I would be interested too in knowing how to solve this problem by pluggin in values..
For now, I can post the algebraic approach, which is quite simple and not time consuming at all.
Let the number of $1 newspaper sold = a
Let the number of $1.25 newspaper sold = b
Let a + b = 100 --------------- (1)
Given:
(P/100)*(a+b) = a --------------- (2)
=> p = a ---------------- (3)
(r/100)(a + 1.25b) = a ------------- (4)
substituting a = p and b = 100 - a = 100 - p in equation (4) we get
r = 100p/(125 - 0.25p)
= 400p/(500 - p) - Option D
