Most of the time, picking numbers is a particularly risky strategy. It's best employed when trying to prove that something ISN'T true, like for instance if all numbers that are divisible by 3 are also divisible by 6. Pick 9 to prove that this statement is incorrect.
This is why I'll stick to that age old method of algebraically solving problems.
So you have that y = k*x - x is a factor of y
x = m*z - x is a multiple of z
Notice that since y = k*x = k*(mz) = km*z, so you have both y and x in terms of z:
x = k*z
y = km*z
I'd use these to eliminate choices:
a. x is divisible by z and z is also divisible by z. This means that the sum x + z will also be divisible by z, so this one is out.
b. y + z = km*z + z = z(km + 1)
(y + z)/x = z(km + 1)/kz = (km + 1)/k = km/k + 1/k = m + 1/k - this one is an integer only if 1/k is also an integer, which happens when k is either 1 or -1. As you can see, this is terribly restrictive, B is the correct answer. If you had this problem on the real thing, just pick B and go on to the next question. However, I will also analyze the other choices, just so you get a clear picture of things.
c. x is divisible by z and y is divisible by z, so their sum will also be divisible by z
d. since both x and y are divisible by z,xy will definitely be divisible by z
e. yz = km*z*z = km*(z^2)
x = k*z
yz/x = km*(z^2)/kz = zm - which will also be an integer.