This set of Switching Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Boolean Algebra – Multiplying Out and Factoring”.
1. The Boolean expression (A + B’).(A + C) when multiplied out results in A + B’.C.
a) True
b) False
View Answer
Explanation: The steps to derive the solution are given below.
Using Distributive law, (X + Y).(X + Z) = X + YZ
= (A + B’).(A + C)
= A + B.C’
2. The Boolean expression (A + B.C’.D) when factored results in (A + B).(A + C’).(A + D).
a) True
b) False
View Answer
Explanation: The steps to derive the solution are given below.
Using Distributive law, (X + Y).(X + Z) = X + YZ
= (A + BC’D)
= (A + B).(A + C’).(A + D)
3. Factor A.B + C.D.
a) (A + C).(B + C).(A + D).(B + D)
b) B.(A + B +C)
c) (A + B).(C + D)
d) A.B.C.D
View Answer
Explanation: The steps to derive the solution are given below.
Using Distributive law, (X + Y).(X + Z) = X + YZ
= A.B + C.D.
= (A.B + C).(A.B + D)
Using Distributive law, (X + Y).(X + Z) = X + YZ
= (A + C).(B + C).(A + D).(B + D)
4. Factor A.B + A.C’ + D’.E’.
a) (A + D’).( A + E’).(B + C’ + D’).(B + C’ + E’)
b) (A + D’).( A + E’).B.C ’
c) (A + D’).( A + E’).(B + C’)
d) A.(B + C’ + D’).(B + C’ + E’)
View Answer
Explanation: The steps to derive the solution are given below.
Using the ordinary distributive law, X.Y + X.Z = X.(Y + Z)
= A.(B + C’) + D’.E’
Using the second distributive law, (X + Y).(X + Z) = X + YZ
= (A. (B + C’) + D’).(A.(B + C’) + E’)
Using the second distributive law, (X + Y).(X + Z) = X + YZ
= (A + D’).( A + E’).(B + C’ + D’).(B + C’ + E’)
5. Factor the boolean expression A + B.C.D.E
a) (A + B).(A + C).(A + D).(A + E)
b) (A + B).(B + D).(C + E)
c) (A + B).(C + D).(E + B)
d) (A + B).(C + D).(D + E)
View Answer
Explanation: The steps to derive the solution are given below.
Using Distributive law, (X + Y).(X + Z) = X + YZ
= A + B.C.D.E
= (A + B).(A + C).(A + D).(A + E)
6. Multiply out the Boolean expression (A’ + B).(A’ + C).(A’ + D).
a) A’ + B.C.D
b) A’ + B.C.D’
c) A’ + B.C + D
d) A’ + B + C + D
View Answer
Explanation: The steps to derive the solution are given below.
Using Distributive law, (X + Y).(X + Z) = X + YZ
= (A’ + B).(A’ + C).(A’ + D)
= A’ + B.C.D
7. Multiply out the Boolean expression (A + B).(A + C).(A + D).
a) A + B.C.D
b) A + B.C.D’
c) A + B.C + D
d) A + B + C + D
View Answer
Explanation: The steps to derive the solution are given below.
Using Distributive law, (X + Y).(X + Z) = X + YZ
= (A + B).(A + C).(A + D)
= A + B.C.D
8. Multiply out the Boolean expression (A + B’).(A + C’).(A + D’).
a) A + B’.C’.D’
b) A + B.C.D
c) A + B.C’ + D’
d) A + B + C + D
View Answer
Explanation: The steps to derive the solution are given below.
Using Distributive law, (X + Y).(X + Z) = X + YZ
= (A + B’).(A + C’).(A + D’)
= A + B’.C’.D’
9. Multiply out the Boolean expression (A + B).(A + C).(A + D’).
a) A + B.C.D’
b) A + B.C.D
c) A + B.C’ + D’
d) A + B + C + D
View Answer
Explanation: The steps to derive the solution are given below.
Using Distributive law, (X + Y).(X + Z) = X + YZ
= (A + B).(A + C).(A + D’)
= A + B.C.D’
10. Multiply out the Boolean expression (A + B).(A + C’).(A + D).
a) A + B.C’.D
b) A + B.C.D
c) A + B.C’ + D’
d) A + B + C + D
View Answer
Explanation: The steps to derive the solution are given below.
Using Distributive law, (X + Y).(X + Z) = X + YZ
= (A + B).(A + C’).(A + D’)
= A + B.C’.D
Sanfoundry Global Education & Learning Series – Switching Circuits.
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