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# For any numbers x and y, x#y = xy - x - y if x#y=1 which

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CEO
Joined: 21 Jan 2007
Posts: 2694
Location: New York City
For any numbers x and y, x#y = xy - x - y if x#y=1 which  [#permalink]

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Updated on: 24 Dec 2013, 01:44
5
18
00:00

Difficulty:

55% (hard)

Question Stats:

60% (01:30) correct 40% (01:21) wrong based on 639 sessions

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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

M23-33

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.
Math Expert
Joined: 02 Sep 2009
Posts: 47157
Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which  [#permalink]

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24 Dec 2013, 01:45
2
4
bmwhype2 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

M23-33

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.

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CEO
Joined: 17 Nov 2007
Posts: 3484
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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21 Dec 2007, 08:10
2
D

xy-x-y=1 ==> y(x-1)-x=1 ==> y(x-1)=(x+1) ==> y=(x+1)/(x-1) ==>

y=(x-1+2)/(x-1) ==> y=1+2/(x-1)

2/(x-1)<>0 ==> y<>1
Director
Joined: 03 Sep 2006
Posts: 839

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21 Dec 2007, 21:52
1
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)/(X-1)

Now substitute the choices, For Y=1,

X+1 = X-1

which is not possible, that means Y=1 is not possible.
Retired Moderator
Joined: 14 Dec 2013
Posts: 3195
Location: Germany
Schools: HHL Leipzig
GMAT 1: 780 Q50 V47
WE: Corporate Finance (Pharmaceuticals and Biotech)

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24 Dec 2013, 01:56
1
1
walker wrote:
D

xy-x-y=1 ==> y(x-1)-x=1 ==> y(x-1)=(x+1) ==> y=(x+1)/(x-1) ==>

y=(x-1+2)/(x-1) ==> y=1+2/(x-1)

2/(x-1)<>0 ==> y<>1

similarly x= (y+1)/(y-1)...denominator cannot be 0, hence y cannot be 1.
Senior Manager
Joined: 25 Mar 2013
Posts: 265
Location: United States
Concentration: Entrepreneurship, Marketing
GPA: 3.5
Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which  [#permalink]

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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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Intern
Joined: 02 Mar 2010
Posts: 19
Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which  [#permalink]

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12 Nov 2014, 21:28
Bunuel wrote:
bmwhype2 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

M23-33

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.

Hey Bunuel,
What are the guidelines to factor $$xy-x-y=1$$ as $$(1-x)(1-y)-1$$? Can you please provide a link of study or similar problems? I'm unable to understand.
Thanks,
Intern
Joined: 05 Aug 2015
Posts: 46
Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which  [#permalink]

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01 Dec 2015, 22:42
prsnt11 wrote:
Bunuel wrote:
bmwhype2 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

M23-33

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.

Hey Bunuel,
What are the guidelines to factor $$xy-x-y=1$$ as $$(1-x)(1-y)-1$$? 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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Senior Manager
Joined: 20 Aug 2015
Posts: 392
Location: India
GMAT 1: 760 Q50 V44
Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which  [#permalink]

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02 Dec 2015, 02:20
happyface101 wrote:
prsnt11 wrote:

Hey Bunuel,
What are the guidelines to factor $$xy-x-y=1$$ as $$(1-x)(1-y)-1$$? 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 $$xy-x-y=1$$ as $$(1-x)(1-y)-1$$
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 (1-x)(1-y) = 0. This means either x = 1 or y =1

By bringing the equation in the form $$(1-x)(1-y)$$ = 2
We have proved that (1-x)(1-y) ≠ 0'

Hence y cannot take the value 1

Does this help?
Intern
Joined: 05 Aug 2015
Posts: 46
Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which  [#permalink]

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02 Dec 2015, 21:28
TeamGMATIFY wrote:
happyface101 wrote:
prsnt11 wrote:

Hey Bunuel,
What are the guidelines to factor $$xy-x-y=1$$ as $$(1-x)(1-y)-1$$? 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 $$xy-x-y=1$$ as $$(1-x)(1-y)-1$$
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 (1-x)(1-y) = 0. This means either x = 1 or y =1

By bringing the equation in the form $$(1-x)(1-y)$$ = 2
We have proved that (1-x)(1-y) ≠ 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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Retired Moderator
Joined: 14 Dec 2013
Posts: 3195
Location: Germany
Schools: HHL Leipzig
GMAT 1: 780 Q50 V47
WE: Corporate Finance (Pharmaceuticals and Biotech)
Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which  [#permalink]

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03 Dec 2015, 15:47
1
1
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 second-order 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.
Director
Joined: 04 Jun 2016
Posts: 606
GMAT 1: 750 Q49 V43
For any numbers x and y, x#y = xy - x - y if x#y=1 which  [#permalink]

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Updated on: 26 Jul 2016, 08:26
2
bmwhype2 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
M23-33

1 = xy - x - y
take y to LHS and take 1 to RHS
y = xy - x -1
y=x(y-1)-1
Pay attention to the term (y-1) ; if we take the value of y = 1 as given in option D then the term (y-1) will become 0
y=x(y-1)-1 {take y=1}
1=x(1-1)-1
1=x(0)-1
1=-1 THIS DOESNT MAKE SENSE.
SO Y = 1 cannot be the value of y

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Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly.
FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.

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.
Intern
Joined: 13 May 2014
Posts: 11
Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which  [#permalink]

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26 Jul 2016, 06:42
3
1
bmwhype2 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

M23-33

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Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which  [#permalink]

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Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which &nbs [#permalink] 14 Aug 2017, 05:20
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