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Boolean Algebra and Logic Simplification

Q1:

For the SOP expression , how many 0s are in the truth table's output column?

A zero

B 1

C 4

D 5

ANS:C - 4

AB+B'C
= AB.1+B'C.1
= AB(C+C')+B'C(A+A') {Complement element}
= ABC+ABC'+AB'C+A'B'C {Distributive law}

As 4 numbers of output is present so 4 "1"s are present in the output. As variables are A,B,C, therefore n=3. So total outputs are 8 (2^n). number of "0"s are 8-4 = 4.



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