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