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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 Source: GMAT Club Tests  hardest GMAT questions OA is D. How can y = 0 if xy = 1?



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Re: m23 #33 [#permalink]
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05 Feb 2010, 07:55
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Given x # y = xy  x  y If x # y = 1 => xy  x  y = 1 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/PrincetonNew Jersey. If you are interested in joining me, please contact me at vshrivastava@hotmail.com.
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Re: m23 #33 [#permalink]
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15 Aug 2008, 11:37
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jallenmorris wrote: It's still a good question.
I plugged some numbers.
x =2, finds that y = 1
x = 3, finds that y = 2
so y = x+1
If y = 1, then x = 0 and you do not get 1 from x#y. Another way to do it: 1 = xy  x  y 1 = x*(y1)  y y + 1 = x*(y1) If y = 1, this equation does not hold.



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Re: m23 #33 [#permalink]
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11 Jan 2013, 06:42
zoinnk wrote: 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 Source: GMAT Club Tests  hardest GMAT questions OA is D. How can y = 0 if xy = 1? For any numbers \(x\) and \(y\), \(x@y=xyxy\). 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 \(xyxy=1\), which is the same as \((1x)(1y)1=1\) or \((1x)(1y)=2\). Now, if \(y=1\) then \((1x)(11)=0\neq{2}\), so in order the given equation to hold true \(y\) cannot equal to 1. Answer; D.
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Re: m23 #33 [#permalink]
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15 Aug 2008, 10:02
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It's still a good question. I plugged some numbers. x =2, finds that y = 1 x = 3, finds that y = 2 so y = x+1 If y = 1, then x = 0 and you do not get 1 from x#y.
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Re: m23 #33 [#permalink]
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05 Feb 2010, 09:41
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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 ?
(C) 2008 GMAT Club  m23#33
A 2 B 1 C 0 D 1 E 2
My way: xyxy=1 > x(y1)y=1 >x(y1)=1+y >x=(1+y)/(y1) > y1 should be 0 so y cannot be 1.



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Re: m23 #33 [#permalink]
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05 Feb 2010, 13:35
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IMO D Another way xyxy = 1 => y = \(\frac{(x+1)}{(x1)}\) clearly y =1 is not possible as x+1 can never be equal to x1
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Re: m23 #33 [#permalink]
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13 Mar 2010, 17:53
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xyxy = 1 and if y=1, the equation will result in 1 = +1...not possible...so Ans is D.



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Re: m23 #33 [#permalink]
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21 Apr 2014, 05:45
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Plug in the numbers and see how: A) 2xx+2=1>x=1/3 B) xx+1=1> x=0 C) x=1 D) xx1=1=> 1=1 (ten ten!)
So choose D



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Re: m23 #33 [#permalink]
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05 Feb 2010, 08:17
pretty easy :: xy(x+y)=1 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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Re: m23 #33 [#permalink]
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08 Feb 2010, 15:22
x#y =1 => xyxy =1 => x(y1) = 1+y => x = (1+y)/(y1)
denominator cannot be zero hence y != 1



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Re: m23 #33 [#permalink]
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19 Mar 2010, 08:23
zoinnk wrote: 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 Source: GMAT Club Tests  hardest GMAT questions OA is D. How can y = 0 if xy = 1? xyxy =1 => x(y1) = 1+y => x = (1+y)/(y1) definitely y cannot 1 as this will make x indefinite. so d
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Re: m23 #33 [#permalink]
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10 Feb 2011, 06:23
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)



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Re: m23 #33 [#permalink]
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10 Feb 2011, 10:58
another way of finding the answer; check all the options and eliminate the odd one. option 1. y=2 then, x#y=1=xyxy x=1/3 option 2.y=1 then, x#y=1=xyxy x=0 option 3.y=0 then, x#y=1=xyxy x=1 option 4.y=1 then, x#y=1=xyxy ==>1=1, which is wrong so, D cant be an option. option 5,y=2 then, x#y=1=xyxy x=3 hence option D is correct.
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Re: m23 #33 [#permalink]
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14 Feb 2011, 09:55
@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 xyxy = 1 , it would give 1 = 1 , thus Y != 0



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Re: m23 #33 [#permalink]
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14 Feb 2012, 07:53
raulsy wrote: @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 xyxy = 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 \(x#y=xyxy\). So # does not represent any arithmetic operation: +, , /, or *. Hope it's clear.
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Re: m23 #33 [#permalink]
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11 Jan 2013, 06:38
zoinnk wrote: 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 Source: GMAT Club Tests  hardest GMAT questions OA is D. How can y = 0 if xy = 1? Racked brains and got to D, only to find B marked wrongly in the inmail question of the day. PFA screenshot. Good question.
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Re: m23 #33 [#permalink]
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11 Jan 2013, 06:46
Bunuel wrote: zoinnk wrote: 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 Source: GMAT Club Tests  hardest GMAT questions OA is D. How can y = 0 if xy = 1? For any numbers \(x\) and \(y\), \(x@y=xyxy\). 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 \(xyxy=1\), which is the same as \((1x)(1y)1=1\) or \((1x)(1y)=2\). Now, if \(y=1\) then \((1x)(11)=0\neq{2}\), so in order the given equation to hold true \(y\) cannot equal to 1. Answer; D. Bunuel! Great work.
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Re: m23 #33 [#permalink]
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13 Jan 2013, 04:14



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Re: m23 #33 [#permalink]
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01 Feb 2014, 02:03
By plugging we can get that y cannot be equal to 1











