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Director
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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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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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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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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/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, 01:15, edited 1 time in total.
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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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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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Y cannot be 1 so ans is D
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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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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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Senior Manager
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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? 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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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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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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@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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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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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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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? 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. Answer; D. Bunuel! Great work.
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Someone should change the answer to the question in the GMAT question of the day. The answer displayed is B whereas the answer discussed in the forum is D.
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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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