xy = ?

[aru4912]

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

[lsr]

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.

[aru4912]

Got it .. excellent explanation .. thanks lsr.

[Juggler]

Originally Posted by lsr

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.

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.........

[Juggler]

Aru,

Do you have any more questions like these.......

great explanation lsr!

[karta]

great explanation lsr

thanks

[bads123]

Y can also be zero hence cannot be cancelled

[rrgre]

was a little tricky problm.

[adityanlal]

Great Explanation. I was wondering, if the equation is solved then why do we need the other two equations. After explanation, it was clear. Thanks lsr.