IMO = B
I consider, the options provided by you are as follows:-
A. (x+z)/z
B. (y+z)/x
C. (x+y)/z
D. (xy)/z
E. (yz)/x
FIRST APPROACH
Lets assume X = 9 (Don't take prime number, as X is multiple of Z)
so Z could be equal to 3
and as X is factor of Y, so we can take any value of Y that is completely divisible by 9.
Lets say Y = 45
so X = 9, Y = 45 and Z = 3
Now plug these values one by in each equation. Only for choice (B) the results comes into fraction. Hence (B) is the right answer.
SECOND APPROACH
A. (x+z)/z
For A, x is multiple of z. So we add one more z to x and then divide by z, it would result into the integer. (For example, if z = 2 and x = 6, so if add 2 to 6 and divide by 2, it would result into the integer. Simple thing: Its comes under the same table.)
B. (y+z)/x
For B, x is the factor of y but not of z. So, if we add z to y and try to divide it by x, then obviously it would result z as the reminder.
C. (x+y)/z
For C, x is multiple of z. It means z is factor of x. Its already given that x is factor of y. So by these two facts, its clear that z is also factor of y. So, when we add two number which are having common factor, then dividing by common factor would result into an integer.
D. (xy)/z
For D, same reason as I have stated for C. (only the operator symbol is changed which unaffected by the fact stated in the option C)
E. (yz)/x
For E, as x is factor of y, so when we divide y by x, it would result an integer. And then multiplying an integer with another integer would result an integer.
Choice is yours. Which ever you find easiest can blindly follow that.