From the question stem we know that each of the numbers in question is either 0 or 1.

The first stamement can be simplified to 8w + 4x + 2y + z = 11
The only possibility of plugging in zeroes and ones for the values of these numbers and getting a sum of 11 is:

8(1) + 4(0) + 2(1) + 1(1).
Therefore, w = 1, x = 0, y = 1 and z = 1. w + x + y + z = 1 + 0 + 1 + 1 = 3. Sufficient.

Similarly Statement 2 can be simplified to 27w + 9x + 3y + z = 31
The only possibility of plugging in zeroes and ones for the values of these numbers and getting a sum of 31 is:

27(1) + 9(0) + 3(1) + 1(1).

Again, w = 1, x = 0, y = 1 and z = 1. w + x + y + z = 1 + 0 + 1 + 1 = 3. Sufficient.

Hence D

both stmts are suff

Stmt 1:
8w+4x+2y+z/16 = 11/16
8w+4x+2y+z = 11
place in values w,x,y,z is either 0 or 1
so x = 0 then the equation holds true so w+x+y+z = 3 - STMT 1 suff

similarly solve for stmt 2 - you will get 27w+9x+2y+z = 11
so w+x+y+z = 2 - SUFF