 1264 Location: Madrid Followers: 29 Kudos [?]: 298 [0], given: 0 ### Show Tags 09 Oct 2006, 14:16 asaf wrote: If X$Y = (x^2 - y^2)/ (x-y) then what is the value of X$Y ? 1. X + Y = 3 2. X - Y = 2 X$Y= (x+y) * (x-y)/(x-y)
= x + y
A.

What did you wrongly assume in your last step?
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09 Oct 2006, 14:56
Thanks for the attempt kevin the OA says its C

here is the OE

We can simplify the stem. x$y = (x + y) * (x - y ) / (x-y) We can simplify it further and will end up with x$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.
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10 Oct 2006, 03:30
Again, x=y=0 or in this case even just x=y.
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10 Oct 2006, 05:59
kevincan wrote:
asaf wrote:
If X$Y = (x^2 - y^2)/ (x-y) then what is the value of X$Y ?

1. X + Y = 3

2. X - Y = 2

X$Y= (x+y) * (x-y)/(x-y) = x + y A. What did you wrongly assume in your last step?[/quote] I honestly dont see anything wrong. Please do correct me. Intern Joined: 05 Jul 2006 Posts: 1 Followers: 0 Kudos [?]: 0 [0], given: 0 ### Show Tags 10 Oct 2006, 07:05 We cannot assume (x-y) as anything - what if it is 0, then we cannot divide, we need it to be a non zero number which B gives. Senior Manager Joined: 28 Aug 2006 Posts: 306 Followers: 13 Kudos [?]: 150 [0], given: 0 ### Show Tags 10 Oct 2006, 09:00 It must be C. Don't fall into the trap that x-y can be cancelled in both Numerator and Denominator...... We can do this provided x-y !=0. That's nowhere mentioned in 1. So it must be C _________________ SVP Joined: 01 May 2006 Posts: 1797 Followers: 9 Kudos [?]: 149 [0], given: 0 ### Show Tags 10 Oct 2006, 10:12 Normally, the question will already contain x != y. In addition, the equation exists and is given: so we cannot have x=y. It is not as we are searching the solution and we are transforming an equation. In my eyes, the answer is more A than C. Senior Manager Joined: 28 Aug 2006 Posts: 306 Followers: 13 Kudos [?]: 150 [0], given: 0 ### Show Tags 10 Oct 2006, 10:30 Hey fig, don't u think that u are violating the basic rule of Data Sufficiency. IT is given that x$y = (x^2-y^2)/(x-y).
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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10 Oct 2006, 11:11
cicerone wrote:
Hey fig, don't u think that u are violating the basic rule of Data Sufficiency.

IT is given that x$y = (x^2-y^2)/(x-y). That's it ............... We know that the above function is difined if x!=y. But how can we consider that the given statment implies this................. Well, I would like to say u no An equation is definied and implies by its existence that x!=y ... In my eyes, it's widely enough Senior Manager Joined: 28 Aug 2006 Posts: 306 Followers: 13 Kudos [?]: 150 [0], given: 0 ### Show Tags 10 Oct 2006, 11:17 Fig wrote: cicerone wrote: Hey fig, don't u think that u are violating the basic rule of Data Sufficiency. IT is given that x$y = (x^2-y^2)/(x-y).
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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10 Oct 2006, 11:32
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$y = (x^2-y^2)/(x-y). That's it ............... We know that the above function is difined if x!=y. But how can we consider that the given statment implies this................. Well, I would like to say u no An equation is definied and implies by its existence that x!=y ... In my eyes, it's widely enough Hey fig u mean to say that when x=y the function is not defined. Now 1 says x+y =3 I will take x=1.5 and y=1.5............... Then where is the function to calculate with the above values , i mean when they are same the function is not defined. So i say that 1 is not sufficient Yes... To me, we have a system of equations : (1) x$y = (x^2-y^2)/(x-y) (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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10 Oct 2006, 12:02
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$y = (x^2-y^2)/(x-y). That's it ............... We know that the above function is difined if x!=y. But how can we consider that the given statment implies this................. Well, I would like to say u no An equation is definied and implies by its existence that x!=y ... In my eyes, it's widely enough Hey fig u mean to say that when x=y the function is not defined. Now 1 says x+y =3 I will take x=1.5 and y=1.5............... Then where is the function to calculate with the above values , i mean when they are same the function is not defined. So i say that 1 is not sufficient Yes... To me, we have a system of equations : (1) x$y = (x^2-y^2)/(x-y) (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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10 Oct 2006, 12:15
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, 13:32, edited 2 times in total.
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10 Oct 2006, 13:24
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.

x-y !=0 is not stated in this question and cannot be assumed.

HongHu any takes on this one ?
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10 Oct 2006, 18:45
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 xy-y=0, or y(x-1)=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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10 Oct 2006, 22:33
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
10 Oct 2006, 22:33
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