Tough one

288 = 2^5.3^2
since x is divisible by 6 we will say x = 2*3*m
so
sqrt(288kx) = sqrt(2^5*3^2*k*2*3*m)
==>sqrt(2^6*3^2*k*3*m)
= 24sqrt(3*k*m)
from here you can easily eliminate 4 of the 5 choices and left with "B"

not sure how to eliminate C & D. Can you elaborate?

sure
d. 24 sqrt(6k)
our equation = 24sqrt(3*k*m)
as per question k and x are +ve integers, and x is divisible by 3
this implies x = (any int)*6
we have already taken care of 6,
so to satisfy d we can pick a = 2

e. 72 sqrt(k)
following above reasoning if you pick a =3
= 24sqrt(3*k*3)
= 72sqrt(k)
c) m =1
a) m = k
b) to satisfy b we have to select m = 1/3 which will be a fraction but as per question stem our variable "m" should an integer.

Hope this helps

Hey amitabhprasad,
How did you eliminate A? K is not in the square root.

umaa wrote:Hey amitabhprasad,
How did you eliminate A? K is not in the square root.

It can be eliminated by assuming m=k, hence, sqrt(k^2)=K, we're left with 24k sqrt(3)

Right.. I thought abt it. thx..