|
|
#1 (permalink) |
|
TestMagic Guru-in-Training
![]() ![]() ![]() Join Date: Jun 2006
Location: New York
Posts: 724
![]() |
xy = ?
If 6xy = x2y + 9y, what is the value of xy?
(1) y – x = 3 (2) x3 < 0 Friends, I solved this problem by the foll. approach: 6xy = x2y + 9y Divide by Y =>6X = x^2 + 9 => (X-3)^2 = 0 => X = 3 So 1 alone is suff. But thats not the OA. I think I am wrong somewhere in the above approach.I would appreciate if someone can point the mistake that I have made. Thanks, Arun B |
|
|
|
|
|
#2 (permalink) |
|
TestMagic Guru-in-Training
![]() ![]() ![]() Join Date: Jan 2005
Location: Toronto
Posts: 897
![]() |
Arun, you cannot divide by 'y' since y might equal zero. (in dividing by 'y' you are "missing out" on one of the roots).
Question stem: 6xy=(x^2)*y+9y 0=y*[(x^2)-6x+9] 0=y*(x-3)^2 y=0 and x can be any number. xy=0 OR x=3 and y can be any number. xy=3y First statement: y-x=3 If x=3; y=6 and xy=18 If y=0; x=-3 and xy=0 Not sufficient. Second statement: x^3<0 x<0 Therefore x does not equal 3; so y must equal zero and xy=0. Sufficient. Answer is B. |
|
|
|
|
|
#4 (permalink) | |
|
I JUST got here.
Join Date: Sep 2007
Posts: 15
![]() |
Quote:
HIii ISR, The explanation was wonderful.... Can i have some more questions like these if you have to get a beetter understanding of this type......... |
|
|
|
|
Contact TestMagic TestMagic Forums Archive Privacy Statement
TestMagic Locations
Legal
Privacy
SEO by vBSEO 3.2.0
Copyright © 2009 TestMagic
Ad Management by RedTyger