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For any numbers x and y, x#y = xy  x  y if x#y=1 which
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Updated on: 24 Dec 2013, 01:44
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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 M2333
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Originally posted by bmwhype2 on 21 Dec 2007, 07:14.
Last edited by Bunuel on 24 Dec 2013, 01:44, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.




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Re: For any numbers x and y, x#y = xy  x  y if x#y=1 which
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24 Dec 2013, 01:45
bmwhype2 wrote: 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
M2333 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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D
xyxy=1 ==> y(x1)x=1 ==> y(x1)=(x+1) ==> y=(x+1)/(x1) ==>
y=(x1+2)/(x1) ==> y=1+2/(x1)
2/(x1)<>0 ==> y<>1



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21 Dec 2007, 21:52
bmwhype2 wrote: For any numbers x and y, x#y = xy  x  y if x#y=1 which cannot be the value of y?
2 1 0 1 2
Y= (X+1)/(X1)
Now substitute the choices, For Y=1,
X+1 = X1
which is not possible, that means Y=1 is not possible.



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walker wrote: D
xyxy=1 ==> y(x1)x=1 ==> y(x1)=(x+1) ==> y=(x+1)/(x1) ==>
y=(x1+2)/(x1) ==> y=1+2/(x1)
2/(x1)<>0 ==> y<>1 similarly x= (y+1)/(y1)...denominator cannot be 0, hence y cannot be 1.



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Re: For any numbers x and y, x#y = xy  x  y if x#y=1 which
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05 Jan 2014, 07:35
D x#y = xy  x  y where x#y = 1 so, choosing x and y values from option x= 0 and x = 1 then substitute 1 = 0* 1  0  1 = 1 , where equation is # if y = 1 , seeing with x= 0 and y = 1 ,then yes 1 = 1, same as other too..
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Re: For any numbers x and y, x#y = xy  x  y if x#y=1 which
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12 Nov 2014, 21:28
Bunuel wrote: bmwhype2 wrote: 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
M2333 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. Hey Bunuel, What are the guidelines to factor \(xyxy=1\) as \((1x)(1y)1\)? Can you please provide a link of study or similar problems? I'm unable to understand. Thanks,



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Re: For any numbers x and y, x#y = xy  x  y if x#y=1 which
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01 Dec 2015, 22:42
prsnt11 wrote: Bunuel wrote: bmwhype2 wrote: 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
M2333 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. Hey Bunuel, What are the guidelines to factor \(xyxy=1\) as \((1x)(1y)1\)? Can you please provide a link of study or similar problems? I'm unable to understand. Thanks, I second this question  can someone please provide some guidance? Thanks!!
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Re: For any numbers x and y, x#y = xy  x  y if x#y=1 which
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02 Dec 2015, 02:20
happyface101 wrote: prsnt11 wrote: Hey Bunuel, What are the guidelines to factor \(xyxy=1\) as \((1x)(1y)1\)? Can you please provide a link of study or similar problems? I'm unable to understand. Thanks,
I second this question  can someone please provide some guidance? Thanks!! There are no guidelines as such to factorization of \(xyxy=1\) as \((1x)(1y)1\) Since the question is asking us to find the values that y cannot take, we are trying to make an equation in terms of y. If (1x)(1y) = 0. This means either x = 1 or y =1 By bringing the equation in the form \((1x)(1y)\) = 2 We have proved that (1x)(1y) ≠ 0' Hence y cannot take the value 1 Does this help?



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Re: For any numbers x and y, x#y = xy  x  y if x#y=1 which
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02 Dec 2015, 21:28
TeamGMATIFY wrote: happyface101 wrote: prsnt11 wrote: Hey Bunuel, What are the guidelines to factor \(xyxy=1\) as \((1x)(1y)1\)? Can you please provide a link of study or similar problems? I'm unable to understand. Thanks,
I second this question  can someone please provide some guidance? Thanks!! There are no guidelines as such to factorization of \(xyxy=1\) as \((1x)(1y)1\) Since the question is asking us to find the values that y cannot take, we are trying to make an equation in terms of y. If (1x)(1y) = 0. This means either x = 1 or y =1 By bringing the equation in the form \((1x)(1y)\) = 2 We have proved that (1x)(1y) ≠ 0' Hence y cannot take the value 1 Does this help? This doesn't help bc I still don't know the thought process of turning xy−x−y=1 into (1−x)(1−y)−1. Everything else after that make sense to me.
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Re: For any numbers x and y, x#y = xy  x  y if x#y=1 which
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03 Dec 2015, 15:47
happyface101 wrote: This doesn't help bc I still don't know the thought process of turning xy−x−y=1 into (1−x)(1−y)−1. Everything else after that make sense to me.
There are practically four different methods to factorise secondorder expressions: 1. Take out common factor 2. Grouping 3. Difference of two squares 4. Middle term breaking This particular example falls under category 2. If you would like further detailing on these 4 methods, you may either refer to a Maths text book of class 8 or write back to me.



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For any numbers x and y, x#y = xy  x  y if x#y=1 which
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Updated on: 26 Jul 2016, 08:26
bmwhype2 wrote: 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 M2333 1 = xy  x  y take y to LHS and take 1 to RHS y = xy  x 1 y=x(y1)1 Pay attention to the term (y1) ; if we take the value of y = 1 as given in option D then the term (y1) will become 0 y=x(y1)1 {take y=1} 1=x(11)1 1=x(0)1 1=1 THIS DOESNT MAKE SENSE. SO Y = 1 cannot be the value of y ANSWER IS D
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Originally posted by LogicGuru1 on 26 Jul 2016, 05:28.
Last edited by LogicGuru1 on 26 Jul 2016, 08:26, edited 1 time in total.



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Re: For any numbers x and y, x#y = xy  x  y if x#y=1 which
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26 Jul 2016, 06:42
bmwhype2 wrote: 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
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Re: For any numbers x and y, x#y = xy  x  y if x#y=1 which
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