OG - Quant Review DS - #80

[greenpepper]

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

According to the book,
(2) Substituting y for x in (x-1)*(y-1)=y gives (y-1)*(y-1)=1 or thus only that y^2-2y+1=1; this cannot be solved uniquely for y; NOT sufficient.

This equation gives y=o or y=2. Since x=y and xy>0, both x and y cannot be 0. Therefore, y=2. I think the statement 2 is also sufficient. Am I wrong? Thank you.

[funkmurthy]

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

Given: (x-1) * (y-1) = 1;
xy - (x+y) + 1 = 1
xy - (x+y) = 0; .......EQ-1

1.x+y = xy ; then
EQ-1 ; xy-xy =0 ; True : Sufficient.

2.x=y ;
EQ-1 : y.y - (y + y) =0
y^2-2y !=0 ; Insufficient.
Ans:A

[christiancryan]

Hi greenpepper,

I hate to say it, but yes, you're wrong -- you've assumed the QUESTION is true! It's very easy to do this with yes/no questions written as equations or inequalities; a good technique to avoid this error is to always write a question mark with the question or its rephrasings (yes, I love to boldly split my infinitives):

If you substitute statement (2) into the question, you still get a QUESTION: "IS y^2 - 2y + 1 = 1?" Rephrasing this question, you wind up with "IS y^2 - 2y = 0?", then "IS y = 2 OR 0?" and finally, ruling out y=0 from the given information that xy > 0, you get "IS y = 2?"

Since you don't know whether y = 2, the answer is "I don't know."

Hope this is helpful.