Minimization of Boolean Expression MCQ Quiz - Objective Question with Answer for Minimization of Boolean Expression - Download Free PDF
Last updated on Jun 6, 2025
Latest Minimization of Boolean Expression MCQ Objective Questions
Minimization of Boolean Expression Question 1:
Consider the following min term expression for F.
F (P, Q, R, S) = ∑ 0, 2, 5, 7, 8, 10, 13, 15
The min-terms 2, 7, 8 and 13 are ‘do not care’ terms. The minimal sum-of-products form for F isAnswer (Detailed Solution Below)
Minimization of Boolean Expression Question 1 Detailed Solution
F (P, Q, R, S) = ∑ 0, 2, 5, 7, 8, 10, 13, 15
Don’t care min terms are 2, 7, 8, 13
By plotting the K-map, the minimal SOP (sum of products) can be found.
Explanation –
While putting the terms to k-map following things happen,
- 3rd and 4th columns are swapped
- 3rd and 4th rows.
- term 2 is going to (0, 3) column instead of (0, 2)
- 8 is going to (3, 0) instead of (2,0)
Solving, the above K-map, we get Q̅S̅ + QS
Minimization of Boolean Expression Question 2:
What is the result of applying the complement law A + A' in Boolean algebra?
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 2 Detailed Solution
Explanation:
To understand why the complement law holds, we need to analyze the truth table for the expression A + A'. A truth table lists all possible values of the variables involved and the resulting value of the expression for each combination of variable values. In this case, we are dealing with a single variable A, which can either be 0 or 1.
A | A' | A + A' |
---|---|---|
0 | 1 | 1 |
1 | 0 | 1 |
Name |
AND Form |
OR Form |
Identity law |
1.A=A |
0+A=A |
Null Law |
0.A=0 |
1+A=1 |
Idempotent Law |
A.A=A |
A+A=A |
Inverse Law |
AA’=0 |
A+A’=1 |
Commutative Law |
AB=BA |
A+B=B+A |
Associative Law |
(AB)C |
(A+B)+C = A+(B+C) |
Distributive Law |
A+BC=(A+B)(A+C) |
A(B+C)=AB+AC |
Absorption Law |
A(A+B)=A |
A+AB=A |
De Morgan’s Law |
(AB)’=A’+B’ |
(A+B)’=A’B’ |
Minimization of Boolean Expression Question 3:
Let \(\mathrm{f}(\mathrm{~A}, \mathrm{~B})=\overline{\mathrm{A}}+\mathrm{B}\), Simplified expression for function f (f (x y, y), z) is :
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 3 Detailed Solution
Concept:
We are given a Boolean function in the nested form: \( f(f(xy, y), z) \). To simplify this, we need to understand how the function \( f \) behaves and use Boolean algebra rules.
Step 1: Assume definition of function
Let us assume \( f(a, b) = a + b \), a common definition used in Boolean simplification unless otherwise specified.
So,
\( f(xy, y) = xy + y \)
Using the identity: \( xy + y = y \) (Since \( y(x + 1) = y \))
Step 2: Substitute into outer function
Now substitute into the outer function:
\( f(y, z) = y + z \)
Hence, the simplified expression is:
\( y + z \)
Compare with Options:
- Option 1: \( \bar{x} + z \) ❌
- Option 2: \( xyz \) ❌
- Option 3: \( xy̅ + z \) ❌
- Option 4: None of the options ✅
Final Answer: ✅ None of the options
Minimization of Boolean Expression Question 4:
What happens when a bit string is XORed with itself n times as shown below ?
[B ⊕ (B ⊕ (B ⊕ (B ..........n times]
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 4 Detailed Solution
The correct answer is Remains unchanged when n is even.
Key Points
- When a bit string is XORed (exclusive OR) with itself, the result is always 0.
- For example, if you have a bit string B, XORing B with B will result in 0 (B ⊕ B = 0).
- XOR operation is both associative and commutative, meaning the order of operations does not change the result.
- When the bit string is XORed with itself an even number of times, the result remains unchanged.
- Mathematically, B ⊕ B ⊕ B ⊕ B (n times, where n is even) simplifies to 0 ⊕ 0 = 0.
Additional Information
- The XOR operation is widely used in various fields such as cryptography, error detection, and correction algorithms.
- In computer science, XOR is used in binary addition without carrying.
- When n is odd, XORing the bit string with itself results in the bit string itself (B ⊕ B ⊕ B = B).
- XOR can be used to swap values of two variables without using a temporary variable.
Minimization of Boolean Expression Question 5:
Which of the following is the complement of the boolean function.
A'B + CD' + A'B + CD'
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 5 Detailed Solution
The correct answer is Option 3) (A + B')(C' + D).
Key Points
Given Boolean function: A'B + CD' + A'B + CD'
- First simplify the given expression:
- A'B + A'B = A'B (redundant term)
- CD' + CD' = CD' (redundant term)
- So, the simplified function = A'B + CD'
Now, we are asked to find the complement of this function:
Let F = A'B + CD'
Then the complement is: F' = (A'B + CD')'
Apply De Morgan’s Law:
- (A'B + CD')' = (A'B)' · (CD')'
- (A'B)' = A + B'
- (CD')' = C' + D
Therefore, F' = (A + B')(C' + D)
Hence, the correct answer is: Option 3) (A + B')(C' + D)
Top Minimization of Boolean Expression MCQ Objective Questions
Boolean algebra obeys
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 6 Detailed Solution
Download Solution PDF
Name |
AND Form |
OR Form |
Identity law |
1.A = A |
0 + A = A |
Null Law |
0.A = 0 |
1 + A = 1 |
Idempotent Law |
A.A = A |
A + A = A |
Inverse Law |
AA’ = 0 |
A + A’ = 1 |
Commutative Law |
AB = BA |
A + B = B + A |
Associative Law |
(AB)C |
(A + B) + C = A + (B + C) |
Distributive Law |
A + BC = (A + B)(A + C) |
A(B + C) = AB + AC |
Absorption Law |
A(A + B) = A |
A + AB = A |
De Morgan’s Law |
(AB)’ = A’ + B’ |
(A + B)’ = A’B’ |
What would be yielded after simplifying the following Boolean expression:
Y = \(\rm\overline{(A+\bar{B}+C)+(B+\bar{C})}\)
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 7 Detailed Solution
Download Solution PDFConcept
1.) De-morgan's law: \(\overline{A+B}=\overline{A}\space\overline{B}\)
2.) \(\overline{A}{A}=0\)
3.) \(\overline{A}+{A}=1\)
Explanation
\(Y=\rm\overline{(A+\bar{B}+C)+(B+\bar{C})}\)
Using De-morgan's Law:
\(Y=\overline{(A+\overline{B}+C)}\space \overline{(B+\overline{C})}\)
\(Y=(\overline{A}B\overline{C})\space (\overline{B}C)\)
\(Y=\overline{A}(B\overline {B})(C\overline {C})\)
Y = 0
Considering X and Y as binary variables, the Boolean expression X + Y + 1 is equivalent to
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 8 Detailed Solution
Download Solution PDFFROM laws of Boolean algebra
1 + any variable = 1
Name |
AND Form |
OR Form |
Identity law |
1.A=A |
0+A=A |
Null Law |
0.A=0 |
1+A=1 |
Idempotent Law |
A.A=A |
A+A=A |
Inverse Law |
AA’=0 |
A+A’=1 |
Commutative Law |
AB=BA |
A+B=B+A |
Associative Law |
(AB)C |
(A+B)+C = A+(B+C) |
Distributive Law |
A+BC=(A+B)(A+C) |
A(B+C)=AB+AC |
Absorption Law |
A(A+B)=A |
A+AB=A |
De Morgan’s Law |
(AB)’=A’+B’ |
(A+B)’=A’B’ |
Determine the minimised expression of Boolean function
F = X̅ Z̅ + Y̅ Z̅ + Y Z̅ + XYZAnswer (Detailed Solution Below)
Minimization of Boolean Expression Question 9 Detailed Solution
Download Solution PDFConcept:
All Boolean algebra laws are shown below
Name |
AND Form |
OR Form |
Identity law |
1.A = A |
0 + A = A |
Null Law |
0.A = 0 |
1 + A = 1 |
Idempotent Law |
A. A = A |
A + A = A |
Inverse Law |
AA’ = 0 |
A + A’ = 1 |
Commutative Law |
AB = BA |
A + B = B + A |
Associative Law |
(AB)C |
(A + B) + C = A + (B + C) |
Distributive Law |
A + BC = (A + B) (A + C) |
A (B + C) = AB + AC |
Absorption Law |
A (A + B) = A |
A + AB = A |
De Morgan’s Law |
(AB)’ = A’ + B’ |
(A + B)’ = A’B’ |
Calculation:
F = X̅ Z̅ + Y̅ Z̅ + Y Z̅ + XYZ
= X̅ Z̅ + Z̅ (Y̅ + Y) + XYZ
= X̅ Z̅ + Z̅ + XYZ
= Z̅ (1 + X̅) + XYZ
= Z̅ + XYZ
Now using Distributive Law
= (Z̅ + Z)(Z̅ + XY)
= Z̅ + XY
Which Boolean function does the following Karnaugh map represent?
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 10 Detailed Solution
Download Solution PDFThe correct answer is option 1.
Concept:
The given K-Map is,
F = a'b'c+a'bc'+ab'c'+abc
F= a'(b'c+bc')+a (b'c'+bc)
F=a'(b⊕c)+a(b⊙c)
F=a'(b⊕c)+a(b⊕c)'
F=a⊕b⊕c
Hence the correct answer is Exclusive OR.
Simplified expression/s for following Boolean function F(A, B, C, D) = ∑ (0, 1, 2, 3, 6, 12, 13, 14, 15) is/are
(A) A'B' + AB + A'C'D'
(B) A'B' + AB + A'CD'
(C) A'B' + AB + BC'D'
(D) A'B' + AB + BCD'
Choose the correct answer from the options given below:
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 11 Detailed Solution
Download Solution PDFThe correct answer is option 4
K-maps
F(A, B, C, D) = ∑ (0, 1, 2, 3, 6, 12, 13, 14, 15)
Two K-Maps can be constructed from the given boolean function
The expression for K-Map 1 is AB + A'B' + A'CD'
The expression for K-Map 2 is AB +A'B' + BCD'
A ’literal’ in Boolean Algebra means
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 12 Detailed Solution
Download Solution PDF- A variable is a symbol that may take on the value 0 or 1.
- A literal is the use of a variable or its complement in an expression.
- A term is an expression formed by literals and operations at one level.
For example, the following function:
F1 = xy + xy'z + x'yz
Has 3 variables (x,y,z),
8 literals (x,y,x,y',z,x',y,z), and
4 terms (xy, xy'z, x'yz, and the OR term that combines the first level AND terms).
The number of essential prime implicants for the Function Y = A'B'C'D + A'BCD' + ABC'D' is given by ______
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 13 Detailed Solution
Download Solution PDFConcept:
Implicants: Every min-term in SOP form or max-term in POS form in a Boolean function is termed as an implicant.
For example,
F = AB + AC
AB and AC are called implicants.
Prime Implicants: All pairs that cannot be a part of any quad or all quads that cannot be a part of any octet in a K-map are termed as prime implicants.
Essential Prime Implicants: Those prime implicants that cover at least one min-term that can’t be covered by any other prime implicant are called essential prime implicants.
Calculation:
Given the Boolean function,
F (A, B, C, D) = A'B'C'D + A'BCD' + ABC'D'
For the above Boolean function, the K – map representation is:
Hence we can see there three minterms in the given function.
Essential prime implicants are also three as they are not covered by any other prime implicant.
Hence option (3) is the correct answer.
Consider the Karnaugh map given below. Where X represents “don’t care” and blank represents 0.
Assume for all inputs (a, b, c, d), the respective complements (a̅, b̅, c̅, d̅) are also available. The above logic is implemented using 2-input NOR gates only. The minimum number of gates required is ________.
Answer (Detailed Solution Below) 1
Minimization of Boolean Expression Question 14 Detailed Solution
Download Solution PDFConcept:
Draw the K- map, convert the K-map into a SOP (sum of product) or POS (product of sum) form. While reducing the K-map in these forms, a don’t care will be needed only when with the use of don’t cares we can reduce the term size.
Diagram: K – Map
From the K-map simplification:
F(a, b, c, d) = a̅.c
F(a, b, c, d) = \(\overline {\left( {a + \bar c} \right)}\)
Diagram:
Therefore, only one NOR gate is needed to implement the minimized function
For the given Boolean function F (x, y, z) = Σ(0, 2, 4, 5, 6) simplified output will be-
Answer (Detailed Solution Below)
Minimization of Boolean Expression Question 15 Detailed Solution
Download Solution PDFConcept:
We can simplify the given boolean function with the help of K-Map.
The K-map is a systematic way of simplifying Boolean expressions. With the help of the K-map method, we can find the simplest POS and SOP expression, which is known as the minimum expression.
Just like the truth table, a K-map contains all the possible values of input variables and their corresponding output values.
The K-map method is used for expressions containing 2, 3, 4, and 5 variables.
Calculation:
Given ; F (x, y, z) = Σ(0, 2, 4, 5, 6)
3 variable K-map:
The grouping of cells has shown below
The expression obtained from the K-Map → F = z' + xy'
Additional Information Note 1 − If outputs are not defined for some combination of inputs, then those output values will be represented with the don’t care symbol ‘x’. That means, we can consider them as either ‘0’ or ‘1’.
Note 2 − If don’t care terms are also present, then place doesn’t care ‘x’ in the respective cells of the K-map. Consider only the don’t care ‘x’ that are helpful for grouping the maximum number of adjacent ones. In those cases, treat the don’t care value as ‘1’.