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If xy > 0, does (x-1)(y-1) = 1?

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If xy > 0, does (x-1)(y-1) = 1? [#permalink] New post 03 Dec 2010, 21:47
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A
B
C
D
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Question Stats:

69% (01:39) correct 31% (00:27) wrong based on 35 sessions
If xy > 0, does (x-1)(y-1) = 1?

(1) x + y = xy
(2) x = y

[Reveal] Spoiler:
I choose the answer D, but OG says option 2 is not sufficient. Could you please let me know where I am going wrong?

OG Explanation: (2) Substituting y for x in (x-1)(y-1) = 1 gives (y-1)(y-1) = 1 or thus only that y2-2y+1 = 1; this cannot be solved uniquely for y; NOT sufficient.

I was able to solve this as,
y2-2y+1 = 1
y(y-2)=0 --by subtracting 1 from both sides.

Implies y=0 or y-2 = 0
Given xy > 0, which implies neither x or y not equal to 0. Therefore y = 0 is not possible.
Thus y-2 = 0; y = 2. Thus I think option 2 is sufficient.
[Reveal] Spoiler: OA

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Re: Quantitative review - Official guide - DS Question 80 - help [#permalink] New post 04 Dec 2010, 01:56
Expert's post
abelnazareth wrote:
I choose the answer D, but OG says option 2 is not sufficient. Could you please let me know where I am going wrong?

80. If xy > 0, does (x-1)(y-1) = 1?
(1) x + y = xy
(2) x = y

OG Explanation: (2) Substituting y for x in (x-1)(y-1) = 1 gives (y-1)(y-1) = 1 or thus only that y2-2y+1 = 1; this cannot be solved uniquely for y; NOT sufficient.

I was able to solve this as,
y2-2y+1 = 1
y(y-2)=0 --by subtracting 1 from both sides.

Implies y=0 or y-2 = 0
Given xy > 0, which implies neither x or y not equal to 0. Therefore y = 0 is not possible.
Thus y-2 = 0; y = 2. Thus I think option 2 is sufficient.


If xy>0 does (x-1)(y-1)=1?
(1) x + y = xy
(2) x=y

xy>0 means that either both x and y are positive or both are negative (so neither of unknowns equals to zero: x\neq{0} and y\neq{0}).

Question: is (x-1)(y-1)=1? --> is xy-x-y+1=1? is x+y=xy?

(1) x+y=xy --> directly gives us the answer YES. Sufficient.

(2) x=y --> question becomes: is x+x=x^2? --> is x(x-2)=0? --> is x=0 or x=2? --> as given that x\neq{0}, then the question becomes is x=2? We don't know that (or in other words for x=y and xy>0 equation (x-1)(y-1)=1 holds true for x=y=2, but we don't know whether that's true, thus we can answer whether (x-1)(y-1)=1 holds true), hence this statement is not sufficient.

Answer: A.

P.S.

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Re: Quantitative review - Official guide - DS Question 80 - help [#permalink] New post 05 Dec 2010, 20:40
Statement 2 says x=y
If x*y>0 & x=y
x*y could be (-2)*(-2)=4 or 2*2=4

x*y=x+y
-2*-2=-2+-2
4 = -4 . so x*y=x+y

if x*y=x+y
2*2=2+2
4=4. so x*y=x+y
two different answers , so statement 2 is insufficient
Re: Quantitative review - Official guide - DS Question 80 - help   [#permalink] 05 Dec 2010, 20:40
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