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My answer :- RS 60 :)

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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