My answer :- RS 60 :)
Given F= AD'+A'C'+A'D'C'+A'B'C
From K-map, Minimized F will be F=A'B'+A'C'+AD'
Cost of each minterm
A'B'(2 input AND gate and 2 inverters)=10+5+5=20,
A'C'(2 input AND gate and 2 inverters)=10+5+5=20,
AD'(2 input AND gate and 1 inverter)=10+5=15,
F(3 input OR gate)=3*5=15
The total cost is 20+20+15+15=Rs 70
I am still of the opinion that we cannot take only one inverter, since inverter is an input to the two different the AND gates A'C' and A'B'. Though, it is the same inverter that is used, but it is used twice with different combination of another input to the gate. In the question it is mentioned that all gates cost Rs.5 per input. You are just calculating the number of inverters and number of gates(AND/OR) but not checking from where the input is coming/going to.