This set of Discrete Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Minimization of Boolean Functions”.
1. Find the simplified expression A’BC’+AC’.
a) B
b) A+C
c) (A+B)C’
d) B’C
View Answer
Explanation: Given: A’BC’ + AC’
= C’(A’B + A)
= C’(A’B + (A + AB)) // Absorption law
= C’(A + A’B + AB)
= C’(A + (A’ + A)B)
= C’(A + 1B)
= C’(A + B).
2. Evaluate the expression: (X + Z)(X + XZ’) + XY + Y.
a) XY+Z’
b) Y+XZ’+Y’Z
c) X’Z+Y
d) X+Y
View Answer
Explanation: (X + Z)(X + XZ’) + XY + Y [Original Expression]
= (x + z)X(1 + Z’) + XY + Y [Distributive]
= (X + Z)X + XY + Y [Complement, Identity]
= (X+Z)X + Y(X+1) [ Distributive]
= (X+Z)X + Y [Idempotent]
= XX + XZ + Y [Distributive]
= X + XZ + Y [Identity]
= X(1+Z) + Y
= X + Y [Idempotent].
3. Simplify the expression: A’(A + BC) + (AC + B’C).
a) (AB’C+BC’)
b) (A’B+C’)
c) (A+ BC)
d) C
View Answer
Explanation: Given: A’(A + BC) + (AC + B’C)
= A’A + A’BC + AC + B’C
= A’BC + C(A + B’)
= C(A’B + A + B’)
= C(A + B + B’)
= C(A + 1)
= C.
4. What is the simplification value of MN(M + N’) + M(N + N’)?
a) M
b) MN+M’N’
c) (1+M)
d) M+N’
View Answer
Explanation: Given: MN(M + N’) + M(N + N’)
= MN(M+N’) + M.1
= MNM + MNN’ + M
= MN + 0 +M
= M(N + 1)
= M
5. Simplify the expression XZ’ + (Y + Y’Z) + XY.
a) (1+XY’)
b) YZ + XY’ + Z’
c) (X + Y +Z)
d) XY’+ Z’
View Answer
Explanation: Given: X Z’ + (Y + Y’Z) + XY
= XZ’ + (Y + Z) + XY
= XZ’ + Y + Z + XY
= (XZ’ + Z) + (Y + XY)
= (X + Z) + Y (1 + X)
= X + Y + Z.
6. Find the simplified term Y’ (X’ + Y’) (X + X’Y)?
a) XY’
b) X’Y
c) X + Y
d) X’Y’
View Answer
Explanation: Given: Y’ (X’ + Y’) (X + X’Y)
= Y’(X’ + Y’)(X + Y)
= (X’Y’ + Y’)(X + Y)
= (XX’Y’ + X’Y’Y + XY’ + YY’)
= XY’.
7. If an expression is given that x+x’y’z=x+y’z, find the minimal expression of the function F(x,y,z) = x+x’y’z+yz?
a) y’ + z
b) xz + y’
c) x + z
d) x’ + y
View Answer
Explanation: We have, x+x’y’z+yz
= x+y’z+yz [since, x+x’y’z=x+y’z]
= x+z(y’+y)
= x + z.
8. Simplify the expression: XY’ + X’ + Y’X’.
a) X’ + Y
b) XY’
c) (XY)’
d) Y’ + X
View Answer
Explanation: Given XY’+X’+Y’X’ = Y’(X+X’) + X’ = Y’.1 + X’ = X’ + Y’ = (XY)’ [De Morgan’s law].
9. Minimize the Boolean expression using Boolean identities: A′B+ABC′+BC’+A’BC.
a) B(AC)’
b) AC’ + B’
c) ABC + B’
d) BC’ + A’B
View Answer
Explanation: Rearrange the given expression: A′B+ABC′+BC’+A’BC into:
= (A’B + A’BC) + (ABC′ + BC’)
= A’B(1 + C) + BC’(A + 1)
= A’B + BC’
= B(A’ + C’)
= B(AC)’.
10. Minimize the following Boolean expression using Boolean identities.
F(A,B,C) = (A+BC’)(AB’+C)
a) A + B + C’
b) AC’ + B
c) B + AC
d) A(B’ + C)
View Answer
Explanation: Given, F(A,B,C) = (A+BC’)(AB’+C)
= (AAB’ + BC’AB’ + AC + BC’C)
= (AB’ + 0 + AC + 0)
= A(B’ + C).
Sanfoundry Global Education & Learning Series – Discrete Mathematics.
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