y + 5 = adding $5 to the full retail price
\(\frac{100-z}{100}\) is the proportion of the price after the discount, which will be less than 1.

We can interpret Quantity A as:
\((y+5)(\frac{{100-z}}{{100}}) = y(\frac{{100-z}}{{100}}) + 5(\frac{{100-z}}{{100}})\)

and Quantity B as:
\(y(\frac{{100-z}}{{100}}) + 5\)

Subtract \(y(\frac{{100-z}}{{100}})\) from both, and you're left with
\(5(\frac{{100-z}}{{100}})\) vs. \(5\)

Since \((\frac{{100-z}}{{100}})\) will be less than 1, \(5(\frac{{100-z}}{{100}}) < 5\)
Thus, B > A


To demo with actual numbers, let y = 100 and z = 20.
A = \((100+5)(\frac{{100-20}}{{100}}) = 100(\frac{{100-20}}{{100}}) + 5(\frac{{100-20}}{{100}}) = 100(.8) + 5(.8) = 84 \)
B = \(100(\frac{{100-20}}{{100}}) + 5 = 100(.8) + 5 = 85\)
85 > 84