Register    Login    Search    Rss Feeds

 Page 1 of 1 [ 3 posts ] 



 
Author Message
 Post subject: OG - Quant Review DS - #80
 Post Posted: Sat Aug 18, 2007 1:33 pm 
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.


Top 
 Post subject:
 Post Posted: Sun Aug 19, 2007 12:38 am 
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


Top 
 Post subject:
 Post Posted: Tue Aug 21, 2007 2:57 am 
Offline
ManhattanGMAT Staff


Posts: 79
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.


Top 
Display posts from previous:  Sort by  
 
 Page 1 of 1 [ 3 posts ] 





Who is online

Users browsing this forum: No registered users and 0 guests

 
 

 
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to: