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Solving for the value of x: xy - x - y = 1 => x (y - 1) = (y + 1) => x = (y + 1) / ( y - 1)

So, y - 1 cannot be 0; therefore, y cannot be 1.

Answer is D.

===================================================== Dear All: I'm looking for a study partner. I live in Plainsboro/Princeton-New Jersey. If you are interested in joining me, please contact me at vshrivastava@hotmail.com.

Last edited by vshrivastava on 22 Feb 2013, 00:15, edited 1 time in total.

from options:: put y=1 x-(1+x)=1 to satisfy this... -1=1 not possible so ans is for y=1 ----------------- rest all the values would give sm ans .. --------------------------
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" What [i] do is not beyond anybody else's competence"- warren buffett My Gmat experience -http://gmatclub.com/forum/gmat-710-q-47-v-41-tips-for-non-natives-107086.html

Just plug in the answer choices to find the correct answer. Kaplan teaches us to try options D and B first because 60% of the times, the correct answer is either of them. (esp true for questions that can be solved by plugging values from answers choices)

xy - x - y = 1 xy - x = 1 + y

From D, y=1 x.1 - x = 1 + 1 x - x = 2 This cannot be true for no matter what value x takes, hence we have our correct answer. (luckily without looking at other options -- thanks Kaplan)

@all I think we all are overlooking a fact out here that in the question its given , X#Y = 1 , how i approached this questn was to find out a value of X by putting in values of Y from the options , taking into account a feasible operator, and then putting the values of Both X & Y in the second eqn to see if it results in 1

for say if Y = 0 , X has to be +1 and #(operator) should be ADDITION /SUBTRACTION , but if now we put X= 1 and Y= 0 in xy-x-y = 1 , it would give -1 = 1 , thus Y != 0

@all I think we all are overlooking a fact out here that in the question its given , X#Y = 1 , how i approached this questn was to find out a value of X by putting in values of Y from the options , taking into account a feasible operator, and then putting the values of Both X & Y in the second eqn to see if it results in 1

for say if Y = 0 , X has to be +1 and #(operator) should be ADDITION /SUBTRACTION , but if now we put X= 1 and Y= 0 in xy-x-y = 1 , it would give -1 = 1 , thus Y != 0

Welcome to GMAT Club.

The point is that # represents some functional relationship between \(x\) and \(y\) described as \(x#y=xy-x-y\). So # does not represent any arithmetic operation: +, -, /, or *.

For any numbers \(x\) and \(y\), \(x@y=xy-x-y\). If \(x@y=1\), which of the following cannot be the value of \(y\) ?

A. -2 B. -1 C. 0 D. 1 E. 2

Given \(xy-x-y=1\), which is the same as \((1-x)(1-y)-1=1\) or \((1-x)(1-y)=2\). Now, if \(y=1\) then \((1-x)(1-1)=0\neq{2}\), so in order the given equation to hold true \(y\) cannot equal to 1.

For any numbers \(x\) and \(y\), \(x@y=xy-x-y\). If \(x@y=1\), which of the following cannot be the value of \(y\) ?

A. -2 B. -1 C. 0 D. 1 E. 2

Given \(xy-x-y=1\), which is the same as \((1-x)(1-y)-1=1\) or \((1-x)(1-y)=2\). Now, if \(y=1\) then \((1-x)(1-1)=0\neq{2}\), so in order the given equation to hold true \(y\) cannot equal to 1.