gmatnmein2010 wrote:If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y - x = 3
(2) x^3< 0

6*x*y = x^2*y + 9*y ==>
(6*x*y - x^2*y - 9*y) =0
y(x^2-6x+9) =0
y(x-3)^2 =0

y =0
or x=3

1.) y - x = 3 ; [ two cases possible y=0 and x =3; in the case of y=0 x = -3 and xy =0 and in the case of x=3 y=6 and xy =18]
Insufficient

2.) x^3<0
x<0
x is not 3; y has to be zero, xy =0

Sufficient

B

ajith wrote:

gmatnmein2010 wrote:If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y - x = 3
(2) x^3< 0

6*x*y = x^2*y + 9*y ==>
(6*x*y - x^2*y - 9*y) =0
y(x^2-6x+9) =0
y(x-3)^2 =0

y =0
or x=3

1.) y - x = 3 ; [ two cases possible y=0 and x =3; in the case of y=0 x = -3 and xy =0 and in the case of x=3 y=6 and xy =18]
Insufficient

2.) x^3<0
x<0
x is not 3; y has to be zero, xy =0

Sufficient

B

@Ajith

I agree that x=3 or y=0 { frm the stem}

Now in statement 2, it says x^3<0.... How is this possible???

Already we have derived value of X=3 and this st 2 contradicts that value..isnt it??


Even with out the help of st 2 we can say value of xy =0 as y=0..

I am not able to understand st 2 clearly.

gmatmachoman wrote:
@Ajith

I agree that x=3 or y=0 { frm the stem}

Now in statement 2, it says x^3<0.... How is this possible???

Already we have derived value of X=3 and this st 2 contradicts that value..isnt it??


Even with out the help of st 2 we can say value of xy =0 as y=0..

I am not able to understand st 2 clearly.

see from the statements given in the question we conclude that either x =3 or y=0
now it is possible that (y =0 and x =- 3 )and (x =3 and y =6)

y is not always zero, x is not always 3 ...
when y is not zero x has to be 3
and when x is not 3 y has to be 0