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# m23 #33

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15 Aug 2008, 09:22
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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

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

OA is D. How can y = 0 if xy = 1?
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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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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*(y-1) - y
y + 1 = x*(y-1)

If y = 1, this equation does not hold.
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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.

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Last edited by vshrivastava on 22 Feb 2013, 01:15, edited 1 time in total.
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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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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:
xy-x-y=1
-> x(y-1)-y=1
->x(y-1)=1+y
->x=(1+y)/(y-1)
-----> y-1 should be 0
so y cannot be 1.
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05 Feb 2010, 13:35
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IMO D

Another way

xy-x-y = 1

=> y = $$\frac{(x+1)}{(x-1)}$$

clearly y =1 is not possible as x+1 can never be equal to x-1
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08 Feb 2010, 15:22
x#y =1 => xy-x-y =1
=> x(y-1) = 1+y
=> x = (1+y)/(y-1)

denominator cannot be zero hence y != 1
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13 Mar 2010, 17:53
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xy-x-y = 1 and if y=1, the equation will result in -1 = +1...not possible...so Ans is D.
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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

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

OA is D. How can y = 0 if xy = 1?

xy-x-y =1 => x(y-1) = 1+y => x = (1+y)/(y-1)
definitely y cannot 1 as this will make x indefinite.
so d
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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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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=xy-x-y
x=1/3
option 2.y=-1
then, x#y=1=xy-x-y
x=0
option 3.y=0
then, x#y=1=xy-x-y
x=-1
option 4.y=1
then, x#y=1=xy-x-y ==>1=-1, which is wrong
so, D cant be an option.
option 5,y=2
then, x#y=1=xy-x-y
x=3

hence option D is correct.
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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 xy-x-y = 1 , it would give -1 = 1 , thus Y != 0
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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 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 *.

Hope it's clear.
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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

[Reveal] Spoiler: OA
D

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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11 Jan 2013, 06:42
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Expert's post
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

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

OA is D. How can y = 0 if xy = 1?

For any numbers $$x$$ and $$y$$, $$x@y=xy-x-y$$. 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.

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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

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

OA is D. How can y = 0 if xy = 1?

For any numbers $$x$$ and $$y$$, $$x@y=xy-x-y$$. 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.

Bunuel! Great work.
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13 Jan 2013, 04:14
californiaNY wrote:
since I posted that screenshot, my daily GMAT questions have stopped! Am I assuming things or are we getting a bug check?

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01 Feb 2014, 02:03
By plugging we can get that y cannot be equal to 1
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21 Apr 2014, 05:45
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Plug in the numbers and see how:
A) -2x-x+2=1->x=1/3
B) -x-x+1=1-> x=0
C) -x=1
D) x-x-1=1=> -1=1 (ten ten!)

So choose D
Re: m23 #33   [#permalink] 21 Apr 2014, 05:45
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# m23 #33

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